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On Kalman’s functor for bounded hemi-implicative semilattices and hemi-implicative lattices

On Kalman’s functor for bounded hemi-implicative semilattices and hemi-implicative lattices Abstract Hemi-implicative semilattices (lattices), originally defined under the name of weak implicative semilattices (lattices), were introduced by the second author of the present article. A hemi-implicative semilattice is an algebra $$(H,{\wedge},{\rightarrow},1)$$ of type $$(2,2,0)$$ such that $$(H,{\wedge})$$ is a meet semilattice, $$1$$ is the greatest element with respect to the order, $$a{\rightarrow} a = 1$$ for every $$a\in H$$ and for every $$a$$, $$b$$, $$c\in H$$, if $$a\leq b{\rightarrow} c$$ then $$a{\wedge} b \leq c$$. A bounded hemi-implicative semilattice is an algebra $$(H,{\wedge},{\rightarrow},0,1)$$ of type $$(2,2,0,0)$$ such that $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice and $$0$$ is the first element with respect to the order. A hemi-implicative lattice is an algebra $$(H,{\wedge},\vee,{\rightarrow},0,1)$$ of type $$(2,2,2,0,0)$$ such that $$(H,{\wedge},\vee,0,1)$$ is a bounded distributive lattice and the reduct algebra $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice. In this article, we introduce an equivalence for the categories of bounded hemi-implicative semilattices and hemi-implicative lattices, respectively, which is motivated by an old construction due J. Kalman that relates bounded distributive lattices and Kleene algebras. 1 Introduction Inspired by results due to J. Kalman relating to lattices [15], R. Cignoli proved in [8] that a construction of J. Kalman can be extended to a functor $${\mathrm{K}}$$ from the category of bounded distributive lattices to the category of Kleene algebras and that this functor has a left adjoint [8, Theorem 1.7]. He also showed that there exists an equivalence between the category of bounded distributive lattices and the full subcategory of centered Kleene algebras whose objects satisfy a condition called interpolation property [8, Theorem 2.4]. Moreover, R. Cignoli also proved that there exists an equivalence between the category of Heyting algebras and the category of centered Nelson algebras [8, Theorem 3.14]. These results were extended by J.L. Castiglioni, R. Lewin, M. Menni and M. Sagastume in the context of residuated lattices [3, 4]. On the other hand, the original Kalman’s construction was also extended in [5] by J.L. Castiglioni, S. Celani and the second author of the present article to the framework of algebras with implication $$(H,{\wedge},\vee, {\rightarrow}, 0,1)$$ which satisfy the following additional condition: for every $$a,b,c\in H$$, if $$a\leq b{\rightarrow} c$$ then $$a{\wedge} b \leq c$$. Algebras with implication were introduced by S. Celani in [6]. A generalization of Heyting algebras is provided by the notion of hemi-implicative semilattice (lattice), introduced in [21] under the name weak implicative semilattices (lattices). An algebra $$(H,{\wedge},{\rightarrow},1)$$ of type $$(2,2,0)$$ is said to be a hemi-implicative semilattice if $$(H,{\wedge},1)$$ is an upper bounded semilattice1, $$a{\rightarrow} a = 1$$ for every $$a\in H$$ and for every $$a,b,c\in H$$, if $$a\leq b{\rightarrow} c$$ then $$a{\wedge} b \leq c$$. A bounded hemi-implicative semilattice is an algebra $$(H,{\wedge},{\rightarrow},0,1)$$ of type $$(2,2,0,0)$$ such that $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice and $$0$$ is the first element with respect to the order. A hemi-implicative lattice is an algebra $$(H,{\wedge},\vee,{\rightarrow},0,1)$$ of type $$(2,2,2,0,0)$$ such that $$(H,{\wedge},\vee,0,1)$$ is a bounded distributive lattice and the reduct algebra $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice. Implicative semilattices [17] and Hilbert algebras with infimum [12] are examples of hemi-implicative semilattices. Semi-Heyting algebras [20] and some algebras studied in [5] are examples of hemi-implicative lattices. For instance, the RWH-algebras, introduced and studied by S. Celani and the first author of this article in [7], are examples of hemi-implicative lattices. The applications of Kalman’s construction given in [8] suggest that it is potentially fruitful to understand Kalman’s work in the context of bounded hemi-implicative semilattices and hemi-implicative lattices. We do this in the present article. The main goal of the article is to introduce and study an equivalence for the categories of bounded hemi-implicative semilattices and hemi-implicative lattices, respectively, and for some of its full subcategories. The article is organized as follows. In Section 2, we give some results about Kalman’s functor for bounded distributive lattices and Heyting algebras. In Section 3 we generalize Kalman’s functor for the category whose objects are posets with first element and whose morphisms are maps which preserve finite existing infima and the first element (note that the morphisms of this category are in particular order-preserving maps). Moreover, we apply the mentioned equivalence to build up an equivalence for the category whose objects are bounded semilattices and whose morphisms are the corresponding algebra homomorphisms. In Section 4, we recall definitions and properties about hemi-implicative semilattices (lattices) [21], Hilbert algebras with infimum [12], implicative semilattices [17] and semi-Heyting algebras [20]. In Section 5 we employ results of Sections 3 and 4 to establish equivalences, following the original Kalman’s construction, for the categories of bounded hemi-implicative semilattices, bounded Hilbert algebras with infimum, bounded implicative semilattices, hemi-implicative lattices, respectively, and the category of semi-Heyting algebras. Finally, in Section 6 we introduce and study the notion of well-behaved congruences for the objects corresponding to the categories introduced in Section 5. We give a table with some of the categories we shall consider in this article: $$\mathbf{Category}$$  $$\mathbf{Objects}$$  $$\mathbf{Morphisms}$$  $${\mathsf{BDL}}$$  Bounded distributive lattices  Algebra homomorphisms  $${\mathsf{KA_{{\mathrm{c}}}}}$$  Centered Kleene algebras  Algebra homomorphisms  $${\mathsf{HA}}$$  Heyting algebras  Algebra homomorphisms  $${\mathsf{NA_{{\mathrm{c}}}}}$$  Centered Nelson algebras  Algebra homomorphisms  $${\mathsf{NL_{{\mathrm{c}}}}}$$  Centered Nelson lattices  Algebra homomorphisms  $${\mathsf{P_0}}$$  Posets with bottom  Certain order morphisms  $${\mathsf{KP}}$$  Kleene posets  Certain order morphisms  $${\mathsf{MS}}$$  Bounded semilattices  Algebra homomorphisms  $${\mathsf{KMS}}$$  Certain objects of $${\mathsf{KP}}$$  Morphisms of $${\mathsf{KP}}$$  $${\mathsf{hIS_{0}}}$$  Bounded hemi-implicative semilattices  Algebra homomorphisms  $${\mathsf{hBDL}}$$  Hemi-implicative lattices  Algebra homomorphisms  $${\mathsf{Hil_{0}}}$$  Bounded Hilbert algebras with infimum  Algebra homomorphisms  $${\mathsf{IS_{0}}}$$  Bounded implicative semilattices  Algebra homomorphisms  $${\mathsf{SH}}$$  Semi-Heyting algebras  Algebra homomorphisms  $${\mathsf{KhIS_{0}}}$$  Objects of $${\mathsf{KMS}}$$ with an additional  Certain morphisms of $${\mathsf{KMS}}$$    operation    $${\mathsf{KHil_{0}}}$$  Certain objects of $${\mathsf{KhIS_{0}}}$$  Morphisms of $${\mathsf{KhIS_{0}}}$$  $${\mathsf{KIS_{0}}}$$  Certain objects of $${\mathsf{KhIS_{0}}}$$  Morphisms of $${\mathsf{KhIS_{0}}}$$  $${\mathsf{KhBDL}}$$  Objects of $${\mathsf{KA_{{\mathrm{c}}}}}$$ with an additional  Certain morphisms of $${\mathsf{KA_{{\mathrm{c}}}}}$$    operation    $${\mathsf{KSH}}$$  Certain objects of $${\mathsf{KhBDL}}$$  Morphisms of $${\mathsf{KhBDL}}$$  $$\mathbf{Category}$$  $$\mathbf{Objects}$$  $$\mathbf{Morphisms}$$  $${\mathsf{BDL}}$$  Bounded distributive lattices  Algebra homomorphisms  $${\mathsf{KA_{{\mathrm{c}}}}}$$  Centered Kleene algebras  Algebra homomorphisms  $${\mathsf{HA}}$$  Heyting algebras  Algebra homomorphisms  $${\mathsf{NA_{{\mathrm{c}}}}}$$  Centered Nelson algebras  Algebra homomorphisms  $${\mathsf{NL_{{\mathrm{c}}}}}$$  Centered Nelson lattices  Algebra homomorphisms  $${\mathsf{P_0}}$$  Posets with bottom  Certain order morphisms  $${\mathsf{KP}}$$  Kleene posets  Certain order morphisms  $${\mathsf{MS}}$$  Bounded semilattices  Algebra homomorphisms  $${\mathsf{KMS}}$$  Certain objects of $${\mathsf{KP}}$$  Morphisms of $${\mathsf{KP}}$$  $${\mathsf{hIS_{0}}}$$  Bounded hemi-implicative semilattices  Algebra homomorphisms  $${\mathsf{hBDL}}$$  Hemi-implicative lattices  Algebra homomorphisms  $${\mathsf{Hil_{0}}}$$  Bounded Hilbert algebras with infimum  Algebra homomorphisms  $${\mathsf{IS_{0}}}$$  Bounded implicative semilattices  Algebra homomorphisms  $${\mathsf{SH}}$$  Semi-Heyting algebras  Algebra homomorphisms  $${\mathsf{KhIS_{0}}}$$  Objects of $${\mathsf{KMS}}$$ with an additional  Certain morphisms of $${\mathsf{KMS}}$$    operation    $${\mathsf{KHil_{0}}}$$  Certain objects of $${\mathsf{KhIS_{0}}}$$  Morphisms of $${\mathsf{KhIS_{0}}}$$  $${\mathsf{KIS_{0}}}$$  Certain objects of $${\mathsf{KhIS_{0}}}$$  Morphisms of $${\mathsf{KhIS_{0}}}$$  $${\mathsf{KhBDL}}$$  Objects of $${\mathsf{KA_{{\mathrm{c}}}}}$$ with an additional  Certain morphisms of $${\mathsf{KA_{{\mathrm{c}}}}}$$    operation    $${\mathsf{KSH}}$$  Certain objects of $${\mathsf{KhBDL}}$$  Morphisms of $${\mathsf{KhBDL}}$$  If $$\mathrm{A}$$ is one of the categories $${\mathsf{KA_{{\mathrm{c}}}}}$$, $${\mathsf{KP}}$$, $${\mathsf{KMS}}$$, $${\mathsf{KhIS_{0}}}$$, $${\mathsf{KHil_{0}}}$$ and $${\mathsf{KhBDL}}$$, then we write $$\mathrm{A}^{{\mathrm{CK}}}$$ to denote the full subcategory of $$\mathrm{A}$$ whose objects satisfy the condition $$({\mathrm{CK}})$$, that will be defined later. The results we expound in the present article are motivated by the abstraction of ideas coming from different varieties of algebras related to some constructive logics, as Heyting algebras and Nelson algebras, and in particular by the existent categorical equivalence between the category of Heyting algebras and the category of centered Nelson algebras (see [8]) combined with the fact that the variety of centered Nelson algebras is term equivalent to the variety of centered Nelson lattices, as it is shown in [22] (see also [2]). In this article, we introduce and study categories which are closely connected with the category of centered Nelson lattices, as for instance the category $${\mathsf{KP}}$$ of Kleene posets (of which centered Nelson lattices can be seen as particular cases) and the category $${\mathsf{KhBDL}}$$ of centered Kleene algebras endowed with a binary operation which generalizes the implication of Nelson lattices. We consider that the study of the above mentioned categories is interesting in itself. We also think that the categorical equivalences and some related properties studied in this article can be of interest for future work concerning the understanding of the categories of bounded hemi-implicative semilattices and hemi-implicative lattices, respectively. 2 Basic results The definition of the functor from the category of Kleene algebras to the category of bounded distributive lattices given by R. Cignoli [8] is based on Priestley duality, and the interpolation property for Kleene algebras considered by Cignoli in establishing the equivalence is stated in topological terms. On the other hand, M. Sagastume proved in an unpublished manuscript [19] that in centered Kleene algebras the interpolation property is equivalent to an algebraic condition called (CK), that we will state later on. Moreover, she presented an equivalence between the category of bounded distributive lattices and the category of centered Kleene algebras that satisfy (CK), but using a different (purely algebraic) construction to that given by R. Cignoli in [8]. In what follows we describe this equivalence whose details can be found in [5]. We assume the reader is familiar with bounded distributive lattices and Heyting algebras [1]. A De Morgan algebra is an algebra $$(H,{\wedge},\vee,{\sim},0,1)$$ of type $$(2,2,1,0,0)$$ such that $$(H,{\wedge}, \vee,0,1)$$ is a bounded distributive lattice and $${\sim}$$ fulfills the equations   \[ {\sim} {\sim} x = x \quad {\text{and}} \quad {\sim}(x \vee y) = {\sim} x {\wedge} {\sim} y. \] An operation $${\sim}$$ which satisfies the previous two equations is called De Morgan involution. A Kleene algebra is a De Morgan algebra in which the inequality   \[ x{\wedge} {\sim} x \leq y \vee {\sim} y \] holds. A centered Kleene algebra is an algebra $$(H,{\wedge},\vee,{\sim},c, 0,1)$$ where the algebra $$(H,{\wedge},\vee,$$$${\sim},0,1)$$ is a Kleene algebra and $${\mathrm{c}}$$ is an element such that $${\mathrm{c}} = {\sim} {\mathrm{c}}.$$ It is immediate to see that $${\mathrm{c}}$$ is necessarily unique. The element $${\mathrm{c}}$$ is called center. We write $${\mathsf{BDL}}$$ for the category of bounded distributive lattices and $${\mathsf{KA_{{\mathrm{c}}}}}$$ for the category of centered Kleene algebras. In both cases the morphisms are the corresponding algebra homomorphisms. It is interesting to note that if $$T$$ and $$U$$ are centered Kleene algebras and $$f:T{\rightarrow} U$$ is a morphism of Kleene algebras then $$f$$ preserves necessarily the centre, i.e., $$f({\mathrm{c}}) = {\mathrm{c}}$$. The functor $${\mathrm{K}}$$ from the category $${\mathsf{BDL}}$$ to the category $${\mathsf{KA_{{\mathrm{c}}}}}$$ is defined as follows. For an object $$H\in {\mathsf{BDL}}$$ we let   \[ {\mathrm{K}}(H): =\{(a,b) \in H\times H: a{\wedge} b = 0\}. \] This set is endowed with the operations and the distinguished elements defined by:   \begin{eqnarray*} (a,b)\vee (d,e) & := & (a\vee d,b{\wedge} e)\\ (a,b){\wedge} (d,e)& := & (a{\wedge} d,b\vee e)\\ {\sim} (a,b)& := & (b,a)\\ 0 & := & (0,1)\\ 1 & := & (1,0)\\ {\mathrm{c}} & := & (0,0). \end{eqnarray*} We have that $$({\mathrm{K}}(H),{\wedge},\vee,\sim,{\mathrm{c}},0,1)\in {\mathsf{KA_{{\mathrm{c}}}}}$$. For a morphism $$f:H {\rightarrow} G \in {\mathsf{BDL}}$$, the map $${\mathrm{K}}(f):{\mathrm{K}}(H) {\rightarrow} {\mathrm{K}}(G)$$ defined by   $${\mathrm{K}}(f)(a,b) = (f(a),f(b))$$ is a morphism in $${\mathsf{KA_{{\mathrm{c}}}}}$$. Hence, $${\mathrm{K}}$$ is a functor from $${\mathsf{BDL}}$$ to $${\mathsf{KA_{{\mathrm{c}}}}}$$. Let $$(T,{\wedge},\vee,{\sim},{\mathrm{c}},0,1)\in {\mathsf{KA_{{\mathrm{c}}}}}$$. The set   $${\mathrm{C}}(T):=\{x\in T:x\geq {\mathrm{c}}\}$$ is the universe of a subalgebra of $$(T,{\wedge},\vee,{\mathrm{c}},1)$$ and $$({\mathrm{C}}(T),{\wedge},\vee,{\mathrm{c}},1) \in {\mathsf{BDL}}$$. Moreover, if $$g:T{\rightarrow} U$$ is a morphism in $${\mathsf{KA_{{\mathrm{c}}}}}$$, then the map $${\mathrm{C}}(g): {\mathrm{C}}(T) {\rightarrow} {\mathrm{C}}(U)$$, given by $${\mathrm{C}}(g)(x) = g(x)$$, is a morphism in $${\mathsf{BDL}}$$. Thus, $${\mathrm{C}}$$ is a functor from $${\mathsf{KA_{{\mathrm{c}}}}}$$ to $${\mathsf{BDL}}$$. Let $$H \in {\mathsf{BDL}}$$. The map $$\alpha_{H}: H {\rightarrow} {\mathrm{C}}({\mathrm{K}}(H))$$ given by $$\alpha_{H}(a) = (a,0)$$ is an isomorphism in $${\mathsf{BDL}}$$. If $$T \in {\mathsf{KA_{{\mathrm{c}}}}}$$, then the map $$\beta_T: T{\rightarrow} {\mathrm{K}}({\mathrm{C}}(T))$$ given by $$\beta_T(x) = (x\vee {\mathrm{c}}, {\sim} x \vee {\mathrm{c}})$$ is injective and a morphism in $${\mathsf{KA_{{\mathrm{c}}}}}$$. It is not difficult to show that the functor $${\mathrm{K}}: {\mathsf{BDL}} {\rightarrow} {\mathsf{KA_{{\mathrm{c}}}}}$$ has as left adjoint the functor $${\mathrm{C}}: {\mathsf{KA_{{\mathrm{c}}}}} {\rightarrow} {\mathsf{BDL}}$$ with unit $$\beta$$ and counit $$\alpha^{-1}$$. We are interested though in an equivalence between $${\mathsf{BDL}}$$ and the full subcategory of $${\mathsf{KA_{{\mathrm{c}}}}}$$ whose objects satisfy the condition (CK) we proceed to state. Let $$T\in {\mathsf{KA_{{\mathrm{c}}}}}$$. We consider the algebraic condition:   \begin{equation} \label{eq-CK} (\forall x, y \geq c)(x{\wedge} y = {\mathrm{c}} \ \longrightarrow \ (\exists z)(z\vee {\mathrm{c}} = x \ \& \ {\sim} z \vee {\mathrm{c}} = y)). \end{equation} (CK) This condition characterizes the surjectivity of $$\beta_T$$, i.e., for every $$T\in {\mathsf{KA_{{\mathrm{c}}}}}$$, $$T$$ satisfies (CK) if and only if $$\beta_T$$ is a surjective map, as shown in [19]. The condition (CK) is not necessarily verified in every centered Kleene algebra (see [5]). We write $${\mathsf{KA_{{\mathrm{c}}}^{CK}}}$$ for the full subcategory of $${\mathsf{KA_{{\mathrm{c}}}}}$$ whose objects satisfy (CK). The functor $${\mathrm{K}}$$ can then be seen as a functor from $${\mathsf{BDL}}$$ to $${\mathsf{KA_{{\mathrm{c}}}^{CK}}}$$. The next theorem was proved by M. Sagastume in [19]. A complete proof of it can be also found in [5]. Theorem 1 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{BDL}}$$ and $${\mathsf{KA_{{\mathrm{c}}}^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Let $$T\in {\mathsf{KA_{{\mathrm{c}}}}}$$. We know that $$\beta_T$$ is not necessarily a surjective map. However we will prove that $$\beta_T$$ is an epimorphism. Before, we need a lemma that is interesting in its own right. It tells us that the morphisms in $${\mathsf{KA_{{\mathrm{c}}}}}$$ are determined by their behaviour on the elements greater than or equal to the center. Lemma 2 If $$f:T{\rightarrow} U$$ and $$g:T{\rightarrow} U$$ are morphisms in $${\mathsf{KA_{{\mathrm{c}}}}}$$ and $$f(x) = g(x)$$ whenever $$x\in {\mathrm{C}}(T)$$, then $$f(x) = g(x)$$ for every $$x\in T$$. Proof. Suppose that $$f(x) = g(x)$$ whenever $$x\in {\mathrm{C}}(T)$$. Let $$x$$ be an arbitrary element of $$T$$. Then   \[ \begin{array} [c]{lllll} f(x) \vee {\mathrm{c}} & = & f(x) \vee f({\mathrm{c}}) & & \\ & = & f(x\vee {\mathrm{c}})& & \\ & = & g(x\vee {\mathrm{c}})& & \\ & = & g(x) \vee g({\mathrm{c}})& & \\ & = & g(x) \vee {\mathrm{c}},& & \end{array} \] so we obtain that $$f(x) \vee {\mathrm{c}} = g(x) \vee {\mathrm{c}}$$. Similarly we can prove that $${\sim} f(x) \vee {\mathrm{c}} = {\sim} g(x) \vee {\mathrm{c}}$$, which is equivalent to $$f(x) {\wedge} {\mathrm{c}} = g(x) {\wedge} {\mathrm{c}}$$. Hence, it follows from the distributivity of the underlying lattice of $$U$$ that $$f(x) = g(x)$$. ■ Proposition 3 Let $$T\in {\mathsf{KA_{{\mathrm{c}}}}}$$. Then $$\beta_T$$ is an epimorphism. Proof. Let $$f: {\mathrm{K}}({\mathrm{C}}(T)) {\rightarrow} U$$ and $$g:{\mathrm{K}}({\mathrm{C}}(T)){\rightarrow} U$$ be morphisms in $${\mathsf{KA_{{\mathrm{c}}}}}$$ such that $$f \circ \beta_T = g \circ \beta_T$$, where $$\circ$$ denotes the composition of functions. We will prove that $$f = g$$. Let $$(x,y) \in {\mathrm{C}}({\mathrm{K}}({\mathrm{C}}(T)))$$, i.e., $$x{\wedge} y = {\mathrm{c}}$$, $$x\geq {\mathrm{c}}$$, $$y \geq {\mathrm{c}}$$ and $$({\mathrm{c}},{\mathrm{c}}) \leq (x,y)$$, where we also write $$\leq$$ for the order associated with the underlying lattice of $${\mathrm{K}}({\mathrm{C}}(T))$$. In particular we have that $$y\leq {\mathrm{c}}$$, so $$y = {\mathrm{c}}$$. Then $$(x,y) = (x,{\mathrm{c}})$$. Besides, since $$x\geq {\mathrm{c}}$$ we have that $$\beta_{T}(x) = (x,{\mathrm{c}})$$. Then   \[ \begin{array} [c]{lllll} f(x,y) & = & f(x,{\mathrm{c}}) & & \\ & = & (f \circ \beta_{T})(x)& & \\ & = & (g \circ \beta_{T})(x)& & \\ & = & g(x, {\mathrm{c}})& & \\ & = & g(x,y).& & \end{array} \] Hence, $$f(x,y) = g(x,y)$$ whenever $$(x,y) \in {\mathrm{C}}({\mathrm{K}}({\mathrm{C}}(T)))$$. Therefore, it follows from Lemma 2 that $$f(x,y) = g(x,y)$$ for every $$(x,y) \in {\mathrm{K}}({\mathrm{C}}(T))$$, which was our aim. ■ Let $$H \in {\mathsf{BDL}}$$ and $$a$$, $$b\in H$$. If the relative pseudocomplement of $$a$$ with respect to $$b$$ exists, then we denote it by $$a {\rightarrow}_{{\mathsf{HA}}} b$$. Recall that a Nelson algebra [8] is a Kleene algebra such that for each pair $$x$$, $$y$$ there exists the binary operation $${\Rightarrow}$$ given by $$x{\Rightarrow} y: = x {\rightarrow}_{{\mathsf{HA}}} ({\sim x} \vee y)$$ and for every $$x,y,z$$ it holds that $$(x {\wedge} y)\Rightarrow z = x \Rightarrow (y\Rightarrow z)$$. The binary operation $$\Rightarrow$$ so defined is called the weak implication. We denote by $${\mathsf{HA}}$$ the category of Heyting algebras. M. Fidel [11] and D. Vakarelov [23] proved independently that if $$H\in {\mathsf{HA}}$$, then the Kleene algebra $${\mathrm{K}}(H)$$ is a Nelson algebra, in which the weak implication is defined for pairs $$(a, b)$$ and $$(d, e$$) in $${\mathrm{K}}(H)$$ as follows:   \begin{equation} \label{ic} (a,b)\Rightarrow(d,e):= (a{\rightarrow} d, a{\wedge} e). \end{equation} (1) We say that an algebra $$(T,{\wedge},\vee,\Rightarrow, {\sim},{\mathrm{c}},0,1)$$ is a centered Nelson algebra if the reduct $$(T,{\wedge},\vee,\Rightarrow, {\sim},0,1)$$ is a Nelson algebra and $${\mathrm{c}}$$ satisfies $${\sim} {\mathrm{c}} = {\mathrm{c}}$$. We write $${\mathsf{NA_{{\mathrm{c}}}}}$$ for the category of centered Nelson algebras. The following result appears in [3, Proposition 3.7] and is a reformulation of [8, Theorem 3.14]. Theorem 4 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{HA}}$$ and $${\mathsf{NA_{{\mathrm{c}}}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. We assume the reader is familiar with commutative residuated lattices [13]. An involutive residuated lattice is a bounded, integral and commutative residuated lattice $$(T,{\wedge}, \vee, \ast,{\rightarrow}, 0, 1)$$ such that for every $$x\in T$$ it holds that $$\neg \neg x = x$$, where $$\neg x: = x{\rightarrow} 0$$ and $$0$$ is the first element of $$T$$ [2]. In an involutive residuated lattice it holds that $$x \ast y = \neg (x {\rightarrow} \neg y)$$ and $$x{\rightarrow} y = \neg (x \ast \neg y)$$. A Nelson lattice [2] is an involutive residuated lattice $$(T,{\wedge},\vee, *,{\rightarrow},0,1)$$ which satisfies the additional inequality $$(x^2 {\rightarrow} y){\wedge} ((\neg y)^2 {\rightarrow} \neg x) \leq x{\rightarrow} y$$, where $$x^2:=x\ast x$$. See also [23]. Remark 5 Let $$(T,{\wedge}, \vee, \Rightarrow,{\sim}, 0,1)$$ be a Nelson algebra. We define on $$T$$ the binary operations $$*$$ and $${\rightarrow}$$ by   \begin{eqnarray*} x*y:=& {\sim} (x \Rightarrow {\sim} y) \vee {\sim} (y \Rightarrow {\sim} x), \hspace{1cm} x {\rightarrow} y :=& (x \Rightarrow y) {\wedge} ({\sim} y\Rightarrow {\sim} x). \end{eqnarray*} Then Theorem 3.1 of [2] says that $$(T,{\wedge}, \vee, {\rightarrow},*, 0,1)$$ is a Nelson lattice. Moreover, $${\sim} x = \neg x = x{\rightarrow} 0$$. Let $$(T,{\wedge},\vee,*,{\rightarrow},0,1)$$ be a Nelson lattice. We define on $$T$$ a binary operation $$\Rightarrow$$ and a unary operation $$\sim$$ by   \begin{eqnarray*} x \Rightarrow y:=& x^2 {\rightarrow} y, \hspace{1cm} {\sim} x:=& \neg x, \end{eqnarray*} where $$x^2 = x*x$$. Then Theorem 3.6 of [2] says that the algebra $$(T,{\wedge}, \vee,\Rightarrow,{\sim},0,1)$$ is a Nelson algebra. In [2, Theorem 3.11] it was also proved that the category of Nelson algebras and the category of Nelson lattices are isomorphic. Taking into account the construction of this isomorphism in [2] we have that the variety of Nelson algebras and the variety of Nelson lattices are term equivalent and the term equivalence is given by the operations we have defined before. The results from [2] about the connections between Nelson algebras and Nelson lattices mentioned in Remark 5 are based on results from Spinks and Veroff [22]. In particular, the term equivalence of the varieties of Nelson algebras and Nelson lattices was discovered by Spinks and Veroff in [22]. A centered Nelson lattice is an algebra $$(T,{\wedge},\vee,*,{\rightarrow},{\mathrm{c}},0,1)$$, where the reduct $$(T,{\wedge},\vee,*,{\rightarrow},0,1)$$ is a Nelson lattice and $${\mathrm{c}}$$ is an element such that $$\neg {\mathrm{c}} = {\mathrm{c}}$$. It follows from Remark 5 that the variety of centered Nelson algebras and the variety of centered Nelson lattices are term equivalent. We write $${\mathsf{NL_{{\mathrm{c}}}}}$$ for the category of centered Nelson lattices. Remark 6 Let $$(H,{\wedge}, \vee,{\rightarrow},0,1) \in {\mathsf{HA}}$$. Then $$({\mathrm{K}}(H),{\wedge},\vee,{\Rightarrow},{\sim},{\mathrm{c}},0,1) \in {\mathsf{NA_{{\mathrm{c}}}}}$$. Hence it follows from Remark 5 that $$({\mathrm{K}}(H),{\wedge},\vee,*,{\rightarrow},{\mathrm{c}},0,1) \in {\mathsf{NL_{{\mathrm{c}}}}}$$, where for $$(a,b)$$ and $$(d,e)$$ in $${\mathrm{K}}(H)$$ the operations $$\ast$$ and $${\rightarrow}$$ take the form   \begin{eqnarray*} (a,b) \ * \ (d,e) = (a{\wedge} d, (a{\rightarrow} e) \ {\wedge} \ (d{\rightarrow} b)),\\ (a,b) {\rightarrow} (d,e) = ((a{\rightarrow} d){\wedge} (e{\rightarrow} b), a {\wedge} e). \end{eqnarray*} We write $${\rightarrow}$$ both for the implication in $$H$$ as for the implication in $${\mathrm{K}}(H)$$. It follows from Theorem 4 and Remark 5 that there is a categorical equivalence between $${\mathsf{HA}}$$ and $${\mathsf{NL_{{\mathrm{c}}}}}$$, as it was also mentioned in [5, Corollary 2.11]. In what follows we will make explicit a construction of this equivalence. Proposition 7 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{HA}}$$ and $${\mathsf{NL_{{\mathrm{c}}}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Proof. Let $$H \in {\mathsf{HA}}$$. Then the centered Kleene algebra $$({\mathrm{K}}(H),{\wedge},\vee,{\sim},{\mathrm{c}},0,1)$$ endowed with the two operations given in Remark 6 is a centered Nelson lattice. It is immediate that if $$f$$ is a morphism in $${\mathsf{HA}}$$, then $${\mathrm{K}}(f)$$ is a morphism in $${\mathsf{NL_{{\mathrm{c}}}}}$$. Let $$(T,{\wedge},\vee,*,{\rightarrow}, {\mathrm{c}},0,1)\in {\mathsf{NL_{{\mathrm{c}}}}}$$. Taking into account Remark 5 we deduce that $$(T,{\wedge}, \vee,{\Rightarrow}, {\sim}, {\mathrm{c}},0,1)\in {\mathsf{NA_{{\mathrm{c}}}}}$$, where $$x {\Rightarrow} y = x^{2} {\rightarrow} y$$. Moreover,   \begin{equation} \label{nanl1} x {\rightarrow} y = (x {\Rightarrow} y) {\wedge} ({\sim} y {\Rightarrow} {\sim x}). \end{equation} (2) Let $$x$$, $$y\geq {\mathrm{c}}$$. We will prove that $$x {\rightarrow} y = x {\rightarrow}_{{\mathsf{HA}}} y$$. To show it, note that straightforward computations show that   \begin{equation} \label{nanl2} x{\Rightarrow} y = x {\rightarrow}_{{\mathsf{HA}}} \ y. \end{equation} (3) Besides, $${\sim} y \Rightarrow {\sim} x = {\sim} y {\rightarrow}_{{\mathsf{HA}}} (y \vee {\sim} x)$$. Since $$y\geq {\mathrm{c}}$$ and $${\sim} x \leq {\mathrm{c}}$$, then $$y \vee {\sim} x = y$$. Then $${\sim} y \Rightarrow {\sim} x = {\sim} y {\rightarrow}_{{\mathsf{HA}}} y$$. Hence, it follows from (2) and (3) that   $$ x{\rightarrow} y = (x{\rightarrow}_{{\mathsf{HA}}} y) {\wedge} ({\sim} y {\rightarrow}_{{\mathsf{HA}}} y).$$ Note that $$x{\rightarrow} y = x{\rightarrow}_{{\mathsf{HA}}} y$$ if and only if $$x{\rightarrow}_{{\mathsf{HA}}} y \leq {\sim} y {\rightarrow} y$$, which is equivalent to $${\sim} y {\wedge} (x{\rightarrow}_{{\mathsf{HA}}} y) \leq y$$. But $${\sim} y \leq {\mathrm{c}}$$ and $$x {\rightarrow}_{{\mathsf{HA}}} y \geq {\mathrm{c}}$$, so $${\sim} y {\wedge} (x{\rightarrow}_{{\mathsf{HA}}} y) = {\sim} y$$. Hence, $${\sim} y {\wedge} (x{\rightarrow}_{{\mathsf{HA}}} y) \leq y$$ if and only if $${\sim} y \leq y$$. Since $$y\geq {\mathrm{c}}$$, then $${\sim} y \leq {\mathrm{c}}$$, so $${\sim} y \leq y$$. Then we have that $$x{\rightarrow} y = x {\rightarrow}_{{\mathsf{HA}}} y$$. Thus, $$({\mathrm{C}}(T), {\wedge}, \vee,{\rightarrow},{\mathrm{c}},1) \in {\mathsf{HA}}$$. Straightforward computations show that if $$g$$ is a morphism in $${\mathsf{NL_{{\mathrm{c}}}}}$$, then $${\mathrm{C}}(g)$$ is a morphism in $${\mathsf{HA}}$$. It is also immediate that if $$H\in {\mathsf{HA}}$$ then $$\alpha_H$$ is an isomorphism in $${\mathsf{HA}}$$. Let $$(T,{\wedge}, \vee,*,{\rightarrow}, {\mathrm{c}},0,1)\in {\mathsf{NL_{{\mathrm{c}}}}}$$. It follows from Theorem 4 and Remark 5 that $$\beta_T$$ preserves $${\rightarrow}$$. Therefore, $$\beta_T$$ is an isomorphism in $${\mathsf{NL_{{\mathrm{c}}}}}$$. ■ The main goal of this article is to find a generalization of Proposition 7 replacing the categories of Heyting algebras and centered Nelson lattices by the categories of bounded hemi-implicative semilattices and hemi-implicative lattices, respectively. To make it possible, we start studying an equivalence for a particular category of posets with first element. Then we employ it to obtain an equivalence for the category of bounded semilattices. Finally, taking into account the last mentioned equivalence, we build up an equivalence for the categories of bounded hemi-implicative semilattices and hemi-implicative lattices, respectively, and for some of its full subcategories. 3 Kalman’s functor for posets with bottom and for bounded semilattices In this section, we generalize the equivalence given for the category of bounded distributive lattices but replacing this category by the category whose objects are posets with first element and whose morphisms are maps which preserve finite existing infima and the first element. Then we apply this equivalence to establish an equivalence for the category whose objects are bounded semilattices and whose morphisms are the corresponding algebra homomorphisms. We start with some preliminary definitions and properties. Let $$(P,\leq,0)$$ be a poset with first element, and let $$(P \times P,\preceq)$$ be the poset with universe the cartesian product $$P\times P$$ where the order $$\preceq$$ is given by   \[ (a,b) \preceq (d,e)\; \text{if and only if}\; a\leq d \ \text{and} \ e \leq b. \] In other words, $$(P \times P,\preceq)$$ is the direct product of $$(P,\leq)$$ with its dual. Let $$(P,\leq)$$ and $$(Q,\leq)$$ be posets. Let $$f:(P,\leq) {\rightarrow} (Q,\leq)$$ be a function. We say that $$f$$preserves finite existing infima if for every $$a$$, $$b\in P$$ such that $$a{\wedge} b$$ exists in $$P$$ then $$f(a){\wedge} f(b)$$ exists in $$Q$$ and $$f(a{\wedge} b) = f(a){\wedge} f(b)$$. Definition 8 The category $${\mathsf{P_0}}$$ has as objects the posets with first element and has as morphisms the maps between posets with first element which preserve the finite existing infima and the first element. Note that every morphism in $${\mathsf{P_0}}$$ preserves the order. It follows from the fact that morphisms preserve the finite existing infima. Let $$P \in {\mathsf{P_0}}$$. We define the following set:   \begin{equation} {\mathrm{K}}(P): = \{(a,b)\in P\times P: a{\wedge} b \ \text{exists and} \ a{\wedge} b = 0\}. \end{equation} (4) This set is the natural one to associate with the poset $$P$$ if we aim to generalize the original Kalman’s construction given for bounded distributive lattices [15]. To attain the generalization we first order $${\mathrm{K}}(P)$$ with the order induced by the poset $$(P \times P,\preceq)$$ defined above. It is immediate from the definition that $${\mathrm{K}}(P)$$ is closed under the unary operation $${\sim}$$ on $$P \times P$$ given by $${\sim}(a, b) = (b, a)$$, and that the element $${\mathrm{c}} = (0, 0)$$ belongs to $${\mathrm{K}}(P)$$. Thus we obtain the structure $${\mathrm{K}}(P) := ({\mathrm{K}}(P),\preceq,{\sim}, {\mathrm{c}}).$$ The following elemental lemma plays a fundamental role in some proofs of this section. Lemma 9 Let $$(b,d) \in {\mathrm{K}}(P)$$. The following conditions hold: (1) For every $$a \in P$$, $$(a,0) {\wedge} (b,d)$$ exists in $${\mathrm{K}}(P)$$ if and only if $$a{\wedge} b$$ exists in $$P$$. If these conditions hold, then $$(a,0) {\wedge} (b, d) = (a{\wedge} b, d)$$. (2) $$(b,d) {\wedge} (0,0)$$ exists in $${\mathrm{K}}(P)$$ and $$(b,d) {\wedge} (0,0) = (0,d)$$. (3) $$(b,d) \vee (0,0)$$ exists in $${\mathrm{K}}(P)$$ and $$(b,d) \vee (0,0) = (b,0).