A categorical equivalence between semi-Heyting algebras and centered semi-Nelson algebras

A categorical equivalence between semi-Heyting algebras and centered semi-Nelson algebras Abstract Motivated by a construction due to R. Cignoli that relates Heyting algebras and centered Nelson algebras, in this paper we prove that there exists an equivalence between the category of semi-Heyting algebras and the category of centered semi-Nelson algebras. 1 Introduction Inspired by results from [17] due to J. Kalman relating to lattices, R. Cignoli proved in [11, Theorem 2.4] that there exists an equivalence between the category of bounded distributive lattices and a particular full subcategory of centered Kleene algebras. Moreover, he also proved that there exists an equivalence between the category of Heyting algebras and the category of centered Nelson algebras [11, Theorem 3.14] (see also [10, 16]). In this paper we extend the previous result in order to show that there is an equivalence between the category of semi-Heyting algebras [20] and the category of semi-Nelson algebras [12] with center. In the process of our research on the topic of the present paper, we have found useful to place our problems in the following general context. We assume that the reader is familiar with bounded distributive lattices and Heyting algebras [5]. A De Morgan algebra is an algebra (H, ∧, ∨, ∼, 0, 1) of type (2, 2, 1, 0, 0) such that (H, ∧, ∨, 0, 1) is a bounded distributive lattice and ∼ fulfills the equations ∼∼x = x and ∼(x ∨ y) = ∼x ∧ ∼y. An operation ∼ which satisfies the previous two equations is called involution. A Kleene algebra is a De Morgan algebra in which the inequality x ∧ ∼x ≤ y ∨ ∼y holds. We say that an algebra (H, ∧, ∨, ∼, c, 0, 1) of type (2, 2, 1, 0, 0, 0) is a centered Kleene algebra if (H, ∧, ∨, ∼, 0, 1) is a Kleene algebra and c is such that c = ∼c (this element is called center). It is immediate to see that c is necessarily unique. We write 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 center, i.e. f(c) = c. For an object H ∈ BDL we define   $$ \mathrm{K}(H): =\{(a,b) \in H\times H: a\wedge b = 0\}. $$This set could be endowed with the operations and the distinguished elements defined by:   \begin{align*} (a,b)\vee (d,e) & := (a\vee d,b\wedge e)\\[-2pt] (a,b)\wedge (d,e)& := (a\wedge d,b\vee e)\\[-2pt] {\sim} (a,b)& := (b,a)\\[-2pt] 0 & := (0,1)\\[-2pt] 1 & := (1,0)\\[-2pt] \mathrm{c} & := (0,0). \end{align*} In particular, $$(\mathrm{K}(H),\wedge ,\vee , \sim ,\mathrm{c}, 0,1)\in \mathsf{KA_{\mathrm{c}}}$$. For a morphism $$f:H \rightarrow G \in \mathsf{BDL}$$, the map $$\mathrm{K}(\,f):\mathrm{K}(H) \rightarrow \mathrm{K}(G)$$, defined by K(f)(a, b) = (f(a), f(b)), is a morphism in $$\mathsf{KA_{\mathrm{c}}}$$. Moreover, K is a functor from 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, ∧, ∨, c, 1) and (C(T), ∧, ∨, c, 1) ∈ BDL. For a morphism $$g:T\rightarrow U \in \mathsf{KA_{\mathrm{c}}}$$, the map $$\mathrm{C}(g): \mathrm{C}(T) \rightarrow \mathrm{C}(U)$$, given by C(g)(x) = g(x), is a morphism in BDL. Moreover, C is a functor from $$\mathsf{KA_{\mathrm{c}}}$$ to BDL. Let H ∈ BDL. The map $$\alpha _{H}: H \rightarrow \mathrm{C}(\mathrm{K}(H))$$ given by $$\alpha _{H}(a) = (a,0)$$ is an isomorphism in BDL. If $$T \in \mathsf{KA_{\mathrm{c}}}$$, 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 it is a morphism in $$\mathsf{KA_{\mathrm{c}}}$$, but it is not necessarily surjective (see [16]). Let $$T\in \mathsf{KA_{\mathrm{c}}}$$. Consider the following algebraic condition:   \begin{align}(\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{align} (CK)This condition characterizes the surjectivity of $$\beta _{T}$$, that is, for