$$ Proof. In general, if $$y = (e, u) \in {\mathrm{K}}(P)$$, we write $$\pi_1(y)$$ for the first coordinate and $$\pi_2(y)$$ for the second coordinate (i.e. $$\pi_1(y) = e$$ and $$\pi_2(y) = u$$). Let $$(b,d) \in {\mathrm{K}}(P)$$. We proceed to the proof. (1) Suppose that $$a \in P$$ and $$(a,0) {\wedge} (b,d)$$ exists in $${\mathrm{K}}(P)$$. Let $$x := (a,0) {\wedge} (b,d)$$. We have that $$x \preceq (a,0)$$ and $$x\preceq (b,d)$$, so by the definition of $$\preceq$$ we obtain that $$\pi_1(x) \leq a$$ and $$\pi_1(x) \leq b$$. Thus $$\pi_1(x)$$ is a lower bound of the set $$\{a,b\}$$. Let now $$e$$ be a lower bound of $$\{a,b\}$$, i.e., $$e\leq a$$ and $$e\leq b$$. The fact that $$(b,d) \in {\mathrm{K}}(P)$$ implies that $$(e,d) \in {\mathrm{K}}(P)$$. Since $$(e,d)\preceq (a,0)$$ and $$(e,d) \preceq (b,0)$$, then $$(e,d)\preceq x$$. Hence, $$e\leq \pi_1(x)$$. Thus, $$\pi_1(x) = a{\wedge} b$$. Conversely, suppose that $$a{\wedge} b$$ exists in $$P$$. We have that $$(a{\wedge} b,d) \in {\mathrm{K}}(P)$$, $$(a{\wedge} b,d) \preceq (a,0)$$, and $$(a{\wedge} b,d) \preceq (b,d)$$. Let $$(e, u) \in {\mathrm{K}}(P)$$ such that $$(e,u)\preceq (a,0)$$ and $$(e,u) \preceq (b,d)$$, i.e., $$e\leq a$$, $$e\leq b$$ and $$d\leq u$$. Since $$e \leq a{\wedge} b$$, then $$(e,u) \preceq (a{\wedge} b, d)$$. Hence we obtain that $$(a,0){\wedge} (b,d)$$ exists in $${\mathrm{K}}(P)$$, and moreover $$(a,0){\wedge} (b,d) = (a{\wedge} b,d)$$. (2) To prove that $$(b,d) {\wedge} (0,0)$$ exists in $${\mathrm{K}}(P)$$, let us see that $$(0,d)$$ is the infimum of $$(b,d)$$ and $$(0,0)$$. We have that $$(0,d)\preceq (b,d)$$ and $$(0,d)\preceq (0,0)$$, so $$(0,d)$$ is a lower bound of $$\{(b,d), (0,0)\}$$. Let $$(e,u) \in {\mathrm{K}}(P)$$ be a lower bound of $$\{(b,d), (0,0)\}$$, so $$(e,u) \preceq (b,d)$$ and $$(e,u) \preceq (0,0)$$. Hence, $$e = 0$$ and $$d\leq u$$, and so $$(e,u) \leq (0,d)$$. Therefore we obtain that $$(b,d) {\wedge} (0,0) = (0,d)$$. (3) In a similar way it can be proved that $$(b,d) \vee (0,0)$$ exists in $${\mathrm{K}}(P)$$ and is $$(b,0)$$. ■ Motivated by properties of $${\mathrm{K}}(P)$$ we give the following definition. Definition 10 A structure $$(T,\leq,{\sim},{\mathrm{c}})$$ is a Kleene poset if the following conditions hold: (1) $$(T,\leq)$$ is a poset. (2) $${\sim}$$ is an unary operation on $$T$$ which is an involution, i.e., $${\sim} {\sim} x = x$$ for every $$x\in T$$ and is order reversing, i.e., for every $$x,y\in T$$, if $$x\leq y$$, then $${\sim} y \leq {\sim} x$$. (3) $${\mathrm{c}} = {\sim} {\mathrm{c}}$$. (4) $$x\vee {\mathrm{c}}$$ exists, for every $$x \in T$$. (5) For every $$x\in T$$, $$(x\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}})$$ exists and $$(x\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}) = {\mathrm{c}}$$. (6) For every $$x,y \in T$$, if $$x{\wedge} {\mathrm{c}} \leq y {\wedge} {\mathrm{c}}$$ and $$x\vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$$, then $$x \leq y$$. The element $${\mathrm{c}}$$ of the previous definition will be also called center. The next lemma justifies the use of $${\wedge}$$ in the statement of the condition 6. Lemma 11 Let $$(T, \leq)$$ be a poset satisfying $$2$$., $$3$$., $$4$$. and $$5$$. of Definition 10. (1) Let $$x,y \in T$$. If $$x{\wedge} y$$ exists, then $${\sim} x \vee {\sim} y$$ exists and $${\sim} x \vee {\sim} y = {\sim} (x{\wedge} y)$$. Analogously, if $$x\vee y$$ exists, then $${\sim} x {\wedge}dge {\sim} y$$ exists and $${\sim} x {\wedge}dge {\sim} y = {\sim} (x\vee y)$$. (2) For every $$x\in T$$, $$x{\wedge} {\mathrm{c}}$$ exists and $$x{\wedge} {\mathrm{c}} = {\sim} ({\sim} x \vee {\mathrm{c}})$$. (3) The element $${\mathrm{c}}$$ is unique. Proof. Straightforward computations show the first two assertions. To prove that the centre is unique, let $${\mathrm{c}}$$ and $${\mathrm{c}}'$$ be centres. Then $${\mathrm{c}} = ({\mathrm{c}}' \vee {\mathrm{c}}){\wedge} ({\sim} {\mathrm{c}}' \vee {\mathrm{c}})$$. Since $${\sim} {\mathrm{c}}' = {\mathrm{c}}'$$, then $${\mathrm{c}} = {\mathrm{c}}' \vee {\mathrm{c}}$$. Hence, $${\mathrm{c}}'\leq {\mathrm{c}}$$. Analogously, $${\mathrm{c}}\leq {\mathrm{c}}'$$. Thus, $${\mathrm{c}} = {\mathrm{c}}'$$. ■ In what follows we introduce the category of Kleene posets. Definition 12 We denote by $${\mathsf{KP}}$$ the category whose objects are the Kleene posets and whose morphisms are the maps $$g$$ between Kleene posets that preserve the order, the involution and the finite existing infima over elements greater than or equal to the center. Note that if $$g:T{\rightarrow} U$$ is a morphism in $${\mathsf{KP}}$$, $${\mathrm{c}}$$ is the center of $$T$$ and $${\mathrm{c}}'$$ is the center of $$U$$, then $$g({\mathrm{c}}) = {\mathrm{c}}'$$. It follows from the fact that $$g({\mathrm{c}}) = {\sim} g({\mathrm{c}})$$ and the fact that the center is unique. If $$T\in {\mathsf{KP}}$$, we define $${\mathrm{C}}(T)$$ as in the case of centered Kleene algebras. If $$f$$ is a morphism in $${\mathsf{P_0}}$$ and $$g$$ is a morphism in $${\mathsf{KP}}$$ we define $${\mathrm{K}}(f)$$ and $${\mathrm{C}}(g)$$ as in Section 1, respectively. Lemma 13 (a) If $$(P,\leq,0) \in {\mathsf{P_0}}$$, then $$({\mathrm{K}}(P),\preceq,{\sim}, {\mathrm{c}}) \in {\mathsf{KP}}$$. (b) If $$f \in {\mathsf{P_0}}$$, then $${\mathrm{K}}(f) \in {\mathsf{KP}}$$. Proof. It follows from straightforward computations based on Lemma 9 that if $$(P,\leq,0) \in {\mathsf{P_0}}$$, then $$({\mathrm{K}}(P),\preceq,{\sim},$$$$ {\mathrm{c}}) \in {\mathsf{KP}}$$. In what follows we will prove that if $$f:P {\rightarrow} Q$$ is a morphism in $${\mathsf{P_0}}$$, then $${\mathrm{K}}(f):{\mathrm{K}}(P){\rightarrow} {\mathrm{K}}(Q)$$ is a morphism in $${\mathsf{KP}}$$. By the definition of $${\mathsf{P_0}}$$ we have that if $$(a,b) \in {\mathrm{K}}(P)$$, then $$f(a) {\wedge}dge f(b)$$ exists in $$Q$$ and is $$f(a{\wedge} b) = f(0)$$, which is $$0$$ in $$Q$$. Thus, $$(f(a),f(b))\in {\mathrm{K}}(Q)$$. Therefore, $${\mathrm{K}}(f)$$ is indeed a map from $${\mathrm{K}}(P)$$ to $${\mathrm{K}}(Q)$$. Since $$f$$ preserves the order, then $${\mathrm{K}}(f)$$ preserves the order. It is immediate that $${\mathrm{K}}(f)$$ preserves the involution. Let $$(a,0)$$ and $$(b,0)$$ be elements such that $$(a,0) {\wedge} (b,0)$$ exists. It follows from Lemma 9 that $$a{\wedge} b$$ exists and $$(a,0) {\wedge} (b,0) = (a{\wedge} b,0)$$. Then, $$f(a){\wedge} f(b)$$ exists and $$f(a{\wedge} b) = f(a){\wedge} f(b)$$. Again by Lemma 9 we obtain that $$(f(a),0){\wedge} (f(b),0)$$ exists and $$(f(a),0){\wedge} (f(b),0) = (f(a){\wedge} f(b),0)$$. Hence, $${\mathrm{K}}(f)((a,0){\wedge} (b,0)) = {\mathrm{K}}(f)(a,0) {\wedge} {\mathrm{K}}(f)(b,0)$$. Therefore, $${\mathrm{K}}(f)$$ is a morphism in $${\mathsf{KP}}$$. ■ Using the previous lemma, it is immediate to see that $${\mathrm{K}}$$ defines a functor from $${\mathsf{P_0}}$$ to $${\mathsf{KP}}$$. Let $$P \in {\mathsf{P_0}}$$. The map $$\alpha_P:P {\rightarrow} {\mathrm{C}}({\mathrm{K}}(P))$$ given by $$\alpha_{P}(a,b) = (a,0)$$ is easily seen to be an isomorphism in $${\mathsf{P_0}}$$. The fact that $$\alpha_P$$ is morphism is a consequence of Lemma 9. The proof of the following lemma is immediate. It easily follows from it that $${\mathrm{C}}: {\mathsf{KP}} {\rightarrow} {\mathsf{P_0}}$$ is a functor. Lemma 14 (a) If $$(T, \leq, {\sim},{\mathrm{c}}) \in {\mathsf{KP}}$$ then $$({\mathrm{C}}(T),\leq,{\mathrm{c}}) \in {\mathsf{P_0}}$$. (b) If $$g \in {\mathsf{KP}}$$ then $${\mathrm{C}}(g) \in {\mathsf{P_0}}$$. Definition 15 For $$T\in {\mathsf{KP}}$$ we also name $$({\mathrm{CK}})$$ to the following condition   \begin{equation} \label{eq-CK-2} (\forall x, y \geq c)(\text{if } x{\wedge} y \text{ exists and } x{\wedge} y = {\mathrm{c}}, \text{ then } (\exists z)(z\vee {\mathrm{c}} = x \ \& \ {\sim} z \vee {\mathrm{c}} = y)). \end{equation} (CK) Remark 16 In Section 2, the condition $$({\mathrm{CK}})$$ was defined for centered Kleene algebras. Notice that if $$(T,{\wedge},\vee,{\sim},{\mathrm{c}},0,1)$$ is a centered Kleene algebra, then the structure $$(T,\leq,{\sim},{\mathrm{c}})$$ is a Kleene poset, where $$\leq$$ is the order associated with the lattice $$(T,{\wedge},\vee)$$. In particular, we have that $$(T,{\wedge},\vee,{\sim},{\mathrm{c}},0,1)$$ satisfies the condition $$({\mathrm{CK}})$$ given in Section 2 if and only if $$(T,\leq,{\sim},{\mathrm{c}})$$ satisfies the condition $$({\mathrm{CK}})$$ given in Definition 15. This fact justifies the use of the same label for both conditions. As in the case of bounded distributive lattices, if $$(P,\leq,0) \in {\mathsf{P_0}}$$, then the structure $$({\mathrm{K}}(P),\preceq,{\sim}, {\mathrm{c}})$$ satisfies $$({\mathrm{CK}})$$. Remark 17 For $$T\in {\mathsf{KP}}$$ and $$x\in T$$ we have $$(x\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}) = {\mathrm{c}}$$, which shows that the map $$\beta_T:T {\rightarrow} {\mathrm{K}}({\mathrm{C}}(T))$$ defined by $$\beta_{T}(x) = (x\vee {\mathrm{c}},{\sim} x \vee {\mathrm{c}})$$ is a well-defined map. We also have that $$T$$ satisfies $$({\mathrm{CK}})$$ if and only if $$\beta_T$$ is surjective. Let $$f:(P,\leq) {\rightarrow} (Q,\leq)$$ be an order isomorphism, i.e., a bijective map such that for every $$a$$, $$b\in P$$, $$a\leq b$$ if and only if $$f(a)\leq f(b)$$. Let $$a$$, $$b\in P$$ such that $$a{\wedge} b$$ exists. Then $$f(a) {\wedge} f(b)$$ exists and $$f(a) {\wedge} f(b) = f(a{\wedge} b)$$. Straightforward computations prove the following remark. Remark 18 Let $$f:T{\rightarrow} U$$ be a morphism in $${\mathsf{KP}}$$. If $$f$$ is an order isomorphism, then $$f$$ is an isomorphism in $${\mathsf{KP}}$$. Lemma 19 If $$T \in {\mathsf{KP}}$$, then $$\beta_T$$ is an injective morphism in $${\mathsf{KP}}$$. Moreover, if $$T$$ satisfies $$({\mathrm{CK}})$$, then $$\beta_T$$ is an isomorphism in $${\mathsf{KP}}$$. Proof. To show that $$\beta_T$$ preserves the order, let $$x,y \in T$$ such that $$x\leq y$$. Then $$x\vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$$ and $${\sim} x \vee {\mathrm{c}} \geq {\sim} y \vee {\mathrm{c}}$$, which means that $$\beta_T(x) \leq \beta_T(y)$$. Thus, $$\beta_T$$ preserves the order. It is immediate that $$\beta_T$$ preserves the involution. Let now $$x,y \in T$$ such that $$x,y \geq {\mathrm{c}}$$. Assume that $$x{\wedge} y$$ exists. So $$\beta_T(x{\wedge} y) = (x{\wedge} y, {\mathrm{c}})$$. Moreover, we have that $$\beta_T(x) = (x,{\mathrm{c}})$$ and $$\beta_T(y) = (y,{\mathrm{c}})$$. Thus, it follows from Lemma 9 that $$(x,c) {\wedge} (y,c)$$ exists and $$(x,{\mathrm{c}}) {\wedge} (y,{\mathrm{c}}) = (x{\wedge} y, {\mathrm{c}})$$. Then $$\beta_T(x {\wedge} y) = \beta_T(x) {\wedge} \beta_T(y)$$. Hence, $$\beta_T$$ is a morphism in $${\mathsf{KP}}$$. Now we will prove that for every $$x,y \in T$$, $$x\leq y$$ if and only if $$\beta_T(x) \leq \beta_T(y)$$. The fact that if $$x\leq y$$, then $$\beta_T(x) \leq \beta_T(y)$$ was proved before. To prove the converse, suppose that $$\beta_T(x) \leq \beta_T(y)$$, i.e., $$x\vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$$ and $$x {\wedge} {\mathrm{c}} \leq y {\wedge} {\mathrm{c}}$$. So, by the definition of Kleene poset we have $$x\leq y$$. In particular, $$\beta_T$$ is an injective map. Finally, assume that $$T$$ satisfies $$({\mathrm{CK}})$$. It follows from remarks 17 and 18 that $$\beta_T$$ is an isomorphism in $${\mathsf{KP}}$$. ■ Straightforward calculations prove that if $$f:P{\rightarrow} Q$$ is a morphism in $${\mathsf{P_0}}$$ then $$ ({\mathrm{C}} \circ {\mathrm{K}})(f) \circ \alpha_{P} = \alpha_{Q} \circ f$$, and if $$g:T {\rightarrow} U$$ is a morphism in $${\mathsf{KP}}$$ then $$ ({\mathrm{K}} \circ {\mathrm{C}})(g) \circ \beta_{T} = \beta_U \circ g$$. Theorem 20 Let $${\mathsf{KP^{CK}}}$$ be the full subcategory of $${\mathsf{KP}}$$ whose objects satisfy the condition $$({\mathrm{CK}})$$. The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{P_0}}$$ and $${\mathsf{KP^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Let $${\mathsf{MS}}$$ be the category whose objects are bounded semilattices and whose morphisms are the algebra homomorphisms. Definition 21 We write $${\mathsf{KMS}}$$ to denote the category whose objects are the structures $$(T, \leq, {\sim},{\mathrm{c}}, 0,1)$$ which satisfy the following conditions: $${\mathrm{(KM1)}}$$$$(T, \leq, {\sim},{\mathrm{c}}) \in {\mathsf{KP}}$$. $${\mathrm{(KM2)}}$$$$0$$ is the first element of $$(T,\leq)$$ and $$1$$ is the greatest element of $$(T,\leq)$$. $${\mathrm{(KM3)}}$$ For every $$x,y \in T$$, if $$x\geq {\mathrm{c}}$$, then $$x{\wedge} y$$ exists. $${\mathrm{(KM4)}}$$ For every $$x,y \in T$$, if $$x\geq {\mathrm{c}}$$, then $$(x{\wedge} y)\vee {\mathrm{c}} = x{\wedge} (y\vee {\mathrm{c}})$$. The morphisms of $${\mathsf{KMS}}$$ are maps $$g$$ between objects of $${\mathsf{KMS}}$$ which preserve the order, the involution and such that for every $$x,y\geq {\mathrm{c}}$$, $$g(x{\wedge} y) = g(x) {\wedge} g(y)$$. We write $${\mathsf{KMS^{CK}}}$$ to denote the full subcategory of $${\mathsf{KMS}}$$ whose objects satisfy the condition $$({\mathrm{CK}})$$. Note that in presence of the condition $${\mathrm{(KM1)}}$$ we can replace the condition $${\mathrm{(KM3)}}$$ by the following condition: $$(x\vee {\mathrm{c}}){\wedge} y$$ exists for every $$x$$, $$y$$. Also note that in presence of the conditions $${\mathrm{(KM1)}}$$ and $${\mathrm{(KM3)}}$$, the condition $${\mathrm{(KM4)}}$$ can be replaced by the condition $$((x\vee {\mathrm{c}}){\wedge} y)\vee {\mathrm{c}} = (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}})$$ for every $$x$$, $$y$$. Recall that $${\mathsf{MS}}$$ is the category whose objects are bounded semilattices and whose morphisms are the algebra homomorphisms between them. Corollary 22 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{MS}}$$ and $${\mathsf{KMS^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Proof. Let $$H\in {\mathsf{MS}}$$. The condition $${\mathrm{(KM1)}}$$ for $${\mathrm{K}}(H)$$ follows from Theorem 20. We also have that $$(0,1)$$ is the first element of $${\mathrm{K}}(H)$$ and $$(1,0)$$ is the last element of $${\mathrm{K}}(H)$$, i.e., we have the condition $${\mathrm{(KM2)}}$$. Let $$x,y \in {\mathrm{K}}(H)$$ with $$x\geq {\mathrm{c}}$$. Then there are $$a,b,d\in H$$ such that $$x = (a,0)$$, $$y = (b,d)$$ and $$b{\wedge} d = 0$$. Since in particular $$a{\wedge} b$$ exists, then it follows from Lemma 9 that $$x {\wedge} y$$ exists and $$x{\wedge} y = (a {\wedge} b,d)$$. Then we have proved $${\mathrm{(KM3)}}$$. Again taking into account Lemma 9 we deduce that $$((a,0){\wedge} (b,d))\vee (0,0) = (a{\wedge} b,0)$$. The mentioned lemma also implies that $$(b,d) \vee (0,0) = (b,0)$$ and $$(a,0) {\wedge} (b,0) = (a{\wedge} b,0)$$. Thus, $$(x{\wedge} y)\vee {\mathrm{c}} = x{\wedge} (y\vee {\mathrm{c}})$$, which is $${\mathrm{(KM4)}}$$. Therefore $${\mathrm{K}}(H) \in {\mathsf{KMS}}$$. It is immediate that if $$T\in {\mathsf{KMS}}$$, then $${\mathrm{C}}(T) \in {\mathsf{MS}}$$. The rest of the proof follows from Theorem 20. ■ 4 The variety of hemi-implicative semilattices (lattices) In this section, we recall definitions and properties about the algebras we will consider later: hemi-implicative semilattices (lattices) [21], Hilbert algebras with infimum [12], implicative semilattices [17] and semi-Heyting algebras [20]. Definition 23 A hemi-implicative semilattice is an algebra $$(H,{\wedge},{\rightarrow}, 1)$$ of type $$(2,2,0)$$ which satisfies the following conditions: $${\mathrm{(W1)}}$$$$(H,{\wedge},1)$$ is an upper bounded semilattice, $${\mathrm{(W2)}}$$ for every $$a,b,d \in H$$, if $$a\leq b{\rightarrow} d$$ then $$a {\wedge} b \leq d$$, $${\mathrm{(W3)}}$$$$a{\rightarrow} a = 1$$ for every $$a\in H$$. A bounded hemi-implicative semilattice is an algebra $$(H,{\wedge},{\rightarrow},0,1)$$ of type $$(2,2,0,0)$$ such that $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice and $$0$$ is the first element with respect to the order. A hemi-implicative lattice is an algebra $$(H,{\wedge},\vee,{\rightarrow},0,1)$$ of type $$(2,2,2,0,0)$$ such that $$(H,{\wedge},\vee,0,1)\in {\mathsf{BDL}}$$ and $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice. Hemi-implicative semilattices were called weak implicative semilattices in [21]. We write $${\mathsf{hIS_{0}}}$$ for the category of bounded hemi-implicative semilattices and $${\mathsf{hBDL}}$$ for the category of hemi-implicative lattices. Remark 24 If $$(H,{\wedge})$$ is a semilattice and $${\rightarrow}$$ a binary operation, then $$H$$ satisfies $${\mathrm{(W2)}}$$ if and only if for every $$a,b \in A$$, $$a{\wedge} (a{\rightarrow} b)\leq b$$. Therefore, the class of hemi-implicative semilattices (lattices) is a variety [21]. The variety of Hilbert algebras is the algebraic counterpart of the implicative fragment of Intuitionistic Propositional Logic. These algebras were introduced in the early 50s by Henkin and Skolem for some investigations on the implication in intuitionistic logic and other non-classical logics [18]. In the 1960s, they were studied especially by Horn and Diego [10]. Definition 25 A Hilbert algebra is an algebra $$(H,{\rightarrow}, 1)$$ of type $$(2,0)$$ that satisfies the following conditions: (1) $$a{\rightarrow} (b{\rightarrow} a) = 1$$. (2) $$a{\rightarrow} (b{\rightarrow} d) = (a{\rightarrow} b){\rightarrow} (a{\rightarrow} d)$$. (3) If $$a{\rightarrow} b = b{\rightarrow} a = 1$$, then $$a = b$$. It is a well-known fact that Hilbert algebras form a variety. In every Hilbert algebra, we have the partial order defined by $$a\leq b$$ if and only if $$a{\rightarrow} b = 1$$. In particular, $$a{\rightarrow} a = 1$$ for every $$a$$. Example 26 In any poset $$(H,\leq)$$ with last element $$1$$ it is possible to define the following binary operation:   \[ a {\rightarrow} b = \begin{cases} 1, &\text{if $a \leq b$;}\\ b, &\text{if $a\nleq b$.} \end{cases} \] The structure $$(H,{\rightarrow} ,1)$$ is a Hilbert algebra. For the following definition see [12]. Definition 27 An algebra $$(H,{\wedge},{\rightarrow},1)$$ is a Hilbert algebra with infimum if the following conditions hold: (1) $$(H,{\rightarrow},1)$$ is a Hilbert algebra, (2) $$(H,{\wedge},1)$$ is an upper bounded semilattice, (3) For every $$a,b \in H$$, $$a\leq b$$ if and only if $$a{\rightarrow} b = 1$$, where $$\leq$$ is the semilattice order. An algebra $$(H,{\wedge},{\rightarrow},0,1)$$ of type $$(2,2,0,0)$$ is a bounded Hilbert algebra with infimum if $$(H,{\wedge},{\rightarrow},1)$$ is a Hilbert algebra with infimum and $$0$$ is the first element with respect to the induced order. In [12] it is proved that the class of Hilbert algebras with infimum is a variety. We note that this result also follows from the results given by P. M. Idziak in [14] for BCK-algebras with lattice operations. The following proposition can be found in [12, Theorem 2.1]. Proposition 28 Let $$(H,{\wedge},{\rightarrow},1)$$ be an algebra of type $$(2,2,0)$$. Then $$(H,{\wedge},{\rightarrow},1)$$ is a Hilbert algebra with infimum if and only if for every $$a,b,d\in H$$ the following conditions hold: (a) $$(H,{\rightarrow},1)$$ is a Hilbert algebra, (b) $$(H,{\wedge},1)$$ is an upper bounded semilattice, (c) $$a{\wedge} (a{\rightarrow} b) = a{\wedge} b$$, (d) $$a {\rightarrow} (b{\wedge} d) \leq (a{\rightarrow} b) {\wedge} (a{\rightarrow} d)$$. In every Hilbert algebra with infimum we have $$a{\rightarrow} a = 1$$ and $$a{\wedge} (a{\rightarrow} b) \leq b$$, so the variety of Hilbert algebras with infimum is a subvariety of the variety of hemi-implicative semilattices. We will write $${\mathsf{Hil_{0}}}$$ for the category whose objects are bounded Hilbert algebras with infimum and whose morphisms are the corresponding algebra homomorphisms. Clearly $${\mathsf{Hil_{0}}}$$ is a full subcategory of $${\mathsf{hIS_{0}}}$$. Definition 29 An implicative semilattice is an algebra $$(H, {\wedge}, {\rightarrow})$$ of type $$(2,2)$$ such that $$(H,{\wedge})$$ is a semilattice, and for every $$a,b,d\in H$$ we have that $$a{\wedge} b \leq d$$ if and only if $$a\leq b{\rightarrow} d$$. Implicative semilattices have a greatest element, denoted by $$1$$. In this article we shall include the constant $$1$$ in the language of the algebras. Implicative semilattices are the algebraic models of the implication-conjunction fragment of Intuitionistic Propositional Logic. For more details about these algebras see [9]. An algebra $$(H,{\wedge},{\rightarrow},0,1)$$ of type $$(2,2,0,0)$$ is a bounded implicative semilattice if $$(H,{\wedge},{\rightarrow},1)$$ is an implicative semilattice and $$0$$ is the first element with respect to the order. We write $${\mathsf{IS_{0}}}$$ for the category whose objects are bounded implicative semilattices and whose morphisms are the corresponding algebra homomorphisms. We have that $${\mathsf{IS_{0}}}$$ is a full subcategory of $${\mathsf{hIS_{0}}}$$. It is part of the folklore of the subject that the class of implicative semilattices is a variety. There are many ways to axiomatize the variety of implicative semilattices. In the following lemma we propose a possible axiomatization that will play an important role in the next section. Lemma 30 Let $$(H,{\wedge},1)$$ be an upper bounded semilattice and $${\rightarrow}$$ a binary operation on $$H$$. The following conditions are equivalent: (a) For every $$a,b,d \in H$$, $$a\leq b{\rightarrow} d$$ if and only if $$a{\wedge} b \leq d$$. (b) For every $$a,b,d \in H$$ the following conditions hold: (1) $$a{\wedge} (a{\rightarrow} b) \leq b$$, (2) $$a{\rightarrow} a = 1$$, (3) $$a{\rightarrow} (b{\wedge} d) = (a{\rightarrow} b) {\wedge} (a{\rightarrow} d)$$, (4) $$a\leq b {\rightarrow} (a{\wedge} b)$$. Proof. Assume the conditions (1), (2), (3) and (4) of (b). It follows from (1) that if $$a\leq b{\rightarrow} d$$, then $$a{\wedge} b \leq d$$. Suppose now that $$a{\wedge} b \leq d$$. It follows from (3) that $$b{\rightarrow} (a{\wedge} b) \leq b{\rightarrow} d$$. But by (4) we have that $$a\leq b {\rightarrow} (a{\wedge} b)$$, so $$a\leq b{\rightarrow} d$$. Then, $$a\leq b{\rightarrow} d$$ if and only if $$a{\wedge} b \leq d$$. For the converse of this property see [17]. ■ Remark 31 A moment of reflection shows that implicative semilattices are Hilbert algebras with infimum where the implication is the right residuum of the infimum, or equivalently, where the following equation holds [12]: $$a\leq b {\rightarrow} (a{\wedge} b)$$. Alternatively, it follows from Lemma 30 that an implicative semilattice is a hemi-implicative semilattice which satisfies $$a{\rightarrow} (b{\wedge} d) = (a{\rightarrow} b) {\wedge} (a{\rightarrow} d)$$ and $$a\leq b {\rightarrow} (a{\wedge} b)$$ for every $$a,b,d$$. Semi-Heyting algebras were introduced by H. P. Sankappanavar in [20] as an abstraction of Heyting algebras. These algebras share with Heyting algebras the following properties: they are pseudocomplemented and distributive lattices and their congruences are determined by the lattice filters. Definition 32 An algebra $$(H, {\wedge}, \vee, {\rightarrow}, 0, 1)$$ of type $$(2,2,2,0,0)$$ is a semi-Heyting algebra if the following conditions hold for every $$a,b,d$$ in $$H$$: $${\mathrm{(SH1)}}$$$$(H, {\wedge}, \vee, 0, 1)\in {\mathsf{BDL}}$$, $${\mathrm{(SH2)}}$$$$a{\wedge} (a{\rightarrow} b) = a {\wedge} b$$, $${\mathrm{(SH3)}}$$$$a{\wedge} (b{\rightarrow} d) = a {\wedge} ((a{\wedge} b) {\rightarrow} (a{\wedge} d))$$, $${\mathrm{(SH4)}}$$$$a{\rightarrow} a = 1$$. We write $${\mathsf{SH}}$$ for the category of semi-Heyting algebras. A semi-Heyting algebra can be seen as a hemi-implicative lattice which satisfies $${\mathrm{(SH2)}}$$ and $${\mathrm{(SH3)}}$$. Therefore, $${\mathsf{SH}}$$ is a full subcategory of $${\mathsf{hBDL}}$$. Remark 33 Implicative semilattices satisfy the inequality $$a\leq b {\rightarrow} (a{\wedge} b)$$ because $$a{\wedge} b \leq a{\wedge} b$$, or simply by Lemma 30. Semi-Heyting algebras also satisfy the inequality $$a\leq b {\rightarrow} (a{\wedge} b)$$. This fact follows from $${\mathrm{(SH3)}}$$ and $${\mathrm{(SH4)}}$$ in the following way:   \[ \begin{array} [c]{lllll} a {\wedge} (b {\rightarrow} (a{\wedge} b)) & = & a{\wedge} ((a{\wedge} b) {\rightarrow} (a{\wedge} b)) & & \\ & = & a{\wedge} 1& & \\ & = & a,& & \end{array} \] which means that $$a\leq b {\rightarrow} (a{\wedge} b)$$. In the following example we will show the following facts: $${\mathsf{Hil_{0}}}$$ is a proper subvariety of $${\mathsf{hIS_{0}}}$$ and $${\mathsf{SH}}$$ is a proper subvariety of $${\mathsf{hBDL}}$$. Example 34 Let $$H$$ be the chain of three elements with $$0<a<1$$. We define on $$H$$ the following binary operation:   \[ \begin{array} [c]{c|ccc} {\rightarrow} & 0 & a & 1\\\hline 0 & 1 & a & 1\\ a & 0 & 1 & 1\\ 1 & 0 & 0 & 1 \end{array} \] Straightforward computations show that $$(H, {\wedge}, \vee,{\rightarrow}, 0,1) \in {\mathsf{hBDL}}$$. In particular, $$(H,{\wedge},{\rightarrow},0,1) \in {\mathsf{hIS_{0}}}$$. However, $$(H,{\wedge},{\rightarrow},0,1) \notin {\mathsf{Hil_{0}}}$$ and $$(H,{\wedge},\vee,{\rightarrow},0,1) \notin {\mathsf{SH}}$$ because $$1{\wedge} (1{\rightarrow} a) \neq 1{\wedge} a$$. It is a known fact that the variety $${\mathsf{IS_{0}}}$$ is properly included in $${\mathsf{Hil_{0}}}$$. We give an example that shows it. Let $$H$$ be the universe of the boolean lattice of four elements, where $$a$$ and $$b$$ are the atoms. Then $$(H,{\wedge},{\rightarrow},0,1) \in {\mathsf{Hil_{0}}}$$, where $${\rightarrow}$$ is the operation defined in Example 26. Since $$a{\rightarrow} 0 = 0$$ and $$0\neq b$$, then $$(H,{\wedge},{\rightarrow},0,1) \notin {\mathsf{IS_{0}}}$$. It is also a known fact that $${\mathsf{HA}}$$ is a proper subvariety of $${\mathsf{SH}}$$. We also provide an example that shows it. Consider the chain of two elements. We define the following binary operation:   \[ \begin{array} [c]{c|cc} {\rightarrow} & 0 & 1\\\hline 0 & 1 & 0\\ 1 & 0 & 1 \end{array} \] Then $$(H,{\wedge},\vee,{\rightarrow},0,1)\in {\mathsf{SH}}$$. Since $$0{\rightarrow} 1 = 0$$ and $$0 \neq 1$$, then $$(H,{\wedge},\vee,{\rightarrow},0,1)$$ is not a Heyting algebra. The following diagrams show the relations among the categories defined in this section:   \begin{equation*} \begin{array}{cc} {\mathsf{hIS_{0}}} & {\mathsf{hBDL}}\\ {\Large{|}} & {\Large{|}}\\ {\mathsf{Hil_{0}}} & {\mathsf{SH}}\\ {\Large{|}} & {\Large{|}}\\ {\mathsf{IS_{0}}} & {\mathsf{HA}}\\ \end{array} \end{equation*} In [5] an extension of Kalman’s functor was studied for the variety of algebras with implication $$(H,{\wedge},\vee, {\rightarrow}, 0,1)$$ which satisfy $$a{\wedge} (a{\rightarrow} b)\leq b$$ for every $$a$$, $$b$$. 5 Kalman’s construction for $${\mathsf{hIS_{0}}}$$ and $${\mathsf{hBDL}}$$ The fact that Kalman’s construction can be extended consistently to Heyting algebras led us to believe that some of the picture could be lifted to the varieties $${\mathsf{hIS_{0}}}$$ and $${\mathsf{hBDL}}$$. More precisely, it arises the natural question of whether is it possible to find some category $${\mathsf{KhIS_{0}}}$$ to obtain an equivalence between $${\mathsf{hIS_{0}}}$$ and some full subcategory of $${\mathsf{KhIS_{0}}}$$, making the following diagram commute: Similarly, it arises the question of whether is it possible to find some category $${\mathsf{KhBDL}}$$ to obtain an equivalence between $${\mathsf{hBDL}}$$ and certain full subcategory of $${\mathsf{KhBDL}}$$, making the following diagram commute: In this section, we answer these questions in the positive. Moreover, we extend Kalman’s functor to the categories $${\mathsf{Hil_{0}}}$$, $${\mathsf{IS_{0}}}$$ and $${\mathsf{SH}}$$. The aim of Section 3 was to obtain a categorical equivalence between $${\mathsf{MS}}$$ and $${\mathsf{KMS^{CK}}}$$ (Corollary 22) to be applied in the present section to the category $${\mathsf{hIS_{0}}}$$. For the case of the category $${\mathsf{hBDL}}$$ we will also use Theorem 1, which establishes an equivalence between $${\mathsf{BDL}}$$ and $${\mathsf{KA_{{\mathrm{c}}}^{CK}}}$$. 5.1 Kalman’s construction for $${\mathsf{hIS_{0}}}$$ Let $$H\in {\mathsf{hIS_{0}}}$$. We write $${\rightarrow}$$ for the implication of $$H$$ and define a binary operation on $${\mathrm{K}}(H)$$ (also denoted $${\rightarrow}$$) by   \begin{equation} \label{eqi1} (a,b) {\rightarrow} (d,e): = ((a{\rightarrow} d) {\wedge} (e{\rightarrow} b), a{\wedge} e). \end{equation} (5) This definition is motivated by Remark 6. Note that since $$a{\wedge} b = d{\wedge} e = 0$$, then $$(a{\rightarrow} d){\wedge} (e{\rightarrow} b) {\wedge} a{\wedge} e = 0$$ because $$a{\wedge} (a{\rightarrow} d) \leq d$$ and $$d{\wedge} e = 0$$. Hence, $$(a,b) {\rightarrow} (d,e) \in {\mathrm{K}}(H)$$. The next definition is motivated by the original Kalman’s construction. Definition 35 We denote by $${\mathsf{KhIS_{0}}}$$ the category whose objects are the structures $$(T, \leq, {\sim}, {\rightarrow}, {\mathrm{c}}, 0,1)$$ such that $$(T, \leq, {\sim},{\mathrm{c}}, 0,1)\in {\mathsf{KMS}}$$ and $${\rightarrow}$$ is a binary operation on $$T$$ which satisfies the following conditions for every $$x,y \in T$$: $${\mathrm{(K1)}}$$$${\mathrm{c}} \leq x {\rightarrow} (y\vee {\mathrm{c}})$$, $${\mathrm{(K2)}}$$$$x {\wedge} ((x\vee {\mathrm{c}}){\rightarrow} (y\vee {\mathrm{c}}))\leq y \vee {\mathrm{c}}$$, $${\mathrm{(K3)}}$$$$x {\rightarrow} x = 1$$, $${\mathrm{(K4)}}$$$$(x{\rightarrow} y){\wedge} {\mathrm{c}} = ({\sim} x {\wedge} {\mathrm{c}}) \vee (y{\wedge} {\mathrm{c}})$$, $${\mathrm{(K5)}}$$$$(x{\rightarrow} {\sim} y) \vee {\mathrm{c}} = ((x\vee {\mathrm{c}}) {\rightarrow}({\sim} y \vee {\mathrm{c}})) {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} ({\sim} x \vee {\mathrm{c}}))$$. The morphisms of $${\mathsf{KhIS_{0}}}$$ are the morphisms $$g$$ of $${\mathsf{KMS}}$$ which satisfy the condition $$g(x{\rightarrow} y) = g(x){\rightarrow} g(y)$$ for every $$x$$, $$y$$. In what follows we will prove that if $$H \in {\mathsf{hIS_{0}}}$$, then $${\mathrm{K}}(H) \in {\mathsf{KhIS_{0}}}$$, where the binary operation $${\rightarrow}$$ in $${\mathrm{K}}(H)$$ is that defined in (5). Proposition 36 Let $$H \in {\mathsf{hIS_{0}}}$$. Then $${\mathrm{K}}(H) \in {\mathsf{KhIS_{0}}}$$. Furthermore, $${\mathrm{K}}$$ extends to a functor from $${\mathsf{hIS_{0}}}$$ to $${\mathsf{KhIS_{0}}}$$, which we also denote by $${\mathrm{K}}$$. Proof. Throughout this proof we use Lemmas 9 and 11. Recall that $$c = (0,0)$$. Let $$(a,b), (d,e) \in {\mathrm{K}}(H)$$. In particular, $$(d,e) \vee {\mathrm{c}} = (d,0)$$. Then   \[ \begin{array} [c]{lllll} (a,b){\rightarrow} ((d,e) \vee {\mathrm{c}})& = & (a,b){\rightarrow} (d,0)& & \\ & = & ((a{\rightarrow} d){\wedge} (0{\rightarrow} b),0)& & \\ & \succeq & {\mathrm{c}}.& & \end{array} \] Thus we have proved the condition $${\mathrm{(K1)}}$$. To prove $${\mathrm{(K2)}}$$ we make the following computation:   \[ \begin{array} [c]{lllll} (a,b) {\wedge} (((a,b) \vee {\mathrm{c}}) {\rightarrow} ((d,e) \vee {\mathrm{c}}))& = &(a,b) {\wedge} ((a,0){\rightarrow} (d,0))& & \\ & = & (a,b) {\wedge} (a{\rightarrow} d, 0)& & \\ & = & (a{\wedge} (a{\rightarrow} d),b) & &\\ & \preceq & (d,b) & & \\ & \preceq & (d,0) & & \\ & = & (d,e) \vee {\mathrm{c}}. & & \\ \end{array} \] The proof of the condition $${\mathrm{(K3)}}$$ is immediate. To prove $${\mathrm{(K4)}}$$, note that $$((a,b){\rightarrow} (d,e)){\wedge} {\mathrm{c}} = (0,a{\wedge} e)$$ and   \[ \begin{array} [c]{lllll} ({\sim} (a,b){\wedge} {\mathrm{c}}) \vee ((d,e) {\wedge} {\mathrm{c}})& = &(0,a) \vee (0,e)& & \\ & = & (0,a{\wedge} e).& & \end{array} \] Hence, we have that   \[ ((a,b){\rightarrow} (d,e)){\wedge} {\mathrm{c}} = ({\sim} (a,b){\wedge} {\mathrm{c}}) \vee ((d,e) {\wedge} {\mathrm{c}}). \] Finally we shall prove $${\mathrm{(K5)}}$$. First note that   \[ \begin{array} [c]{lllll} ((a,b) {\rightarrow} {\sim}(d,e)) \vee {\mathrm{c}}& = &((a,b) {\rightarrow} (e,d))\vee {\mathrm{c}}& & \\ & = & ((a{\rightarrow} e){\wedge} (d{\rightarrow} b),0).& & \end{array} \] Then,   \begin{equation} \label{K5on} ((a,b) {\rightarrow} {\sim}(d,e)) \vee {\mathrm{c}} = ((a{\rightarrow} e){\wedge} (d{\rightarrow} b),0). \end{equation} (6) On the other hand,   \[ \begin{array} [c]{lllll} ((a,0){\rightarrow} (e,0)){\wedge} ((d,0){\rightarrow} (b,0)) & = &(a{\rightarrow} e,0) {\wedge} (d{\rightarrow} b,0)& & \\ & = & ((a{\rightarrow} e){\wedge} (d{\rightarrow} b),0).& & \end{array} \] Hence,   \begin{equation} \label{K5tw} ((a,0){\rightarrow} (e,0)){\wedge} ((d,0){\rightarrow} (b,0)) = ((a{\rightarrow} e){\wedge} (d{\rightarrow} b),0). \end{equation} (7) Since $$(a,b) \vee {\mathrm{c}} = (a,0)$$, $${\sim}(d,e) \vee {\mathrm{c}} = (e,0)$$, $$(d,e) \vee {\mathrm{c}} = (d,0)$$, and $${\sim}(a,b) \vee {\mathrm{c}} = (b,0)$$, then it follows from (6) and (7) that the condition $${\mathrm{(K5)}}$$ holds. Thus, $${\mathrm{K}}(H)\in {\mathsf{KhIS_{0}}}$$. Let $$f:H{\rightarrow} G$$ be a morphism in $${\mathsf{hIS_{0}}}$$. Straightforward computations show that $${\mathrm{K}}(f)$$ preserves the implication operation, which implies that $${\mathrm{K}}(f)$$ is a morphism in $${\mathsf{KhIS_{0}}}$$. ■ Proposition 37 If $$(T, \leq, {\sim}, {\rightarrow}, {\mathrm{c}}, 0,1)\in {\mathsf{KhIS_{0}}}$$, then $$({\mathrm{C}}(T),{\wedge},{\rightarrow},{\mathrm{c}},1) \in {\mathsf{hIS_{0}}}$$. Furthermore, $${\mathrm{C}}$$ extends to a functor from $${\mathsf{KhIS_{0}}}$$ to $${\mathsf{hIS_{0}}}$$, which we also denote by $${\mathrm{C}}$$. Proof. We have that $${\mathrm{C}}(T)$$ is closed under the operation $${\rightarrow}$$. To prove it, let $$x,y \geq {\mathrm{c}}$$. By $${\mathrm{(K1)}}$$ we have that   \[ \begin{array} [c]{lllll} {\mathrm{c}} & \leq &(x\vee {\mathrm{c}}) {\rightarrow} (y\vee {\mathrm{c}})& & \\ & = & x {\rightarrow} y,& & \end{array} \] so $$x{\rightarrow} y \in {\mathrm{C}}(T)$$. Thus, the restriction of $${\rightarrow}$$ to $${\mathrm{C}}(T)$$ is indeed an operation on $${\mathrm{C}}(T)$$. Let $$x,y \geq {\mathrm{c}}$$. It follows from $${\mathrm{(K2)}}$$ that $$x{\wedge} (x{\rightarrow} y)\leq y$$ and it follows from $${\mathrm{(K3)}}$$ that $$x{\rightarrow} x = 1$$. Then $$({\mathrm{C}}(T),{\wedge},{\rightarrow},{\mathrm{c}},1) \in {\mathsf{hIS_{0}}}$$. The rest of the proof is immediate. ■ Remark 38 For $$H\in {\mathsf{hIS_{0}}}$$ we have that $$\alpha_{H}(a{\rightarrow} b) = \alpha_{H}(a) {\rightarrow} \alpha_{H}(b)$$ for every $$a$$, $$b\in H$$. Moreover, $$\alpha_H$$ is an isomorphism in $${\mathsf{hIS_{0}}}$$. For the case of $$T\in {\mathsf{KhIS_{0}}}$$ we will prove that $$\beta_T$$ preserves the implication. Lemma 39 Let $$T\in {\mathsf{KhIS_{0}}}$$. Then $$\beta_T$$ is injective and a morphism in $${\mathsf{KhIS_{0}}}$$. Moreover, if $$T$$ satisfies $$({\mathrm{CK}})$$ then $$\beta_T$$ is an isomorphism in $${\mathsf{KhIS_{0}}}$$. Proof. We need to prove that $$\beta_T(x{\rightarrow} y) = \beta_T(x) {\rightarrow} \beta_T(y)$$ for every $$x,y$$. In an equivalent way, we need to prove that $$\beta_T(x{\rightarrow} {\sim} y) = \beta_T(x) {\rightarrow} \beta_T({\sim} y)$$ for every $$x,y$$. It follows from $${\mathrm{(K4)}}$$ and $${\mathrm{(K5)}}$$ that   \[ \begin{array} [c]{lllll} \beta_T(x{\rightarrow} {\sim} y) & = & ((x{\rightarrow} {\sim} y) \vee {\mathrm{c}},{\sim} (x{\rightarrow} {\sim} y) \vee {\mathrm{c}})) & & \\ & = & (((x\vee {\mathrm{c}}) {\rightarrow}({\sim} y \vee {\mathrm{c}})) {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} ({\sim} x \vee {\mathrm{c}})), (x\vee {\mathrm{c}}) {\wedge} (y \vee {\mathrm{c}})) & & \\ & = & \beta_T(x) {\rightarrow} \beta_T({\sim} y).& & \end{array} \] ■ We write $${\mathsf{KhIS_{0}^{CK}}}$$ for the full subcategory of $${\mathsf{KhIS_{0}}}$$ whose objects satisfy $$({\mathrm{CK}})$$. The proof of the following theorem follows from Corollary 22, Proposition 36, Proposition 37, Remark 38 and Lemma 39. Theorem 40 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{hIS_{0}}}$$ and $${\mathsf{KhIS_{0}^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Let $$T \in {\mathsf{KhIS_{0}}}$$. We define the following condition for every $$x,y \in T$$: $${\mathrm{(K6)}}$$$$x \leq (y\vee {\mathrm{c}}) {\rightarrow} ((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}))$$. The next lemma is the motivation to consider the condition $${\mathrm{(K6)}}$$. Lemma 41 If $$H\in {\mathsf{hIS_{0}}}$$ satisfies the inequality $$a\leq b {\rightarrow} (a{\wedge} b)$$ for every $$a, b$$, then $${\mathrm{K}}(H)$$ satisfies $${\mathrm{(K6)}}$$. Proof. First note that for every $$(a,b) \in {\mathrm{K}}(H)$$, $$(a,b) \vee {\mathrm{c}} = (a,0)$$. To prove $${\mathrm{(K6)}}$$ we make the following computation:   \[ \begin{array} [c]{lllll} (d,0) {\rightarrow} ((a,0){\wedge} (d,0)) & = & (d,0) {\rightarrow} (a{\wedge} d,0) & & \\ & = & (d {\rightarrow} (a{\wedge} d),0)& & \\ & \succeq & (a,b).& & \end{array} \] Hence, we obtain $${\mathrm{(K6)}}$$. ■ The following lemma will play an important role in this article. Lemma 42 If $$T \in {\mathsf{KhIS_{0}}}$$ satisfies $${\mathrm{(K6)}}$$, then $$T$$ satisfies $$({\mathrm{CK}})$$. Proof. Let $$x,y \geq {\mathrm{c}}$$ such that $$x{\wedge} y = {\mathrm{c}}$$. Taking into account $${\mathrm{(KM3)}}$$ we can define $$z = (y {\rightarrow} {\sim} y) {\wedge} x$$. It follows from $${\mathrm{(K4)}}$$ that   \[ \begin{array} [c]{lllll} z{\wedge} {\mathrm{c}} & = & ((y{\rightarrow} {\sim} y) {\wedge} {\mathrm{c}}){\wedge} x & & \\ & = & (({\sim} y {\wedge} {\mathrm{c}}) \vee ({\sim} y {\wedge} {\mathrm{c}})) {\wedge} x& & \\ & = & ({\sim} y {\wedge} {\mathrm{c}}) {\wedge} x& & \\ & = & {\sim} y {\wedge} x& & \\ & = & {\sim} y.& & \end{array} \] Hence, $${\sim} z \vee {\mathrm{c}} = y$$. In order to prove that $$z\vee {\mathrm{c}} = x$$, we use the conditions $${\mathrm{(KM3)}}$$, $${\mathrm{(KM4)}}$$, $${\mathrm{(K5)}}$$, $${\mathrm{(K6)}}$$ and the fact that $$x{\wedge} y = {\mathrm{c}}$$ as follows:   \[ \begin{array} [c]{lllll} z\vee {\mathrm{c}} & = & (x{\wedge} (y{\rightarrow} {\sim} y))\vee {\mathrm{c}} & & \\ & = & x{\wedge} ((y{\rightarrow} {\sim} y) \vee {\mathrm{c}}) & &\\ & = & ((y\vee {\mathrm{c}}) {\rightarrow} ({\sim} y \vee {\mathrm{c}})){\wedge} x& & \\ & = & (x\vee {\mathrm{c}}) {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} {\mathrm{c}})& & \\ & = & (x \vee {\mathrm{c}}) {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} (x{\wedge} y)) & & \\ & = & (x \vee {\mathrm{c}}) {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} ((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}))) & &\\ & = & x \vee {\mathrm{c}} & &\\ & = & x. & & \end{array} \] Therefore, $$z\vee {\mathrm{c}} = x$$. ■ 5.2 Kalman’s construction for $${\mathsf{Hil_{0}}}$$ We write $${\mathsf{KHil_{0}}}$$ for the full subcategory of $${\mathsf{KhIS_{0}}}$$ whose objects satisfy the following conditions for every $$x,y,z$$: $${\mathrm{(KHil1)}}$$$$(x\vee {\mathrm{c}}){\rightarrow} (y{\rightarrow} (x\vee {\mathrm{c}})) = 1$$, $${\mathrm{(KHil2)}}$$$$x {\rightarrow} ((y\vee {\mathrm{c}}) {\rightarrow} (z\vee {\mathrm{c}})) = (x {\rightarrow} (y\vee {\mathrm{c}})) {\rightarrow} (x{\rightarrow} (z \vee {\mathrm{c}}))$$, $${\mathrm{(KHil3)}}$$ If $$x{\rightarrow} y = y {\rightarrow} x = 1$$, then $$x = y$$, $${\mathrm{(KHil4)}}$$$$x{\wedge} ((x\vee {\mathrm{c}}){\rightarrow} (y\vee {\mathrm{c}})) = x{\wedge} (y\vee {\mathrm{c}})$$, $${\mathrm{(KHil5)}}$$$$x{\rightarrow} ((y\vee {\mathrm{c}}) {\wedge} (z\vee {\mathrm{c}})) \leq (x{\rightarrow} (y\vee {\mathrm{c}})){\wedge} (x{\rightarrow} (z\vee {\mathrm{c}}))$$. Example 43 In every centered Kleene algebra $$(T,{\wedge},\vee,{\sim},{\mathrm{c}},0,1)$$ it is possible to define a binary operation, that we denote by $${\rightarrow}$$, as follows:   \[ x {\rightarrow} y = \begin{cases} 1, &\text{if $x \vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$ and $x{\wedge} {\mathrm{c}} \leq y{\wedge} {\mathrm{c}}$;}\\ {\sim} x \vee (y{\wedge} {\mathrm{c}}), &\text{if $x \vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$ and $x{\wedge} {\mathrm{c}} \nleq y{\wedge} {\mathrm{c}} $;}\\ y \vee ({\sim} x {\wedge} {\mathrm{c}}), &\text{if $x \vee {\mathrm{c}} \nleq y \vee {\mathrm{c}}$ and $x{\wedge} {\mathrm{c}} \leq y{\wedge} {\mathrm{c}}$;}\\ ((y\vee {\mathrm{c}}){\wedge} {\sim} x) \vee (({\sim} x \vee {\mathrm{c}}){\wedge} y), &\text{if $x\vee {\mathrm{c}} \nleq y\vee {\mathrm{c}}$ and $x{\wedge} {\mathrm{c}} \nleq y{\wedge} {\mathrm{c}}$.} \end{cases} \] It is possible to prove that $$(T,\leq, {\sim},{\mathrm{c}},0,1)\in {\mathsf{KMS}}$$, and it is not difficult to see that $$(T,{\sim},{\rightarrow},{\mathrm{c}},0,1) \in {\mathsf{KHil_{0}}}$$. By endowing the centered Kleene algebra given in [5, Example 2.5] with the binary operation $${\rightarrow}$$ just defined we obtain an example of an object of $${\mathsf{KHil_{0}}}$$ which does not satisfy the condition $$({\mathrm{CK}})$$. Lemma 44 (a) If $$H\in {\mathsf{Hil_{0}}}$$, then $${\mathrm{K}}(H) \in {\mathsf{KHil_{0}}}$$. (b) If $$T\in {\mathsf{KHil_{0}}}$$, then $${\mathrm{C}}(T) \in {\mathsf{Hil_{0}}}$$. Proof. Let $$H\in {\mathsf{Hil_{0}}}$$ and take $$(a,b)$$, $$(d,e)$$ and $$(f,g)$$ in $${\mathrm{K}}(H)$$. In what follows we will use Proposition 28. Taking into account that $$a{\rightarrow} (d{\rightarrow} a) = 1$$ we obtain   \[ \begin{array} [c]{lllll} (a,0) {\rightarrow} ((d,e) {\rightarrow} (a,0)) & = & (a,0) {\rightarrow} (d{\rightarrow} a,0) & & \\ & = & (a {\rightarrow} (d{\rightarrow} a) ,0)& & \\ & = & (1,0),& & \end{array} \] which is the condition $${\mathrm{(KHil1)}}$$. Since $$a{\rightarrow} (d{\rightarrow} f) = (a{\rightarrow} d) {\rightarrow} (a{\rightarrow} f)$$, then   \[ \begin{array} [c]{lllll} (a,b) {\rightarrow} ((d,0) {\rightarrow} (f,0)) & = & (a,b) {\rightarrow} (d{\rightarrow} f,0) & & \\ & = & (a{\rightarrow} (d{\rightarrow} f), 0)& & \\ & = & ((a{\rightarrow} d) {\rightarrow} (a {\rightarrow} f), 0)& & \\ & = & (a{\rightarrow} d,0) {\rightarrow} (a {\rightarrow} f), 0)& & \\ & = & ((a,b) {\rightarrow} (d,0)) {\rightarrow} ((a,b){\rightarrow} (f,0)).& & \end{array} \] Hence, we have proved $${\mathrm{(KHil2)}}$$. In order to prove $${\mathrm{(KHil3)}}$$ suppose that $$(a,b) {\rightarrow} (d,e) = (d,e) {\rightarrow} (a,b) = (1,0)$$, so $$a{\rightarrow} d = d{\rightarrow} a = 1$$ and $$b{\rightarrow} e = e {\rightarrow} b = 1$$. Then $$a = d$$ and $$b = e$$, i.e., $$(a,b) = (d,e)$$, which was our aim. The condition $${\mathrm{(KHil4)}}$$ is a consequence of the equality $$a{\wedge} (a{\rightarrow} d) = a{\wedge} d$$. Indeed,   \[ \begin{array} [c]{lllll} (a,b) {\wedge} ((a,0) {\rightarrow} (d,0)) & = & (a,b) {\wedge} (a{\rightarrow} d,0) & & \\ & = & (a{\wedge} (a{\rightarrow} d), b)& & \\ & = & (a{\wedge} d,b)& & \\ & = & (a,b) {\wedge} (d,0).& & \end{array} \] Finally, we will prove $${\mathrm{(KHil5)}}$$. By the condition $$a{\rightarrow} (d{\wedge} f) \leq (a{\rightarrow} d){\wedge} (a{\rightarrow} f)$$ we have that   \[ \begin{array} [c]{lllll} (a,b) {\rightarrow} ((d,0){\wedge} (f,0)) & = & (a,b) {\rightarrow} (d{\wedge} f,0) & & \\ & = & (a {\rightarrow} (d{\wedge} f), 0)& & \\ & \preceq & ((a{\rightarrow} d){\wedge} (a{\rightarrow} f), 0)& & \\ & = & (a{\rightarrow} d, 0) {\wedge} (a{\rightarrow} f,0)& & \\ & = & ((a,b) {\rightarrow} (d,0)) {\wedge} ((a,b) {\rightarrow} (f,0)).& & \end{array} \] Then $${\mathrm{K}}(H) \in {\mathsf{KHil_{0}}}$$. Finally, it follows from Proposition 28 that if $$T\in {\mathsf{KHil_{0}}}$$, then $${\mathrm{C}}(T) \in {\mathsf{Hil_{0}}}$$ ■ We write $${\mathsf{KHil_{0}^{CK}}}$$ for the full subcategory of $${\mathsf{KHil_{0}}}$$ whose objects satisfy $$({\mathrm{CK}})$$. The following corollary follows from Theorem 40 and Lemma 44. Corollary 45 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{Hil_{0}}}$$ and $${\mathsf{KHil_{0}^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. 5.3 Kalman’s construction for $${\mathsf{IS_{0}}}$$ We write $${\mathsf{KIS_{0}}}$$ for the full subcategory of $${\mathsf{KhIS_{0}}}$$ whose objects $$T$$ satisfy the condition $${\mathrm{(K6)}}$$ and the following additional condition for every $$x,y \in T$$: $${\mathrm{(K7)}}$$$$x {\rightarrow} ((y\vee {\mathrm{c}}) {\wedge} (z\vee {\mathrm{c}})) = (x {\rightarrow} (y\vee {\mathrm{c}})){\wedge} (x {\rightarrow} (z\vee {\mathrm{c}}))$$. Lemma 46 (a) If $$H\in {\mathsf{IS_{0}}}$$, then $${\mathrm{K}}(H) \in {\mathsf{KIS_{0}}}$$. (b) If $$T\in {\mathsf{KIS_{0}}}$$, then $${\mathrm{C}}(T) \in {\mathsf{IS_{0}}}$$. Proof. Let $$H\in {\mathsf{IS_{0}}}$$. The fact that $${\mathrm{K}}(H)$$ satisfies $${\mathrm{(K6)}}$$ follows from Lemmas 30 and 41. On the other hand, it follows from Lemma 30 that   \[ \begin{array} [c]{lllll} (a,b){\rightarrow} ((d,0){\wedge} (f,0)) & = & (a,b) {\rightarrow} (d{\wedge} f,0) & & \\ & = & (a {\rightarrow} (d{\wedge} f),0)& & \\ & = & ((a{\rightarrow} d){\wedge} (a{\rightarrow} f),0)& & \\ & = & ((a,b){\rightarrow} (d,0)){\wedge} ((a,b){\rightarrow} (f,0)).& & \end{array} \] Thus, we have the condition $${\mathrm{(K7)}}$$. Then $${\mathrm{K}}(H) \in {\mathsf{KIS_{0}}}$$. The fact that if $$T\in {\mathsf{KIS_{0}}}$$, then $${\mathrm{C}}(T) \in {\mathsf{IS_{0}}}$$ is also consequence of Lemma 30. ■ The following corollary follows from Theorem 40, Lemma 42 and Lemma 46. Corollary 47 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{IS_{0}}}$$ and $${\mathsf{KIS_{0}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Since $${\mathsf{IS_{0}}}$$ is a full subcategory of $${\mathsf{Hil_{0}}}$$, it follows from corollaries 45 and 47 that $${\mathsf{KIS_{0}}}$$ is a full subcategory of $${\mathsf{KHil_{0}^{CK}}}$$. 5.4 Kalman’s construction for $${\mathsf{hBDL}}$$ In what follows we define a category which will be related with the category $${\mathsf{hBDL}}$$. Definition 48 We write $${\mathsf{KhBDL}}$$ for the category whose objects are the algebras $$(T,{\wedge},\vee,$$$${\rightarrow}, {\sim},{\mathrm{c}},0,1)$$ of type $$(2,2,2,1,0,0,0)$$ such that $$(T,{\wedge},\vee, {\sim},{\mathrm{c}},0,1)\in {\mathsf{KA_{{\mathrm{c}}}}}$$ and the conditions $${\mathrm{(K1)}}$$, $${\mathrm{(K2)}}$$, $${\mathrm{(K3)}}$$, $${\mathrm{(K4)}}$$ and $${\mathrm{(K5)}}$$ are satisfied. The morphisms of the category are the corresponding algebra homomorphisms. By the Example 43, in every centered Kleene algebra $$(T,{\wedge},\vee,0,{\mathrm{c}},0,1)$$ we can define a binary operation $${\rightarrow}$$ such that $$(T,{\wedge},\vee,{\rightarrow},{\mathrm{c}},0,1) \in {\mathsf{KhBDL}}$$. In particular, if $$(T,{\wedge},\vee,0, {\mathrm{c}},0,1)$$ is the centered Kleene algebra given in [5, Example 2.5], then $$(T,{\wedge},\vee,{\rightarrow},{\mathrm{c}},0,1)\in {\mathsf{KhBDL}}$$, where $${\rightarrow}$$ is the implication considered in Example 43. It is immediate that $$(T,{\wedge},\vee,{\mathrm{c}},0,1)$$ does not satisfy the condition ($${\mathrm{CK}}$$). Note that $$(H,{\wedge},\vee,{\rightarrow}, 0,1) \in {\mathsf{hBDL}}$$ if and only if $$(H,{\wedge},\vee,0,1) \in {\mathsf{BDL}}$$ and $$(H,{\wedge},{\rightarrow},0,1) \in {\mathsf{hIS_{0}}}$$. Also note that $$(T,{\wedge},\vee,{\rightarrow}, {\sim},{\mathrm{c}},0,1) \in {\mathsf{KhBDL}}$$ if and only if $$(T,{\wedge},\vee, {\sim},{\mathrm{c}},0,1)\in {\mathsf{KA_{{\mathrm{c}}}}}$$ and $$(T, \leq, {\sim}, {\rightarrow}, {\mathrm{c}}, 0,1) \in {\mathsf{KhIS_{0}}}$$. We write $${\mathsf{KhBDL^{CK}}}$$ for the full subcategory of $${\mathsf{KhBDL}}$$ whose objects satisfy $$({\mathrm{CK}})$$. Theorem 49 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{hBDL}}$$ and $${\mathsf{KhBDL^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Proof. It follows from Theorem 1 and Theorem 40. ■ 5.5 Kalman’s construction for $${\mathsf{SH}}$$ We write $${\mathsf{KSH}}$$ for the full subcategory of $${\mathsf{KhBDL}}$$ whose objects satisfy the condition $${\mathrm{(KHil4)}}$$ and the following additional condition: $${\mathrm{(KSH3)}}$$$$x {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} (z\vee {\mathrm{c}})) = x {\wedge}(((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}})){\rightarrow} ((x\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}})))$$. Lemma 50 (a) If $$H\in {\mathsf{SH}}$$, then $${\mathrm{K}}(H) \in {\mathsf{KSH}}$$. (b) If $$T\in {\mathsf{KSH}}$$, then $${\mathrm{C}}(T) \in {\mathsf{SH}}$$. (c) If $$T \in {\mathsf{KSH}}$$, then $$T$$ satisfies $${\mathrm{(K6)}}$$. In particular, $$T$$ satisfies $$({\mathrm{CK}})$$. Proof. Let $$H\in {\mathsf{SH}}$$. The condition $${\mathrm{(KHil4)}}$$ follows from $${\mathrm{(SH2)}}$$ (see proof of Lemma 44). Let $$(a,b)$$, $$(d,e)$$ and $$(f,g)$$ in $${\mathrm{K}}(H)$$. Taking into account $${\mathrm{(SH3)}}$$ we have that   \[ \begin{array} [c]{lllll} (a,b){\wedge} ((d,0){\rightarrow} (f,0)) & = & (a,b) {\wedge} (d{\rightarrow} f,0) & & \\ & = & (a {\wedge} (d{\rightarrow} f),b)& & \\ & = & (a {\wedge} ((a{\wedge} d){\rightarrow} (a{\wedge} f)),b)& & \\ & = & (a,b) {\wedge} ((a{\wedge} d){\rightarrow} (a{\wedge} f),0)& & \\ & = & (a,b) {\wedge} ((a{\wedge} d,0) {\rightarrow} (a{\wedge} f,0))& & \\ & = & (a,b) {\wedge} (((a,0){\wedge} (d,0)) {\rightarrow} ((a,0){\wedge} (f,0))),& & \end{array} \] which is the condition $${\mathrm{(KSH3)}}$$. Then $${\mathrm{K}}(H) \in {\mathsf{KSH}}$$. It is immediate that if $$T\in {\mathsf{KSH}}$$ then $${\mathrm{C}}(T) \in {\mathsf{SH}}$$. To prove that $$T$$ satisfies $${\mathrm{(K6)}}$$ we will use $${\mathrm{(K3)}}$$ and $${\mathrm{(KSH3)}}$$ as follows:   \[ \begin{array} [c]{lllll} x{\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} ((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}))) & = & x{\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} (((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}})) \vee {\mathrm{c}}) & & \\ & = & x {\wedge} (((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}})) {\rightarrow} ((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}}))) & & \\ & = & x{\wedge} 1& & \\ & = & x.& & \end{array} \] Then $$x\leq (y\vee {\mathrm{c}}) {\rightarrow} ((x\vee {\mathrm{c}}) {\rightarrow} (y\vee {\mathrm{c}}))$$, i.e., the condition $${\mathrm{(K6)}}$$. Therefore, it follows from Lemma 42 that $$T$$ satisfies ($${\mathrm{CK}}$$). ■ Theorem 51 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{SH}}$$ and $${\mathsf{KSH}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Proof. It follows from Theorem 49 and Lemma 50. ■ 6 Well-behaved congruences in $${\mathsf{KhIS_{0}}}$$ and congruences in $${\mathsf{KhBDL}}$$ In this section, we introduce the concept of the well-behaved congruences over objects of $${\mathsf{KhIS_{0}}}$$. They are equivalence relations with some additional properties. We will prove that if $$T\in {\mathsf{KhIS_{0}}}$$ and $$\theta$$ is a well-behaved congruence on $$T$$, then it is possible to define on the quotient $$T/\theta$$ a partial order and operations so that $$T/\theta \in {\mathsf{KhIS_{0}}}$$. For $$T\in {\mathsf{KhIS_{0}}}$$ we study the relation between the well-behaved congruences of $$T$$ and the congruences of $${\mathrm{C}}(T)$$, and in particular for the cases where $$T\in {\mathsf{KHil_{0}}}$$ or $$T \in {\mathsf{KIS_{0}}}$$. For $$T\in {\mathsf{KhBDL}}$$ or $$T\in {\mathsf{KSH}}$$ we also study the relation between the congruences of $$T$$ and the congruences of $${\mathrm{C}}(T)$$. Finally, we study the principal well-behaved congruences of the objects in $${\mathsf{KhIS_{0}}}$$, $${\mathsf{KHil_{0}}}$$ and $${\mathsf{KIS_{0}}}$$ and the principal congruences of the objects in $${\mathsf{KhBDL}}$$ and $${\mathsf{KSH}}$$. We start by fixing notation and giving some useful definitions. Let $$X$$ be a set, $$x \in X$$ and $$\theta$$ an equivalence relation on $$X$$. We write $$x/\theta$$ to indicate the equivalence class of $$x$$ associated with the equivalence relation $$\theta$$, and $$X/\theta$$ to indicate the quotient set of $$X$$ associated with $$\theta$$ (i.e. the set of equivalence classes). If $$T$$ is an algebra, we write $${\mathrm{Con}}(T)$$ to denote the set of as well as the lattice of congruences of $$T$$. Definition 52 Let $$T\in {\mathsf{KhIS_{0}}}$$. We say that an equivalence relation $$\theta$$ of $$T$$ is a well-behaved congruence of $$T$$ if it satisfies the following conditions: $${\mathrm{(C1)}}$$$$\theta \in {\mathrm{Con}}((T,{\rightarrow},{\sim}))$$. $${\mathrm{(C2)}}$$ For $$x,y \in T$$, $$(x,y) \in \theta$$ if and only if $$(x\vee {\mathrm{c}},y\vee {\mathrm{c}}) \in \theta$$ and $$({\sim} x \vee {\mathrm{c}}, {\sim} y \vee {\mathrm{c}}) \in \theta$$. $${\mathrm{(C3)}}$$ For $$x$$, $$y$$, $$z$$ and $$w$$ in $${\mathrm{C}}(T)$$, if $$(x,y)\in \theta$$ and $$(z,w) \in \theta$$, then $$(x{\wedge} z,y{\wedge} w) \in \theta$$. Note that the intersection of any family of well-behaved congruences of $$T\in {\mathsf{KhIS_{0}}}$$ is a well-behaved congruence; therefore the set of well-behaved congruences of $$T$$ ordered by the inclusion relation is a complete lattice. Remark 53 The definition of well-behaved congruence can be also given for algebras of $${\mathsf{KhBDL}}$$. In this case, if $$T\in {\mathsf{KhBDL}}$$, then every congruence of $$T$$ is a well-behaved congruence. In what follows we define a binary relation in $$T/\theta$$, where $$T\in {\mathsf{KhIS_{0}}}$$ and $$\theta$$ is a well-behaved congruence of $$T$$. Definition 54 Let $$T \in {\mathsf{KhIS_{0}}}$$. If $$\theta$$ is a well-behaved congruence of $$T$$, then we define in $$T/\theta$$ the following binary relation $$\ll_{\theta}$$ by:   \[ x/\theta \ll_{\theta} y/\theta\; \textrm{if and only if}\; ((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta\; \textrm{and}\; (({\sim} y \vee {\mathrm{c}}){\wedge} ({\sim} x \vee {\mathrm{c}}), {\sim} y \vee {\mathrm{c}}) \in \theta. \] If there is no ambiguity, we write $$\ll$$ in place of $$\ll_{\theta}$$. Note that the definition given is good, in the sense that it is independent of the elements selected as representativess of the equivalence classes. To show it, suppose that $$x/\theta \ll y/\theta$$. Let $$z\in x/\theta$$ and $$w\in y/\theta$$. Then by $${\mathrm{(C2)}}$$ we have that $$(x\vee {\mathrm{c}}, z\vee {\mathrm{c}}) \in \theta$$, $$({\sim} x\vee {\mathrm{c}}, {\sim} z\vee {\mathrm{c}}) \in \theta$$, $$(y\vee {\mathrm{c}}, w\vee {\mathrm{c}}) \in \theta$$, and $$({\sim} z\vee {\mathrm{c}}, {\sim} w\vee {\mathrm{c}}) \in \theta$$. Hence it follows from $${\mathrm{(C3)}}$$ that   \[ ((z\vee {\mathrm{c}}){\wedge} (w\vee {\mathrm{c}}), (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}})) \in \theta. \] Since, by the assumption, $$((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta$$, and $$(x\vee {\mathrm{c}},z\vee {\mathrm{c}}) \in \theta$$, then   \[ ((z\vee {\mathrm{c}}){\wedge} (w\vee {\mathrm{c}}), z\vee {\mathrm{c}}) \in \theta. \] In a similar way it can be proved that $$(({\sim} z\vee {\mathrm{c}}){\wedge} ({\sim} w\vee {\mathrm{c}}), \sim w\vee {\mathrm{c}}) \in \theta$$. Remark 55 Let $$T\in {\mathsf{KA_{{\mathrm{c}}}}}$$ and $$\theta \in {\mathrm{Con}}(T)$$. Since the class of centered Kleene algebras is a variety, then $$T/\theta \in {\mathsf{KA_{{\mathrm{c}}}}}$$. In particular, the lattice order $$\leq$$ of $$T/\theta$$ is given by $$x/\theta \leq y/\theta$$ if and only if $$x/\theta = (x{\wedge} y)/\theta$$. In this framework the relation $$\ll$$ given in Definition 54 coincides with the relation $$\leq$$, i.e.,   \[ x/\theta \leq y/\theta\;\textrm{if and only if}\; x/\theta \ll y/\theta. \] To prove it note first that from the distributivity of the underlying lattice of $$T$$ it follows that $$x/\theta \leq y/\theta$$ if and only if $$(x \vee {\mathrm{c}}, (x{\wedge} y)\vee {\mathrm{c}}) \in \theta$$ and $$(x{\wedge} {\mathrm{c}}, (x{\wedge} y){\wedge} {\mathrm{c}}) \in \theta$$. Besides, we have that $$(x{\wedge} y) \vee {\mathrm{c}} = (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}})$$. Since $$\theta$$ preserves the involution, then $$(x{\wedge} {\mathrm{c}}, (x{\wedge} y) {\wedge} {\mathrm{c}}) \in \theta$$ if and only if $$({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\sim} y \vee {\mathrm{c}}) \in \theta$$. Therefore   \begin{equation} \label{oi1} x/\theta \leq y/\theta \ \textrm{if and only if} \ ((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta \ \textrm{and} \ ({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\sim} y \vee {\mathrm{c}}) \in \theta. \end{equation} (8) We also have   \begin{equation} \label{oi2} ({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\sim} y \vee {\mathrm{c}}) \in \theta \ \textrm{if and only if} \ (({\sim} x \vee {\mathrm{c}}) {\wedge} ({\sim} y \vee {\mathrm{c}}), {\sim y} \vee {\mathrm{c}})\in \theta. \end{equation} (9) To prove (9), suppose that $$\;({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\sim} y \vee {\mathrm{c}}) \in \theta$$. Since $$({\sim} y \vee {\mathrm{c}},{\sim} y \vee {\mathrm{c}})\in \theta$$, then taking $${\wedge}$$ we obtain that $$(({\sim} x \vee {\mathrm{c}}) {\wedge} ({\sim} y \vee {\mathrm{c}}), {\sim y} \vee {\mathrm{c}})\in \theta$$. Conversely, assume that $$(({\sim} x \vee {\mathrm{c}}) {\wedge} ({\sim} y \vee {\mathrm{c}}), {\sim y} \vee {\mathrm{c}})\in \theta$$. Since $$({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\mathrm{c}}) \in \theta$$, then taking $$\vee$$ we obtain that $$({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\sim} y \vee {\mathrm{c}}) \in \theta$$, so $$({\sim} x \vee {\sim} y \vee {\mathrm{c}}, {\sim} y \vee {\mathrm{c}}) \in \theta$$. Then we have proved (9). Therefore, it follows from (8) and (9) that $$x/\theta \leq y/\theta$$ if and only if $$x/\theta \ll y/\theta$$. Lemma 56 Let $$T \in {\mathsf{KhIS_{0}}}$$ and $$\theta$$ a well-behaved congruence of $$T$$. Then $$(T,\ll)$$ is a poset. Proof. Let $$\theta$$ be a well-behaved congruence of $$T$$. The reflexivity of $$\theta$$ implies the reflexivity of $$\ll$$. To prove that $$\ll$$ is antisymmetric, let $$x, y\in T$$ be such that $$x/\theta \ll y/ \theta$$ and $$y/\theta \ll x/\theta$$, which means that   \[ ((x\vee {\mathrm{c}}) {\wedge} (y{\wedge} {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta, \]   \[ (({\sim} y \vee {\mathrm{c}}){\wedge} ({\sim} x \vee {\mathrm{c}}), {\sim} y \vee {\mathrm{c}}) \in \theta, \]   \[ ((y\vee {\mathrm{c}}) {\wedge} (x\vee {\mathrm{c}}), y\vee {\mathrm{c}}) \in \theta, \]   \[ (({\sim} x \vee {\mathrm{c}}){\wedge} ({\sim} y \vee {\mathrm{c}}), {\sim} x \vee {\mathrm{c}}) \in \theta. \] Since $$(x\vee {\mathrm{c}}, (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}})) \in \theta$$ and $$((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), y\vee {\mathrm{c}}) \in \theta$$, then $$(x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \theta$$. Analogously we have that $$({\sim} x \vee {\mathrm{c}}, {\sim} y \vee {\mathrm{c}}) \in \theta$$. Hence, it follows from $${\mathrm{(C2)}}$$ that $$(x,y)\in \theta$$, i.e., $$x/\theta = y/\theta$$. We conclude that $$\ll$$ is antisymmetric. Finally we will prove that $$\ll$$ is transitive. Let $$x$$, $$y$$ and $$z$$ be elements of $$T$$ such that $$x/\theta \ll y/\theta$$ and $$y/\theta \ll z/\theta$$. In particular,   \begin{equation} \label{EQ1} ((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}}), x\vee {\mathrm{c}})\in \theta, \end{equation} (10)  \begin{equation} \label{EQ2} ((y\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}}), y\vee {\mathrm{c}}) \in \theta. \end{equation} (11) It follows from (10) and $${\mathrm{(C3)}}$$ that   \begin{equation} \label{EQ3} ((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}}), (x\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}})) \in \theta, \end{equation} (12) and it follows from (11) and $${\mathrm{(C3)}}$$ that   \begin{equation} \label{EQ4} ((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}}), (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}})) \in \theta. \end{equation} (13) Hence, by (12) and (13) we obtain that $$((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), (x\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}})) \in \theta$$. Thus, taking into account (10) we have $$((x\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta$$. Similarly we can show that $$(({\sim} z \vee {\mathrm{c}}){\wedge} ({\sim} x \vee {\mathrm{c}}), {\sim} z \vee {\mathrm{c}}) \in \theta$$. Thus, $$x/\theta \ll z/\theta$$. Hence, $$\ll$$ is transitive. ■ Lemma 57 Let $$T\in {\mathsf{KhIS_{0}}}$$ and $$x,y \in T$$. If $$x\leq y$$, then $$x/\theta \ll y/\theta$$. Proof. Let $$x\leq y$$. Then $${\sim} y \leq {\sim} x$$. Hence, we have $$x\vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$$ and $${\sim} y \vee {\mathrm{c}} \leq {\sim} x \vee {\mathrm{c}}$$, i.e., $$(x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}) = x\vee {\mathrm{c}}$$ and $$({\sim} y \vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}) = {\sim} y \vee {\mathrm{c}}$$. Since $$\theta$$ is a reflexive relation, then $$((x\vee {\mathrm{c}}) {\wedge} (y{\wedge} {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta$$ and $$(({\sim} y \vee {\mathrm{c}}){\wedge} ({\sim} x \vee {\mathrm{c}}), {\sim} y \vee {\mathrm{c}}) \in \theta$$, i.e., $$x/\theta \ll y/\theta$$. ■ For $$T\in {\mathsf{KhIS_{0}}}$$ and $$\theta$$ a well-behaved congruence of $$T$$, we have in particular that $$\theta$$ is a congruence of $$(T,{\sim},{\rightarrow})$$. Let us use also the symbols $${\sim}$$ and $${\rightarrow}$$ to refer to the respective induced operations on $$T/\theta$$. Proposition 58 Let $$(T,\leq,{\sim},{\rightarrow},{\mathrm{c}},0,1)\in {\mathsf{KhIS_{0}}}$$ and $$\theta$$ a well-behaved congruence of $$T$$. Then $$(T/\theta,\ll,{\sim},{\rightarrow}, {\mathrm{c}}/\theta, 0/\theta, 1/\theta) \in {\mathsf{KhIS_{0}}}$$. Proof. Step 1. $$(T/\theta,\ll,{\sim},{\mathrm{c}}/\theta) \in {\mathsf{KP}}$$. It follows from Lemma 56 that $$(T/\theta,\ll)$$ is a poset. It is immediate that $${\sim}$$ is an involution in $$(T/\theta,\ll)$$ which is order reversing and that $${\sim}{\mathrm{c}}/\theta = {\mathrm{c}}/\theta$$. Let $$x\in T$$. In what follows we will prove that the supremum of $$x/\theta$$ and $${\mathrm{c}}/\theta$$ with respect to the order $$\ll$$ exists in $$T/\theta$$, and we will denote it by $$x/\theta \vee {\mathrm{c}}/\theta$$. Moreover, we will prove that $$x/\theta \vee {\mathrm{c}}/\theta = (x\vee {\mathrm{c}})/\theta$$. First note that if $$y\in x/\theta$$, then it follows from $${\mathrm{(C2)}}$$ that $$(x\vee {\mathrm{c}},y\vee {\mathrm{c}}) \in \theta$$, i.e., that $$(x\vee {\mathrm{c}})/\theta = (y\vee {\mathrm{c}})/\theta$$. Now we will show that $$x/\theta \vee {\mathrm{c}}/\theta$$ exists. Since $$x\leq x\vee {\mathrm{c}}$$ and $${\mathrm{c}}\leq x \vee {\mathrm{c}}$$, it follows from Lemma 57 that $$x/\theta \ll (x\vee {\mathrm{c}})/\theta$$ and $${\mathrm{c}}/\theta \ll (x\vee {\mathrm{c}})/\theta$$. Let $$z\in T$$ be such that $$x/\theta \ll z/\theta$$ and $${\mathrm{c}}/\theta\ll z/\theta$$. Then $$((x\vee {\mathrm{c}}){\wedge}(z\vee {\mathrm{c}}), x\vee {\mathrm{c}})\in \theta$$, $$(({\sim}z \vee {\mathrm{c}}){\wedge}({\sim}x \vee {\mathrm{c}}), {\sim}z \vee {\mathrm{c}}) \in \theta$$ and $$(c,{\sim} z \vee {\mathrm{c}})\in \theta$$. We need to prove that $$(x\vee {\mathrm{c}})/\theta \ll z/\theta$$. By the previous assertions we have in particular that   \begin{equation} \label{sup1} (((x\vee {\mathrm{c}})\vee {\mathrm{c}}){\wedge}(z\vee {\mathrm{c}}), (x\vee {\mathrm{c}}) \vee {\mathrm{c}})\in \theta. \end{equation} (14) On the other hand,   \[ ({\sim}z \vee {\mathrm{c}}){\wedge} ({\sim}(x\vee {\mathrm{c}})\vee {\mathrm{c}}) = {\mathrm{c}}. \] But $$(c,{\sim}z \vee {\mathrm{c}}) \in \theta$$, so   \begin{equation} \label{sup2} (({\sim}z \vee {\mathrm{c}}){\wedge} ({\sim}(x\vee {\mathrm{c}})\vee {\mathrm{c}}), {\sim}z \vee {\mathrm{c}}) \in \theta. \end{equation} (15) Hence, it follows from (14) and (15) that $$(x\vee {\mathrm{c}})/\theta \ll z/\theta$$. Thus, $$x/\theta \vee {\mathrm{c}}/\theta$$ exists and $$x/\theta \vee {\mathrm{c}}/\theta = (x\vee {\mathrm{c}})/\theta$$. In what follows we will prove that for every $$x,y \in T$$,   \[ (x/\theta\vee {\mathrm{c}}/\theta){\wedge} ({\sim}x/\theta \vee {\mathrm{c}}/\theta) = {\mathrm{c}}/\theta, \] or, equivalently, that   \begin{equation} \label{center} (x\vee {\mathrm{c}})/\theta {\wedge} ({\sim} x \vee {\mathrm{c}})/\theta = {\mathrm{c}}/\theta, \end{equation} (16) where we also use $${\wedge}$$ for the infimum with respect to $$\ll$$. To prove (16), note that it follows from Lemma 57 that $$c/\theta \ll (x\vee {\mathrm{c}})/\theta$$ and $${\mathrm{c}}/\theta \ll ({\sim} x \vee {\mathrm{c}})/\theta$$. Let $$z\in T$$ such that $$z/\theta \ll (x\vee {\mathrm{c}})/\theta$$ and $$z/\theta \ll ({\sim}x \vee {\mathrm{c}})/\theta$$. In particular,   \begin{equation} \label{sup3} ((z\vee {\mathrm{c}}) {\wedge} (x\vee {\mathrm{c}}), z \vee {\mathrm{c}}) \in \theta, \end{equation} (17)  \begin{equation} \label{sup4} ((z\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}), z\vee {\mathrm{c}}) \in \theta. \end{equation} (18) It follows from $${\mathrm{(C3)}}$$, (17) and (18) that   \begin{equation} ((z\vee {\mathrm{c}}) {\wedge} (x\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}) , z \vee {\mathrm{c}}) \in \theta. \end{equation} (19) Since $$(x\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}) = {\mathrm{c}}$$, then $$({\mathrm{c}}, z\vee {\mathrm{c}}) \in \theta$$, i.e., $$z/\theta \ll {\mathrm{c}}/\theta$$. Therefore, $$(x/\theta\vee {\mathrm{c}}/\theta){\wedge} ({\sim}x/\theta \vee {\mathrm{c}}/\theta) = {\mathrm{c}}/\theta$$. For $$x,y\in T$$ assume that $$(x\vee {\mathrm{c}})/\theta \ll (y\vee {\mathrm{c}})/\theta$$ and $$(x{\wedge} {\mathrm{c}})/ \theta \ll (y{\wedge} {\mathrm{c}})/\theta$$. It is immediate that $$x/\theta \ll y/\theta$$. Then we conclude that $$(T/\theta,\ll,{\sim},{\mathrm{c}}/\theta) \in {\mathsf{KP}}$$. Step 2. $$(T/\theta,\ll,{\sim},{\mathrm{c}}/\theta,0/\theta, 1/\theta) \in {\mathsf{KMS}}$$. Since for every $$x\in T$$ we have $$0\leq x \leq 1$$, it follows from Lemma 57 that $$0/\theta \ll x/\theta \ll 1/\theta$$, i.e., $$0/\theta$$ is the first element of $$(T/\theta,\ll)$$ and $$1/\theta$$ is the last element of $$(T/\theta,\ll)$$. Let $$x$$ and $$y$$ be elements of $$T$$. Recall that it follows from $${\mathrm{(KM3)}}$$ that $$(x\vee {\mathrm{c}}) {\wedge} y$$ exists. We will prove that $$(x \vee {\mathrm{c}})/\theta {\wedge} y/\theta$$ exists and is $$((x\vee {\mathrm{c}}){\wedge} y)/\theta$$. To do it, we will prove first that if $$(x,z) \in \theta$$ and $$(y,w) \in \theta$$, then $$((x\vee {\mathrm{c}}){\wedge} y)/\theta = ((z\vee {\mathrm{c}}){\wedge} w)/\theta$$. Let $$(x,z) \in \theta$$ and $$(y,w) \in \theta$$. It follows from $${\mathrm{(C2)}}$$ that $$(x\vee {\mathrm{c}},z\vee {\mathrm{c}}) \in \theta$$ and $$(y\vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta$$. By $${\mathrm{(C3)}}$$ we have that   \begin{equation} \label{EQ5} ((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}}),(z\vee {\mathrm{c}}) {\wedge} (w\vee {\mathrm{c}}))\in \theta. \end{equation} (20) Taking into account $${\mathrm{(KM4)}}$$ we also have   \begin{equation} \label{EQ6} ((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}} = (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), \end{equation} (21)  \begin{equation} \label{EQ7} ((z\vee {\mathrm{c}}){\wedge} w) \vee {\mathrm{c}} = (z\vee {\mathrm{c}}){\wedge} (w\vee {\mathrm{c}}). \end{equation} (22) Hence, it follows from (20), (21) and (22) that   \begin{equation} \label{EQ7-1} (((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}, ((z\vee {\mathrm{c}}){\wedge} w) \vee {\mathrm{c}})\in \theta. \end{equation} (23) In a similar way, taking into account that $$({\sim} x\vee {\mathrm{c}},{\sim} z\vee {\mathrm{c}}) \in \theta$$ and $$({\sim} y\vee {\mathrm{c}},{\sim} w\vee {\mathrm{c}}) \in \theta$$ we have   \begin{equation} \label{EQ7-2} ((({\sim} x\vee {\mathrm{c}}){\wedge} {\sim} y) \vee {\mathrm{c}}, (({\sim} z\vee {\mathrm{c}}){\wedge} {\sim} w) \vee {\mathrm{c}})\in \theta. \end{equation} (24) Then it follows from (23), (24) and $${\mathrm{(C2)}}$$ that   \[ ((x\vee {\mathrm{c}}){\wedge} y, (z\vee {\mathrm{c}}){\wedge} w)\in \theta. \] Now we will prove that $$(x/\theta \vee {\mathrm{c}}/\theta) {\wedge} y/\theta$$ exists and is $$((x\vee {\mathrm{c}}){\wedge} y)/\theta$$. This is equivalent to prove that $$(x \vee {\mathrm{c}})/\theta {\wedge} y/\theta$$ exists and is $$((x\vee {\mathrm{c}}){\wedge} y)/\theta$$. Since $$(x\vee {\mathrm{c}}){\wedge} y \leq x\vee {\mathrm{c}}$$ and $$(x\vee {\mathrm{c}}){\wedge} y \leq y$$, then it follows from Lemma 57 that $$((x\vee {\mathrm{c}}){\wedge} y)/\theta \ll (x\vee {\mathrm{c}})/\theta$$ and $$((x\vee {\mathrm{c}}){\wedge} y)/\theta \ll y/\theta$$. Let $$z\in T$$ be such that $$z/\theta \ll (x\vee {\mathrm{c}})/\theta$$ and $$z/\theta \ll y/\theta$$. In particular,   \begin{equation} \label{EQ8} ((z\vee {\mathrm{c}}){\wedge} (x\vee {\mathrm{c}}), z\vee {\mathrm{c}})\in \theta, \end{equation} (25)  \begin{equation} \label{EQ9} ((z\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), z\vee {\mathrm{c}})\in \theta, \end{equation} (26)  \begin{equation} \label{EQ10} (({\sim} y\vee {\mathrm{c}}){\wedge} ({\sim} z\vee {\mathrm{c}}), {\sim} y\vee {\mathrm{c}})\in \theta. \end{equation} (27) We need to prove that $$z/\theta \ll ((x\vee {\mathrm{c}}){\wedge} y)/\theta$$, which means that   \begin{equation} \label{EQ11} ((z\vee {\mathrm{c}}){\wedge} (((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}), z\vee {\mathrm{c}}) \in \theta \end{equation} (28) and   \begin{equation} \label{EQ12} ({\sim}((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}) {\wedge} ({\sim z} \vee {\mathrm{c}}), {\sim}((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}) \in \theta. \end{equation} (29) It follows from $${\mathrm{(KM4)}}$$ that   \begin{equation} \label{EQ12-1} (z\vee {\mathrm{c}}){\wedge} (((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}) = (z\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}){\wedge} (x\vee {\mathrm{c}}), \end{equation} (30) and it follows from (25) and $${\mathrm{(C3)}}$$ that   \begin{equation} \label{EQ12-2} ((z\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}){\wedge} (x\vee {\mathrm{c}}), (z\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}))\in \theta. \end{equation} (31) Thus, by (26), (30) and (31) we obtain (28). It is immediate that the condition (29) is equal to the condition (27) because   \[ {\sim}((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}) {\wedge} ({\sim z} \vee {\mathrm{c}}) = ({\sim} y\vee {\mathrm{c}}){\wedge} ({\sim} z\vee {\mathrm{c}}), \]   \[ {\sim}((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}} = {\sim} y\vee {\mathrm{c}}. \] Then $$(T/\theta,\ll,{\sim},{\mathrm{c}}/\theta,0/\theta, 1/\theta)$$ satisfies $${\mathrm{(KM3)}}$$. The condition $${\mathrm{(KM4)}}$$ follows from the previous steps and from the same condition on $$T$$. In consequence, we obtain that $$(T/\theta,\ll,{\sim},{\mathrm{c}}/\theta,0/\theta, 1/\theta)\in {\mathsf{KMS}}$$. Step 3. $$(T/\theta,\ll,{\sim},{\rightarrow},{\mathrm{c}}/\theta,0/\theta, 1/\theta) \in {\mathsf{KhIS_{0}}}$$. The other conditions to be an object of $${\mathsf{KhIS_{0}}}$$ follow from the previous steps, the fact that $$T\in {\mathsf{KhIS_{0}}}$$ and Lemma 57. ■ In what follows we will study the lattice of well-behaved congruences of any object of $${\mathsf{KhIS_{0}}}$$. We start with some preliminary definitions. Let $$T\in {\mathsf{KhIS_{0}}}$$. Recall that it follows from previous results of this article that $${\mathrm{C}}(T) \in {\mathsf{hIS_{0}}}$$. Note that $$T$$ does not necessarily satisfy the condition ($${\mathrm{CK}}$$). We write $${\mathrm{Con_{wb}}}(T)$$ to refer both to the set and to the lattice of well-behaved congruences of $$T$$. For $$\theta \in {\mathrm{Con_{wb}}}(T)$$ we define the binary relation $$\Gamma(\theta)$$ on $${\mathrm{C}}(T)$$ as the restriction of $$\theta$$ to $${\mathrm{C}}(T) \times {\mathrm{C}}(T)$$. For $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$ we define the relation $$\Sigma{(\tau)} \subseteq T \times T$$ in the following way:   \[ \text{$(x,y) \in \Sigma(\tau)$ if and only if $(x\vee {\mathrm{c}},y\vee {\mathrm{c}}) \in \tau$ and $({\sim} x\vee {\mathrm{c}},{\sim} y\vee {\mathrm{c}}) \in \tau$.} \] We prove that $$\Sigma(\tau)$$ is a well behaved congruence of $$T$$. Lemma 59 Let $$T\in {\mathsf{KhIS_{0}}}$$ and $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$. Then $$\Sigma(\tau) \in {\mathrm{Con_{wb}}}(T)$$. Proof. Let $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$. Straightforward computations show that $$\Sigma(\tau)$$ satisfies $${\mathrm{(C2)}}$$. To show that $$\Sigma(\tau)$$ satisfies $${\mathrm{(C3)}}$$, let $$x$$, $$y$$, $$z$$ and $$w$$ in $${\mathrm{C}}(T)$$ be such that $$(x,y) \in \Sigma(\tau)$$ and $$(z,w) \in \Sigma(\tau)$$, which means that $$(x,y) \in \tau$$ and $$(z,w) \in \tau$$. Then $$(x{\wedge} z,y{\wedge} w)\in \tau$$, because $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$. But $$(x{\wedge} z) \vee {\mathrm{c}} = x{\wedge} z$$ and $$(y{\wedge} w) \vee {\mathrm{c}} = y{\wedge} w$$. Thus,   \[ ((x{\wedge} y) \vee {\mathrm{c}}, (z{\wedge} w)\vee {\mathrm{c}}) \in \tau. \] On the other hand, since $${\sim} (x{\wedge} z) \vee {\mathrm{c}} = {\mathrm{c}}$$ and $${\sim} (y{\wedge} w) \vee {\mathrm{c}} = {\mathrm{c}}$$, then   \[ ({\sim} (x{\wedge} z) \vee {\mathrm{c}}, {\sim} (y{\wedge} w) \vee {\mathrm{c}}) \in \tau. \] Hence, $$(x{\wedge} z, z{\wedge} w) \in \Sigma(\tau)$$, so the condition $${\mathrm{(C3)}}$$ holds. Now we show the condition $${\mathrm{(C1)}}$$. It is immediate that $$\Sigma(\tau)$$ is congruence with respect to $${\sim}$$. To prove that $$\Sigma(\tau)$$ is congruence with respect to $${\rightarrow}$$, let $$(x,y) \in \Sigma(\tau)$$ and $$(z,w) \in \Sigma(\tau)$$, so   \begin{equation} \label{eq1} (x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \tau, \end{equation} (32)  \begin{equation} \label{eq2} (z\vee {\mathrm{c}}, w\vee {\mathrm{c}}) \in \tau, \end{equation} (33)  \begin{equation} \label{eq3} ({\sim} x\vee {\mathrm{c}}, {\sim} y\vee {\mathrm{c}}) \in \tau, \end{equation} (34)  \begin{equation} \label{eq4} ({\sim} z\vee {\mathrm{c}}, {\sim} w \vee {\mathrm{c}}) \in \tau. \end{equation} (35) Then taking $${\wedge}$$ in (32) and (35) we have   \begin{equation} \label{eq5} ((x\vee {\mathrm{c}}) {\wedge} ({\sim} z \vee {\mathrm{c}}), (y\vee {\mathrm{c}}){\wedge} ({\sim} w \vee {\mathrm{c}})) \in \tau. \end{equation} (36) But it follows from $${\mathrm{(K4)}}$$ that $${\sim}(x{\rightarrow} z) \vee {\mathrm{c}} = (x\vee {\mathrm{c}}) {\wedge} ({\sim} z \vee {\mathrm{c}})$$ and $${\sim}(y {\rightarrow} w) \vee {\mathrm{c}} = (y\vee {\mathrm{c}}){\wedge} ({\sim} w \vee {\mathrm{c}})$$. So by (36) we obtain that   \begin{equation}\label{eq6} ({\sim}(x{\rightarrow} z) \vee {\mathrm{c}}, {\sim}(y{\rightarrow} w) \vee {\mathrm{c}}) \in \tau. \end{equation} (37) On the other hand, taking $${\rightarrow}$$ between (32) and (33) we have that   \begin{equation}\label{eq7} ((x\vee {\mathrm{c}}) {\rightarrow} (z\vee {\mathrm{c}}), (y\vee {\mathrm{c}}) {\rightarrow} (w\vee {\mathrm{c}}))\in \tau, \end{equation} (38) and taking $${\rightarrow}$$ between (35) and (34) we obtain   \begin{equation}\label{eq8} (({\sim} z\vee {\mathrm{c}}) {\rightarrow} ({\sim} x\vee {\mathrm{c}}), ({\sim} w\vee {\mathrm{c}}) {\rightarrow} ({\sim} y\vee {\mathrm{c}}))\in \tau. \end{equation} (39) Define now the following elements:   \[ t := ((x\vee {\mathrm{c}}){\rightarrow} (z\vee {\mathrm{c}})){\wedge} (({\sim} z\vee {\mathrm{c}}) {\rightarrow} ({\sim} x \vee {\mathrm{c}})), \]   \[ u := ((y\vee {\mathrm{c}}){\rightarrow} (w\vee {\mathrm{c}})){\wedge} (({\sim} w\vee {\mathrm{c}}){\rightarrow} ({\sim} y \vee {\mathrm{c}})). \] Taking $${\wedge}$$ in (38) and (39) we obtain that   \begin{equation}\label{eq9} (t,u)\in \tau. \end{equation} (40) Besides, it follows from $${\mathrm{(K5)}}$$ that   \begin{equation} \label{eq10} (x{\rightarrow} z) \vee {\mathrm{c}} = ((x\vee {\mathrm{c}}) {\rightarrow} (z\vee {\mathrm{c}})) {\wedge} (({\sim} z \vee {\mathrm{c}}) {\rightarrow} ({\sim} x \vee {\mathrm{c}})), \end{equation} (41)  \begin{equation}\label{eq11} (y{\rightarrow} w) \vee z = ((y\vee {\mathrm{c}}){\rightarrow} (w\vee {\mathrm{c}})) {\wedge} (({\sim} w \vee {\mathrm{c}}){\rightarrow} ({\sim} y \vee {\mathrm{c}})). \end{equation} (42) Taking into account (40), (41) and (42) we have   \begin{equation} \label{eq12} ((x{\rightarrow} z)\vee {\mathrm{c}}, (y{\rightarrow} w)\vee {\mathrm{c}}) \in \tau. \end{equation} (43) Thus, by (37) and (43) the condition $$(x{\rightarrow} z, y{\rightarrow} w) \in \Sigma(\tau)$$ is satisfied. This implies that $$\Sigma(\tau) \in {\mathrm{Con_{wb}}}(T)$$. ■ Proposition 60 Let $$T\in {\mathsf{KhIS_{0}}}$$. There exists an isomorphism between $${\mathrm{Con_{wb}}}(T)$$ and $${\mathrm{Con}}({\mathrm{C}}(T))$$, which is established via the assignments $$\theta \mapsto \Gamma(\theta)$$ and $$\tau \mapsto \Sigma(\tau)$$. Proof. Let $$\theta\in {\mathrm{Con_{wb}}}(T)$$. It follows from $${\mathrm{(C1)}}$$ and $${\mathrm{(C3)}}$$ that $$\Gamma(\theta) \in {\mathrm{Con}} ({\mathrm{C}}(T))$$. Suppose now that $$\theta\in {\mathrm{Con_{wb}}}(T)$$, $$\sigma \in {\mathrm{Con_{wb}}}(T)$$ and $$\Gamma(\theta) = \Gamma(\sigma)$$. Let $$(x,y) \in \theta$$. Then by $${\mathrm{(C2)}}$$ we have $$(x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \theta$$ and $$({\sim} x \vee {\mathrm{c}}, {\sim} y \vee {\mathrm{c}}) \in \theta$$, so $$(x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \Gamma(\theta)$$ and $$({\sim}x \vee {\mathrm{c}}, {\sim}y \vee {\mathrm{c}}) \in \Gamma(\theta)$$. Since $$\Gamma(\theta) = \Gamma(\sigma)$$, $$(x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \sigma$$ and $$({\sim}x \vee {\mathrm{c}}, {\sim} y \vee {\mathrm{c}}) \in \sigma$$. Hence, it follows from $${\mathrm{(C2)}}$$ again that $$(x,y) \in \sigma$$. Thus, $$\theta \subseteq \sigma$$. For the same reason we have the other inclusion, so $$\theta = \sigma$$. Lemma 59 shows that if $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$, then $$\Sigma(\tau)\in {\mathrm{Con_{wb}}}(T)$$. Besides it is immediate that $$\Gamma(\Sigma(\tau)) = \tau$$. We also have that for $$\theta\in {\mathrm{Con_{wb}}}(T)$$ and $$\sigma \in {\mathrm{Con_{wb}}}(T)$$, $$\theta \subseteq \sigma$$ if and only if $$\Sigma(\theta) \subseteq \Sigma (\sigma)$$. Therefore, we obtain an isomorphism between $${\mathrm{Con_{wb}}}(T)$$ and $${\mathrm{Con}}({\mathrm{C}}(T))$$. ■ Let $$T\in {\mathsf{KhBDL}}$$. If $$\theta \in {\mathrm{Con}}(T)$$ and $$\tau \in {\mathrm{C}}(T)$$, we define $$\Gamma(\theta)$$ and $$\Sigma(\tau)$$ as for the case of $${\mathsf{KhIS_{0}}}$$. If $$\theta \in {\mathrm{Con}}(T)$$, then $$\theta$$ satisfies $${\mathrm{(C1)}}$$, $${\mathrm{(C2)}}$$, and $${\mathrm{(C3)}}$$. Let $$\tau \in {\mathrm{C}}(T)$$. The distributivity of the underlying lattice of $$T$$ proves that $$\Sigma(\tau)$$ preserves $${\wedge}$$ and $$\vee$$. Then from the proof of Proposition 60 the next result follows. Proposition 61 Let $$T\in {\mathsf{KhBDL}}$$. There exists an isomorphism between $${\mathrm{Con}}(T)$$ and $${\mathrm{Con}}({\mathrm{C}}(T))$$, which is established via the assignments $$\theta \mapsto \Gamma(\theta)$$ and $$\tau \mapsto \Sigma(\tau)$$. Let $$H \in {\mathsf{hIS_{0}}}$$ or $$H \in {\mathsf{hBDL}}$$. Let $$\theta \in {\mathrm{Con}}(H)$$ and $$\tau \in {\mathrm{Con}}({\mathrm{C}}({\mathrm{K}}(H)))$$. Since the map $$\alpha:H{\rightarrow} {\mathrm{C}}(K(H))$$ given by $$\alpha(a) = (a,0)$$ is an isomorphism, we have that the binary relation $$\alpha(\theta) = \{(\alpha(a),\alpha(b)): (a,b) \in \theta\}$$ in $${\mathrm{C}}({\mathrm{K}}(H))$$ is a congruence of $${\mathrm{C}}({\mathrm{K}}(H))$$. Moreover, the relation $$\alpha^{-1}(\tau)$$ in $$H$$ given by $$(a,b) \in \alpha^{-1}(\tau)$$ if and only if $$((a,0), (b,0)) \in \tau$$ is a congruence of $$H$$. Then the following result follows from propositions 60 and 61. Corollary 62 (a) Let $$H\in {\mathsf{hIS_{0}}}$$. There exists an isomorphism between $${\mathrm{Con}}(H)$$ and $${\mathrm{Con_{wb}}}({\mathrm{K}}(H))$$, which is established via the assignments $$\theta \mapsto \Sigma(\alpha(\theta))$$ and $$\tau \mapsto \alpha^{-1}(\Gamma(\tau))$$. (b) Let $$H\in {\mathsf{hBDL}}$$. There exists an isomorphism between $${\mathrm{Con}}(H)$$ and $${\mathrm{Con}}({\mathrm{K}}(H))$$, which is established via the assignments $$\theta \mapsto \Sigma(\alpha(\theta))$$ and $$\tau \mapsto \alpha^{-1}(\Gamma(\tau))$$. Remark 63 Let $$H\in {\mathsf{hIS_{0}}}$$, $$\theta \in {\mathrm{Con}}(H)$$ and $$\tau \in {\mathrm{Con_{wb}}}({\mathrm{C}}({\mathrm{K}}(H)))$$. Then   \[ ((a,b),(d,e)) \in \Sigma(\alpha(\theta))\; \textrm{if and only if} \; (a,d) \in \theta\; \textrm{and}\; (b,e) \in \theta, \]   \[ (a,b) \in \alpha^{-1}(\Gamma(\tau))\; \textrm{if and only if}\; ((a,0),(b,0)) \in \tau. \] Similarly for $$H\in {\mathsf{hBDL}}$$. 6.1 Well-behaved congruences and congruences: the relation with some family of filters and some applications We start by recalling some facts about congruences in $${\mathsf{hIS_{0}}}$$ and congruences in $${\mathsf{hBDL}}$$ [21]. Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$. As usual, we say that $$F$$ is a filter if it is a nonempty subset of $$H$$ which satisfies the following conditions: (1) If $$a\in F$$ and $$b\in F$$ then $$a{\wedge} b \in F$$. (2) If $$a\in F$$ and $$a\leq b$$ then $$b\in F$$. We also consider the binary relation associated with $$F \subseteq H$$  \[ \Theta(F) = \{(a,b) \in H\times H: a{\wedge} f = b {\wedge} f\;\text{for some}\; f\in F\}. \] Note that if $$H$$ is an upper bounded semilattice and $$F$$ is a filter, then $$\Theta(F)$$ is a congruence. Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$. For $$a,b,f\in H$$ we define the following element of $$H$$:   $$t(a,b,f): = (a {\rightarrow} b) {\leftrightarrow} ((a{\wedge} f) {\rightarrow} (b{\wedge} f)),$$ where $$a{\leftrightarrow} b:= (a{\rightarrow} b){\wedge} (b{\rightarrow} a)$$. The next definition was introduced in [21]. Definition 64 Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$, and let $$F$$ be a filter of $$H$$. We say that $$F$$ is a congruent filter if $$t(a,b,f)\in F$$ whenever $$a$$, $$b \in H$$ and $$f\in F$$. Note that the set of all congruent filters of $$H\in {\mathsf{hIS_{0}}}$$ or of $$H\in {\mathsf{hBDL}}$$ is closed under arbitrary intersections and therefore for every $$X \subseteq H$$ the congruent filter generated by $$X$$ exists. Remark 65 Let $$F$$ be a congruent filter of a hemi-implicative semilattice (lattice). We will see that $$(a,b) \in \Theta(F)$$ if and only if $$a{\leftrightarrow} b \in F$$. To show it, suppose that $$a{\leftrightarrow} b \in F$$. Since $$a{\wedge} (a{\leftrightarrow} b) = b {\wedge}(b{\leftrightarrow} a)$$, then $$(a,b) \in \Theta(F)$$. Conversely, assume that $$(a,b) \in \Theta(F)$$, i.e., $$a{\wedge} f = b{\wedge} f$$ for some $$f\in F$$. Since $$t(a,b,f) \in F$$ and $$t(a,b,f) = (a{\rightarrow} b) {\leftrightarrow} 1$$, then $$1{\rightarrow} (a{\rightarrow} b) \in F$$ because $$(a{\rightarrow} b) {\leftrightarrow} 1 \leq 1{\rightarrow} (a{\rightarrow} b)$$. Since $$1{\rightarrow} (a{\rightarrow} b) \leq a{\rightarrow} b$$, then $$a{\rightarrow} b\in F$$. In a similar way we can show that $$b{\rightarrow} a\in F$$. Hence, $$a{\leftrightarrow} b\in F$$. Thus,   \[ \Theta(F) = \{(a,b)\in H\times H: a{\leftrightarrow} b \in F\}. \] The following result was proved in [21]. Theorem 66 Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$. There exists an isomorphism between $${\mathrm{Con}}(H)$$ and the lattice of congruent filters of $$H$$, which is established via the assignments $$\theta \mapsto 1/\theta$$ and $$F\mapsto \Theta(F)$$. Taking into account Theorem 66, it is possible to show that Proposition 61 can be seen as a corollary of Proposition 60. To show this assertion, let $$T_1 = (T,{\wedge},\vee,{\rightarrow},{\sim},{\mathrm{c}},0,1) \in {\mathsf{hBDL}}$$. Then we write $$T_2 = (T,\leq,{\sim},{\rightarrow},{\mathrm{c}},0,1)$$ for the corresponding object of $${\mathsf{KhIS_{0}}}$$. Since the set of congruent filters of $${\mathrm{C}}(T_1)$$ is equal to the set of congruent filters of $${\mathrm{C}}(T_2)$$, then it follows from Theorem 66 that $${\mathrm{Con}}({\mathrm{C}}(T_1)) = {\mathrm{Con}}({\mathrm{C}}(T_2))$$. In what follows we will see that $${\mathrm{Con}}(T_1) = {\mathrm{Con_{wb}}}(T_2)$$. It is immediate that $${\mathrm{Con}}(T_1) \subseteq {\mathrm{Con_{wb}}}(T_2)$$. Conversely, let $$\theta \in {\mathrm{Con_{wb}}}(T_2)$$. We will prove that $$\theta$$ preserves $${\wedge}$$ and $$\vee$$. Let $$(x,y)\in \theta$$ and $$(z,w) \in \theta$$. Then it follows from $${\mathrm{(C2)}}$$ that $$(x\vee {\mathrm{c}}, z\vee {\mathrm{c}}) \in \theta$$ and $$(y \vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta$$. Then by $${\mathrm{(C3)}}$$ we have that $$((x\vee {\mathrm{c}}){\wedge} (z \vee {\mathrm{c}}), (y\vee {\mathrm{c}}){\wedge} (w\vee {\mathrm{c}}))\in \theta$$. But by the distributivity of the underlying lattice of $$T_1$$ we deduce that $$(x{\wedge} z) \vee {\mathrm{c}} = (x\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}})$$ and $$(y{\wedge} w) \vee {\mathrm{c}} = (y\vee {\mathrm{c}}){\wedge} (w\vee {\mathrm{c}})$$. Thus,   \begin{equation} \label{cwbc0} ((x{\wedge} z)\vee {\mathrm{c}}, (y{\wedge} w)\vee {\mathrm{c}} \in \theta. \end{equation} (44) Besides, since $$(x,y)\in \theta$$ and $$(z,w)\in \theta$$, then it follows from the condition $${\mathrm{(C2)}}$$ that $$({\sim} x \vee {\mathrm{c}},{\sim} y \vee {\mathrm{c}}) \in \theta$$ and $$({\sim} y \vee {\mathrm{c}},{\sim} w \vee {\mathrm{c}}) \in \theta$$. Equivalently, we have that   \begin{equation} \label{cwbc1} ({\sim} x \vee {\mathrm{c}},{\sim} y \vee {\mathrm{c}}) \in \Gamma(\theta), \end{equation} (45)  \begin{equation} \label{cwbc2} ({\sim} y \vee {\mathrm{c}},{\sim} w \vee {\mathrm{c}}) \in \Gamma(\theta). \end{equation} (46) Since $$\Gamma(\theta) \in {\mathrm{Con}}({\mathrm{C}}(T_2))$$ and $${\mathrm{Con}}({\mathrm{C}}(T_1)) = {\mathrm{Con}}({\mathrm{C}}(T_2))$$, then taking $$\vee$$ in (45) and (46) we obtain $$({\sim} x \vee {\sim} y \vee {\mathrm{c}}, {\sim} z \vee {\sim} w \vee {\mathrm{c}}) \in \theta$$, i.e.,   \begin{equation} \label{cwbc4} ({\sim}(x{\wedge} z)\vee {\mathrm{c}}, {\sim}(z{\wedge} w) \vee {\mathrm{c}})\in \theta. \end{equation} (47) Then it follows from $${\mathrm{(C2)}}$$, (44) and (47) that $$(x{\wedge} z, y{\wedge} w)\in \theta$$. The same argument combined with $${\mathrm{(C1)}}$$ proves that $$({\sim} x {\wedge} {\sim} z, {\sim} y {\wedge} {\sim} w) \in \theta$$, so $$(x\vee z,y\vee w)\in \theta$$. Hence, $$\theta$$ preserves $${\wedge}$$ and $$\vee$$, which implies that $$\theta \in {\mathrm{Con}}(T_1)$$. Then $${\mathrm{Con}}(T_1) = {\mathrm{Con_{wb}}}(T_2)$$. Therefore, since $${\mathrm{Con}}(T_1) = {\mathrm{Con_{wb}}}(T_2)$$ and $${\mathrm{Con}}({\mathrm{C}}(T_1)) = {\mathrm{Con}}({\mathrm{C}}(T_2))$$, we deduce that Proposition 61 can be seen as a corollary of Proposition 60. Corollary 67 Let $$T\in {\mathsf{KhIS_{0}}}$$. There exists an isomorphism between $${\mathrm{Con_{wb}}}(T)$$ and the lattice of congruent filters of $${\mathrm{C}}(T)$$, which is established via the assignments $$\theta \mapsto 1/\Gamma(\theta)$$ and $$F\mapsto \Sigma(\Theta(F))$$. Proof. It follows from Proposition 60 and Theorem 66. ■ Similarly, the following result follows from Proposition 61 and Theorem 66. Corollary 68 Let $$T\in {\mathsf{KhBDL}}$$. There exists an isomorphism between $${\mathrm{Con}}(T)$$ and the lattice of congruent filters of $${\mathrm{C}}(T)$$, which is established via the assignments $$\theta \mapsto 1/\Gamma(\theta)$$ and $$F\mapsto \Sigma(\Theta(F))$$. For implicative semilattices Corollary 67 can be simplified, and for semi-Heyting algebras Corollary 68 also can be simplified. More precisely: if $$H \in {\mathsf{IS_{0}}}$$ or $$H\in {\mathsf{SH}}$$ then the congruent filters of $$H$$ are all the filters of $$H$$ [21]. Let $$H\in {\mathsf{Hil_{0}}}$$ and $$F\subseteq H$$. Recall that $$F$$ is said to be a deductive system [10] if the following conditions are satisfied: (a) $$1\in F$$, (b) if $$a\in F$$ and $$a{\rightarrow} b\in F$$ then $$b\in F$$. Also recall that a deductive system $$F$$ is said to be absorbent [12] if $$a {\rightarrow} (a{\wedge} b)\in F$$ whenever $$a\in F$$. It follows from Theorem 66 and [12, Lemma 3.3] that the congruent filters of $$H$$ are the absorbent deductive systems of $$H$$. Definition 69 Let $$A$$ be an algebra and $$a_1,b_1,\dots,a_n,b_n$$ elements of $$A$$. We write $$\theta_{A}((a_1,b_1),\dots,(a_n,b_n))$$ for the congruence generated by $$(a_1,b_1),\dots,(a_n,b_n)$$. If $$T\in {\mathsf{KhIS_{0}}}$$ we also write $$\theta_{T}((a_1,b_1),\dots,(a_n,b_n))$$ for the well-behaved congruence generated by $$(a_1,b_1),\dots,(a_n,b_n)$$. Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$, and let $$a \in H$$. We refer by $$F^{c}(a)$$ to the congruent filter generated by $$\{a\}$$. In [21] the following assertions were proved: (1) if $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$, then $$(d,e) \in \theta_{H}(a,b)$$ if and only if $$d{\leftrightarrow} e \in F^{c}(a{\leftrightarrow} b)$$; (2) if $$H\in {\mathsf{IS_{0}}}$$ or $$H\in {\mathsf{SH}}$$ then $$(d,e) \in \theta_{H}(a,b)$$ if and only if $$a{\leftrightarrow} b \leq d{\leftrightarrow} e$$. The following remark will be used later. Remark 70 Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$. Let $$\tau \in {\mathrm{Con}}(H)$$. Then $$(a,b) \in \tau$$ if and only if $$a{\leftrightarrow} b \in 1/\tau$$. Moreover, $$(a,b)$$, $$(d,e) \in \tau$$ if and only if $$(a{\leftrightarrow} b){\wedge} (d{\leftrightarrow} e) \in 1/\tau$$. In what follows we describe some aspects of the principal well-behaved congruences of the objects of $${\mathsf{KhIS_{0}}}$$ and some aspects of the principal congruences of the algebras in $${\mathsf{KhBDL}}$$. Let $$T\in {\mathsf{KhIS_{0}}}$$ or $$T\in {\mathsf{KhBDL}}$$. For $$x$$ and $$y$$ elements of $$T$$ we also write $$x{\leftrightarrow} y$$ for the element $$(x{\rightarrow} y){\wedge} (y{\rightarrow} x)$$. Lemma 71 Let $$T\in {\mathsf{KhIS_{0}}}$$ or $$T\in {\mathsf{KhBDL}}$$. Let $$x$$, $$y$$, $$z$$, $$w \in T$$. Then (a) $$(z,w) \in \theta_{T}(x,y)$$ if and only if $$(z\vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}}))$$ and $$(\sim z\vee {\mathrm{c}},\sim w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}}))$$. (b) If $$x$$, $$y$$, $$z$$ and $$w$$ are in $${\mathrm{C}}(T)$$, then   \[ 1/\theta_{{\mathrm{C}}(T)}((x,y),(z,w)) = F^{c}((x{\leftrightarrow} y) {\wedge} (z{\leftrightarrow} w)). \] Proof. We consider $$T\in {\mathsf{KhIS_{0}}}$$ (the proof for $$T\in {\mathsf{KhBDL}}$$ is analogous). First we prove a). Let $$(z,w) \in \theta_{T}(x,y)$$. Then $$(z,w) \in \theta$$ for every $$\theta \in {\mathrm{Con_{wb}}}(T)$$ such that $$(x,y) \in \theta$$. Now we see that   \[ (z\vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})), \]   \[ (\sim z\vee {\mathrm{c}},\sim w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})). \] Let $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$ such that $$(x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \tau$$ and $$(\sim x \vee {\mathrm{c}}, \sim y \vee {\mathrm{c}}) \in \tau$$. It follows from Proposition 60 that $$\Sigma(\tau) \in {\mathrm{Con_{wb}}}(T)$$. We also have that $$(x,y) \in \Sigma(\tau)$$. Then by hypothesis we obtain that $$(z,w) \in \Sigma(\tau)$$. Hence, $$(z\vee {\mathrm{c}}, w\vee {\mathrm{c}}) \in \tau$$ and $$(\sim z \vee {\mathrm{c}}, \sim w \vee {\mathrm{c}}) \in \tau$$, which was our aim. Conversely, assume that $$(z\vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}}))$$ and $$(\sim z\vee {\mathrm{c}},\sim w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}}))$$. Let $$\theta \in {\mathrm{Con_{wb}}}(T)$$ be such that $$(x,y) \in \theta$$. It follows from Proposition 60 that $$\Gamma(\theta) \in {\mathrm{Con}}({\mathrm{C}}(T))$$. Moreover, $$(x\vee {\mathrm{c}},y\vee {\mathrm{c}}) \in \Gamma(\theta)$$ and $$(\sim x\vee {\mathrm{c}},\sim y\vee {\mathrm{c}}) \in \Gamma(\theta)$$. Thus by hypothesis we have that $$(z\vee {\mathrm{c}}, w\vee {\mathrm{c}}) \in \Gamma(\theta)$$ and $$(\sim z\vee {\mathrm{c}}, \sim w\vee {\mathrm{c}}) \in \Gamma(\theta)$$, i.e., $$(z\vee {\mathrm{c}}, w\vee {\mathrm{c}}) \in \theta$$ and $$(\sim z\vee {\mathrm{c}}, \sim w\vee {\mathrm{c}}) \in \theta$$. Then it follows from $${\mathrm{(C2)}}$$ that $$(z,w) \in \theta$$. Thus, $$(z,w) \in \theta_{T}(x,y)$$. Finally, we prove b). Let $$H\in {\mathsf{hIS_{0}}}$$. We write $$\tau$$ for an arbitrary well-behaved congruence of $$H$$. Then   \[ \theta_{H}((x,y),(z,w)) = \bigcap\{\tau \in {\mathrm{Con_{wb}}}(H): (x,y),(z,w) \in \tau\}. \] Hence,   \[ 1/\theta_{H}((x,y),(z,w)) = \bigcap\{1/\tau: \tau \in {\mathrm{Con_{wb}}}(H) \text{ and } (x,y), (z,w) \in \tau\}. \] Then it follows from Remark 70 that   \[ 1/\theta_{H}((x,y),(z,w)) = \bigcap\{ 1/\tau: \tau \in {\mathrm{Con_{wb}}}(H) \text{ and } (x{\leftrightarrow} y){\wedge} (z{\leftrightarrow} w) \in 1/\tau\}. \] Thus, by Theorem 66 we have that   \[ 1/\theta_{H}((x,y),(z,w)) = F^{c}((x{\leftrightarrow} y){\wedge} (z{\leftrightarrow} w)). \] In particular, the last assertion holds for $$H = {\mathrm{C}}(T)$$. ■ Remark 72 The proof of item (b) of Lemma 71 can be done in a different way. Let $$\theta$$ be a congruence of an algebra $$H\in {\mathsf{hIS_{0}}}$$, and let $$a,b\in H$$. Straightforward computations show that $$F^{c}(a) \vee F^{c}(b) = F^{c}(a {\wedge} b)$$, where $$\vee$$ is the supremum in the lattice of congruent filters of $$H$$. On the other hand, it follows from general results from universal algebra that $$\theta_{H}((x,y),(z,w)) = \theta_{H}(x,y) \vee \theta_H(z,w)$$, where $$\vee$$ is the supremum in the lattice of congruences of $$H$$ [20]. In [21] it was proved that $$1/\theta_{H}(x,y) = F^{c}(x{\leftrightarrow} y)$$. Then   \[ \begin{array} [c]{lllll} 1/\theta_{H}((x,y),(z,w)) & = & 1/\theta_{H}(x,y) \vee 1/\theta_{H}(z,w) & & \\ & = & F^{c}(x{\leftrightarrow} y) \vee F^{c}(z{\leftrightarrow} w)& & \\ & = & F^{c}((x{\leftrightarrow} y) {\wedge} (z{\leftrightarrow} w)).& & \end{array} \] Let $$T\in {\mathsf{KhIS_{0}}}$$ or $$T\in {\mathsf{KhBDL}}$$. For every $$x$$, $$y\in T$$ we define the following binary term:   \[ q(x,y) = ((x\vee {\mathrm{c}}) {\leftrightarrow} (y\vee {\mathrm{c}})){\wedge} ((\sim x\vee {\mathrm{c}}) {\leftrightarrow} (\sim y\vee {\mathrm{c}})). \] In the proof of the following corollary we will use Remark 70 and Lemma 71. Corollary 73 Let $$T \in {\mathsf{KhIS_{0}}}$$ or $$T\in {\mathsf{KhBDL}}$$. Let $$x$$, $$y$$, $$z$$, $$w \in T$$. (a) $$(z,w) \in \theta_{T}(x,y)$$ if and only if $$q(z,w) \in F^{c}(q(x,y))$$. (b) If $$T\in {\mathsf{KIS_{0}}}$$ or $$T \in {\mathsf{KSH}}$$ then $$(z,w) \in \theta_{T}(x,y)$$ if and only if $$q(x,y) \leq q(z,w)$$. Proof. The condition $$(z,w) \in \theta_{T}(x,y)$$ is equivalent to the conditions   \[ (z\vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})), \]   \[ (\sim z\vee {\mathrm{c}},\sim w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})), \] which are equivalent to   \[ (z\vee {\mathrm{c}}) {\leftrightarrow} (w\vee {\mathrm{c}}) \in 1/\theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})), \]   \[ ({\sim} z\vee {\mathrm{c}}) {\leftrightarrow} ({\sim} w\vee {\mathrm{c}}) \in 1/\theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})), \] which happens if and only if   \[ q(z,w) \in 1/\theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})). \] But this last fact is equivalent to say that $$q(z,w) \in F^{c}(q(x,y))$$. If $$T \in {\mathsf{KIS_{0}}}$$ or $$T\in {\mathsf{KSH}}$$, then $$F^{c}(q(x,y))$$ is equal to the filter generated by $$\{q(x,y)\}$$, so $$(z,w) \in \theta_{T}(x,y)$$ if and only if $$q(x,y) \leq q(z,w)$$. ■ Acknowledgements This project has received funding from the European Union’s Horizon 2020 research and innovation programme under the Marie Sklodowska-Curie grant agreement [No 689176]. R.J. was also partially supported by the research grant 2014 SGR 788 from the government of Catalonia and by the research projects MTM2011-25747 and MTM2016-74892-P from the government of Spain, which include feder funds from the European Union. H.J.S.M. was also supported by CONICET Project PIP 112-201501-00412, and he thanks Marta Sagastume for several conversations concerning the matter of this article. Footnotes 1Let $$(H,\leq)$$ be a poset. If any two elements $$a$$, $$b \in H$$ have a greatest lower bound (i.e. an infimum), which is denoted by $$a {\wedge} b$$, then the algebra $$(H,{\wedge})$$ is called a meet semilattice. Throughout this article we write semilattice in place of meet semilattice. A semilattice $$(H,{\wedge})$$ is said to be upper bounded if it has a greatest element; in this case we write $$(H,{\wedge},1)$$, where $$1$$ is the last element of $$(H,\leq)$$. A bounded semilattice is an algebra $$(H,{\wedge},0,1)$$ of type $$(2,0,0)$$ such that $$(H,{\wedge},1)$$ is an upper bounded semilattice and $$0$$ is the first element of $$(H,\leq)$$. Frequently in the literature what we call upper bounded semilattice is known as bounded semilattice. References [1] Balbes R. and Dwinger. P. Distributive Lattices . University of Missouri Press, 1974. [2] Busaniche M. and Cignoli. R. Constructive logic with strong negation as a substructural logic. Journal of Logic and Computation , 20, 761– 793, 2010. Google Scholar CrossRef Search ADS   [3] Castiglioni J. L. Menni M. and Sagastume. M. On some categories of involutive centered residuated lattices. Studia Logica , 90, 93– 124, 2008. Google Scholar CrossRef Search ADS   [4] Castiglioni J. L. Lewin R. and Sagastume. M. On a definition of a variety of monadic l-groups. Studia Logica , 102, 67– 92, 2014. Google Scholar CrossRef Search ADS   [5] Castiglioni J. L. Celani S. and San Martín. H. J. Kleene algebras with implication. Algebra Universalis , 77, 375– 393, 2017. Google Scholar CrossRef Search ADS   [6] Celani. S. Bounded distributive lattices with fusion and implication. Southeast Asian Bulletin Mathematics , 27, 1– 10, 2003. [7] Celani S. A. and Jansana. R. Bounded distributive lattices with strict implication. Mathematical Logic Quarterly , 51, 219– 246, 2005. Google Scholar CrossRef Search ADS   [8] Cignoli. R. The Class of Kleene Algebras satisfying an interpolation property and Nelson algebras. Algebra Universalis , 23, 262– 292, 1986. [9] Curry. H. B. Foundations of Mathematical Logic . McGraw-Hill, 1963. [10] Diego. A. Sobre Algebras de Hilbert . Notas de Lógica Matemática. Instituto de Matemática, Universidad Nacional del Sur, 1965. [11] Fidel. M. M. An algebraic study of a propositional system of Nelson. Mathematical Logic, In Proceedings of the First Brazilian Conference. Lectures in Pure and Applied Mathematics , Vol. 39, Arruda A. I., Da Costa N. C. A., Chuaqui R., eds., pp. 99– 117. Marcel Dekker, 1978. [12] Figallo A. V. Ramon G. and Saad. S. A note on the Hilbert algebras with infimum. Mathematica Contemporanea , 24, 23– 37, 2003. [13] Hart J. Raftery L. and Tsinakis. C. The structure of commutative residuated lattices. Internat. J. Algebra Comput. , 12, 509– 524, 2002. Google Scholar CrossRef Search ADS   [14] Idziak. P. M. Lattice operations in BCK-algebras. Mathematica Japonica , 29, 839– 846, 1984. [15] Kalman. J. A. Lattices with involution. Transaction of the American Mathematica Society , 87, 485– 491, 1958. Google Scholar CrossRef Search ADS   [16] Monteiro. A. Construction des Algèbres de Nelson Finies. Bulletin of the polish Academy of Science , 11, 359– 362, 1963. [17] Nemitz. W. Implicative semi-lattices. Transactions of the American Mathematica Society , 117, 128– 142, 1965. Google Scholar CrossRef Search ADS   [18] Rasiowa. H. An algebraic approach to non-classical logics. In Studies in logic and the Foundations of Mathematics , 78. Nort-Holland and PNN, 1974. [19] Sagastume. M. Categorical equivalence between centered Kleene algebras with condition (CK) and bounded distributive lattices . Unpublished manuscript ( 2004). [20] Sankappanavar H.P., Semi-Heyting algebras: an abstraction from Heyting algebras. In Proceedings of the 9th Congreso ‘Dr. Antonio A. R.’ , pp. 33– 66, Actas Congr. ‘Dr. Antonio A. R. Monteiro’, Univ. Nac. del Sur, 2008. [21] San Martín. H. J. On congruences in weak implicative semi-lattices. Soft Computing , 21, 3167– 3176, 2017. Google Scholar CrossRef Search ADS   [22] Spinks M. and Veroff. M. Constructive logic with strong negation is a substructural logic. I. Studia Logica , 88, 325– 348, 2008. Google Scholar CrossRef Search ADS   [23] Vakarelov. D. Notes on N-lattices and constructive logic with strong negation. Studia Logica , 34, 109– 125, 1977. Google Scholar CrossRef Search ADS   © The Author 2017. Published by Oxford University Press. All rights reserved. For Permissions, please email: journals.permissions@oup.com http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Logic Journal of the IGPL Oxford University Press

On Kalman’s functor for bounded hemi-implicative semilattices and hemi-implicative lattices

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Abstract

Abstract Hemi-implicative semilattices (lattices), originally defined under the name of weak implicative semilattices (lattices), were introduced by the second author of the present article. A hemi-implicative semilattice is an algebra $$(H,{\wedge},{\rightarrow},1)$$ of type $$(2,2,0)$$ such that $$(H,{\wedge})$$ is a meet semilattice, $$1$$ is the greatest element with respect to the order, $$a{\rightarrow} a = 1$$ for every $$a\in H$$ and for every $$a$$, $$b$$, $$c\in H$$, if $$a\leq b{\rightarrow} c$$ then $$a{\wedge} b \leq c$$. A bounded hemi-implicative semilattice is an algebra $$(H,{\wedge},{\rightarrow},0,1)$$ of type $$(2,2,0,0)$$ such that $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice and $$0$$ is the first element with respect to the order. A hemi-implicative lattice is an algebra $$(H,{\wedge},\vee,{\rightarrow},0,1)$$ of type $$(2,2,2,0,0)$$ such that $$(H,{\wedge},\vee,0,1)$$ is a bounded distributive lattice and the reduct algebra $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice. In this article, we introduce an equivalence for the categories of bounded hemi-implicative semilattices and hemi-implicative lattices, respectively, which is motivated by an old construction due J. Kalman that relates bounded distributive lattices and Kleene algebras. 1 Introduction Inspired by results due to J. Kalman relating to lattices [15], R. Cignoli proved in [8] that a construction of J. Kalman can be extended to a functor $${\mathrm{K}}$$ from the category of bounded distributive lattices to the category of Kleene algebras and that this functor has a left adjoint [8, Theorem 1.7]. He also showed that there exists an equivalence between the category of bounded distributive lattices and the full subcategory of centered Kleene algebras whose objects satisfy a condition called interpolation property [8, Theorem 2.4]. Moreover, R. Cignoli also proved that there exists an equivalence between the category of Heyting algebras and the category of centered Nelson algebras [8, Theorem 3.14]. These results were extended by J.L. Castiglioni, R. Lewin, M. Menni and M. Sagastume in the context of residuated lattices [3, 4]. On the other hand, the original Kalman’s construction was also extended in [5] by J.L. Castiglioni, S. Celani and the second author of the present article to the framework of algebras with implication $$(H,{\wedge},\vee, {\rightarrow}, 0,1)$$ which satisfy the following additional condition: for every $$a,b,c\in H$$, if $$a\leq b{\rightarrow} c$$ then $$a{\wedge} b \leq c$$. Algebras with implication were introduced by S. Celani in [6]. A generalization of Heyting algebras is provided by the notion of hemi-implicative semilattice (lattice), introduced in [21] under the name weak implicative semilattices (lattices). An algebra $$(H,{\wedge},{\rightarrow},1)$$ of type $$(2,2,0)$$ is said to be a hemi-implicative semilattice if $$(H,{\wedge},1)$$ is an upper bounded semilattice1, $$a{\rightarrow} a = 1$$ for every $$a\in H$$ and for every $$a,b,c\in H$$, if $$a\leq b{\rightarrow} c$$ then $$a{\wedge} b \leq c$$. A bounded hemi-implicative semilattice is an algebra $$(H,{\wedge},{\rightarrow},0,1)$$ of type $$(2,2,0,0)$$ such that $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice and $$0$$ is the first element with respect to the order. A hemi-implicative lattice is an algebra $$(H,{\wedge},\vee,{\rightarrow},0,1)$$ of type $$(2,2,2,0,0)$$ such that $$(H,{\wedge},\vee,0,1)$$ is a bounded distributive lattice and the reduct algebra $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice. Implicative semilattices [17] and Hilbert algebras with infimum [12] are examples of hemi-implicative semilattices. Semi-Heyting algebras [20] and some algebras studied in [5] are examples of hemi-implicative lattices. For instance, the RWH-algebras, introduced and studied by S. Celani and the first author of this article in [7], are examples of hemi-implicative lattices. The applications of Kalman’s construction given in [8] suggest that it is potentially fruitful to understand Kalman’s work in the context of bounded hemi-implicative semilattices and hemi-implicative lattices. We do this in the present article. The main goal of the article is to introduce and study an equivalence for the categories of bounded hemi-implicative semilattices and hemi-implicative lattices, respectively, and for some of its full subcategories. The article is organized as follows. In Section 2, we give some results about Kalman’s functor for bounded distributive lattices and Heyting algebras. In Section 3 we generalize Kalman’s functor for the category whose objects are posets with first element and whose morphisms are maps which preserve finite existing infima and the first element (note that the morphisms of this category are in particular order-preserving maps). Moreover, we apply the mentioned equivalence to build up an equivalence for the category whose objects are bounded semilattices and whose morphisms are the corresponding algebra homomorphisms. In Section 4, we recall definitions and properties about hemi-implicative semilattices (lattices) [21], Hilbert algebras with infimum [12], implicative semilattices [17] and semi-Heyting algebras [20]. In Section 5 we employ results of Sections 3 and 4 to establish equivalences, following the original Kalman’s construction, for the categories of bounded hemi-implicative semilattices, bounded Hilbert algebras with infimum, bounded implicative semilattices, hemi-implicative lattices, respectively, and the category of semi-Heyting algebras. Finally, in Section 6 we introduce and study the notion of well-behaved congruences for the objects corresponding to the categories introduced in Section 5. We give a table with some of the categories we shall consider in this article: $$\mathbf{Category}$$  $$\mathbf{Objects}$$  $$\mathbf{Morphisms}$$  $${\mathsf{BDL}}$$  Bounded distributive lattices  Algebra homomorphisms  $${\mathsf{KA_{{\mathrm{c}}}}}$$  Centered Kleene algebras  Algebra homomorphisms  $${\mathsf{HA}}$$  Heyting algebras  Algebra homomorphisms  $${\mathsf{NA_{{\mathrm{c}}}}}$$  Centered Nelson algebras  Algebra homomorphisms  $${\mathsf{NL_{{\mathrm{c}}}}}$$  Centered Nelson lattices  Algebra homomorphisms  $${\mathsf{P_0}}$$  Posets with bottom  Certain order morphisms  $${\mathsf{KP}}$$  Kleene posets  Certain order morphisms  $${\mathsf{MS}}$$  Bounded semilattices  Algebra homomorphisms  $${\mathsf{KMS}}$$  Certain objects of $${\mathsf{KP}}$$  Morphisms of $${\mathsf{KP}}$$  $${\mathsf{hIS_{0}}}$$  Bounded hemi-implicative semilattices  Algebra homomorphisms  $${\mathsf{hBDL}}$$  Hemi-implicative lattices  Algebra homomorphisms  $${\mathsf{Hil_{0}}}$$  Bounded Hilbert algebras with infimum  Algebra homomorphisms  $${\mathsf{IS_{0}}}$$  Bounded implicative semilattices  Algebra homomorphisms  $${\mathsf{SH}}$$  Semi-Heyting algebras  Algebra homomorphisms  $${\mathsf{KhIS_{0}}}$$  Objects of $${\mathsf{KMS}}$$ with an additional  Certain morphisms of $${\mathsf{KMS}}$$    operation    $${\mathsf{KHil_{0}}}$$  Certain objects of $${\mathsf{KhIS_{0}}}$$  Morphisms of $${\mathsf{KhIS_{0}}}$$  $${\mathsf{KIS_{0}}}$$  Certain objects of $${\mathsf{KhIS_{0}}}$$  Morphisms of $${\mathsf{KhIS_{0}}}$$  $${\mathsf{KhBDL}}$$  Objects of $${\mathsf{KA_{{\mathrm{c}}}}}$$ with an additional  Certain morphisms of $${\mathsf{KA_{{\mathrm{c}}}}}$$    operation    $${\mathsf{KSH}}$$  Certain objects of $${\mathsf{KhBDL}}$$  Morphisms of $${\mathsf{KhBDL}}$$  $$\mathbf{Category}$$  $$\mathbf{Objects}$$  $$\mathbf{Morphisms}$$  $${\mathsf{BDL}}$$  Bounded distributive lattices  Algebra homomorphisms  $${\mathsf{KA_{{\mathrm{c}}}}}$$  Centered Kleene algebras  Algebra homomorphisms  $${\mathsf{HA}}$$  Heyting algebras  Algebra homomorphisms  $${\mathsf{NA_{{\mathrm{c}}}}}$$  Centered Nelson algebras  Algebra homomorphisms  $${\mathsf{NL_{{\mathrm{c}}}}}$$  Centered Nelson lattices  Algebra homomorphisms  $${\mathsf{P_0}}$$  Posets with bottom  Certain order morphisms  $${\mathsf{KP}}$$  Kleene posets  Certain order morphisms  $${\mathsf{MS}}$$  Bounded semilattices  Algebra homomorphisms  $${\mathsf{KMS}}$$  Certain objects of $${\mathsf{KP}}$$  Morphisms of $${\mathsf{KP}}$$  $${\mathsf{hIS_{0}}}$$  Bounded hemi-implicative semilattices  Algebra homomorphisms  $${\mathsf{hBDL}}$$  Hemi-implicative lattices  Algebra homomorphisms  $${\mathsf{Hil_{0}}}$$  Bounded Hilbert algebras with infimum  Algebra homomorphisms  $${\mathsf{IS_{0}}}$$  Bounded implicative semilattices  Algebra homomorphisms  $${\mathsf{SH}}$$  Semi-Heyting algebras  Algebra homomorphisms  $${\mathsf{KhIS_{0}}}$$  Objects of $${\mathsf{KMS}}$$ with an additional  Certain morphisms of $${\mathsf{KMS}}$$    operation    $${\mathsf{KHil_{0}}}$$  Certain objects of $${\mathsf{KhIS_{0}}}$$  Morphisms of $${\mathsf{KhIS_{0}}}$$  $${\mathsf{KIS_{0}}}$$  Certain objects of $${\mathsf{KhIS_{0}}}$$  Morphisms of $${\mathsf{KhIS_{0}}}$$  $${\mathsf{KhBDL}}$$  Objects of $${\mathsf{KA_{{\mathrm{c}}}}}$$ with an additional  Certain morphisms of $${\mathsf{KA_{{\mathrm{c}}}}}$$    operation    $${\mathsf{KSH}}$$  Certain objects of $${\mathsf{KhBDL}}$$  Morphisms of $${\mathsf{KhBDL}}$$  If $$\mathrm{A}$$ is one of the categories $${\mathsf{KA_{{\mathrm{c}}}}}$$, $${\mathsf{KP}}$$, $${\mathsf{KMS}}$$, $${\mathsf{KhIS_{0}}}$$, $${\mathsf{KHil_{0}}}$$ and $${\mathsf{KhBDL}}$$, then we write $$\mathrm{A}^{{\mathrm{CK}}}$$ to denote the full subcategory of $$\mathrm{A}$$ whose objects satisfy the condition $$({\mathrm{CK}})$$, that will be defined later. The results we expound in the present article are motivated by the abstraction of ideas coming from different varieties of algebras related to some constructive logics, as Heyting algebras and Nelson algebras, and in particular by the existent categorical equivalence between the category of Heyting algebras and the category of centered Nelson algebras (see [8]) combined with the fact that the variety of centered Nelson algebras is term equivalent to the variety of centered Nelson lattices, as it is shown in [22] (see also [2]). In this article, we introduce and study categories which are closely connected with the category of centered Nelson lattices, as for instance the category $${\mathsf{KP}}$$ of Kleene posets (of which centered Nelson lattices can be seen as particular cases) and the category $${\mathsf{KhBDL}}$$ of centered Kleene algebras endowed with a binary operation which generalizes the implication of Nelson lattices. We consider that the study of the above mentioned categories is interesting in itself. We also think that the categorical equivalences and some related properties studied in this article can be of interest for future work concerning the understanding of the categories of bounded hemi-implicative semilattices and hemi-implicative lattices, respectively. 2 Basic results The definition of the functor from the category of Kleene algebras to the category of bounded distributive lattices given by R. Cignoli [8] is based on Priestley duality, and the interpolation property for Kleene algebras considered by Cignoli in establishing the equivalence is stated in topological terms. On the other hand, M. Sagastume proved in an unpublished manuscript [19] that in centered Kleene algebras the interpolation property is equivalent to an algebraic condition called (CK), that we will state later on. Moreover, she presented an equivalence between the category of bounded distributive lattices and the category of centered Kleene algebras that satisfy (CK), but using a different (purely algebraic) construction to that given by R. Cignoli in [8]. In what follows we describe this equivalence whose details can be found in [5]. We assume the reader is familiar with bounded distributive lattices and Heyting algebras [1]. A De Morgan algebra is an algebra $$(H,{\wedge},\vee,{\sim},0,1)$$ of type $$(2,2,1,0,0)$$ such that $$(H,{\wedge}, \vee,0,1)$$ is a bounded distributive lattice and $${\sim}$$ fulfills the equations   \[ {\sim} {\sim} x = x \quad {\text{and}} \quad {\sim}(x \vee y) = {\sim} x {\wedge} {\sim} y. \] An operation $${\sim}$$ which satisfies the previous two equations is called De Morgan involution. A Kleene algebra is a De Morgan algebra in which the inequality   \[ x{\wedge} {\sim} x \leq y \vee {\sim} y \] holds. A centered Kleene algebra is an algebra $$(H,{\wedge},\vee,{\sim},c, 0,1)$$ where the algebra $$(H,{\wedge},\vee,$$$${\sim},0,1)$$ is a Kleene algebra and $${\mathrm{c}}$$ is an element such that $${\mathrm{c}} = {\sim} {\mathrm{c}}.$$ It is immediate to see that $${\mathrm{c}}$$ is necessarily unique. The element $${\mathrm{c}}$$ is called center. We write $${\mathsf{BDL}}$$ for the category of bounded distributive lattices and $${\mathsf{KA_{{\mathrm{c}}}}}$$ for the category of centered Kleene algebras. In both cases the morphisms are the corresponding algebra homomorphisms. It is interesting to note that if $$T$$ and $$U$$ are centered Kleene algebras and $$f:T{\rightarrow} U$$ is a morphism of Kleene algebras then $$f$$ preserves necessarily the centre, i.e., $$f({\mathrm{c}}) = {\mathrm{c}}$$. The functor $${\mathrm{K}}$$ from the category $${\mathsf{BDL}}$$ to the category $${\mathsf{KA_{{\mathrm{c}}}}}$$ is defined as follows. For an object $$H\in {\mathsf{BDL}}$$ we let   \[ {\mathrm{K}}(H): =\{(a,b) \in H\times H: a{\wedge} b = 0\}. \] This set is endowed with the operations and the distinguished elements defined by:   \begin{eqnarray*} (a,b)\vee (d,e) & := & (a\vee d,b{\wedge} e)\\ (a,b){\wedge} (d,e)& := & (a{\wedge} d,b\vee e)\\ {\sim} (a,b)& := & (b,a)\\ 0 & := & (0,1)\\ 1 & := & (1,0)\\ {\mathrm{c}} & := & (0,0). \end{eqnarray*} We have that $$({\mathrm{K}}(H),{\wedge},\vee,\sim,{\mathrm{c}},0,1)\in {\mathsf{KA_{{\mathrm{c}}}}}$$. For a morphism $$f:H {\rightarrow} G \in {\mathsf{BDL}}$$, the map $${\mathrm{K}}(f):{\mathrm{K}}(H) {\rightarrow} {\mathrm{K}}(G)$$ defined by   $${\mathrm{K}}(f)(a,b) = (f(a),f(b))$$ is a morphism in $${\mathsf{KA_{{\mathrm{c}}}}}$$. Hence, $${\mathrm{K}}$$ is a functor from $${\mathsf{BDL}}$$ to $${\mathsf{KA_{{\mathrm{c}}}}}$$. Let $$(T,{\wedge},\vee,{\sim},{\mathrm{c}},0,1)\in {\mathsf{KA_{{\mathrm{c}}}}}$$. The set   $${\mathrm{C}}(T):=\{x\in T:x\geq {\mathrm{c}}\}$$ is the universe of a subalgebra of $$(T,{\wedge},\vee,{\mathrm{c}},1)$$ and $$({\mathrm{C}}(T),{\wedge},\vee,{\mathrm{c}},1) \in {\mathsf{BDL}}$$. Moreover, if $$g:T{\rightarrow} U$$ is a morphism in $${\mathsf{KA_{{\mathrm{c}}}}}$$, then the map $${\mathrm{C}}(g): {\mathrm{C}}(T) {\rightarrow} {\mathrm{C}}(U)$$, given by $${\mathrm{C}}(g)(x) = g(x)$$, is a morphism in $${\mathsf{BDL}}$$. Thus, $${\mathrm{C}}$$ is a functor from $${\mathsf{KA_{{\mathrm{c}}}}}$$ to $${\mathsf{BDL}}$$. Let $$H \in {\mathsf{BDL}}$$. The map $$\alpha_{H}: H {\rightarrow} {\mathrm{C}}({\mathrm{K}}(H))$$ given by $$\alpha_{H}(a) = (a,0)$$ is an isomorphism in $${\mathsf{BDL}}$$. If $$T \in {\mathsf{KA_{{\mathrm{c}}}}}$$, then the map $$\beta_T: T{\rightarrow} {\mathrm{K}}({\mathrm{C}}(T))$$ given by $$\beta_T(x) = (x\vee {\mathrm{c}}, {\sim} x \vee {\mathrm{c}})$$ is injective and a morphism in $${\mathsf{KA_{{\mathrm{c}}}}}$$. It is not difficult to show that the functor $${\mathrm{K}}: {\mathsf{BDL}} {\rightarrow} {\mathsf{KA_{{\mathrm{c}}}}}$$ has as left adjoint the functor $${\mathrm{C}}: {\mathsf{KA_{{\mathrm{c}}}}} {\rightarrow} {\mathsf{BDL}}$$ with unit $$\beta$$ and counit $$\alpha^{-1}$$. We are interested though in an equivalence between $${\mathsf{BDL}}$$ and the full subcategory of $${\mathsf{KA_{{\mathrm{c}}}}}$$ whose objects satisfy the condition (CK) we proceed to state. Let $$T\in {\mathsf{KA_{{\mathrm{c}}}}}$$. We consider the algebraic condition:   \begin{equation} \label{eq-CK} (\forall x, y \geq c)(x{\wedge} y = {\mathrm{c}} \ \longrightarrow \ (\exists z)(z\vee {\mathrm{c}} = x \ \& \ {\sim} z \vee {\mathrm{c}} = y)). \end{equation} (CK) This condition characterizes the surjectivity of $$\beta_T$$, i.e., for every $$T\in {\mathsf{KA_{{\mathrm{c}}}}}$$, $$T$$ satisfies (CK) if and only if $$\beta_T$$ is a surjective map, as shown in [19]. The condition (CK) is not necessarily verified in every centered Kleene algebra (see [5]). We write $${\mathsf{KA_{{\mathrm{c}}}^{CK}}}$$ for the full subcategory of $${\mathsf{KA_{{\mathrm{c}}}}}$$ whose objects satisfy (CK). The functor $${\mathrm{K}}$$ can then be seen as a functor from $${\mathsf{BDL}}$$ to $${\mathsf{KA_{{\mathrm{c}}}^{CK}}}$$. The next theorem was proved by M. Sagastume in [19]. A complete proof of it can be also found in [5]. Theorem 1 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{BDL}}$$ and $${\mathsf{KA_{{\mathrm{c}}}^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Let $$T\in {\mathsf{KA_{{\mathrm{c}}}}}$$. We know that $$\beta_T$$ is not necessarily a surjective map. However we will prove that $$\beta_T$$ is an epimorphism. Before, we need a lemma that is interesting in its own right. It tells us that the morphisms in $${\mathsf{KA_{{\mathrm{c}}}}}$$ are determined by their behaviour on the elements greater than or equal to the center. Lemma 2 If $$f:T{\rightarrow} U$$ and $$g:T{\rightarrow} U$$ are morphisms in $${\mathsf{KA_{{\mathrm{c}}}}}$$ and $$f(x) = g(x)$$ whenever $$x\in {\mathrm{C}}(T)$$, then $$f(x) = g(x)$$ for every $$x\in T$$. Proof. Suppose that $$f(x) = g(x)$$ whenever $$x\in {\mathrm{C}}(T)$$. Let $$x$$ be an arbitrary element of $$T$$. Then   \[ \begin{array} [c]{lllll} f(x) \vee {\mathrm{c}} & = & f(x) \vee f({\mathrm{c}}) & & \\ & = & f(x\vee {\mathrm{c}})& & \\ & = & g(x\vee {\mathrm{c}})& & \\ & = & g(x) \vee g({\mathrm{c}})& & \\ & = & g(x) \vee {\mathrm{c}},& & \end{array} \] so we obtain that $$f(x) \vee {\mathrm{c}} = g(x) \vee {\mathrm{c}}$$. Similarly we can prove that $${\sim} f(x) \vee {\mathrm{c}} = {\sim} g(x) \vee {\mathrm{c}}$$, which is equivalent to $$f(x) {\wedge} {\mathrm{c}} = g(x) {\wedge} {\mathrm{c}}$$. Hence, it follows from the distributivity of the underlying lattice of $$U$$ that $$f(x) = g(x)$$. ■ Proposition 3 Let $$T\in {\mathsf{KA_{{\mathrm{c}}}}}$$. Then $$\beta_T$$ is an epimorphism. Proof. Let $$f: {\mathrm{K}}({\mathrm{C}}(T)) {\rightarrow} U$$ and $$g:{\mathrm{K}}({\mathrm{C}}(T)){\rightarrow} U$$ be morphisms in $${\mathsf{KA_{{\mathrm{c}}}}}$$ such that $$f \circ \beta_T = g \circ \beta_T$$, where $$\circ$$ denotes the composition of functions. We will prove that $$f = g$$. Let $$(x,y) \in {\mathrm{C}}({\mathrm{K}}({\mathrm{C}}(T)))$$, i.e., $$x{\wedge} y = {\mathrm{c}}$$, $$x\geq {\mathrm{c}}$$, $$y \geq {\mathrm{c}}$$ and $$({\mathrm{c}},{\mathrm{c}}) \leq (x,y)$$, where we also write $$\leq$$ for the order associated with the underlying lattice of $${\mathrm{K}}({\mathrm{C}}(T))$$. In particular we have that $$y\leq {\mathrm{c}}$$, so $$y = {\mathrm{c}}$$. Then $$(x,y) = (x,{\mathrm{c}})$$. Besides, since $$x\geq {\mathrm{c}}$$ we have that $$\beta_{T}(x) = (x,{\mathrm{c}})$$. Then   \[ \begin{array} [c]{lllll} f(x,y) & = & f(x,{\mathrm{c}}) & & \\ & = & (f \circ \beta_{T})(x)& & \\ & = & (g \circ \beta_{T})(x)& & \\ & = & g(x, {\mathrm{c}})& & \\ & = & g(x,y).& & \end{array} \] Hence, $$f(x,y) = g(x,y)$$ whenever $$(x,y) \in {\mathrm{C}}({\mathrm{K}}({\mathrm{C}}(T)))$$. Therefore, it follows from Lemma 2 that $$f(x,y) = g(x,y)$$ for every $$(x,y) \in {\mathrm{K}}({\mathrm{C}}(T))$$, which was our aim. ■ Let $$H \in {\mathsf{BDL}}$$ and $$a$$, $$b\in H$$. If the relative pseudocomplement of $$a$$ with respect to $$b$$ exists, then we denote it by $$a {\rightarrow}_{{\mathsf{HA}}} b$$. Recall that a Nelson algebra [8] is a Kleene algebra such that for each pair $$x$$, $$y$$ there exists the binary operation $${\Rightarrow}$$ given by $$x{\Rightarrow} y: = x {\rightarrow}_{{\mathsf{HA}}} ({\sim x} \vee y)$$ and for every $$x,y,z$$ it holds that $$(x {\wedge} y)\Rightarrow z = x \Rightarrow (y\Rightarrow z)$$. The binary operation $$\Rightarrow$$ so defined is called the weak implication. We denote by $${\mathsf{HA}}$$ the category of Heyting algebras. M. Fidel [11] and D. Vakarelov [23] proved independently that if $$H\in {\mathsf{HA}}$$, then the Kleene algebra $${\mathrm{K}}(H)$$ is a Nelson algebra, in which the weak implication is defined for pairs $$(a, b)$$ and $$(d, e$$) in $${\mathrm{K}}(H)$$ as follows:   \begin{equation} \label{ic} (a,b)\Rightarrow(d,e):= (a{\rightarrow} d, a{\wedge} e). \end{equation} (1) We say that an algebra $$(T,{\wedge},\vee,\Rightarrow, {\sim},{\mathrm{c}},0,1)$$ is a centered Nelson algebra if the reduct $$(T,{\wedge},\vee,\Rightarrow, {\sim},0,1)$$ is a Nelson algebra and $${\mathrm{c}}$$ satisfies $${\sim} {\mathrm{c}} = {\mathrm{c}}$$. We write $${\mathsf{NA_{{\mathrm{c}}}}}$$ for the category of centered Nelson algebras. The following result appears in [3, Proposition 3.7] and is a reformulation of [8, Theorem 3.14]. Theorem 4 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{HA}}$$ and $${\mathsf{NA_{{\mathrm{c}}}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. We assume the reader is familiar with commutative residuated lattices [13]. An involutive residuated lattice is a bounded, integral and commutative residuated lattice $$(T,{\wedge}, \vee, \ast,{\rightarrow}, 0, 1)$$ such that for every $$x\in T$$ it holds that $$\neg \neg x = x$$, where $$\neg x: = x{\rightarrow} 0$$ and $$0$$ is the first element of $$T$$ [2]. In an involutive residuated lattice it holds that $$x \ast y = \neg (x {\rightarrow} \neg y)$$ and $$x{\rightarrow} y = \neg (x \ast \neg y)$$. A Nelson lattice [2] is an involutive residuated lattice $$(T,{\wedge},\vee, *,{\rightarrow},0,1)$$ which satisfies the additional inequality $$(x^2 {\rightarrow} y){\wedge} ((\neg y)^2 {\rightarrow} \neg x) \leq x{\rightarrow} y$$, where $$x^2:=x\ast x$$. See also [23]. Remark 5 Let $$(T,{\wedge}, \vee, \Rightarrow,{\sim}, 0,1)$$ be a Nelson algebra. We define on $$T$$ the binary operations $$*$$ and $${\rightarrow}$$ by   \begin{eqnarray*} x*y:=& {\sim} (x \Rightarrow {\sim} y) \vee {\sim} (y \Rightarrow {\sim} x), \hspace{1cm} x {\rightarrow} y :=& (x \Rightarrow y) {\wedge} ({\sim} y\Rightarrow {\sim} x). \end{eqnarray*} Then Theorem 3.1 of [2] says that $$(T,{\wedge}, \vee, {\rightarrow},*, 0,1)$$ is a Nelson lattice. Moreover, $${\sim} x = \neg x = x{\rightarrow} 0$$. Let $$(T,{\wedge},\vee,*,{\rightarrow},0,1)$$ be a Nelson lattice. We define on $$T$$ a binary operation $$\Rightarrow$$ and a unary operation $$\sim$$ by   \begin{eqnarray*} x \Rightarrow y:=& x^2 {\rightarrow} y, \hspace{1cm} {\sim} x:=& \neg x, \end{eqnarray*} where $$x^2 = x*x$$. Then Theorem 3.6 of [2] says that the algebra $$(T,{\wedge}, \vee,\Rightarrow,{\sim},0,1)$$ is a Nelson algebra. In [2, Theorem 3.11] it was also proved that the category of Nelson algebras and the category of Nelson lattices are isomorphic. Taking into account the construction of this isomorphism in [2] we have that the variety of Nelson algebras and the variety of Nelson lattices are term equivalent and the term equivalence is given by the operations we have defined before. The results from [2] about the connections between Nelson algebras and Nelson lattices mentioned in Remark 5 are based on results from Spinks and Veroff [22]. In particular, the term equivalence of the varieties of Nelson algebras and Nelson lattices was discovered by Spinks and Veroff in [22]. A centered Nelson lattice is an algebra $$(T,{\wedge},\vee,*,{\rightarrow},{\mathrm{c}},0,1)$$, where the reduct $$(T,{\wedge},\vee,*,{\rightarrow},0,1)$$ is a Nelson lattice and $${\mathrm{c}}$$ is an element such that $$\neg {\mathrm{c}} = {\mathrm{c}}$$. It follows from Remark 5 that the variety of centered Nelson algebras and the variety of centered Nelson lattices are term equivalent. We write $${\mathsf{NL_{{\mathrm{c}}}}}$$ for the category of centered Nelson lattices. Remark 6 Let $$(H,{\wedge}, \vee,{\rightarrow},0,1) \in {\mathsf{HA}}$$. Then $$({\mathrm{K}}(H),{\wedge},\vee,{\Rightarrow},{\sim},{\mathrm{c}},0,1) \in {\mathsf{NA_{{\mathrm{c}}}}}$$. Hence it follows from Remark 5 that $$({\mathrm{K}}(H),{\wedge},\vee,*,{\rightarrow},{\mathrm{c}},0,1) \in {\mathsf{NL_{{\mathrm{c}}}}}$$, where for $$(a,b)$$ and $$(d,e)$$ in $${\mathrm{K}}(H)$$ the operations $$\ast$$ and $${\rightarrow}$$ take the form   \begin{eqnarray*} (a,b) \ * \ (d,e) = (a{\wedge} d, (a{\rightarrow} e) \ {\wedge} \ (d{\rightarrow} b)),\\ (a,b) {\rightarrow} (d,e) = ((a{\rightarrow} d){\wedge} (e{\rightarrow} b), a {\wedge} e). \end{eqnarray*} We write $${\rightarrow}$$ both for the implication in $$H$$ as for the implication in $${\mathrm{K}}(H)$$. It follows from Theorem 4 and Remark 5 that there is a categorical equivalence between $${\mathsf{HA}}$$ and $${\mathsf{NL_{{\mathrm{c}}}}}$$, as it was also mentioned in [5, Corollary 2.11]. In what follows we will make explicit a construction of this equivalence. Proposition 7 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{HA}}$$ and $${\mathsf{NL_{{\mathrm{c}}}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Proof. Let $$H \in {\mathsf{HA}}$$. Then the centered Kleene algebra $$({\mathrm{K}}(H),{\wedge},\vee,{\sim},{\mathrm{c}},0,1)$$ endowed with the two operations given in Remark 6 is a centered Nelson lattice. It is immediate that if $$f$$ is a morphism in $${\mathsf{HA}}$$, then $${\mathrm{K}}(f)$$ is a morphism in $${\mathsf{NL_{{\mathrm{c}}}}}$$. Let $$(T,{\wedge},\vee,*,{\rightarrow}, {\mathrm{c}},0,1)\in {\mathsf{NL_{{\mathrm{c}}}}}$$. Taking into account Remark 5 we deduce that $$(T,{\wedge}, \vee,{\Rightarrow}, {\sim}, {\mathrm{c}},0,1)\in {\mathsf{NA_{{\mathrm{c}}}}}$$, where $$x {\Rightarrow} y = x^{2} {\rightarrow} y$$. Moreover,   \begin{equation} \label{nanl1} x {\rightarrow} y = (x {\Rightarrow} y) {\wedge} ({\sim} y {\Rightarrow} {\sim x}). \end{equation} (2) Let $$x$$, $$y\geq {\mathrm{c}}$$. We will prove that $$x {\rightarrow} y = x {\rightarrow}_{{\mathsf{HA}}} y$$. To show it, note that straightforward computations show that   \begin{equation} \label{nanl2} x{\Rightarrow} y = x {\rightarrow}_{{\mathsf{HA}}} \ y. \end{equation} (3) Besides, $${\sim} y \Rightarrow {\sim} x = {\sim} y {\rightarrow}_{{\mathsf{HA}}} (y \vee {\sim} x)$$. Since $$y\geq {\mathrm{c}}$$ and $${\sim} x \leq {\mathrm{c}}$$, then $$y \vee {\sim} x = y$$. Then $${\sim} y \Rightarrow {\sim} x = {\sim} y {\rightarrow}_{{\mathsf{HA}}} y$$. Hence, it follows from (2) and (3) that   $$ x{\rightarrow} y = (x{\rightarrow}_{{\mathsf{HA}}} y) {\wedge} ({\sim} y {\rightarrow}_{{\mathsf{HA}}} y).$$ Note that $$x{\rightarrow} y = x{\rightarrow}_{{\mathsf{HA}}} y$$ if and only if $$x{\rightarrow}_{{\mathsf{HA}}} y \leq {\sim} y {\rightarrow} y$$, which is equivalent to $${\sim} y {\wedge} (x{\rightarrow}_{{\mathsf{HA}}} y) \leq y$$. But $${\sim} y \leq {\mathrm{c}}$$ and $$x {\rightarrow}_{{\mathsf{HA}}} y \geq {\mathrm{c}}$$, so $${\sim} y {\wedge} (x{\rightarrow}_{{\mathsf{HA}}} y) = {\sim} y$$. Hence, $${\sim} y {\wedge} (x{\rightarrow}_{{\mathsf{HA}}} y) \leq y$$ if and only if $${\sim} y \leq y$$. Since $$y\geq {\mathrm{c}}$$, then $${\sim} y \leq {\mathrm{c}}$$, so $${\sim} y \leq y$$. Then we have that $$x{\rightarrow} y = x {\rightarrow}_{{\mathsf{HA}}} y$$. Thus, $$({\mathrm{C}}(T), {\wedge}, \vee,{\rightarrow},{\mathrm{c}},1) \in {\mathsf{HA}}$$. Straightforward computations show that if $$g$$ is a morphism in $${\mathsf{NL_{{\mathrm{c}}}}}$$, then $${\mathrm{C}}(g)$$ is a morphism in $${\mathsf{HA}}$$. It is also immediate that if $$H\in {\mathsf{HA}}$$ then $$\alpha_H$$ is an isomorphism in $${\mathsf{HA}}$$. Let $$(T,{\wedge}, \vee,*,{\rightarrow}, {\mathrm{c}},0,1)\in {\mathsf{NL_{{\mathrm{c}}}}}$$. It follows from Theorem 4 and Remark 5 that $$\beta_T$$ preserves $${\rightarrow}$$. Therefore, $$\beta_T$$ is an isomorphism in $${\mathsf{NL_{{\mathrm{c}}}}}$$. ■ The main goal of this article is to find a generalization of Proposition 7 replacing the categories of Heyting algebras and centered Nelson lattices by the categories of bounded hemi-implicative semilattices and hemi-implicative lattices, respectively. To make it possible, we start studying an equivalence for a particular category of posets with first element. Then we employ it to obtain an equivalence for the category of bounded semilattices. Finally, taking into account the last mentioned equivalence, we build up an equivalence for the categories of bounded hemi-implicative semilattices and hemi-implicative lattices, respectively, and for some of its full subcategories. 3 Kalman’s functor for posets with bottom and for bounded semilattices In this section, we generalize the equivalence given for the category of bounded distributive lattices but replacing this category by the category whose objects are posets with first element and whose morphisms are maps which preserve finite existing infima and the first element. Then we apply this equivalence to establish an equivalence for the category whose objects are bounded semilattices and whose morphisms are the corresponding algebra homomorphisms. We start with some preliminary definitions and properties. Let $$(P,\leq,0)$$ be a poset with first element, and let $$(P \times P,\preceq)$$ be the poset with universe the cartesian product $$P\times P$$ where the order $$\preceq$$ is given by   \[ (a,b) \preceq (d,e)\; \text{if and only if}\; a\leq d \ \text{and} \ e \leq b. \] In other words, $$(P \times P,\preceq)$$ is the direct product of $$(P,\leq)$$ with its dual. Let $$(P,\leq)$$ and $$(Q,\leq)$$ be posets. Let $$f:(P,\leq) {\rightarrow} (Q,\leq)$$ be a function. We say that $$f$$preserves finite existing infima if for every $$a$$, $$b\in P$$ such that $$a{\wedge} b$$ exists in $$P$$ then $$f(a){\wedge} f(b)$$ exists in $$Q$$ and $$f(a{\wedge} b) = f(a){\wedge} f(b)$$. Definition 8 The category $${\mathsf{P_0}}$$ has as objects the posets with first element and has as morphisms the maps between posets with first element which preserve the finite existing infima and the first element. Note that every morphism in $${\mathsf{P_0}}$$ preserves the order. It follows from the fact that morphisms preserve the finite existing infima. Let $$P \in {\mathsf{P_0}}$$. We define the following set:   \begin{equation} {\mathrm{K}}(P): = \{(a,b)\in P\times P: a{\wedge} b \ \text{exists and} \ a{\wedge} b = 0\}. \end{equation} (4) This set is the natural one to associate with the poset $$P$$ if we aim to generalize the original Kalman’s construction given for bounded distributive lattices [15]. To attain the generalization we first order $${\mathrm{K}}(P)$$ with the order induced by the poset $$(P \times P,\preceq)$$ defined above. It is immediate from the definition that $${\mathrm{K}}(P)$$ is closed under the unary operation $${\sim}$$ on $$P \times P$$ given by $${\sim}(a, b) = (b, a)$$, and that the element $${\mathrm{c}} = (0, 0)$$ belongs to $${\mathrm{K}}(P)$$. Thus we obtain the structure $${\mathrm{K}}(P) := ({\mathrm{K}}(P),\preceq,{\sim}, {\mathrm{c}}).$$ The following elemental lemma plays a fundamental role in some proofs of this section. Lemma 9 Let $$(b,d) \in {\mathrm{K}}(P)$$. The following conditions hold: (1) For every $$a \in P$$, $$(a,0) {\wedge} (b,d)$$ exists in $${\mathrm{K}}(P)$$ if and only if $$a{\wedge} b$$ exists in $$P$$. If these conditions hold, then $$(a,0) {\wedge} (b, d) = (a{\wedge} b, d)$$. (2) $$(b,d) {\wedge} (0,0)$$ exists in $${\mathrm{K}}(P)$$ and $$(b,d) {\wedge} (0,0) = (0,d)$$. (3) $$(b,d) \vee (0,0)$$ exists in $${\mathrm{K}}(P)$$ and $$(b,d) \vee (0,0) = (b,0).$$ Proof. In general, if $$y = (e, u) \in {\mathrm{K}}(P)$$, we write $$\pi_1(y)$$ for the first coordinate and $$\pi_2(y)$$ for the second coordinate (i.e. $$\pi_1(y) = e$$ and $$\pi_2(y) = u$$). Let $$(b,d) \in {\mathrm{K}}(P)$$. We proceed to the proof. (1) Suppose that $$a \in P$$ and $$(a,0) {\wedge} (b,d)$$ exists in $${\mathrm{K}}(P)$$. Let $$x := (a,0) {\wedge} (b,d)$$. We have that $$x \preceq (a,0)$$ and $$x\preceq (b,d)$$, so by the definition of $$\preceq$$ we obtain that $$\pi_1(x) \leq a$$ and $$\pi_1(x) \leq b$$. Thus $$\pi_1(x)$$ is a lower bound of the set $$\{a,b\}$$. Let now $$e$$ be a lower bound of $$\{a,b\}$$, i.e., $$e\leq a$$ and $$e\leq b$$. The fact that $$(b,d) \in {\mathrm{K}}(P)$$ implies that $$(e,d) \in {\mathrm{K}}(P)$$. Since $$(e,d)\preceq (a,0)$$ and $$(e,d) \preceq (b,0)$$, then $$(e,d)\preceq x$$. Hence, $$e\leq \pi_1(x)$$. Thus, $$\pi_1(x) = a{\wedge} b$$. Conversely, suppose that $$a{\wedge} b$$ exists in $$P$$. We have that $$(a{\wedge} b,d) \in {\mathrm{K}}(P)$$, $$(a{\wedge} b,d) \preceq (a,0)$$, and $$(a{\wedge} b,d) \preceq (b,d)$$. Let $$(e, u) \in {\mathrm{K}}(P)$$ such that $$(e,u)\preceq (a,0)$$ and $$(e,u) \preceq (b,d)$$, i.e., $$e\leq a$$, $$e\leq b$$ and $$d\leq u$$. Since $$e \leq a{\wedge} b$$, then $$(e,u) \preceq (a{\wedge} b, d)$$. Hence we obtain that $$(a,0){\wedge} (b,d)$$ exists in $${\mathrm{K}}(P)$$, and moreover $$(a,0){\wedge} (b,d) = (a{\wedge} b,d)$$. (2) To prove that $$(b,d) {\wedge} (0,0)$$ exists in $${\mathrm{K}}(P)$$, let us see that $$(0,d)$$ is the infimum of $$(b,d)$$ and $$(0,0)$$. We have that $$(0,d)\preceq (b,d)$$ and $$(0,d)\preceq (0,0)$$, so $$(0,d)$$ is a lower bound of $$\{(b,d), (0,0)\}$$. Let $$(e,u) \in {\mathrm{K}}(P)$$ be a lower bound of $$\{(b,d), (0,0)\}$$, so $$(e,u) \preceq (b,d)$$ and $$(e,u) \preceq (0,0)$$. Hence, $$e = 0$$ and $$d\leq u$$, and so $$(e,u) \leq (0,d)$$. Therefore we obtain that $$(b,d) {\wedge} (0,0) = (0,d)$$. (3) In a similar way it can be proved that $$(b,d) \vee (0,0)$$ exists in $${\mathrm{K}}(P)$$ and is $$(b,0)$$. ■ Motivated by properties of $${\mathrm{K}}(P)$$ we give the following definition. Definition 10 A structure $$(T,\leq,{\sim},{\mathrm{c}})$$ is a Kleene poset if the following conditions hold: (1) $$(T,\leq)$$ is a poset. (2) $${\sim}$$ is an unary operation on $$T$$ which is an involution, i.e., $${\sim} {\sim} x = x$$ for every $$x\in T$$ and is order reversing, i.e., for every $$x,y\in T$$, if $$x\leq y$$, then $${\sim} y \leq {\sim} x$$. (3) $${\mathrm{c}} = {\sim} {\mathrm{c}}$$. (4) $$x\vee {\mathrm{c}}$$ exists, for every $$x \in T$$. (5) For every $$x\in T$$, $$(x\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}})$$ exists and $$(x\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}) = {\mathrm{c}}$$. (6) For every $$x,y \in T$$, if $$x{\wedge} {\mathrm{c}} \leq y {\wedge} {\mathrm{c}}$$ and $$x\vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$$, then $$x \leq y$$. The element $${\mathrm{c}}$$ of the previous definition will be also called center. The next lemma justifies the use of $${\wedge}$$ in the statement of the condition 6. Lemma 11 Let $$(T, \leq)$$ be a poset satisfying $$2$$., $$3$$., $$4$$. and $$5$$. of Definition 10. (1) Let $$x,y \in T$$. If $$x{\wedge} y$$ exists, then $${\sim} x \vee {\sim} y$$ exists and $${\sim} x \vee {\sim} y = {\sim} (x{\wedge} y)$$. Analogously, if $$x\vee y$$ exists, then $${\sim} x {\wedge}dge {\sim} y$$ exists and $${\sim} x {\wedge}dge {\sim} y = {\sim} (x\vee y)$$. (2) For every $$x\in T$$, $$x{\wedge} {\mathrm{c}}$$ exists and $$x{\wedge} {\mathrm{c}} = {\sim} ({\sim} x \vee {\mathrm{c}})$$. (3) The element $${\mathrm{c}}$$ is unique. Proof. Straightforward computations show the first two assertions. To prove that the centre is unique, let $${\mathrm{c}}$$ and $${\mathrm{c}}'$$ be centres. Then $${\mathrm{c}} = ({\mathrm{c}}' \vee {\mathrm{c}}){\wedge} ({\sim} {\mathrm{c}}' \vee {\mathrm{c}})$$. Since $${\sim} {\mathrm{c}}' = {\mathrm{c}}'$$, then $${\mathrm{c}} = {\mathrm{c}}' \vee {\mathrm{c}}$$. Hence, $${\mathrm{c}}'\leq {\mathrm{c}}$$. Analogously, $${\mathrm{c}}\leq {\mathrm{c}}'$$. Thus, $${\mathrm{c}} = {\mathrm{c}}'$$. ■ In what follows we introduce the category of Kleene posets. Definition 12 We denote by $${\mathsf{KP}}$$ the category whose objects are the Kleene posets and whose morphisms are the maps $$g$$ between Kleene posets that preserve the order, the involution and the finite existing infima over elements greater than or equal to the center. Note that if $$g:T{\rightarrow} U$$ is a morphism in $${\mathsf{KP}}$$, $${\mathrm{c}}$$ is the center of $$T$$ and $${\mathrm{c}}'$$ is the center of $$U$$, then $$g({\mathrm{c}}) = {\mathrm{c}}'$$. It follows from the fact that $$g({\mathrm{c}}) = {\sim} g({\mathrm{c}})$$ and the fact that the center is unique. If $$T\in {\mathsf{KP}}$$, we define $${\mathrm{C}}(T)$$ as in the case of centered Kleene algebras. If $$f$$ is a morphism in $${\mathsf{P_0}}$$ and $$g$$ is a morphism in $${\mathsf{KP}}$$ we define $${\mathrm{K}}(f)$$ and $${\mathrm{C}}(g)$$ as in Section 1, respectively. Lemma 13 (a) If $$(P,\leq,0) \in {\mathsf{P_0}}$$, then $$({\mathrm{K}}(P),\preceq,{\sim}, {\mathrm{c}}) \in {\mathsf{KP}}$$. (b) If $$f \in {\mathsf{P_0}}$$, then $${\mathrm{K}}(f) \in {\mathsf{KP}}$$. Proof. It follows from straightforward computations based on Lemma 9 that if $$(P,\leq,0) \in {\mathsf{P_0}}$$, then $$({\mathrm{K}}(P),\preceq,{\sim},$$$$ {\mathrm{c}}) \in {\mathsf{KP}}$$. In what follows we will prove that if $$f:P {\rightarrow} Q$$ is a morphism in $${\mathsf{P_0}}$$, then $${\mathrm{K}}(f):{\mathrm{K}}(P){\rightarrow} {\mathrm{K}}(Q)$$ is a morphism in $${\mathsf{KP}}$$. By the definition of $${\mathsf{P_0}}$$ we have that if $$(a,b) \in {\mathrm{K}}(P)$$, then $$f(a) {\wedge}dge f(b)$$ exists in $$Q$$ and is $$f(a{\wedge} b) = f(0)$$, which is $$0$$ in $$Q$$. Thus, $$(f(a),f(b))\in {\mathrm{K}}(Q)$$. Therefore, $${\mathrm{K}}(f)$$ is indeed a map from $${\mathrm{K}}(P)$$ to $${\mathrm{K}}(Q)$$. Since $$f$$ preserves the order, then $${\mathrm{K}}(f)$$ preserves the order. It is immediate that $${\mathrm{K}}(f)$$ preserves the involution. Let $$(a,0)$$ and $$(b,0)$$ be elements such that $$(a,0) {\wedge} (b,0)$$ exists. It follows from Lemma 9 that $$a{\wedge} b$$ exists and $$(a,0) {\wedge} (b,0) = (a{\wedge} b,0)$$. Then, $$f(a){\wedge} f(b)$$ exists and $$f(a{\wedge} b) = f(a){\wedge} f(b)$$. Again by Lemma 9 we obtain that $$(f(a),0){\wedge} (f(b),0)$$ exists and $$(f(a),0){\wedge} (f(b),0) = (f(a){\wedge} f(b),0)$$. Hence, $${\mathrm{K}}(f)((a,0){\wedge} (b,0)) = {\mathrm{K}}(f)(a,0) {\wedge} {\mathrm{K}}(f)(b,0)$$. Therefore, $${\mathrm{K}}(f)$$ is a morphism in $${\mathsf{KP}}$$. ■ Using the previous lemma, it is immediate to see that $${\mathrm{K}}$$ defines a functor from $${\mathsf{P_0}}$$ to $${\mathsf{KP}}$$. Let $$P \in {\mathsf{P_0}}$$. The map $$\alpha_P:P {\rightarrow} {\mathrm{C}}({\mathrm{K}}(P))$$ given by $$\alpha_{P}(a,b) = (a,0)$$ is easily seen to be an isomorphism in $${\mathsf{P_0}}$$. The fact that $$\alpha_P$$ is morphism is a consequence of Lemma 9. The proof of the following lemma is immediate. It easily follows from it that $${\mathrm{C}}: {\mathsf{KP}} {\rightarrow} {\mathsf{P_0}}$$ is a functor. Lemma 14 (a) If $$(T, \leq, {\sim},{\mathrm{c}}) \in {\mathsf{KP}}$$ then $$({\mathrm{C}}(T),\leq,{\mathrm{c}}) \in {\mathsf{P_0}}$$. (b) If $$g \in {\mathsf{KP}}$$ then $${\mathrm{C}}(g) \in {\mathsf{P_0}}$$. Definition 15 For $$T\in {\mathsf{KP}}$$ we also name $$({\mathrm{CK}})$$ to the following condition   \begin{equation} \label{eq-CK-2} (\forall x, y \geq c)(\text{if } x{\wedge} y \text{ exists and } x{\wedge} y = {\mathrm{c}}, \text{ then } (\exists z)(z\vee {\mathrm{c}} = x \ \& \ {\sim} z \vee {\mathrm{c}} = y)). \end{equation} (CK) Remark 16 In Section 2, the condition $$({\mathrm{CK}})$$ was defined for centered Kleene algebras. Notice that if $$(T,{\wedge},\vee,{\sim},{\mathrm{c}},0,1)$$ is a centered Kleene algebra, then the structure $$(T,\leq,{\sim},{\mathrm{c}})$$ is a Kleene poset, where $$\leq$$ is the order associated with the lattice $$(T,{\wedge},\vee)$$. In particular, we have that $$(T,{\wedge},\vee,{\sim},{\mathrm{c}},0,1)$$ satisfies the condition $$({\mathrm{CK}})$$ given in Section 2 if and only if $$(T,\leq,{\sim},{\mathrm{c}})$$ satisfies the condition $$({\mathrm{CK}})$$ given in Definition 15. This fact justifies the use of the same label for both conditions. As in the case of bounded distributive lattices, if $$(P,\leq,0) \in {\mathsf{P_0}}$$, then the structure $$({\mathrm{K}}(P),\preceq,{\sim}, {\mathrm{c}})$$ satisfies $$({\mathrm{CK}})$$. Remark 17 For $$T\in {\mathsf{KP}}$$ and $$x\in T$$ we have $$(x\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}) = {\mathrm{c}}$$, which shows that the map $$\beta_T:T {\rightarrow} {\mathrm{K}}({\mathrm{C}}(T))$$ defined by $$\beta_{T}(x) = (x\vee {\mathrm{c}},{\sim} x \vee {\mathrm{c}})$$ is a well-defined map. We also have that $$T$$ satisfies $$({\mathrm{CK}})$$ if and only if $$\beta_T$$ is surjective. Let $$f:(P,\leq) {\rightarrow} (Q,\leq)$$ be an order isomorphism, i.e., a bijective map such that for every $$a$$, $$b\in P$$, $$a\leq b$$ if and only if $$f(a)\leq f(b)$$. Let $$a$$, $$b\in P$$ such that $$a{\wedge} b$$ exists. Then $$f(a) {\wedge} f(b)$$ exists and $$f(a) {\wedge} f(b) = f(a{\wedge} b)$$. Straightforward computations prove the following remark. Remark 18 Let $$f:T{\rightarrow} U$$ be a morphism in $${\mathsf{KP}}$$. If $$f$$ is an order isomorphism, then $$f$$ is an isomorphism in $${\mathsf{KP}}$$. Lemma 19 If $$T \in {\mathsf{KP}}$$, then $$\beta_T$$ is an injective morphism in $${\mathsf{KP}}$$. Moreover, if $$T$$ satisfies $$({\mathrm{CK}})$$, then $$\beta_T$$ is an isomorphism in $${\mathsf{KP}}$$. Proof. To show that $$\beta_T$$ preserves the order, let $$x,y \in T$$ such that $$x\leq y$$. Then $$x\vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$$ and $${\sim} x \vee {\mathrm{c}} \geq {\sim} y \vee {\mathrm{c}}$$, which means that $$\beta_T(x) \leq \beta_T(y)$$. Thus, $$\beta_T$$ preserves the order. It is immediate that $$\beta_T$$ preserves the involution. Let now $$x,y \in T$$ such that $$x,y \geq {\mathrm{c}}$$. Assume that $$x{\wedge} y$$ exists. So $$\beta_T(x{\wedge} y) = (x{\wedge} y, {\mathrm{c}})$$. Moreover, we have that $$\beta_T(x) = (x,{\mathrm{c}})$$ and $$\beta_T(y) = (y,{\mathrm{c}})$$. Thus, it follows from Lemma 9 that $$(x,c) {\wedge} (y,c)$$ exists and $$(x,{\mathrm{c}}) {\wedge} (y,{\mathrm{c}}) = (x{\wedge} y, {\mathrm{c}})$$. Then $$\beta_T(x {\wedge} y) = \beta_T(x) {\wedge} \beta_T(y)$$. Hence, $$\beta_T$$ is a morphism in $${\mathsf{KP}}$$. Now we will prove that for every $$x,y \in T$$, $$x\leq y$$ if and only if $$\beta_T(x) \leq \beta_T(y)$$. The fact that if $$x\leq y$$, then $$\beta_T(x) \leq \beta_T(y)$$ was proved before. To prove the converse, suppose that $$\beta_T(x) \leq \beta_T(y)$$, i.e., $$x\vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$$ and $$x {\wedge} {\mathrm{c}} \leq y {\wedge} {\mathrm{c}}$$. So, by the definition of Kleene poset we have $$x\leq y$$. In particular, $$\beta_T$$ is an injective map. Finally, assume that $$T$$ satisfies $$({\mathrm{CK}})$$. It follows from remarks 17 and 18 that $$\beta_T$$ is an isomorphism in $${\mathsf{KP}}$$. ■ Straightforward calculations prove that if $$f:P{\rightarrow} Q$$ is a morphism in $${\mathsf{P_0}}$$ then $$ ({\mathrm{C}} \circ {\mathrm{K}})(f) \circ \alpha_{P} = \alpha_{Q} \circ f$$, and if $$g:T {\rightarrow} U$$ is a morphism in $${\mathsf{KP}}$$ then $$ ({\mathrm{K}} \circ {\mathrm{C}})(g) \circ \beta_{T} = \beta_U \circ g$$. Theorem 20 Let $${\mathsf{KP^{CK}}}$$ be the full subcategory of $${\mathsf{KP}}$$ whose objects satisfy the condition $$({\mathrm{CK}})$$. The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{P_0}}$$ and $${\mathsf{KP^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Let $${\mathsf{MS}}$$ be the category whose objects are bounded semilattices and whose morphisms are the algebra homomorphisms. Definition 21 We write $${\mathsf{KMS}}$$ to denote the category whose objects are the structures $$(T, \leq, {\sim},{\mathrm{c}}, 0,1)$$ which satisfy the following conditions: $${\mathrm{(KM1)}}$$$$(T, \leq, {\sim},{\mathrm{c}}) \in {\mathsf{KP}}$$. $${\mathrm{(KM2)}}$$$$0$$ is the first element of $$(T,\leq)$$ and $$1$$ is the greatest element of $$(T,\leq)$$. $${\mathrm{(KM3)}}$$ For every $$x,y \in T$$, if $$x\geq {\mathrm{c}}$$, then $$x{\wedge} y$$ exists. $${\mathrm{(KM4)}}$$ For every $$x,y \in T$$, if $$x\geq {\mathrm{c}}$$, then $$(x{\wedge} y)\vee {\mathrm{c}} = x{\wedge} (y\vee {\mathrm{c}})$$. The morphisms of $${\mathsf{KMS}}$$ are maps $$g$$ between objects of $${\mathsf{KMS}}$$ which preserve the order, the involution and such that for every $$x,y\geq {\mathrm{c}}$$, $$g(x{\wedge} y) = g(x) {\wedge} g(y)$$. We write $${\mathsf{KMS^{CK}}}$$ to denote the full subcategory of $${\mathsf{KMS}}$$ whose objects satisfy the condition $$({\mathrm{CK}})$$. Note that in presence of the condition $${\mathrm{(KM1)}}$$ we can replace the condition $${\mathrm{(KM3)}}$$ by the following condition: $$(x\vee {\mathrm{c}}){\wedge} y$$ exists for every $$x$$, $$y$$. Also note that in presence of the conditions $${\mathrm{(KM1)}}$$ and $${\mathrm{(KM3)}}$$, the condition $${\mathrm{(KM4)}}$$ can be replaced by the condition $$((x\vee {\mathrm{c}}){\wedge} y)\vee {\mathrm{c}} = (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}})$$ for every $$x$$, $$y$$. Recall that $${\mathsf{MS}}$$ is the category whose objects are bounded semilattices and whose morphisms are the algebra homomorphisms between them. Corollary 22 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{MS}}$$ and $${\mathsf{KMS^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Proof. Let $$H\in {\mathsf{MS}}$$. The condition $${\mathrm{(KM1)}}$$ for $${\mathrm{K}}(H)$$ follows from Theorem 20. We also have that $$(0,1)$$ is the first element of $${\mathrm{K}}(H)$$ and $$(1,0)$$ is the last element of $${\mathrm{K}}(H)$$, i.e., we have the condition $${\mathrm{(KM2)}}$$. Let $$x,y \in {\mathrm{K}}(H)$$ with $$x\geq {\mathrm{c}}$$. Then there are $$a,b,d\in H$$ such that $$x = (a,0)$$, $$y = (b,d)$$ and $$b{\wedge} d = 0$$. Since in particular $$a{\wedge} b$$ exists, then it follows from Lemma 9 that $$x {\wedge} y$$ exists and $$x{\wedge} y = (a {\wedge} b,d)$$. Then we have proved $${\mathrm{(KM3)}}$$. Again taking into account Lemma 9 we deduce that $$((a,0){\wedge} (b,d))\vee (0,0) = (a{\wedge} b,0)$$. The mentioned lemma also implies that $$(b,d) \vee (0,0) = (b,0)$$ and $$(a,0) {\wedge} (b,0) = (a{\wedge} b,0)$$. Thus, $$(x{\wedge} y)\vee {\mathrm{c}} = x{\wedge} (y\vee {\mathrm{c}})$$, which is $${\mathrm{(KM4)}}$$. Therefore $${\mathrm{K}}(H) \in {\mathsf{KMS}}$$. It is immediate that if $$T\in {\mathsf{KMS}}$$, then $${\mathrm{C}}(T) \in {\mathsf{MS}}$$. The rest of the proof follows from Theorem 20. ■ 4 The variety of hemi-implicative semilattices (lattices) In this section, we recall definitions and properties about the algebras we will consider later: hemi-implicative semilattices (lattices) [21], Hilbert algebras with infimum [12], implicative semilattices [17] and semi-Heyting algebras [20]. Definition 23 A hemi-implicative semilattice is an algebra $$(H,{\wedge},{\rightarrow}, 1)$$ of type $$(2,2,0)$$ which satisfies the following conditions: $${\mathrm{(W1)}}$$$$(H,{\wedge},1)$$ is an upper bounded semilattice, $${\mathrm{(W2)}}$$ for every $$a,b,d \in H$$, if $$a\leq b{\rightarrow} d$$ then $$a {\wedge} b \leq d$$, $${\mathrm{(W3)}}$$$$a{\rightarrow} a = 1$$ for every $$a\in H$$. A bounded hemi-implicative semilattice is an algebra $$(H,{\wedge},{\rightarrow},0,1)$$ of type $$(2,2,0,0)$$ such that $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice and $$0$$ is the first element with respect to the order. A hemi-implicative lattice is an algebra $$(H,{\wedge},\vee,{\rightarrow},0,1)$$ of type $$(2,2,2,0,0)$$ such that $$(H,{\wedge},\vee,0,1)\in {\mathsf{BDL}}$$ and $$(H,{\wedge},{\rightarrow},1)$$ is a hemi-implicative semilattice. Hemi-implicative semilattices were called weak implicative semilattices in [21]. We write $${\mathsf{hIS_{0}}}$$ for the category of bounded hemi-implicative semilattices and $${\mathsf{hBDL}}$$ for the category of hemi-implicative lattices. Remark 24 If $$(H,{\wedge})$$ is a semilattice and $${\rightarrow}$$ a binary operation, then $$H$$ satisfies $${\mathrm{(W2)}}$$ if and only if for every $$a,b \in A$$, $$a{\wedge} (a{\rightarrow} b)\leq b$$. Therefore, the class of hemi-implicative semilattices (lattices) is a variety [21]. The variety of Hilbert algebras is the algebraic counterpart of the implicative fragment of Intuitionistic Propositional Logic. These algebras were introduced in the early 50s by Henkin and Skolem for some investigations on the implication in intuitionistic logic and other non-classical logics [18]. In the 1960s, they were studied especially by Horn and Diego [10]. Definition 25 A Hilbert algebra is an algebra $$(H,{\rightarrow}, 1)$$ of type $$(2,0)$$ that satisfies the following conditions: (1) $$a{\rightarrow} (b{\rightarrow} a) = 1$$. (2) $$a{\rightarrow} (b{\rightarrow} d) = (a{\rightarrow} b){\rightarrow} (a{\rightarrow} d)$$. (3) If $$a{\rightarrow} b = b{\rightarrow} a = 1$$, then $$a = b$$. It is a well-known fact that Hilbert algebras form a variety. In every Hilbert algebra, we have the partial order defined by $$a\leq b$$ if and only if $$a{\rightarrow} b = 1$$. In particular, $$a{\rightarrow} a = 1$$ for every $$a$$. Example 26 In any poset $$(H,\leq)$$ with last element $$1$$ it is possible to define the following binary operation:   \[ a {\rightarrow} b = \begin{cases} 1, &\text{if $a \leq b$;}\\ b, &\text{if $a\nleq b$.} \end{cases} \] The structure $$(H,{\rightarrow} ,1)$$ is a Hilbert algebra. For the following definition see [12]. Definition 27 An algebra $$(H,{\wedge},{\rightarrow},1)$$ is a Hilbert algebra with infimum if the following conditions hold: (1) $$(H,{\rightarrow},1)$$ is a Hilbert algebra, (2) $$(H,{\wedge},1)$$ is an upper bounded semilattice, (3) For every $$a,b \in H$$, $$a\leq b$$ if and only if $$a{\rightarrow} b = 1$$, where $$\leq$$ is the semilattice order. An algebra $$(H,{\wedge},{\rightarrow},0,1)$$ of type $$(2,2,0,0)$$ is a bounded Hilbert algebra with infimum if $$(H,{\wedge},{\rightarrow},1)$$ is a Hilbert algebra with infimum and $$0$$ is the first element with respect to the induced order. In [12] it is proved that the class of Hilbert algebras with infimum is a variety. We note that this result also follows from the results given by P. M. Idziak in [14] for BCK-algebras with lattice operations. The following proposition can be found in [12, Theorem 2.1]. Proposition 28 Let $$(H,{\wedge},{\rightarrow},1)$$ be an algebra of type $$(2,2,0)$$. Then $$(H,{\wedge},{\rightarrow},1)$$ is a Hilbert algebra with infimum if and only if for every $$a,b,d\in H$$ the following conditions hold: (a) $$(H,{\rightarrow},1)$$ is a Hilbert algebra, (b) $$(H,{\wedge},1)$$ is an upper bounded semilattice, (c) $$a{\wedge} (a{\rightarrow} b) = a{\wedge} b$$, (d) $$a {\rightarrow} (b{\wedge} d) \leq (a{\rightarrow} b) {\wedge} (a{\rightarrow} d)$$. In every Hilbert algebra with infimum we have $$a{\rightarrow} a = 1$$ and $$a{\wedge} (a{\rightarrow} b) \leq b$$, so the variety of Hilbert algebras with infimum is a subvariety of the variety of hemi-implicative semilattices. We will write $${\mathsf{Hil_{0}}}$$ for the category whose objects are bounded Hilbert algebras with infimum and whose morphisms are the corresponding algebra homomorphisms. Clearly $${\mathsf{Hil_{0}}}$$ is a full subcategory of $${\mathsf{hIS_{0}}}$$. Definition 29 An implicative semilattice is an algebra $$(H, {\wedge}, {\rightarrow})$$ of type $$(2,2)$$ such that $$(H,{\wedge})$$ is a semilattice, and for every $$a,b,d\in H$$ we have that $$a{\wedge} b \leq d$$ if and only if $$a\leq b{\rightarrow} d$$. Implicative semilattices have a greatest element, denoted by $$1$$. In this article we shall include the constant $$1$$ in the language of the algebras. Implicative semilattices are the algebraic models of the implication-conjunction fragment of Intuitionistic Propositional Logic. For more details about these algebras see [9]. An algebra $$(H,{\wedge},{\rightarrow},0,1)$$ of type $$(2,2,0,0)$$ is a bounded implicative semilattice if $$(H,{\wedge},{\rightarrow},1)$$ is an implicative semilattice and $$0$$ is the first element with respect to the order. We write $${\mathsf{IS_{0}}}$$ for the category whose objects are bounded implicative semilattices and whose morphisms are the corresponding algebra homomorphisms. We have that $${\mathsf{IS_{0}}}$$ is a full subcategory of $${\mathsf{hIS_{0}}}$$. It is part of the folklore of the subject that the class of implicative semilattices is a variety. There are many ways to axiomatize the variety of implicative semilattices. In the following lemma we propose a possible axiomatization that will play an important role in the next section. Lemma 30 Let $$(H,{\wedge},1)$$ be an upper bounded semilattice and $${\rightarrow}$$ a binary operation on $$H$$. The following conditions are equivalent: (a) For every $$a,b,d \in H$$, $$a\leq b{\rightarrow} d$$ if and only if $$a{\wedge} b \leq d$$. (b) For every $$a,b,d \in H$$ the following conditions hold: (1) $$a{\wedge} (a{\rightarrow} b) \leq b$$, (2) $$a{\rightarrow} a = 1$$, (3) $$a{\rightarrow} (b{\wedge} d) = (a{\rightarrow} b) {\wedge} (a{\rightarrow} d)$$, (4) $$a\leq b {\rightarrow} (a{\wedge} b)$$. Proof. Assume the conditions (1), (2), (3) and (4) of (b). It follows from (1) that if $$a\leq b{\rightarrow} d$$, then $$a{\wedge} b \leq d$$. Suppose now that $$a{\wedge} b \leq d$$. It follows from (3) that $$b{\rightarrow} (a{\wedge} b) \leq b{\rightarrow} d$$. But by (4) we have that $$a\leq b {\rightarrow} (a{\wedge} b)$$, so $$a\leq b{\rightarrow} d$$. Then, $$a\leq b{\rightarrow} d$$ if and only if $$a{\wedge} b \leq d$$. For the converse of this property see [17]. ■ Remark 31 A moment of reflection shows that implicative semilattices are Hilbert algebras with infimum where the implication is the right residuum of the infimum, or equivalently, where the following equation holds [12]: $$a\leq b {\rightarrow} (a{\wedge} b)$$. Alternatively, it follows from Lemma 30 that an implicative semilattice is a hemi-implicative semilattice which satisfies $$a{\rightarrow} (b{\wedge} d) = (a{\rightarrow} b) {\wedge} (a{\rightarrow} d)$$ and $$a\leq b {\rightarrow} (a{\wedge} b)$$ for every $$a,b,d$$. Semi-Heyting algebras were introduced by H. P. Sankappanavar in [20] as an abstraction of Heyting algebras. These algebras share with Heyting algebras the following properties: they are pseudocomplemented and distributive lattices and their congruences are determined by the lattice filters. Definition 32 An algebra $$(H, {\wedge}, \vee, {\rightarrow}, 0, 1)$$ of type $$(2,2,2,0,0)$$ is a semi-Heyting algebra if the following conditions hold for every $$a,b,d$$ in $$H$$: $${\mathrm{(SH1)}}$$$$(H, {\wedge}, \vee, 0, 1)\in {\mathsf{BDL}}$$, $${\mathrm{(SH2)}}$$$$a{\wedge} (a{\rightarrow} b) = a {\wedge} b$$, $${\mathrm{(SH3)}}$$$$a{\wedge} (b{\rightarrow} d) = a {\wedge} ((a{\wedge} b) {\rightarrow} (a{\wedge} d))$$, $${\mathrm{(SH4)}}$$$$a{\rightarrow} a = 1$$. We write $${\mathsf{SH}}$$ for the category of semi-Heyting algebras. A semi-Heyting algebra can be seen as a hemi-implicative lattice which satisfies $${\mathrm{(SH2)}}$$ and $${\mathrm{(SH3)}}$$. Therefore, $${\mathsf{SH}}$$ is a full subcategory of $${\mathsf{hBDL}}$$. Remark 33 Implicative semilattices satisfy the inequality $$a\leq b {\rightarrow} (a{\wedge} b)$$ because $$a{\wedge} b \leq a{\wedge} b$$, or simply by Lemma 30. Semi-Heyting algebras also satisfy the inequality $$a\leq b {\rightarrow} (a{\wedge} b)$$. This fact follows from $${\mathrm{(SH3)}}$$ and $${\mathrm{(SH4)}}$$ in the following way:   \[ \begin{array} [c]{lllll} a {\wedge} (b {\rightarrow} (a{\wedge} b)) & = & a{\wedge} ((a{\wedge} b) {\rightarrow} (a{\wedge} b)) & & \\ & = & a{\wedge} 1& & \\ & = & a,& & \end{array} \] which means that $$a\leq b {\rightarrow} (a{\wedge} b)$$. In the following example we will show the following facts: $${\mathsf{Hil_{0}}}$$ is a proper subvariety of $${\mathsf{hIS_{0}}}$$ and $${\mathsf{SH}}$$ is a proper subvariety of $${\mathsf{hBDL}}$$. Example 34 Let $$H$$ be the chain of three elements with $$0<a<1$$. We define on $$H$$ the following binary operation:   \[ \begin{array} [c]{c|ccc} {\rightarrow} & 0 & a & 1\\\hline 0 & 1 & a & 1\\ a & 0 & 1 & 1\\ 1 & 0 & 0 & 1 \end{array} \] Straightforward computations show that $$(H, {\wedge}, \vee,{\rightarrow}, 0,1) \in {\mathsf{hBDL}}$$. In particular, $$(H,{\wedge},{\rightarrow},0,1) \in {\mathsf{hIS_{0}}}$$. However, $$(H,{\wedge},{\rightarrow},0,1) \notin {\mathsf{Hil_{0}}}$$ and $$(H,{\wedge},\vee,{\rightarrow},0,1) \notin {\mathsf{SH}}$$ because $$1{\wedge} (1{\rightarrow} a) \neq 1{\wedge} a$$. It is a known fact that the variety $${\mathsf{IS_{0}}}$$ is properly included in $${\mathsf{Hil_{0}}}$$. We give an example that shows it. Let $$H$$ be the universe of the boolean lattice of four elements, where $$a$$ and $$b$$ are the atoms. Then $$(H,{\wedge},{\rightarrow},0,1) \in {\mathsf{Hil_{0}}}$$, where $${\rightarrow}$$ is the operation defined in Example 26. Since $$a{\rightarrow} 0 = 0$$ and $$0\neq b$$, then $$(H,{\wedge},{\rightarrow},0,1) \notin {\mathsf{IS_{0}}}$$. It is also a known fact that $${\mathsf{HA}}$$ is a proper subvariety of $${\mathsf{SH}}$$. We also provide an example that shows it. Consider the chain of two elements. We define the following binary operation:   \[ \begin{array} [c]{c|cc} {\rightarrow} & 0 & 1\\\hline 0 & 1 & 0\\ 1 & 0 & 1 \end{array} \] Then $$(H,{\wedge},\vee,{\rightarrow},0,1)\in {\mathsf{SH}}$$. Since $$0{\rightarrow} 1 = 0$$ and $$0 \neq 1$$, then $$(H,{\wedge},\vee,{\rightarrow},0,1)$$ is not a Heyting algebra. The following diagrams show the relations among the categories defined in this section:   \begin{equation*} \begin{array}{cc} {\mathsf{hIS_{0}}} & {\mathsf{hBDL}}\\ {\Large{|}} & {\Large{|}}\\ {\mathsf{Hil_{0}}} & {\mathsf{SH}}\\ {\Large{|}} & {\Large{|}}\\ {\mathsf{IS_{0}}} & {\mathsf{HA}}\\ \end{array} \end{equation*} In [5] an extension of Kalman’s functor was studied for the variety of algebras with implication $$(H,{\wedge},\vee, {\rightarrow}, 0,1)$$ which satisfy $$a{\wedge} (a{\rightarrow} b)\leq b$$ for every $$a$$, $$b$$. 5 Kalman’s construction for $${\mathsf{hIS_{0}}}$$ and $${\mathsf{hBDL}}$$ The fact that Kalman’s construction can be extended consistently to Heyting algebras led us to believe that some of the picture could be lifted to the varieties $${\mathsf{hIS_{0}}}$$ and $${\mathsf{hBDL}}$$. More precisely, it arises the natural question of whether is it possible to find some category $${\mathsf{KhIS_{0}}}$$ to obtain an equivalence between $${\mathsf{hIS_{0}}}$$ and some full subcategory of $${\mathsf{KhIS_{0}}}$$, making the following diagram commute: Similarly, it arises the question of whether is it possible to find some category $${\mathsf{KhBDL}}$$ to obtain an equivalence between $${\mathsf{hBDL}}$$ and certain full subcategory of $${\mathsf{KhBDL}}$$, making the following diagram commute: In this section, we answer these questions in the positive. Moreover, we extend Kalman’s functor to the categories $${\mathsf{Hil_{0}}}$$, $${\mathsf{IS_{0}}}$$ and $${\mathsf{SH}}$$. The aim of Section 3 was to obtain a categorical equivalence between $${\mathsf{MS}}$$ and $${\mathsf{KMS^{CK}}}$$ (Corollary 22) to be applied in the present section to the category $${\mathsf{hIS_{0}}}$$. For the case of the category $${\mathsf{hBDL}}$$ we will also use Theorem 1, which establishes an equivalence between $${\mathsf{BDL}}$$ and $${\mathsf{KA_{{\mathrm{c}}}^{CK}}}$$. 5.1 Kalman’s construction for $${\mathsf{hIS_{0}}}$$ Let $$H\in {\mathsf{hIS_{0}}}$$. We write $${\rightarrow}$$ for the implication of $$H$$ and define a binary operation on $${\mathrm{K}}(H)$$ (also denoted $${\rightarrow}$$) by   \begin{equation} \label{eqi1} (a,b) {\rightarrow} (d,e): = ((a{\rightarrow} d) {\wedge} (e{\rightarrow} b), a{\wedge} e). \end{equation} (5) This definition is motivated by Remark 6. Note that since $$a{\wedge} b = d{\wedge} e = 0$$, then $$(a{\rightarrow} d){\wedge} (e{\rightarrow} b) {\wedge} a{\wedge} e = 0$$ because $$a{\wedge} (a{\rightarrow} d) \leq d$$ and $$d{\wedge} e = 0$$. Hence, $$(a,b) {\rightarrow} (d,e) \in {\mathrm{K}}(H)$$. The next definition is motivated by the original Kalman’s construction. Definition 35 We denote by $${\mathsf{KhIS_{0}}}$$ the category whose objects are the structures $$(T, \leq, {\sim}, {\rightarrow}, {\mathrm{c}}, 0,1)$$ such that $$(T, \leq, {\sim},{\mathrm{c}}, 0,1)\in {\mathsf{KMS}}$$ and $${\rightarrow}$$ is a binary operation on $$T$$ which satisfies the following conditions for every $$x,y \in T$$: $${\mathrm{(K1)}}$$$${\mathrm{c}} \leq x {\rightarrow} (y\vee {\mathrm{c}})$$, $${\mathrm{(K2)}}$$$$x {\wedge} ((x\vee {\mathrm{c}}){\rightarrow} (y\vee {\mathrm{c}}))\leq y \vee {\mathrm{c}}$$, $${\mathrm{(K3)}}$$$$x {\rightarrow} x = 1$$, $${\mathrm{(K4)}}$$$$(x{\rightarrow} y){\wedge} {\mathrm{c}} = ({\sim} x {\wedge} {\mathrm{c}}) \vee (y{\wedge} {\mathrm{c}})$$, $${\mathrm{(K5)}}$$$$(x{\rightarrow} {\sim} y) \vee {\mathrm{c}} = ((x\vee {\mathrm{c}}) {\rightarrow}({\sim} y \vee {\mathrm{c}})) {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} ({\sim} x \vee {\mathrm{c}}))$$. The morphisms of $${\mathsf{KhIS_{0}}}$$ are the morphisms $$g$$ of $${\mathsf{KMS}}$$ which satisfy the condition $$g(x{\rightarrow} y) = g(x){\rightarrow} g(y)$$ for every $$x$$, $$y$$. In what follows we will prove that if $$H \in {\mathsf{hIS_{0}}}$$, then $${\mathrm{K}}(H) \in {\mathsf{KhIS_{0}}}$$, where the binary operation $${\rightarrow}$$ in $${\mathrm{K}}(H)$$ is that defined in (5). Proposition 36 Let $$H \in {\mathsf{hIS_{0}}}$$. Then $${\mathrm{K}}(H) \in {\mathsf{KhIS_{0}}}$$. Furthermore, $${\mathrm{K}}$$ extends to a functor from $${\mathsf{hIS_{0}}}$$ to $${\mathsf{KhIS_{0}}}$$, which we also denote by $${\mathrm{K}}$$. Proof. Throughout this proof we use Lemmas 9 and 11. Recall that $$c = (0,0)$$. Let $$(a,b), (d,e) \in {\mathrm{K}}(H)$$. In particular, $$(d,e) \vee {\mathrm{c}} = (d,0)$$. Then   \[ \begin{array} [c]{lllll} (a,b){\rightarrow} ((d,e) \vee {\mathrm{c}})& = & (a,b){\rightarrow} (d,0)& & \\ & = & ((a{\rightarrow} d){\wedge} (0{\rightarrow} b),0)& & \\ & \succeq & {\mathrm{c}}.& & \end{array} \] Thus we have proved the condition $${\mathrm{(K1)}}$$. To prove $${\mathrm{(K2)}}$$ we make the following computation:   \[ \begin{array} [c]{lllll} (a,b) {\wedge} (((a,b) \vee {\mathrm{c}}) {\rightarrow} ((d,e) \vee {\mathrm{c}}))& = &(a,b) {\wedge} ((a,0){\rightarrow} (d,0))& & \\ & = & (a,b) {\wedge} (a{\rightarrow} d, 0)& & \\ & = & (a{\wedge} (a{\rightarrow} d),b) & &\\ & \preceq & (d,b) & & \\ & \preceq & (d,0) & & \\ & = & (d,e) \vee {\mathrm{c}}. & & \\ \end{array} \] The proof of the condition $${\mathrm{(K3)}}$$ is immediate. To prove $${\mathrm{(K4)}}$$, note that $$((a,b){\rightarrow} (d,e)){\wedge} {\mathrm{c}} = (0,a{\wedge} e)$$ and   \[ \begin{array} [c]{lllll} ({\sim} (a,b){\wedge} {\mathrm{c}}) \vee ((d,e) {\wedge} {\mathrm{c}})& = &(0,a) \vee (0,e)& & \\ & = & (0,a{\wedge} e).& & \end{array} \] Hence, we have that   \[ ((a,b){\rightarrow} (d,e)){\wedge} {\mathrm{c}} = ({\sim} (a,b){\wedge} {\mathrm{c}}) \vee ((d,e) {\wedge} {\mathrm{c}}). \] Finally we shall prove $${\mathrm{(K5)}}$$. First note that   \[ \begin{array} [c]{lllll} ((a,b) {\rightarrow} {\sim}(d,e)) \vee {\mathrm{c}}& = &((a,b) {\rightarrow} (e,d))\vee {\mathrm{c}}& & \\ & = & ((a{\rightarrow} e){\wedge} (d{\rightarrow} b),0).& & \end{array} \] Then,   \begin{equation} \label{K5on} ((a,b) {\rightarrow} {\sim}(d,e)) \vee {\mathrm{c}} = ((a{\rightarrow} e){\wedge} (d{\rightarrow} b),0). \end{equation} (6) On the other hand,   \[ \begin{array} [c]{lllll} ((a,0){\rightarrow} (e,0)){\wedge} ((d,0){\rightarrow} (b,0)) & = &(a{\rightarrow} e,0) {\wedge} (d{\rightarrow} b,0)& & \\ & = & ((a{\rightarrow} e){\wedge} (d{\rightarrow} b),0).& & \end{array} \] Hence,   \begin{equation} \label{K5tw} ((a,0){\rightarrow} (e,0)){\wedge} ((d,0){\rightarrow} (b,0)) = ((a{\rightarrow} e){\wedge} (d{\rightarrow} b),0). \end{equation} (7) Since $$(a,b) \vee {\mathrm{c}} = (a,0)$$, $${\sim}(d,e) \vee {\mathrm{c}} = (e,0)$$, $$(d,e) \vee {\mathrm{c}} = (d,0)$$, and $${\sim}(a,b) \vee {\mathrm{c}} = (b,0)$$, then it follows from (6) and (7) that the condition $${\mathrm{(K5)}}$$ holds. Thus, $${\mathrm{K}}(H)\in {\mathsf{KhIS_{0}}}$$. Let $$f:H{\rightarrow} G$$ be a morphism in $${\mathsf{hIS_{0}}}$$. Straightforward computations show that $${\mathrm{K}}(f)$$ preserves the implication operation, which implies that $${\mathrm{K}}(f)$$ is a morphism in $${\mathsf{KhIS_{0}}}$$. ■ Proposition 37 If $$(T, \leq, {\sim}, {\rightarrow}, {\mathrm{c}}, 0,1)\in {\mathsf{KhIS_{0}}}$$, then $$({\mathrm{C}}(T),{\wedge},{\rightarrow},{\mathrm{c}},1) \in {\mathsf{hIS_{0}}}$$. Furthermore, $${\mathrm{C}}$$ extends to a functor from $${\mathsf{KhIS_{0}}}$$ to $${\mathsf{hIS_{0}}}$$, which we also denote by $${\mathrm{C}}$$. Proof. We have that $${\mathrm{C}}(T)$$ is closed under the operation $${\rightarrow}$$. To prove it, let $$x,y \geq {\mathrm{c}}$$. By $${\mathrm{(K1)}}$$ we have that   \[ \begin{array} [c]{lllll} {\mathrm{c}} & \leq &(x\vee {\mathrm{c}}) {\rightarrow} (y\vee {\mathrm{c}})& & \\ & = & x {\rightarrow} y,& & \end{array} \] so $$x{\rightarrow} y \in {\mathrm{C}}(T)$$. Thus, the restriction of $${\rightarrow}$$ to $${\mathrm{C}}(T)$$ is indeed an operation on $${\mathrm{C}}(T)$$. Let $$x,y \geq {\mathrm{c}}$$. It follows from $${\mathrm{(K2)}}$$ that $$x{\wedge} (x{\rightarrow} y)\leq y$$ and it follows from $${\mathrm{(K3)}}$$ that $$x{\rightarrow} x = 1$$. Then $$({\mathrm{C}}(T),{\wedge},{\rightarrow},{\mathrm{c}},1) \in {\mathsf{hIS_{0}}}$$. The rest of the proof is immediate. ■ Remark 38 For $$H\in {\mathsf{hIS_{0}}}$$ we have that $$\alpha_{H}(a{\rightarrow} b) = \alpha_{H}(a) {\rightarrow} \alpha_{H}(b)$$ for every $$a$$, $$b\in H$$. Moreover, $$\alpha_H$$ is an isomorphism in $${\mathsf{hIS_{0}}}$$. For the case of $$T\in {\mathsf{KhIS_{0}}}$$ we will prove that $$\beta_T$$ preserves the implication. Lemma 39 Let $$T\in {\mathsf{KhIS_{0}}}$$. Then $$\beta_T$$ is injective and a morphism in $${\mathsf{KhIS_{0}}}$$. Moreover, if $$T$$ satisfies $$({\mathrm{CK}})$$ then $$\beta_T$$ is an isomorphism in $${\mathsf{KhIS_{0}}}$$. Proof. We need to prove that $$\beta_T(x{\rightarrow} y) = \beta_T(x) {\rightarrow} \beta_T(y)$$ for every $$x,y$$. In an equivalent way, we need to prove that $$\beta_T(x{\rightarrow} {\sim} y) = \beta_T(x) {\rightarrow} \beta_T({\sim} y)$$ for every $$x,y$$. It follows from $${\mathrm{(K4)}}$$ and $${\mathrm{(K5)}}$$ that   \[ \begin{array} [c]{lllll} \beta_T(x{\rightarrow} {\sim} y) & = & ((x{\rightarrow} {\sim} y) \vee {\mathrm{c}},{\sim} (x{\rightarrow} {\sim} y) \vee {\mathrm{c}})) & & \\ & = & (((x\vee {\mathrm{c}}) {\rightarrow}({\sim} y \vee {\mathrm{c}})) {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} ({\sim} x \vee {\mathrm{c}})), (x\vee {\mathrm{c}}) {\wedge} (y \vee {\mathrm{c}})) & & \\ & = & \beta_T(x) {\rightarrow} \beta_T({\sim} y).& & \end{array} \] ■ We write $${\mathsf{KhIS_{0}^{CK}}}$$ for the full subcategory of $${\mathsf{KhIS_{0}}}$$ whose objects satisfy $$({\mathrm{CK}})$$. The proof of the following theorem follows from Corollary 22, Proposition 36, Proposition 37, Remark 38 and Lemma 39. Theorem 40 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{hIS_{0}}}$$ and $${\mathsf{KhIS_{0}^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Let $$T \in {\mathsf{KhIS_{0}}}$$. We define the following condition for every $$x,y \in T$$: $${\mathrm{(K6)}}$$$$x \leq (y\vee {\mathrm{c}}) {\rightarrow} ((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}))$$. The next lemma is the motivation to consider the condition $${\mathrm{(K6)}}$$. Lemma 41 If $$H\in {\mathsf{hIS_{0}}}$$ satisfies the inequality $$a\leq b {\rightarrow} (a{\wedge} b)$$ for every $$a, b$$, then $${\mathrm{K}}(H)$$ satisfies $${\mathrm{(K6)}}$$. Proof. First note that for every $$(a,b) \in {\mathrm{K}}(H)$$, $$(a,b) \vee {\mathrm{c}} = (a,0)$$. To prove $${\mathrm{(K6)}}$$ we make the following computation:   \[ \begin{array} [c]{lllll} (d,0) {\rightarrow} ((a,0){\wedge} (d,0)) & = & (d,0) {\rightarrow} (a{\wedge} d,0) & & \\ & = & (d {\rightarrow} (a{\wedge} d),0)& & \\ & \succeq & (a,b).& & \end{array} \] Hence, we obtain $${\mathrm{(K6)}}$$. ■ The following lemma will play an important role in this article. Lemma 42 If $$T \in {\mathsf{KhIS_{0}}}$$ satisfies $${\mathrm{(K6)}}$$, then $$T$$ satisfies $$({\mathrm{CK}})$$. Proof. Let $$x,y \geq {\mathrm{c}}$$ such that $$x{\wedge} y = {\mathrm{c}}$$. Taking into account $${\mathrm{(KM3)}}$$ we can define $$z = (y {\rightarrow} {\sim} y) {\wedge} x$$. It follows from $${\mathrm{(K4)}}$$ that   \[ \begin{array} [c]{lllll} z{\wedge} {\mathrm{c}} & = & ((y{\rightarrow} {\sim} y) {\wedge} {\mathrm{c}}){\wedge} x & & \\ & = & (({\sim} y {\wedge} {\mathrm{c}}) \vee ({\sim} y {\wedge} {\mathrm{c}})) {\wedge} x& & \\ & = & ({\sim} y {\wedge} {\mathrm{c}}) {\wedge} x& & \\ & = & {\sim} y {\wedge} x& & \\ & = & {\sim} y.& & \end{array} \] Hence, $${\sim} z \vee {\mathrm{c}} = y$$. In order to prove that $$z\vee {\mathrm{c}} = x$$, we use the conditions $${\mathrm{(KM3)}}$$, $${\mathrm{(KM4)}}$$, $${\mathrm{(K5)}}$$, $${\mathrm{(K6)}}$$ and the fact that $$x{\wedge} y = {\mathrm{c}}$$ as follows:   \[ \begin{array} [c]{lllll} z\vee {\mathrm{c}} & = & (x{\wedge} (y{\rightarrow} {\sim} y))\vee {\mathrm{c}} & & \\ & = & x{\wedge} ((y{\rightarrow} {\sim} y) \vee {\mathrm{c}}) & &\\ & = & ((y\vee {\mathrm{c}}) {\rightarrow} ({\sim} y \vee {\mathrm{c}})){\wedge} x& & \\ & = & (x\vee {\mathrm{c}}) {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} {\mathrm{c}})& & \\ & = & (x \vee {\mathrm{c}}) {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} (x{\wedge} y)) & & \\ & = & (x \vee {\mathrm{c}}) {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} ((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}))) & &\\ & = & x \vee {\mathrm{c}} & &\\ & = & x. & & \end{array} \] Therefore, $$z\vee {\mathrm{c}} = x$$. ■ 5.2 Kalman’s construction for $${\mathsf{Hil_{0}}}$$ We write $${\mathsf{KHil_{0}}}$$ for the full subcategory of $${\mathsf{KhIS_{0}}}$$ whose objects satisfy the following conditions for every $$x,y,z$$: $${\mathrm{(KHil1)}}$$$$(x\vee {\mathrm{c}}){\rightarrow} (y{\rightarrow} (x\vee {\mathrm{c}})) = 1$$, $${\mathrm{(KHil2)}}$$$$x {\rightarrow} ((y\vee {\mathrm{c}}) {\rightarrow} (z\vee {\mathrm{c}})) = (x {\rightarrow} (y\vee {\mathrm{c}})) {\rightarrow} (x{\rightarrow} (z \vee {\mathrm{c}}))$$, $${\mathrm{(KHil3)}}$$ If $$x{\rightarrow} y = y {\rightarrow} x = 1$$, then $$x = y$$, $${\mathrm{(KHil4)}}$$$$x{\wedge} ((x\vee {\mathrm{c}}){\rightarrow} (y\vee {\mathrm{c}})) = x{\wedge} (y\vee {\mathrm{c}})$$, $${\mathrm{(KHil5)}}$$$$x{\rightarrow} ((y\vee {\mathrm{c}}) {\wedge} (z\vee {\mathrm{c}})) \leq (x{\rightarrow} (y\vee {\mathrm{c}})){\wedge} (x{\rightarrow} (z\vee {\mathrm{c}}))$$. Example 43 In every centered Kleene algebra $$(T,{\wedge},\vee,{\sim},{\mathrm{c}},0,1)$$ it is possible to define a binary operation, that we denote by $${\rightarrow}$$, as follows:   \[ x {\rightarrow} y = \begin{cases} 1, &\text{if $x \vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$ and $x{\wedge} {\mathrm{c}} \leq y{\wedge} {\mathrm{c}}$;}\\ {\sim} x \vee (y{\wedge} {\mathrm{c}}), &\text{if $x \vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$ and $x{\wedge} {\mathrm{c}} \nleq y{\wedge} {\mathrm{c}} $;}\\ y \vee ({\sim} x {\wedge} {\mathrm{c}}), &\text{if $x \vee {\mathrm{c}} \nleq y \vee {\mathrm{c}}$ and $x{\wedge} {\mathrm{c}} \leq y{\wedge} {\mathrm{c}}$;}\\ ((y\vee {\mathrm{c}}){\wedge} {\sim} x) \vee (({\sim} x \vee {\mathrm{c}}){\wedge} y), &\text{if $x\vee {\mathrm{c}} \nleq y\vee {\mathrm{c}}$ and $x{\wedge} {\mathrm{c}} \nleq y{\wedge} {\mathrm{c}}$.} \end{cases} \] It is possible to prove that $$(T,\leq, {\sim},{\mathrm{c}},0,1)\in {\mathsf{KMS}}$$, and it is not difficult to see that $$(T,{\sim},{\rightarrow},{\mathrm{c}},0,1) \in {\mathsf{KHil_{0}}}$$. By endowing the centered Kleene algebra given in [5, Example 2.5] with the binary operation $${\rightarrow}$$ just defined we obtain an example of an object of $${\mathsf{KHil_{0}}}$$ which does not satisfy the condition $$({\mathrm{CK}})$$. Lemma 44 (a) If $$H\in {\mathsf{Hil_{0}}}$$, then $${\mathrm{K}}(H) \in {\mathsf{KHil_{0}}}$$. (b) If $$T\in {\mathsf{KHil_{0}}}$$, then $${\mathrm{C}}(T) \in {\mathsf{Hil_{0}}}$$. Proof. Let $$H\in {\mathsf{Hil_{0}}}$$ and take $$(a,b)$$, $$(d,e)$$ and $$(f,g)$$ in $${\mathrm{K}}(H)$$. In what follows we will use Proposition 28. Taking into account that $$a{\rightarrow} (d{\rightarrow} a) = 1$$ we obtain   \[ \begin{array} [c]{lllll} (a,0) {\rightarrow} ((d,e) {\rightarrow} (a,0)) & = & (a,0) {\rightarrow} (d{\rightarrow} a,0) & & \\ & = & (a {\rightarrow} (d{\rightarrow} a) ,0)& & \\ & = & (1,0),& & \end{array} \] which is the condition $${\mathrm{(KHil1)}}$$. Since $$a{\rightarrow} (d{\rightarrow} f) = (a{\rightarrow} d) {\rightarrow} (a{\rightarrow} f)$$, then   \[ \begin{array} [c]{lllll} (a,b) {\rightarrow} ((d,0) {\rightarrow} (f,0)) & = & (a,b) {\rightarrow} (d{\rightarrow} f,0) & & \\ & = & (a{\rightarrow} (d{\rightarrow} f), 0)& & \\ & = & ((a{\rightarrow} d) {\rightarrow} (a {\rightarrow} f), 0)& & \\ & = & (a{\rightarrow} d,0) {\rightarrow} (a {\rightarrow} f), 0)& & \\ & = & ((a,b) {\rightarrow} (d,0)) {\rightarrow} ((a,b){\rightarrow} (f,0)).& & \end{array} \] Hence, we have proved $${\mathrm{(KHil2)}}$$. In order to prove $${\mathrm{(KHil3)}}$$ suppose that $$(a,b) {\rightarrow} (d,e) = (d,e) {\rightarrow} (a,b) = (1,0)$$, so $$a{\rightarrow} d = d{\rightarrow} a = 1$$ and $$b{\rightarrow} e = e {\rightarrow} b = 1$$. Then $$a = d$$ and $$b = e$$, i.e., $$(a,b) = (d,e)$$, which was our aim. The condition $${\mathrm{(KHil4)}}$$ is a consequence of the equality $$a{\wedge} (a{\rightarrow} d) = a{\wedge} d$$. Indeed,   \[ \begin{array} [c]{lllll} (a,b) {\wedge} ((a,0) {\rightarrow} (d,0)) & = & (a,b) {\wedge} (a{\rightarrow} d,0) & & \\ & = & (a{\wedge} (a{\rightarrow} d), b)& & \\ & = & (a{\wedge} d,b)& & \\ & = & (a,b) {\wedge} (d,0).& & \end{array} \] Finally, we will prove $${\mathrm{(KHil5)}}$$. By the condition $$a{\rightarrow} (d{\wedge} f) \leq (a{\rightarrow} d){\wedge} (a{\rightarrow} f)$$ we have that   \[ \begin{array} [c]{lllll} (a,b) {\rightarrow} ((d,0){\wedge} (f,0)) & = & (a,b) {\rightarrow} (d{\wedge} f,0) & & \\ & = & (a {\rightarrow} (d{\wedge} f), 0)& & \\ & \preceq & ((a{\rightarrow} d){\wedge} (a{\rightarrow} f), 0)& & \\ & = & (a{\rightarrow} d, 0) {\wedge} (a{\rightarrow} f,0)& & \\ & = & ((a,b) {\rightarrow} (d,0)) {\wedge} ((a,b) {\rightarrow} (f,0)).& & \end{array} \] Then $${\mathrm{K}}(H) \in {\mathsf{KHil_{0}}}$$. Finally, it follows from Proposition 28 that if $$T\in {\mathsf{KHil_{0}}}$$, then $${\mathrm{C}}(T) \in {\mathsf{Hil_{0}}}$$ ■ We write $${\mathsf{KHil_{0}^{CK}}}$$ for the full subcategory of $${\mathsf{KHil_{0}}}$$ whose objects satisfy $$({\mathrm{CK}})$$. The following corollary follows from Theorem 40 and Lemma 44. Corollary 45 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{Hil_{0}}}$$ and $${\mathsf{KHil_{0}^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. 5.3 Kalman’s construction for $${\mathsf{IS_{0}}}$$ We write $${\mathsf{KIS_{0}}}$$ for the full subcategory of $${\mathsf{KhIS_{0}}}$$ whose objects $$T$$ satisfy the condition $${\mathrm{(K6)}}$$ and the following additional condition for every $$x,y \in T$$: $${\mathrm{(K7)}}$$$$x {\rightarrow} ((y\vee {\mathrm{c}}) {\wedge} (z\vee {\mathrm{c}})) = (x {\rightarrow} (y\vee {\mathrm{c}})){\wedge} (x {\rightarrow} (z\vee {\mathrm{c}}))$$. Lemma 46 (a) If $$H\in {\mathsf{IS_{0}}}$$, then $${\mathrm{K}}(H) \in {\mathsf{KIS_{0}}}$$. (b) If $$T\in {\mathsf{KIS_{0}}}$$, then $${\mathrm{C}}(T) \in {\mathsf{IS_{0}}}$$. Proof. Let $$H\in {\mathsf{IS_{0}}}$$. The fact that $${\mathrm{K}}(H)$$ satisfies $${\mathrm{(K6)}}$$ follows from Lemmas 30 and 41. On the other hand, it follows from Lemma 30 that   \[ \begin{array} [c]{lllll} (a,b){\rightarrow} ((d,0){\wedge} (f,0)) & = & (a,b) {\rightarrow} (d{\wedge} f,0) & & \\ & = & (a {\rightarrow} (d{\wedge} f),0)& & \\ & = & ((a{\rightarrow} d){\wedge} (a{\rightarrow} f),0)& & \\ & = & ((a,b){\rightarrow} (d,0)){\wedge} ((a,b){\rightarrow} (f,0)).& & \end{array} \] Thus, we have the condition $${\mathrm{(K7)}}$$. Then $${\mathrm{K}}(H) \in {\mathsf{KIS_{0}}}$$. The fact that if $$T\in {\mathsf{KIS_{0}}}$$, then $${\mathrm{C}}(T) \in {\mathsf{IS_{0}}}$$ is also consequence of Lemma 30. ■ The following corollary follows from Theorem 40, Lemma 42 and Lemma 46. Corollary 47 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{IS_{0}}}$$ and $${\mathsf{KIS_{0}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Since $${\mathsf{IS_{0}}}$$ is a full subcategory of $${\mathsf{Hil_{0}}}$$, it follows from corollaries 45 and 47 that $${\mathsf{KIS_{0}}}$$ is a full subcategory of $${\mathsf{KHil_{0}^{CK}}}$$. 5.4 Kalman’s construction for $${\mathsf{hBDL}}$$ In what follows we define a category which will be related with the category $${\mathsf{hBDL}}$$. Definition 48 We write $${\mathsf{KhBDL}}$$ for the category whose objects are the algebras $$(T,{\wedge},\vee,$$$${\rightarrow}, {\sim},{\mathrm{c}},0,1)$$ of type $$(2,2,2,1,0,0,0)$$ such that $$(T,{\wedge},\vee, {\sim},{\mathrm{c}},0,1)\in {\mathsf{KA_{{\mathrm{c}}}}}$$ and the conditions $${\mathrm{(K1)}}$$, $${\mathrm{(K2)}}$$, $${\mathrm{(K3)}}$$, $${\mathrm{(K4)}}$$ and $${\mathrm{(K5)}}$$ are satisfied. The morphisms of the category are the corresponding algebra homomorphisms. By the Example 43, in every centered Kleene algebra $$(T,{\wedge},\vee,0,{\mathrm{c}},0,1)$$ we can define a binary operation $${\rightarrow}$$ such that $$(T,{\wedge},\vee,{\rightarrow},{\mathrm{c}},0,1) \in {\mathsf{KhBDL}}$$. In particular, if $$(T,{\wedge},\vee,0, {\mathrm{c}},0,1)$$ is the centered Kleene algebra given in [5, Example 2.5], then $$(T,{\wedge},\vee,{\rightarrow},{\mathrm{c}},0,1)\in {\mathsf{KhBDL}}$$, where $${\rightarrow}$$ is the implication considered in Example 43. It is immediate that $$(T,{\wedge},\vee,{\mathrm{c}},0,1)$$ does not satisfy the condition ($${\mathrm{CK}}$$). Note that $$(H,{\wedge},\vee,{\rightarrow}, 0,1) \in {\mathsf{hBDL}}$$ if and only if $$(H,{\wedge},\vee,0,1) \in {\mathsf{BDL}}$$ and $$(H,{\wedge},{\rightarrow},0,1) \in {\mathsf{hIS_{0}}}$$. Also note that $$(T,{\wedge},\vee,{\rightarrow}, {\sim},{\mathrm{c}},0,1) \in {\mathsf{KhBDL}}$$ if and only if $$(T,{\wedge},\vee, {\sim},{\mathrm{c}},0,1)\in {\mathsf{KA_{{\mathrm{c}}}}}$$ and $$(T, \leq, {\sim}, {\rightarrow}, {\mathrm{c}}, 0,1) \in {\mathsf{KhIS_{0}}}$$. We write $${\mathsf{KhBDL^{CK}}}$$ for the full subcategory of $${\mathsf{KhBDL}}$$ whose objects satisfy $$({\mathrm{CK}})$$. Theorem 49 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{hBDL}}$$ and $${\mathsf{KhBDL^{CK}}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Proof. It follows from Theorem 1 and Theorem 40. ■ 5.5 Kalman’s construction for $${\mathsf{SH}}$$ We write $${\mathsf{KSH}}$$ for the full subcategory of $${\mathsf{KhBDL}}$$ whose objects satisfy the condition $${\mathrm{(KHil4)}}$$ and the following additional condition: $${\mathrm{(KSH3)}}$$$$x {\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} (z\vee {\mathrm{c}})) = x {\wedge}(((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}})){\rightarrow} ((x\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}})))$$. Lemma 50 (a) If $$H\in {\mathsf{SH}}$$, then $${\mathrm{K}}(H) \in {\mathsf{KSH}}$$. (b) If $$T\in {\mathsf{KSH}}$$, then $${\mathrm{C}}(T) \in {\mathsf{SH}}$$. (c) If $$T \in {\mathsf{KSH}}$$, then $$T$$ satisfies $${\mathrm{(K6)}}$$. In particular, $$T$$ satisfies $$({\mathrm{CK}})$$. Proof. Let $$H\in {\mathsf{SH}}$$. The condition $${\mathrm{(KHil4)}}$$ follows from $${\mathrm{(SH2)}}$$ (see proof of Lemma 44). Let $$(a,b)$$, $$(d,e)$$ and $$(f,g)$$ in $${\mathrm{K}}(H)$$. Taking into account $${\mathrm{(SH3)}}$$ we have that   \[ \begin{array} [c]{lllll} (a,b){\wedge} ((d,0){\rightarrow} (f,0)) & = & (a,b) {\wedge} (d{\rightarrow} f,0) & & \\ & = & (a {\wedge} (d{\rightarrow} f),b)& & \\ & = & (a {\wedge} ((a{\wedge} d){\rightarrow} (a{\wedge} f)),b)& & \\ & = & (a,b) {\wedge} ((a{\wedge} d){\rightarrow} (a{\wedge} f),0)& & \\ & = & (a,b) {\wedge} ((a{\wedge} d,0) {\rightarrow} (a{\wedge} f,0))& & \\ & = & (a,b) {\wedge} (((a,0){\wedge} (d,0)) {\rightarrow} ((a,0){\wedge} (f,0))),& & \end{array} \] which is the condition $${\mathrm{(KSH3)}}$$. Then $${\mathrm{K}}(H) \in {\mathsf{KSH}}$$. It is immediate that if $$T\in {\mathsf{KSH}}$$ then $${\mathrm{C}}(T) \in {\mathsf{SH}}$$. To prove that $$T$$ satisfies $${\mathrm{(K6)}}$$ we will use $${\mathrm{(K3)}}$$ and $${\mathrm{(KSH3)}}$$ as follows:   \[ \begin{array} [c]{lllll} x{\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} ((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}))) & = & x{\wedge} ((y\vee {\mathrm{c}}) {\rightarrow} (((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}})) \vee {\mathrm{c}}) & & \\ & = & x {\wedge} (((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}})) {\rightarrow} ((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}}))) & & \\ & = & x{\wedge} 1& & \\ & = & x.& & \end{array} \] Then $$x\leq (y\vee {\mathrm{c}}) {\rightarrow} ((x\vee {\mathrm{c}}) {\rightarrow} (y\vee {\mathrm{c}}))$$, i.e., the condition $${\mathrm{(K6)}}$$. Therefore, it follows from Lemma 42 that $$T$$ satisfies ($${\mathrm{CK}}$$). ■ Theorem 51 The functors $${\mathrm{K}}$$ and $${\mathrm{C}}$$ establish a categorical equivalence between $${\mathsf{SH}}$$ and $${\mathsf{KSH}}$$ with natural isomorphisms $$\alpha$$ and $$\beta$$. Proof. It follows from Theorem 49 and Lemma 50. ■ 6 Well-behaved congruences in $${\mathsf{KhIS_{0}}}$$ and congruences in $${\mathsf{KhBDL}}$$ In this section, we introduce the concept of the well-behaved congruences over objects of $${\mathsf{KhIS_{0}}}$$. They are equivalence relations with some additional properties. We will prove that if $$T\in {\mathsf{KhIS_{0}}}$$ and $$\theta$$ is a well-behaved congruence on $$T$$, then it is possible to define on the quotient $$T/\theta$$ a partial order and operations so that $$T/\theta \in {\mathsf{KhIS_{0}}}$$. For $$T\in {\mathsf{KhIS_{0}}}$$ we study the relation between the well-behaved congruences of $$T$$ and the congruences of $${\mathrm{C}}(T)$$, and in particular for the cases where $$T\in {\mathsf{KHil_{0}}}$$ or $$T \in {\mathsf{KIS_{0}}}$$. For $$T\in {\mathsf{KhBDL}}$$ or $$T\in {\mathsf{KSH}}$$ we also study the relation between the congruences of $$T$$ and the congruences of $${\mathrm{C}}(T)$$. Finally, we study the principal well-behaved congruences of the objects in $${\mathsf{KhIS_{0}}}$$, $${\mathsf{KHil_{0}}}$$ and $${\mathsf{KIS_{0}}}$$ and the principal congruences of the objects in $${\mathsf{KhBDL}}$$ and $${\mathsf{KSH}}$$. We start by fixing notation and giving some useful definitions. Let $$X$$ be a set, $$x \in X$$ and $$\theta$$ an equivalence relation on $$X$$. We write $$x/\theta$$ to indicate the equivalence class of $$x$$ associated with the equivalence relation $$\theta$$, and $$X/\theta$$ to indicate the quotient set of $$X$$ associated with $$\theta$$ (i.e. the set of equivalence classes). If $$T$$ is an algebra, we write $${\mathrm{Con}}(T)$$ to denote the set of as well as the lattice of congruences of $$T$$. Definition 52 Let $$T\in {\mathsf{KhIS_{0}}}$$. We say that an equivalence relation $$\theta$$ of $$T$$ is a well-behaved congruence of $$T$$ if it satisfies the following conditions: $${\mathrm{(C1)}}$$$$\theta \in {\mathrm{Con}}((T,{\rightarrow},{\sim}))$$. $${\mathrm{(C2)}}$$ For $$x,y \in T$$, $$(x,y) \in \theta$$ if and only if $$(x\vee {\mathrm{c}},y\vee {\mathrm{c}}) \in \theta$$ and $$({\sim} x \vee {\mathrm{c}}, {\sim} y \vee {\mathrm{c}}) \in \theta$$. $${\mathrm{(C3)}}$$ For $$x$$, $$y$$, $$z$$ and $$w$$ in $${\mathrm{C}}(T)$$, if $$(x,y)\in \theta$$ and $$(z,w) \in \theta$$, then $$(x{\wedge} z,y{\wedge} w) \in \theta$$. Note that the intersection of any family of well-behaved congruences of $$T\in {\mathsf{KhIS_{0}}}$$ is a well-behaved congruence; therefore the set of well-behaved congruences of $$T$$ ordered by the inclusion relation is a complete lattice. Remark 53 The definition of well-behaved congruence can be also given for algebras of $${\mathsf{KhBDL}}$$. In this case, if $$T\in {\mathsf{KhBDL}}$$, then every congruence of $$T$$ is a well-behaved congruence. In what follows we define a binary relation in $$T/\theta$$, where $$T\in {\mathsf{KhIS_{0}}}$$ and $$\theta$$ is a well-behaved congruence of $$T$$. Definition 54 Let $$T \in {\mathsf{KhIS_{0}}}$$. If $$\theta$$ is a well-behaved congruence of $$T$$, then we define in $$T/\theta$$ the following binary relation $$\ll_{\theta}$$ by:   \[ x/\theta \ll_{\theta} y/\theta\; \textrm{if and only if}\; ((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta\; \textrm{and}\; (({\sim} y \vee {\mathrm{c}}){\wedge} ({\sim} x \vee {\mathrm{c}}), {\sim} y \vee {\mathrm{c}}) \in \theta. \] If there is no ambiguity, we write $$\ll$$ in place of $$\ll_{\theta}$$. Note that the definition given is good, in the sense that it is independent of the elements selected as representativess of the equivalence classes. To show it, suppose that $$x/\theta \ll y/\theta$$. Let $$z\in x/\theta$$ and $$w\in y/\theta$$. Then by $${\mathrm{(C2)}}$$ we have that $$(x\vee {\mathrm{c}}, z\vee {\mathrm{c}}) \in \theta$$, $$({\sim} x\vee {\mathrm{c}}, {\sim} z\vee {\mathrm{c}}) \in \theta$$, $$(y\vee {\mathrm{c}}, w\vee {\mathrm{c}}) \in \theta$$, and $$({\sim} z\vee {\mathrm{c}}, {\sim} w\vee {\mathrm{c}}) \in \theta$$. Hence it follows from $${\mathrm{(C3)}}$$ that   \[ ((z\vee {\mathrm{c}}){\wedge} (w\vee {\mathrm{c}}), (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}})) \in \theta. \] Since, by the assumption, $$((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta$$, and $$(x\vee {\mathrm{c}},z\vee {\mathrm{c}}) \in \theta$$, then   \[ ((z\vee {\mathrm{c}}){\wedge} (w\vee {\mathrm{c}}), z\vee {\mathrm{c}}) \in \theta. \] In a similar way it can be proved that $$(({\sim} z\vee {\mathrm{c}}){\wedge} ({\sim} w\vee {\mathrm{c}}), \sim w\vee {\mathrm{c}}) \in \theta$$. Remark 55 Let $$T\in {\mathsf{KA_{{\mathrm{c}}}}}$$ and $$\theta \in {\mathrm{Con}}(T)$$. Since the class of centered Kleene algebras is a variety, then $$T/\theta \in {\mathsf{KA_{{\mathrm{c}}}}}$$. In particular, the lattice order $$\leq$$ of $$T/\theta$$ is given by $$x/\theta \leq y/\theta$$ if and only if $$x/\theta = (x{\wedge} y)/\theta$$. In this framework the relation $$\ll$$ given in Definition 54 coincides with the relation $$\leq$$, i.e.,   \[ x/\theta \leq y/\theta\;\textrm{if and only if}\; x/\theta \ll y/\theta. \] To prove it note first that from the distributivity of the underlying lattice of $$T$$ it follows that $$x/\theta \leq y/\theta$$ if and only if $$(x \vee {\mathrm{c}}, (x{\wedge} y)\vee {\mathrm{c}}) \in \theta$$ and $$(x{\wedge} {\mathrm{c}}, (x{\wedge} y){\wedge} {\mathrm{c}}) \in \theta$$. Besides, we have that $$(x{\wedge} y) \vee {\mathrm{c}} = (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}})$$. Since $$\theta$$ preserves the involution, then $$(x{\wedge} {\mathrm{c}}, (x{\wedge} y) {\wedge} {\mathrm{c}}) \in \theta$$ if and only if $$({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\sim} y \vee {\mathrm{c}}) \in \theta$$. Therefore   \begin{equation} \label{oi1} x/\theta \leq y/\theta \ \textrm{if and only if} \ ((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta \ \textrm{and} \ ({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\sim} y \vee {\mathrm{c}}) \in \theta. \end{equation} (8) We also have   \begin{equation} \label{oi2} ({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\sim} y \vee {\mathrm{c}}) \in \theta \ \textrm{if and only if} \ (({\sim} x \vee {\mathrm{c}}) {\wedge} ({\sim} y \vee {\mathrm{c}}), {\sim y} \vee {\mathrm{c}})\in \theta. \end{equation} (9) To prove (9), suppose that $$\;({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\sim} y \vee {\mathrm{c}}) \in \theta$$. Since $$({\sim} y \vee {\mathrm{c}},{\sim} y \vee {\mathrm{c}})\in \theta$$, then taking $${\wedge}$$ we obtain that $$(({\sim} x \vee {\mathrm{c}}) {\wedge} ({\sim} y \vee {\mathrm{c}}), {\sim y} \vee {\mathrm{c}})\in \theta$$. Conversely, assume that $$(({\sim} x \vee {\mathrm{c}}) {\wedge} ({\sim} y \vee {\mathrm{c}}), {\sim y} \vee {\mathrm{c}})\in \theta$$. Since $$({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\mathrm{c}}) \in \theta$$, then taking $$\vee$$ we obtain that $$({\sim} x \vee {\mathrm{c}}, {\sim} x \vee {\sim} y \vee {\mathrm{c}}) \in \theta$$, so $$({\sim} x \vee {\sim} y \vee {\mathrm{c}}, {\sim} y \vee {\mathrm{c}}) \in \theta$$. Then we have proved (9). Therefore, it follows from (8) and (9) that $$x/\theta \leq y/\theta$$ if and only if $$x/\theta \ll y/\theta$$. Lemma 56 Let $$T \in {\mathsf{KhIS_{0}}}$$ and $$\theta$$ a well-behaved congruence of $$T$$. Then $$(T,\ll)$$ is a poset. Proof. Let $$\theta$$ be a well-behaved congruence of $$T$$. The reflexivity of $$\theta$$ implies the reflexivity of $$\ll$$. To prove that $$\ll$$ is antisymmetric, let $$x, y\in T$$ be such that $$x/\theta \ll y/ \theta$$ and $$y/\theta \ll x/\theta$$, which means that   \[ ((x\vee {\mathrm{c}}) {\wedge} (y{\wedge} {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta, \]   \[ (({\sim} y \vee {\mathrm{c}}){\wedge} ({\sim} x \vee {\mathrm{c}}), {\sim} y \vee {\mathrm{c}}) \in \theta, \]   \[ ((y\vee {\mathrm{c}}) {\wedge} (x\vee {\mathrm{c}}), y\vee {\mathrm{c}}) \in \theta, \]   \[ (({\sim} x \vee {\mathrm{c}}){\wedge} ({\sim} y \vee {\mathrm{c}}), {\sim} x \vee {\mathrm{c}}) \in \theta. \] Since $$(x\vee {\mathrm{c}}, (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}})) \in \theta$$ and $$((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), y\vee {\mathrm{c}}) \in \theta$$, then $$(x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \theta$$. Analogously we have that $$({\sim} x \vee {\mathrm{c}}, {\sim} y \vee {\mathrm{c}}) \in \theta$$. Hence, it follows from $${\mathrm{(C2)}}$$ that $$(x,y)\in \theta$$, i.e., $$x/\theta = y/\theta$$. We conclude that $$\ll$$ is antisymmetric. Finally we will prove that $$\ll$$ is transitive. Let $$x$$, $$y$$ and $$z$$ be elements of $$T$$ such that $$x/\theta \ll y/\theta$$ and $$y/\theta \ll z/\theta$$. In particular,   \begin{equation} \label{EQ1} ((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}}), x\vee {\mathrm{c}})\in \theta, \end{equation} (10)  \begin{equation} \label{EQ2} ((y\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}}), y\vee {\mathrm{c}}) \in \theta. \end{equation} (11) It follows from (10) and $${\mathrm{(C3)}}$$ that   \begin{equation} \label{EQ3} ((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}}), (x\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}})) \in \theta, \end{equation} (12) and it follows from (11) and $${\mathrm{(C3)}}$$ that   \begin{equation} \label{EQ4} ((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}}), (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}})) \in \theta. \end{equation} (13) Hence, by (12) and (13) we obtain that $$((x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), (x\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}})) \in \theta$$. Thus, taking into account (10) we have $$((x\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta$$. Similarly we can show that $$(({\sim} z \vee {\mathrm{c}}){\wedge} ({\sim} x \vee {\mathrm{c}}), {\sim} z \vee {\mathrm{c}}) \in \theta$$. Thus, $$x/\theta \ll z/\theta$$. Hence, $$\ll$$ is transitive. ■ Lemma 57 Let $$T\in {\mathsf{KhIS_{0}}}$$ and $$x,y \in T$$. If $$x\leq y$$, then $$x/\theta \ll y/\theta$$. Proof. Let $$x\leq y$$. Then $${\sim} y \leq {\sim} x$$. Hence, we have $$x\vee {\mathrm{c}} \leq y \vee {\mathrm{c}}$$ and $${\sim} y \vee {\mathrm{c}} \leq {\sim} x \vee {\mathrm{c}}$$, i.e., $$(x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}) = x\vee {\mathrm{c}}$$ and $$({\sim} y \vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}) = {\sim} y \vee {\mathrm{c}}$$. Since $$\theta$$ is a reflexive relation, then $$((x\vee {\mathrm{c}}) {\wedge} (y{\wedge} {\mathrm{c}}), x\vee {\mathrm{c}}) \in \theta$$ and $$(({\sim} y \vee {\mathrm{c}}){\wedge} ({\sim} x \vee {\mathrm{c}}), {\sim} y \vee {\mathrm{c}}) \in \theta$$, i.e., $$x/\theta \ll y/\theta$$. ■ For $$T\in {\mathsf{KhIS_{0}}}$$ and $$\theta$$ a well-behaved congruence of $$T$$, we have in particular that $$\theta$$ is a congruence of $$(T,{\sim},{\rightarrow})$$. Let us use also the symbols $${\sim}$$ and $${\rightarrow}$$ to refer to the respective induced operations on $$T/\theta$$. Proposition 58 Let $$(T,\leq,{\sim},{\rightarrow},{\mathrm{c}},0,1)\in {\mathsf{KhIS_{0}}}$$ and $$\theta$$ a well-behaved congruence of $$T$$. Then $$(T/\theta,\ll,{\sim},{\rightarrow}, {\mathrm{c}}/\theta, 0/\theta, 1/\theta) \in {\mathsf{KhIS_{0}}}$$. Proof. Step 1. $$(T/\theta,\ll,{\sim},{\mathrm{c}}/\theta) \in {\mathsf{KP}}$$. It follows from Lemma 56 that $$(T/\theta,\ll)$$ is a poset. It is immediate that $${\sim}$$ is an involution in $$(T/\theta,\ll)$$ which is order reversing and that $${\sim}{\mathrm{c}}/\theta = {\mathrm{c}}/\theta$$. Let $$x\in T$$. In what follows we will prove that the supremum of $$x/\theta$$ and $${\mathrm{c}}/\theta$$ with respect to the order $$\ll$$ exists in $$T/\theta$$, and we will denote it by $$x/\theta \vee {\mathrm{c}}/\theta$$. Moreover, we will prove that $$x/\theta \vee {\mathrm{c}}/\theta = (x\vee {\mathrm{c}})/\theta$$. First note that if $$y\in x/\theta$$, then it follows from $${\mathrm{(C2)}}$$ that $$(x\vee {\mathrm{c}},y\vee {\mathrm{c}}) \in \theta$$, i.e., that $$(x\vee {\mathrm{c}})/\theta = (y\vee {\mathrm{c}})/\theta$$. Now we will show that $$x/\theta \vee {\mathrm{c}}/\theta$$ exists. Since $$x\leq x\vee {\mathrm{c}}$$ and $${\mathrm{c}}\leq x \vee {\mathrm{c}}$$, it follows from Lemma 57 that $$x/\theta \ll (x\vee {\mathrm{c}})/\theta$$ and $${\mathrm{c}}/\theta \ll (x\vee {\mathrm{c}})/\theta$$. Let $$z\in T$$ be such that $$x/\theta \ll z/\theta$$ and $${\mathrm{c}}/\theta\ll z/\theta$$. Then $$((x\vee {\mathrm{c}}){\wedge}(z\vee {\mathrm{c}}), x\vee {\mathrm{c}})\in \theta$$, $$(({\sim}z \vee {\mathrm{c}}){\wedge}({\sim}x \vee {\mathrm{c}}), {\sim}z \vee {\mathrm{c}}) \in \theta$$ and $$(c,{\sim} z \vee {\mathrm{c}})\in \theta$$. We need to prove that $$(x\vee {\mathrm{c}})/\theta \ll z/\theta$$. By the previous assertions we have in particular that   \begin{equation} \label{sup1} (((x\vee {\mathrm{c}})\vee {\mathrm{c}}){\wedge}(z\vee {\mathrm{c}}), (x\vee {\mathrm{c}}) \vee {\mathrm{c}})\in \theta. \end{equation} (14) On the other hand,   \[ ({\sim}z \vee {\mathrm{c}}){\wedge} ({\sim}(x\vee {\mathrm{c}})\vee {\mathrm{c}}) = {\mathrm{c}}. \] But $$(c,{\sim}z \vee {\mathrm{c}}) \in \theta$$, so   \begin{equation} \label{sup2} (({\sim}z \vee {\mathrm{c}}){\wedge} ({\sim}(x\vee {\mathrm{c}})\vee {\mathrm{c}}), {\sim}z \vee {\mathrm{c}}) \in \theta. \end{equation} (15) Hence, it follows from (14) and (15) that $$(x\vee {\mathrm{c}})/\theta \ll z/\theta$$. Thus, $$x/\theta \vee {\mathrm{c}}/\theta$$ exists and $$x/\theta \vee {\mathrm{c}}/\theta = (x\vee {\mathrm{c}})/\theta$$. In what follows we will prove that for every $$x,y \in T$$,   \[ (x/\theta\vee {\mathrm{c}}/\theta){\wedge} ({\sim}x/\theta \vee {\mathrm{c}}/\theta) = {\mathrm{c}}/\theta, \] or, equivalently, that   \begin{equation} \label{center} (x\vee {\mathrm{c}})/\theta {\wedge} ({\sim} x \vee {\mathrm{c}})/\theta = {\mathrm{c}}/\theta, \end{equation} (16) where we also use $${\wedge}$$ for the infimum with respect to $$\ll$$. To prove (16), note that it follows from Lemma 57 that $$c/\theta \ll (x\vee {\mathrm{c}})/\theta$$ and $${\mathrm{c}}/\theta \ll ({\sim} x \vee {\mathrm{c}})/\theta$$. Let $$z\in T$$ such that $$z/\theta \ll (x\vee {\mathrm{c}})/\theta$$ and $$z/\theta \ll ({\sim}x \vee {\mathrm{c}})/\theta$$. In particular,   \begin{equation} \label{sup3} ((z\vee {\mathrm{c}}) {\wedge} (x\vee {\mathrm{c}}), z \vee {\mathrm{c}}) \in \theta, \end{equation} (17)  \begin{equation} \label{sup4} ((z\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}), z\vee {\mathrm{c}}) \in \theta. \end{equation} (18) It follows from $${\mathrm{(C3)}}$$, (17) and (18) that   \begin{equation} ((z\vee {\mathrm{c}}) {\wedge} (x\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}) , z \vee {\mathrm{c}}) \in \theta. \end{equation} (19) Since $$(x\vee {\mathrm{c}}) {\wedge} ({\sim} x \vee {\mathrm{c}}) = {\mathrm{c}}$$, then $$({\mathrm{c}}, z\vee {\mathrm{c}}) \in \theta$$, i.e., $$z/\theta \ll {\mathrm{c}}/\theta$$. Therefore, $$(x/\theta\vee {\mathrm{c}}/\theta){\wedge} ({\sim}x/\theta \vee {\mathrm{c}}/\theta) = {\mathrm{c}}/\theta$$. For $$x,y\in T$$ assume that $$(x\vee {\mathrm{c}})/\theta \ll (y\vee {\mathrm{c}})/\theta$$ and $$(x{\wedge} {\mathrm{c}})/ \theta \ll (y{\wedge} {\mathrm{c}})/\theta$$. It is immediate that $$x/\theta \ll y/\theta$$. Then we conclude that $$(T/\theta,\ll,{\sim},{\mathrm{c}}/\theta) \in {\mathsf{KP}}$$. Step 2. $$(T/\theta,\ll,{\sim},{\mathrm{c}}/\theta,0/\theta, 1/\theta) \in {\mathsf{KMS}}$$. Since for every $$x\in T$$ we have $$0\leq x \leq 1$$, it follows from Lemma 57 that $$0/\theta \ll x/\theta \ll 1/\theta$$, i.e., $$0/\theta$$ is the first element of $$(T/\theta,\ll)$$ and $$1/\theta$$ is the last element of $$(T/\theta,\ll)$$. Let $$x$$ and $$y$$ be elements of $$T$$. Recall that it follows from $${\mathrm{(KM3)}}$$ that $$(x\vee {\mathrm{c}}) {\wedge} y$$ exists. We will prove that $$(x \vee {\mathrm{c}})/\theta {\wedge} y/\theta$$ exists and is $$((x\vee {\mathrm{c}}){\wedge} y)/\theta$$. To do it, we will prove first that if $$(x,z) \in \theta$$ and $$(y,w) \in \theta$$, then $$((x\vee {\mathrm{c}}){\wedge} y)/\theta = ((z\vee {\mathrm{c}}){\wedge} w)/\theta$$. Let $$(x,z) \in \theta$$ and $$(y,w) \in \theta$$. It follows from $${\mathrm{(C2)}}$$ that $$(x\vee {\mathrm{c}},z\vee {\mathrm{c}}) \in \theta$$ and $$(y\vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta$$. By $${\mathrm{(C3)}}$$ we have that   \begin{equation} \label{EQ5} ((x\vee {\mathrm{c}}) {\wedge} (y\vee {\mathrm{c}}),(z\vee {\mathrm{c}}) {\wedge} (w\vee {\mathrm{c}}))\in \theta. \end{equation} (20) Taking into account $${\mathrm{(KM4)}}$$ we also have   \begin{equation} \label{EQ6} ((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}} = (x\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), \end{equation} (21)  \begin{equation} \label{EQ7} ((z\vee {\mathrm{c}}){\wedge} w) \vee {\mathrm{c}} = (z\vee {\mathrm{c}}){\wedge} (w\vee {\mathrm{c}}). \end{equation} (22) Hence, it follows from (20), (21) and (22) that   \begin{equation} \label{EQ7-1} (((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}, ((z\vee {\mathrm{c}}){\wedge} w) \vee {\mathrm{c}})\in \theta. \end{equation} (23) In a similar way, taking into account that $$({\sim} x\vee {\mathrm{c}},{\sim} z\vee {\mathrm{c}}) \in \theta$$ and $$({\sim} y\vee {\mathrm{c}},{\sim} w\vee {\mathrm{c}}) \in \theta$$ we have   \begin{equation} \label{EQ7-2} ((({\sim} x\vee {\mathrm{c}}){\wedge} {\sim} y) \vee {\mathrm{c}}, (({\sim} z\vee {\mathrm{c}}){\wedge} {\sim} w) \vee {\mathrm{c}})\in \theta. \end{equation} (24) Then it follows from (23), (24) and $${\mathrm{(C2)}}$$ that   \[ ((x\vee {\mathrm{c}}){\wedge} y, (z\vee {\mathrm{c}}){\wedge} w)\in \theta. \] Now we will prove that $$(x/\theta \vee {\mathrm{c}}/\theta) {\wedge} y/\theta$$ exists and is $$((x\vee {\mathrm{c}}){\wedge} y)/\theta$$. This is equivalent to prove that $$(x \vee {\mathrm{c}})/\theta {\wedge} y/\theta$$ exists and is $$((x\vee {\mathrm{c}}){\wedge} y)/\theta$$. Since $$(x\vee {\mathrm{c}}){\wedge} y \leq x\vee {\mathrm{c}}$$ and $$(x\vee {\mathrm{c}}){\wedge} y \leq y$$, then it follows from Lemma 57 that $$((x\vee {\mathrm{c}}){\wedge} y)/\theta \ll (x\vee {\mathrm{c}})/\theta$$ and $$((x\vee {\mathrm{c}}){\wedge} y)/\theta \ll y/\theta$$. Let $$z\in T$$ be such that $$z/\theta \ll (x\vee {\mathrm{c}})/\theta$$ and $$z/\theta \ll y/\theta$$. In particular,   \begin{equation} \label{EQ8} ((z\vee {\mathrm{c}}){\wedge} (x\vee {\mathrm{c}}), z\vee {\mathrm{c}})\in \theta, \end{equation} (25)  \begin{equation} \label{EQ9} ((z\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}), z\vee {\mathrm{c}})\in \theta, \end{equation} (26)  \begin{equation} \label{EQ10} (({\sim} y\vee {\mathrm{c}}){\wedge} ({\sim} z\vee {\mathrm{c}}), {\sim} y\vee {\mathrm{c}})\in \theta. \end{equation} (27) We need to prove that $$z/\theta \ll ((x\vee {\mathrm{c}}){\wedge} y)/\theta$$, which means that   \begin{equation} \label{EQ11} ((z\vee {\mathrm{c}}){\wedge} (((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}), z\vee {\mathrm{c}}) \in \theta \end{equation} (28) and   \begin{equation} \label{EQ12} ({\sim}((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}) {\wedge} ({\sim z} \vee {\mathrm{c}}), {\sim}((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}) \in \theta. \end{equation} (29) It follows from $${\mathrm{(KM4)}}$$ that   \begin{equation} \label{EQ12-1} (z\vee {\mathrm{c}}){\wedge} (((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}) = (z\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}){\wedge} (x\vee {\mathrm{c}}), \end{equation} (30) and it follows from (25) and $${\mathrm{(C3)}}$$ that   \begin{equation} \label{EQ12-2} ((z\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}){\wedge} (x\vee {\mathrm{c}}), (z\vee {\mathrm{c}}){\wedge} (y\vee {\mathrm{c}}))\in \theta. \end{equation} (31) Thus, by (26), (30) and (31) we obtain (28). It is immediate that the condition (29) is equal to the condition (27) because   \[ {\sim}((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}}) {\wedge} ({\sim z} \vee {\mathrm{c}}) = ({\sim} y\vee {\mathrm{c}}){\wedge} ({\sim} z\vee {\mathrm{c}}), \]   \[ {\sim}((x\vee {\mathrm{c}}){\wedge} y) \vee {\mathrm{c}} = {\sim} y\vee {\mathrm{c}}. \] Then $$(T/\theta,\ll,{\sim},{\mathrm{c}}/\theta,0/\theta, 1/\theta)$$ satisfies $${\mathrm{(KM3)}}$$. The condition $${\mathrm{(KM4)}}$$ follows from the previous steps and from the same condition on $$T$$. In consequence, we obtain that $$(T/\theta,\ll,{\sim},{\mathrm{c}}/\theta,0/\theta, 1/\theta)\in {\mathsf{KMS}}$$. Step 3. $$(T/\theta,\ll,{\sim},{\rightarrow},{\mathrm{c}}/\theta,0/\theta, 1/\theta) \in {\mathsf{KhIS_{0}}}$$. The other conditions to be an object of $${\mathsf{KhIS_{0}}}$$ follow from the previous steps, the fact that $$T\in {\mathsf{KhIS_{0}}}$$ and Lemma 57. ■ In what follows we will study the lattice of well-behaved congruences of any object of $${\mathsf{KhIS_{0}}}$$. We start with some preliminary definitions. Let $$T\in {\mathsf{KhIS_{0}}}$$. Recall that it follows from previous results of this article that $${\mathrm{C}}(T) \in {\mathsf{hIS_{0}}}$$. Note that $$T$$ does not necessarily satisfy the condition ($${\mathrm{CK}}$$). We write $${\mathrm{Con_{wb}}}(T)$$ to refer both to the set and to the lattice of well-behaved congruences of $$T$$. For $$\theta \in {\mathrm{Con_{wb}}}(T)$$ we define the binary relation $$\Gamma(\theta)$$ on $${\mathrm{C}}(T)$$ as the restriction of $$\theta$$ to $${\mathrm{C}}(T) \times {\mathrm{C}}(T)$$. For $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$ we define the relation $$\Sigma{(\tau)} \subseteq T \times T$$ in the following way:   \[ \text{$(x,y) \in \Sigma(\tau)$ if and only if $(x\vee {\mathrm{c}},y\vee {\mathrm{c}}) \in \tau$ and $({\sim} x\vee {\mathrm{c}},{\sim} y\vee {\mathrm{c}}) \in \tau$.} \] We prove that $$\Sigma(\tau)$$ is a well behaved congruence of $$T$$. Lemma 59 Let $$T\in {\mathsf{KhIS_{0}}}$$ and $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$. Then $$\Sigma(\tau) \in {\mathrm{Con_{wb}}}(T)$$. Proof. Let $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$. Straightforward computations show that $$\Sigma(\tau)$$ satisfies $${\mathrm{(C2)}}$$. To show that $$\Sigma(\tau)$$ satisfies $${\mathrm{(C3)}}$$, let $$x$$, $$y$$, $$z$$ and $$w$$ in $${\mathrm{C}}(T)$$ be such that $$(x,y) \in \Sigma(\tau)$$ and $$(z,w) \in \Sigma(\tau)$$, which means that $$(x,y) \in \tau$$ and $$(z,w) \in \tau$$. Then $$(x{\wedge} z,y{\wedge} w)\in \tau$$, because $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$. But $$(x{\wedge} z) \vee {\mathrm{c}} = x{\wedge} z$$ and $$(y{\wedge} w) \vee {\mathrm{c}} = y{\wedge} w$$. Thus,   \[ ((x{\wedge} y) \vee {\mathrm{c}}, (z{\wedge} w)\vee {\mathrm{c}}) \in \tau. \] On the other hand, since $${\sim} (x{\wedge} z) \vee {\mathrm{c}} = {\mathrm{c}}$$ and $${\sim} (y{\wedge} w) \vee {\mathrm{c}} = {\mathrm{c}}$$, then   \[ ({\sim} (x{\wedge} z) \vee {\mathrm{c}}, {\sim} (y{\wedge} w) \vee {\mathrm{c}}) \in \tau. \] Hence, $$(x{\wedge} z, z{\wedge} w) \in \Sigma(\tau)$$, so the condition $${\mathrm{(C3)}}$$ holds. Now we show the condition $${\mathrm{(C1)}}$$. It is immediate that $$\Sigma(\tau)$$ is congruence with respect to $${\sim}$$. To prove that $$\Sigma(\tau)$$ is congruence with respect to $${\rightarrow}$$, let $$(x,y) \in \Sigma(\tau)$$ and $$(z,w) \in \Sigma(\tau)$$, so   \begin{equation} \label{eq1} (x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \tau, \end{equation} (32)  \begin{equation} \label{eq2} (z\vee {\mathrm{c}}, w\vee {\mathrm{c}}) \in \tau, \end{equation} (33)  \begin{equation} \label{eq3} ({\sim} x\vee {\mathrm{c}}, {\sim} y\vee {\mathrm{c}}) \in \tau, \end{equation} (34)  \begin{equation} \label{eq4} ({\sim} z\vee {\mathrm{c}}, {\sim} w \vee {\mathrm{c}}) \in \tau. \end{equation} (35) Then taking $${\wedge}$$ in (32) and (35) we have   \begin{equation} \label{eq5} ((x\vee {\mathrm{c}}) {\wedge} ({\sim} z \vee {\mathrm{c}}), (y\vee {\mathrm{c}}){\wedge} ({\sim} w \vee {\mathrm{c}})) \in \tau. \end{equation} (36) But it follows from $${\mathrm{(K4)}}$$ that $${\sim}(x{\rightarrow} z) \vee {\mathrm{c}} = (x\vee {\mathrm{c}}) {\wedge} ({\sim} z \vee {\mathrm{c}})$$ and $${\sim}(y {\rightarrow} w) \vee {\mathrm{c}} = (y\vee {\mathrm{c}}){\wedge} ({\sim} w \vee {\mathrm{c}})$$. So by (36) we obtain that   \begin{equation}\label{eq6} ({\sim}(x{\rightarrow} z) \vee {\mathrm{c}}, {\sim}(y{\rightarrow} w) \vee {\mathrm{c}}) \in \tau. \end{equation} (37) On the other hand, taking $${\rightarrow}$$ between (32) and (33) we have that   \begin{equation}\label{eq7} ((x\vee {\mathrm{c}}) {\rightarrow} (z\vee {\mathrm{c}}), (y\vee {\mathrm{c}}) {\rightarrow} (w\vee {\mathrm{c}}))\in \tau, \end{equation} (38) and taking $${\rightarrow}$$ between (35) and (34) we obtain   \begin{equation}\label{eq8} (({\sim} z\vee {\mathrm{c}}) {\rightarrow} ({\sim} x\vee {\mathrm{c}}), ({\sim} w\vee {\mathrm{c}}) {\rightarrow} ({\sim} y\vee {\mathrm{c}}))\in \tau. \end{equation} (39) Define now the following elements:   \[ t := ((x\vee {\mathrm{c}}){\rightarrow} (z\vee {\mathrm{c}})){\wedge} (({\sim} z\vee {\mathrm{c}}) {\rightarrow} ({\sim} x \vee {\mathrm{c}})), \]   \[ u := ((y\vee {\mathrm{c}}){\rightarrow} (w\vee {\mathrm{c}})){\wedge} (({\sim} w\vee {\mathrm{c}}){\rightarrow} ({\sim} y \vee {\mathrm{c}})). \] Taking $${\wedge}$$ in (38) and (39) we obtain that   \begin{equation}\label{eq9} (t,u)\in \tau. \end{equation} (40) Besides, it follows from $${\mathrm{(K5)}}$$ that   \begin{equation} \label{eq10} (x{\rightarrow} z) \vee {\mathrm{c}} = ((x\vee {\mathrm{c}}) {\rightarrow} (z\vee {\mathrm{c}})) {\wedge} (({\sim} z \vee {\mathrm{c}}) {\rightarrow} ({\sim} x \vee {\mathrm{c}})), \end{equation} (41)  \begin{equation}\label{eq11} (y{\rightarrow} w) \vee z = ((y\vee {\mathrm{c}}){\rightarrow} (w\vee {\mathrm{c}})) {\wedge} (({\sim} w \vee {\mathrm{c}}){\rightarrow} ({\sim} y \vee {\mathrm{c}})). \end{equation} (42) Taking into account (40), (41) and (42) we have   \begin{equation} \label{eq12} ((x{\rightarrow} z)\vee {\mathrm{c}}, (y{\rightarrow} w)\vee {\mathrm{c}}) \in \tau. \end{equation} (43) Thus, by (37) and (43) the condition $$(x{\rightarrow} z, y{\rightarrow} w) \in \Sigma(\tau)$$ is satisfied. This implies that $$\Sigma(\tau) \in {\mathrm{Con_{wb}}}(T)$$. ■ Proposition 60 Let $$T\in {\mathsf{KhIS_{0}}}$$. There exists an isomorphism between $${\mathrm{Con_{wb}}}(T)$$ and $${\mathrm{Con}}({\mathrm{C}}(T))$$, which is established via the assignments $$\theta \mapsto \Gamma(\theta)$$ and $$\tau \mapsto \Sigma(\tau)$$. Proof. Let $$\theta\in {\mathrm{Con_{wb}}}(T)$$. It follows from $${\mathrm{(C1)}}$$ and $${\mathrm{(C3)}}$$ that $$\Gamma(\theta) \in {\mathrm{Con}} ({\mathrm{C}}(T))$$. Suppose now that $$\theta\in {\mathrm{Con_{wb}}}(T)$$, $$\sigma \in {\mathrm{Con_{wb}}}(T)$$ and $$\Gamma(\theta) = \Gamma(\sigma)$$. Let $$(x,y) \in \theta$$. Then by $${\mathrm{(C2)}}$$ we have $$(x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \theta$$ and $$({\sim} x \vee {\mathrm{c}}, {\sim} y \vee {\mathrm{c}}) \in \theta$$, so $$(x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \Gamma(\theta)$$ and $$({\sim}x \vee {\mathrm{c}}, {\sim}y \vee {\mathrm{c}}) \in \Gamma(\theta)$$. Since $$\Gamma(\theta) = \Gamma(\sigma)$$, $$(x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \sigma$$ and $$({\sim}x \vee {\mathrm{c}}, {\sim} y \vee {\mathrm{c}}) \in \sigma$$. Hence, it follows from $${\mathrm{(C2)}}$$ again that $$(x,y) \in \sigma$$. Thus, $$\theta \subseteq \sigma$$. For the same reason we have the other inclusion, so $$\theta = \sigma$$. Lemma 59 shows that if $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$, then $$\Sigma(\tau)\in {\mathrm{Con_{wb}}}(T)$$. Besides it is immediate that $$\Gamma(\Sigma(\tau)) = \tau$$. We also have that for $$\theta\in {\mathrm{Con_{wb}}}(T)$$ and $$\sigma \in {\mathrm{Con_{wb}}}(T)$$, $$\theta \subseteq \sigma$$ if and only if $$\Sigma(\theta) \subseteq \Sigma (\sigma)$$. Therefore, we obtain an isomorphism between $${\mathrm{Con_{wb}}}(T)$$ and $${\mathrm{Con}}({\mathrm{C}}(T))$$. ■ Let $$T\in {\mathsf{KhBDL}}$$. If $$\theta \in {\mathrm{Con}}(T)$$ and $$\tau \in {\mathrm{C}}(T)$$, we define $$\Gamma(\theta)$$ and $$\Sigma(\tau)$$ as for the case of $${\mathsf{KhIS_{0}}}$$. If $$\theta \in {\mathrm{Con}}(T)$$, then $$\theta$$ satisfies $${\mathrm{(C1)}}$$, $${\mathrm{(C2)}}$$, and $${\mathrm{(C3)}}$$. Let $$\tau \in {\mathrm{C}}(T)$$. The distributivity of the underlying lattice of $$T$$ proves that $$\Sigma(\tau)$$ preserves $${\wedge}$$ and $$\vee$$. Then from the proof of Proposition 60 the next result follows. Proposition 61 Let $$T\in {\mathsf{KhBDL}}$$. There exists an isomorphism between $${\mathrm{Con}}(T)$$ and $${\mathrm{Con}}({\mathrm{C}}(T))$$, which is established via the assignments $$\theta \mapsto \Gamma(\theta)$$ and $$\tau \mapsto \Sigma(\tau)$$. Let $$H \in {\mathsf{hIS_{0}}}$$ or $$H \in {\mathsf{hBDL}}$$. Let $$\theta \in {\mathrm{Con}}(H)$$ and $$\tau \in {\mathrm{Con}}({\mathrm{C}}({\mathrm{K}}(H)))$$. Since the map $$\alpha:H{\rightarrow} {\mathrm{C}}(K(H))$$ given by $$\alpha(a) = (a,0)$$ is an isomorphism, we have that the binary relation $$\alpha(\theta) = \{(\alpha(a),\alpha(b)): (a,b) \in \theta\}$$ in $${\mathrm{C}}({\mathrm{K}}(H))$$ is a congruence of $${\mathrm{C}}({\mathrm{K}}(H))$$. Moreover, the relation $$\alpha^{-1}(\tau)$$ in $$H$$ given by $$(a,b) \in \alpha^{-1}(\tau)$$ if and only if $$((a,0), (b,0)) \in \tau$$ is a congruence of $$H$$. Then the following result follows from propositions 60 and 61. Corollary 62 (a) Let $$H\in {\mathsf{hIS_{0}}}$$. There exists an isomorphism between $${\mathrm{Con}}(H)$$ and $${\mathrm{Con_{wb}}}({\mathrm{K}}(H))$$, which is established via the assignments $$\theta \mapsto \Sigma(\alpha(\theta))$$ and $$\tau \mapsto \alpha^{-1}(\Gamma(\tau))$$. (b) Let $$H\in {\mathsf{hBDL}}$$. There exists an isomorphism between $${\mathrm{Con}}(H)$$ and $${\mathrm{Con}}({\mathrm{K}}(H))$$, which is established via the assignments $$\theta \mapsto \Sigma(\alpha(\theta))$$ and $$\tau \mapsto \alpha^{-1}(\Gamma(\tau))$$. Remark 63 Let $$H\in {\mathsf{hIS_{0}}}$$, $$\theta \in {\mathrm{Con}}(H)$$ and $$\tau \in {\mathrm{Con_{wb}}}({\mathrm{C}}({\mathrm{K}}(H)))$$. Then   \[ ((a,b),(d,e)) \in \Sigma(\alpha(\theta))\; \textrm{if and only if} \; (a,d) \in \theta\; \textrm{and}\; (b,e) \in \theta, \]   \[ (a,b) \in \alpha^{-1}(\Gamma(\tau))\; \textrm{if and only if}\; ((a,0),(b,0)) \in \tau. \] Similarly for $$H\in {\mathsf{hBDL}}$$. 6.1 Well-behaved congruences and congruences: the relation with some family of filters and some applications We start by recalling some facts about congruences in $${\mathsf{hIS_{0}}}$$ and congruences in $${\mathsf{hBDL}}$$ [21]. Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$. As usual, we say that $$F$$ is a filter if it is a nonempty subset of $$H$$ which satisfies the following conditions: (1) If $$a\in F$$ and $$b\in F$$ then $$a{\wedge} b \in F$$. (2) If $$a\in F$$ and $$a\leq b$$ then $$b\in F$$. We also consider the binary relation associated with $$F \subseteq H$$  \[ \Theta(F) = \{(a,b) \in H\times H: a{\wedge} f = b {\wedge} f\;\text{for some}\; f\in F\}. \] Note that if $$H$$ is an upper bounded semilattice and $$F$$ is a filter, then $$\Theta(F)$$ is a congruence. Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$. For $$a,b,f\in H$$ we define the following element of $$H$$:   $$t(a,b,f): = (a {\rightarrow} b) {\leftrightarrow} ((a{\wedge} f) {\rightarrow} (b{\wedge} f)),$$ where $$a{\leftrightarrow} b:= (a{\rightarrow} b){\wedge} (b{\rightarrow} a)$$. The next definition was introduced in [21]. Definition 64 Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$, and let $$F$$ be a filter of $$H$$. We say that $$F$$ is a congruent filter if $$t(a,b,f)\in F$$ whenever $$a$$, $$b \in H$$ and $$f\in F$$. Note that the set of all congruent filters of $$H\in {\mathsf{hIS_{0}}}$$ or of $$H\in {\mathsf{hBDL}}$$ is closed under arbitrary intersections and therefore for every $$X \subseteq H$$ the congruent filter generated by $$X$$ exists. Remark 65 Let $$F$$ be a congruent filter of a hemi-implicative semilattice (lattice). We will see that $$(a,b) \in \Theta(F)$$ if and only if $$a{\leftrightarrow} b \in F$$. To show it, suppose that $$a{\leftrightarrow} b \in F$$. Since $$a{\wedge} (a{\leftrightarrow} b) = b {\wedge}(b{\leftrightarrow} a)$$, then $$(a,b) \in \Theta(F)$$. Conversely, assume that $$(a,b) \in \Theta(F)$$, i.e., $$a{\wedge} f = b{\wedge} f$$ for some $$f\in F$$. Since $$t(a,b,f) \in F$$ and $$t(a,b,f) = (a{\rightarrow} b) {\leftrightarrow} 1$$, then $$1{\rightarrow} (a{\rightarrow} b) \in F$$ because $$(a{\rightarrow} b) {\leftrightarrow} 1 \leq 1{\rightarrow} (a{\rightarrow} b)$$. Since $$1{\rightarrow} (a{\rightarrow} b) \leq a{\rightarrow} b$$, then $$a{\rightarrow} b\in F$$. In a similar way we can show that $$b{\rightarrow} a\in F$$. Hence, $$a{\leftrightarrow} b\in F$$. Thus,   \[ \Theta(F) = \{(a,b)\in H\times H: a{\leftrightarrow} b \in F\}. \] The following result was proved in [21]. Theorem 66 Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$. There exists an isomorphism between $${\mathrm{Con}}(H)$$ and the lattice of congruent filters of $$H$$, which is established via the assignments $$\theta \mapsto 1/\theta$$ and $$F\mapsto \Theta(F)$$. Taking into account Theorem 66, it is possible to show that Proposition 61 can be seen as a corollary of Proposition 60. To show this assertion, let $$T_1 = (T,{\wedge},\vee,{\rightarrow},{\sim},{\mathrm{c}},0,1) \in {\mathsf{hBDL}}$$. Then we write $$T_2 = (T,\leq,{\sim},{\rightarrow},{\mathrm{c}},0,1)$$ for the corresponding object of $${\mathsf{KhIS_{0}}}$$. Since the set of congruent filters of $${\mathrm{C}}(T_1)$$ is equal to the set of congruent filters of $${\mathrm{C}}(T_2)$$, then it follows from Theorem 66 that $${\mathrm{Con}}({\mathrm{C}}(T_1)) = {\mathrm{Con}}({\mathrm{C}}(T_2))$$. In what follows we will see that $${\mathrm{Con}}(T_1) = {\mathrm{Con_{wb}}}(T_2)$$. It is immediate that $${\mathrm{Con}}(T_1) \subseteq {\mathrm{Con_{wb}}}(T_2)$$. Conversely, let $$\theta \in {\mathrm{Con_{wb}}}(T_2)$$. We will prove that $$\theta$$ preserves $${\wedge}$$ and $$\vee$$. Let $$(x,y)\in \theta$$ and $$(z,w) \in \theta$$. Then it follows from $${\mathrm{(C2)}}$$ that $$(x\vee {\mathrm{c}}, z\vee {\mathrm{c}}) \in \theta$$ and $$(y \vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta$$. Then by $${\mathrm{(C3)}}$$ we have that $$((x\vee {\mathrm{c}}){\wedge} (z \vee {\mathrm{c}}), (y\vee {\mathrm{c}}){\wedge} (w\vee {\mathrm{c}}))\in \theta$$. But by the distributivity of the underlying lattice of $$T_1$$ we deduce that $$(x{\wedge} z) \vee {\mathrm{c}} = (x\vee {\mathrm{c}}){\wedge} (z\vee {\mathrm{c}})$$ and $$(y{\wedge} w) \vee {\mathrm{c}} = (y\vee {\mathrm{c}}){\wedge} (w\vee {\mathrm{c}})$$. Thus,   \begin{equation} \label{cwbc0} ((x{\wedge} z)\vee {\mathrm{c}}, (y{\wedge} w)\vee {\mathrm{c}} \in \theta. \end{equation} (44) Besides, since $$(x,y)\in \theta$$ and $$(z,w)\in \theta$$, then it follows from the condition $${\mathrm{(C2)}}$$ that $$({\sim} x \vee {\mathrm{c}},{\sim} y \vee {\mathrm{c}}) \in \theta$$ and $$({\sim} y \vee {\mathrm{c}},{\sim} w \vee {\mathrm{c}}) \in \theta$$. Equivalently, we have that   \begin{equation} \label{cwbc1} ({\sim} x \vee {\mathrm{c}},{\sim} y \vee {\mathrm{c}}) \in \Gamma(\theta), \end{equation} (45)  \begin{equation} \label{cwbc2} ({\sim} y \vee {\mathrm{c}},{\sim} w \vee {\mathrm{c}}) \in \Gamma(\theta). \end{equation} (46) Since $$\Gamma(\theta) \in {\mathrm{Con}}({\mathrm{C}}(T_2))$$ and $${\mathrm{Con}}({\mathrm{C}}(T_1)) = {\mathrm{Con}}({\mathrm{C}}(T_2))$$, then taking $$\vee$$ in (45) and (46) we obtain $$({\sim} x \vee {\sim} y \vee {\mathrm{c}}, {\sim} z \vee {\sim} w \vee {\mathrm{c}}) \in \theta$$, i.e.,   \begin{equation} \label{cwbc4} ({\sim}(x{\wedge} z)\vee {\mathrm{c}}, {\sim}(z{\wedge} w) \vee {\mathrm{c}})\in \theta. \end{equation} (47) Then it follows from $${\mathrm{(C2)}}$$, (44) and (47) that $$(x{\wedge} z, y{\wedge} w)\in \theta$$. The same argument combined with $${\mathrm{(C1)}}$$ proves that $$({\sim} x {\wedge} {\sim} z, {\sim} y {\wedge} {\sim} w) \in \theta$$, so $$(x\vee z,y\vee w)\in \theta$$. Hence, $$\theta$$ preserves $${\wedge}$$ and $$\vee$$, which implies that $$\theta \in {\mathrm{Con}}(T_1)$$. Then $${\mathrm{Con}}(T_1) = {\mathrm{Con_{wb}}}(T_2)$$. Therefore, since $${\mathrm{Con}}(T_1) = {\mathrm{Con_{wb}}}(T_2)$$ and $${\mathrm{Con}}({\mathrm{C}}(T_1)) = {\mathrm{Con}}({\mathrm{C}}(T_2))$$, we deduce that Proposition 61 can be seen as a corollary of Proposition 60. Corollary 67 Let $$T\in {\mathsf{KhIS_{0}}}$$. There exists an isomorphism between $${\mathrm{Con_{wb}}}(T)$$ and the lattice of congruent filters of $${\mathrm{C}}(T)$$, which is established via the assignments $$\theta \mapsto 1/\Gamma(\theta)$$ and $$F\mapsto \Sigma(\Theta(F))$$. Proof. It follows from Proposition 60 and Theorem 66. ■ Similarly, the following result follows from Proposition 61 and Theorem 66. Corollary 68 Let $$T\in {\mathsf{KhBDL}}$$. There exists an isomorphism between $${\mathrm{Con}}(T)$$ and the lattice of congruent filters of $${\mathrm{C}}(T)$$, which is established via the assignments $$\theta \mapsto 1/\Gamma(\theta)$$ and $$F\mapsto \Sigma(\Theta(F))$$. For implicative semilattices Corollary 67 can be simplified, and for semi-Heyting algebras Corollary 68 also can be simplified. More precisely: if $$H \in {\mathsf{IS_{0}}}$$ or $$H\in {\mathsf{SH}}$$ then the congruent filters of $$H$$ are all the filters of $$H$$ [21]. Let $$H\in {\mathsf{Hil_{0}}}$$ and $$F\subseteq H$$. Recall that $$F$$ is said to be a deductive system [10] if the following conditions are satisfied: (a) $$1\in F$$, (b) if $$a\in F$$ and $$a{\rightarrow} b\in F$$ then $$b\in F$$. Also recall that a deductive system $$F$$ is said to be absorbent [12] if $$a {\rightarrow} (a{\wedge} b)\in F$$ whenever $$a\in F$$. It follows from Theorem 66 and [12, Lemma 3.3] that the congruent filters of $$H$$ are the absorbent deductive systems of $$H$$. Definition 69 Let $$A$$ be an algebra and $$a_1,b_1,\dots,a_n,b_n$$ elements of $$A$$. We write $$\theta_{A}((a_1,b_1),\dots,(a_n,b_n))$$ for the congruence generated by $$(a_1,b_1),\dots,(a_n,b_n)$$. If $$T\in {\mathsf{KhIS_{0}}}$$ we also write $$\theta_{T}((a_1,b_1),\dots,(a_n,b_n))$$ for the well-behaved congruence generated by $$(a_1,b_1),\dots,(a_n,b_n)$$. Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$, and let $$a \in H$$. We refer by $$F^{c}(a)$$ to the congruent filter generated by $$\{a\}$$. In [21] the following assertions were proved: (1) if $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$, then $$(d,e) \in \theta_{H}(a,b)$$ if and only if $$d{\leftrightarrow} e \in F^{c}(a{\leftrightarrow} b)$$; (2) if $$H\in {\mathsf{IS_{0}}}$$ or $$H\in {\mathsf{SH}}$$ then $$(d,e) \in \theta_{H}(a,b)$$ if and only if $$a{\leftrightarrow} b \leq d{\leftrightarrow} e$$. The following remark will be used later. Remark 70 Let $$H\in {\mathsf{hIS_{0}}}$$ or $$H\in {\mathsf{hBDL}}$$. Let $$\tau \in {\mathrm{Con}}(H)$$. Then $$(a,b) \in \tau$$ if and only if $$a{\leftrightarrow} b \in 1/\tau$$. Moreover, $$(a,b)$$, $$(d,e) \in \tau$$ if and only if $$(a{\leftrightarrow} b){\wedge} (d{\leftrightarrow} e) \in 1/\tau$$. In what follows we describe some aspects of the principal well-behaved congruences of the objects of $${\mathsf{KhIS_{0}}}$$ and some aspects of the principal congruences of the algebras in $${\mathsf{KhBDL}}$$. Let $$T\in {\mathsf{KhIS_{0}}}$$ or $$T\in {\mathsf{KhBDL}}$$. For $$x$$ and $$y$$ elements of $$T$$ we also write $$x{\leftrightarrow} y$$ for the element $$(x{\rightarrow} y){\wedge} (y{\rightarrow} x)$$. Lemma 71 Let $$T\in {\mathsf{KhIS_{0}}}$$ or $$T\in {\mathsf{KhBDL}}$$. Let $$x$$, $$y$$, $$z$$, $$w \in T$$. Then (a) $$(z,w) \in \theta_{T}(x,y)$$ if and only if $$(z\vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}}))$$ and $$(\sim z\vee {\mathrm{c}},\sim w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}}))$$. (b) If $$x$$, $$y$$, $$z$$ and $$w$$ are in $${\mathrm{C}}(T)$$, then   \[ 1/\theta_{{\mathrm{C}}(T)}((x,y),(z,w)) = F^{c}((x{\leftrightarrow} y) {\wedge} (z{\leftrightarrow} w)). \] Proof. We consider $$T\in {\mathsf{KhIS_{0}}}$$ (the proof for $$T\in {\mathsf{KhBDL}}$$ is analogous). First we prove a). Let $$(z,w) \in \theta_{T}(x,y)$$. Then $$(z,w) \in \theta$$ for every $$\theta \in {\mathrm{Con_{wb}}}(T)$$ such that $$(x,y) \in \theta$$. Now we see that   \[ (z\vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})), \]   \[ (\sim z\vee {\mathrm{c}},\sim w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})). \] Let $$\tau \in {\mathrm{Con}}({\mathrm{C}}(T))$$ such that $$(x\vee {\mathrm{c}}, y\vee {\mathrm{c}}) \in \tau$$ and $$(\sim x \vee {\mathrm{c}}, \sim y \vee {\mathrm{c}}) \in \tau$$. It follows from Proposition 60 that $$\Sigma(\tau) \in {\mathrm{Con_{wb}}}(T)$$. We also have that $$(x,y) \in \Sigma(\tau)$$. Then by hypothesis we obtain that $$(z,w) \in \Sigma(\tau)$$. Hence, $$(z\vee {\mathrm{c}}, w\vee {\mathrm{c}}) \in \tau$$ and $$(\sim z \vee {\mathrm{c}}, \sim w \vee {\mathrm{c}}) \in \tau$$, which was our aim. Conversely, assume that $$(z\vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}}))$$ and $$(\sim z\vee {\mathrm{c}},\sim w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}}))$$. Let $$\theta \in {\mathrm{Con_{wb}}}(T)$$ be such that $$(x,y) \in \theta$$. It follows from Proposition 60 that $$\Gamma(\theta) \in {\mathrm{Con}}({\mathrm{C}}(T))$$. Moreover, $$(x\vee {\mathrm{c}},y\vee {\mathrm{c}}) \in \Gamma(\theta)$$ and $$(\sim x\vee {\mathrm{c}},\sim y\vee {\mathrm{c}}) \in \Gamma(\theta)$$. Thus by hypothesis we have that $$(z\vee {\mathrm{c}}, w\vee {\mathrm{c}}) \in \Gamma(\theta)$$ and $$(\sim z\vee {\mathrm{c}}, \sim w\vee {\mathrm{c}}) \in \Gamma(\theta)$$, i.e., $$(z\vee {\mathrm{c}}, w\vee {\mathrm{c}}) \in \theta$$ and $$(\sim z\vee {\mathrm{c}}, \sim w\vee {\mathrm{c}}) \in \theta$$. Then it follows from $${\mathrm{(C2)}}$$ that $$(z,w) \in \theta$$. Thus, $$(z,w) \in \theta_{T}(x,y)$$. Finally, we prove b). Let $$H\in {\mathsf{hIS_{0}}}$$. We write $$\tau$$ for an arbitrary well-behaved congruence of $$H$$. Then   \[ \theta_{H}((x,y),(z,w)) = \bigcap\{\tau \in {\mathrm{Con_{wb}}}(H): (x,y),(z,w) \in \tau\}. \] Hence,   \[ 1/\theta_{H}((x,y),(z,w)) = \bigcap\{1/\tau: \tau \in {\mathrm{Con_{wb}}}(H) \text{ and } (x,y), (z,w) \in \tau\}. \] Then it follows from Remark 70 that   \[ 1/\theta_{H}((x,y),(z,w)) = \bigcap\{ 1/\tau: \tau \in {\mathrm{Con_{wb}}}(H) \text{ and } (x{\leftrightarrow} y){\wedge} (z{\leftrightarrow} w) \in 1/\tau\}. \] Thus, by Theorem 66 we have that   \[ 1/\theta_{H}((x,y),(z,w)) = F^{c}((x{\leftrightarrow} y){\wedge} (z{\leftrightarrow} w)). \] In particular, the last assertion holds for $$H = {\mathrm{C}}(T)$$. ■ Remark 72 The proof of item (b) of Lemma 71 can be done in a different way. Let $$\theta$$ be a congruence of an algebra $$H\in {\mathsf{hIS_{0}}}$$, and let $$a,b\in H$$. Straightforward computations show that $$F^{c}(a) \vee F^{c}(b) = F^{c}(a {\wedge} b)$$, where $$\vee$$ is the supremum in the lattice of congruent filters of $$H$$. On the other hand, it follows from general results from universal algebra that $$\theta_{H}((x,y),(z,w)) = \theta_{H}(x,y) \vee \theta_H(z,w)$$, where $$\vee$$ is the supremum in the lattice of congruences of $$H$$ [20]. In [21] it was proved that $$1/\theta_{H}(x,y) = F^{c}(x{\leftrightarrow} y)$$. Then   \[ \begin{array} [c]{lllll} 1/\theta_{H}((x,y),(z,w)) & = & 1/\theta_{H}(x,y) \vee 1/\theta_{H}(z,w) & & \\ & = & F^{c}(x{\leftrightarrow} y) \vee F^{c}(z{\leftrightarrow} w)& & \\ & = & F^{c}((x{\leftrightarrow} y) {\wedge} (z{\leftrightarrow} w)).& & \end{array} \] Let $$T\in {\mathsf{KhIS_{0}}}$$ or $$T\in {\mathsf{KhBDL}}$$. For every $$x$$, $$y\in T$$ we define the following binary term:   \[ q(x,y) = ((x\vee {\mathrm{c}}) {\leftrightarrow} (y\vee {\mathrm{c}})){\wedge} ((\sim x\vee {\mathrm{c}}) {\leftrightarrow} (\sim y\vee {\mathrm{c}})). \] In the proof of the following corollary we will use Remark 70 and Lemma 71. Corollary 73 Let $$T \in {\mathsf{KhIS_{0}}}$$ or $$T\in {\mathsf{KhBDL}}$$. Let $$x$$, $$y$$, $$z$$, $$w \in T$$. (a) $$(z,w) \in \theta_{T}(x,y)$$ if and only if $$q(z,w) \in F^{c}(q(x,y))$$. (b) If $$T\in {\mathsf{KIS_{0}}}$$ or $$T \in {\mathsf{KSH}}$$ then $$(z,w) \in \theta_{T}(x,y)$$ if and only if $$q(x,y) \leq q(z,w)$$. Proof. The condition $$(z,w) \in \theta_{T}(x,y)$$ is equivalent to the conditions   \[ (z\vee {\mathrm{c}},w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})), \]   \[ (\sim z\vee {\mathrm{c}},\sim w\vee {\mathrm{c}}) \in \theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})), \] which are equivalent to   \[ (z\vee {\mathrm{c}}) {\leftrightarrow} (w\vee {\mathrm{c}}) \in 1/\theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})), \]   \[ ({\sim} z\vee {\mathrm{c}}) {\leftrightarrow} ({\sim} w\vee {\mathrm{c}}) \in 1/\theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})), \] which happens if and only if   \[ q(z,w) \in 1/\theta_{{\mathrm{C}}(T)}((x\vee {\mathrm{c}}, y\vee {\mathrm{c}}),(\sim x\vee {\mathrm{c}}, \sim y\vee {\mathrm{c}})). \] But this last fact is equivalent to say that $$q(z,w) \in F^{c}(q(x,y))$$. If $$T \in {\mathsf{KIS_{0}}}$$ or $$T\in {\mathsf{KSH}}$$, then $$F^{c}(q(x,y))$$ is equal to the filter generated by $$\{q(x,y)\}$$, so $$(z,w) \in \theta_{T}(x,y)$$ if and only if $$q(x,y) \leq q(z,w)$$. ■ Acknowledgements This project has received funding from the European Union’s Horizon 2020 research and innovation programme under the Marie Sklodowska-Curie grant agreement [No 689176]. R.J. was also partially supported by the research grant 2014 SGR 788 from the government of Catalonia and by the research projects MTM2011-25747 and MTM2016-74892-P from the government of Spain, which include feder funds from the European Union. H.J.S.M. was also supported by CONICET Project PIP 112-201501-00412, and he thanks Marta Sagastume for several conversations concerning the matter of this article. Footnotes 1Let $$(H,\leq)$$ be a poset. 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Logic Journal of the IGPL – Oxford University Press

Published: Feb 1, 2018

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