every $$T\in \mathsf{KA_{\mathrm{c}}}$$ we have that T satisfies (CK) if and only if $$\beta _{T}$$ is a surjective map. The condition (CK) is not necessarily verified in every centered Kleene algebra, see for instance [10, Figure 1]. We write $$\mathsf{KA_{\mathrm{c}}^{CK}}$$ for the full subcategory of $$\mathsf{KA_{\mathrm{c}}}$$ whose objects satisfy (CK). The functor K can then be seen as a functor from BDL to $$\mathsf{KA_{\mathrm{c}}^{CK}}$$. The following result is [10, Theorem 2.7] (see also [11, Theorem 2.4]). Theorem 1.1 The functors K and C establish a categorical equivalence between BDL and $$\mathsf{KA_{\mathrm{c}}^{CK}}$$ with natural isomorphisms $$\alpha $$ and $$\beta $$. Let H ∈ BDL and a, b ∈ H. If the relative pseudocomplement of a with respect to b exists, then we denote it by $$a \rightarrow _{H} b$$. Recall that a Nelson algebra [11, 24] is a Kleene algebra such that for each pair x, y there exists the binary operation ⇒ given by $$x\Rightarrow y: = x \rightarrow _{H} ({\sim x} \vee y)$$ and for every x, y, z it holds that (x ∧ y) ⇒ z = x ⇒ (y ⇒ z). The binary operation ⇒ so defined is called the weak implication. Nelson algebras can be seen as algebras (H, ∧, ∨, ⇒, ∼, 0, 1) of type (2, 2, 2, 1, 0, 0). The class of Nelson algebras is a variety [6, 7, 18]. We say that an algebra (T, ∧, ∨, ⇒, ∼, c, 0, 1) is a centered Nelson algebra if the reduct (T, ∧, ∨, ⇒, ∼, 0, 1) is a Nelson algebra and c satisfies ∼c = c. It is a known fact that centered Nelson algebras satisfy the condition (CK) (see [10]). M. Fidel [14] and D. Vakarelov [23] proved independently that if $$(H,\wedge ,\vee ,\rightarrow ,0,1)$$ is a Heyting algebra then (K(H), ∧, ∨, ⇒, ∼, c, 0, 1) is a centered Nelson algebra, where ⇒ is defined as follows, for pairs (a, b) and (d, e) in K(H):   \begin{align} (a,b)\Rightarrow(d,e):= (a\rightarrow d, a\wedge e). \end{align} (1) Let HA be the category of Heyting algebras and $$\mathsf{NA_{\mathrm{c}}}$$ the category of centered Nelson algebras. The following result is a reformulation of [11, Theorem 3.14] (see also [9]). Theorem 1.2 The functors K and C establish a categorical equivalence between HA and $$\mathsf{NA_{\mathrm{c}}}$$ with natural isomorphisms $$\alpha $$ and $$\beta $$. In what follows we recall the definition of semi-Heyting algebras, which were introduced by H. P. Sankappanavar in [20] as an abstraction of Heyting algebras. Semi-Heyting algebras share with Heyting algebras the following properties: they are pseudocomplemented, distributive lattices and their congruences are determined by the lattice filters. The relationship between the variety of semi-Heyting algebras and the varieties of Heyting algebras (and its expansions) have been studied lately in [1, 4, 19]. Definition 1.3 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: (SH1) (H, ∧, ∨, 0, 1) is a bounded lattice, (SH2) $$a\wedge (a\rightarrow b) = a \wedge b$$, (SH3) $$a\wedge (b\rightarrow d) = a \wedge [(a\wedge b) \rightarrow (a\wedge d)]$$, (SH4) $$a\rightarrow a = 1$$. We write SH for the category of semi-Heyting algebras. It was proved in [20] that the underlying lattice of a semi-Heyting algebra is necessarily distributive. In what follows we recall the definition of pre-semi-Nelson algebra and semi-Nelson algebra respectively. These definitions were introduced in [12]. Definition 1.4 An algebra (T, ∧, ∨, →, ∼, 1) of type (2, 2, 2, 1, 0) is a pre-semi-Nelson algebra if for every x, y, z ∈ H the following conditions are satisfied: (SN1) x ∧ (x ∨ y) = x, (SN2) x ∧ (y ∨ z) = (z ∧ x) ∨ (y ∧ x), (SN3) ∼∼ x = x, (SN4) ∼ (x ∧ y) =∼ x∨ ∼ y, (SN5) x∧ ∼ x = (x∧ ∼ x) ∧ (y∨ ∼ y), (SN6) $$x \wedge (x \to _{N} y) = x \wedge (\sim x \vee y)$$, (SN7) $$x \to _{N} (y \to _{N} z) = (x \wedge y) \to _{N} z$$, (SN8) $$(x \to _{N} y) \to _{N} [(y \to _{N} x) \to _{N} [(x \to z) \to _{N} (y \to z)]] = 1$$, (SN9) $$(x \to _{N} y) \to _{N} [(y \to _{N} x) \to _{N} [(z \to x) \to _{N} (z \to y)]] = 1$$, where $$x \to _{N} y$$ stands for the term x → (x ∧ y). Definition 1.5 A pre-semi-Nelson algebra (T, ∧, ∨, →, ∼, 1) is a semi-Nelson algebra if it also verifies the following conditions for every x, y ∈ H: (SN10) $$(\sim (x \to y)) \to _{N} (x \wedge \sim y) = 1$$, (SN11) $$(x \wedge \sim y) \to _{N} (\sim (x \to y)) = 1$$. We write PSN for the category of pre-semi-Nelson algebras. Notice that the conditions (SN1) and (SN2) are those given by Sholander in [21] which define distributive lattices. In addition, the conditions (SN3), (SN4) and (SN5) define Kleene algebras. Thus, every pre-semi-Nelson algebra is in particular a Kleene algebra. Semi-Nelson algebras were introduced by J. M. Cornejo and I. Viglizzo in [12] as a generalization of Nelson algebras. Moreover, in [12, Corollary 5.2] it was proved that if T is a semi-Nelson algebra then there exists a particular semi-Heyting algebra H such that T is isomorphic to a subalgebra of the semi-Nelson algebra (K(H), ∧, ∨, ⇒, ∼, 1). We write SN for the category of semi-Nelson algebras and $$\mathsf{SN_{c}}$$ for the category of centered semi-Nelson algebras. More precisely, the objects of $$\mathsf{SN_{c}}$$ are defined as algebras $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1)$$ of type (2, 2, 2, 1, 0, 0, 0), where $$(T,\wedge ,\vee ,\rightarrow ,\sim ,1)$$ is a semi-Nelson algebra, 0 =∼ 1 and c satisfies that c =∼ c. The morphisms of $$\mathsf{SN_{c}}$$ are the algebra homomorphisms. 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 SH and $$\mathsf{SN_{c}}$$. More precisely, it arises the natural question if it is possible to prove that there exists an equivalence between SH and $$\mathsf{SN_{c}}$$, making the following diagram commute: where $$i_{1}$$ is the inclusion functor from HA to SH and $$i_{2}$$ is the inclusion functor from $$\mathsf{NA_{\mathrm{c}}}$$ to $$\mathsf{SN_{c}}$$. One might think, since that there exists an equivalence between Heyting algebras and centered Nelson algebras, that the equivalence between semi-Heyting algebras and centered semi-Nelson algebras is a routine exercise to prove. However, the natural construction contains some subtleties and is not entirely straightforward because we need some additional results given in sections 2 and 3 in order to show that the condition (CK) is satisfied in every centered semi-Nelson algebra. We give the following table with the categories we consider in this paper. Category  Objects  BDL  Bounded distributive lattices  $$\mathsf{KA_{\mathrm{c}}}$$  Centered Kleene algebras  HA  Heyting algebras  NA  Nelson algebras  $$\mathsf{NA_{\mathrm{c}}}$$  Centered Nelson algebras  $$\mathsf{NL_{\mathrm{c}}}$$  Centered Nelson lattices  SH  Semi-Heyting algebras  PSN  Pre-semi-Nelson algebras  SN  Semi-Nelson algebras  $$\mathsf{SN_{c}}$$  Centered semi-Nelson algebras  KSH  Centered Kleene algebras endowed with a particular binary operation  Category  Objects  BDL  Bounded distributive lattices  $$\mathsf{KA_{\mathrm{c}}}$$  Centered Kleene algebras  HA  Heyting algebras  NA  Nelson algebras  $$\mathsf{NA_{\mathrm{c}}}$$  Centered Nelson algebras  $$\mathsf{NL_{\mathrm{c}}}$$  Centered Nelson lattices  SH  Semi-Heyting algebras  PSN  Pre-semi-Nelson algebras  SN  Semi-Nelson algebras  $$\mathsf{SN_{c}}$$  Centered semi-Nelson algebras  KSH  Centered Kleene algebras endowed with a particular binary operation  Category  Objects  BDL  Bounded distributive lattices  $$\mathsf{KA_{\mathrm{c}}}$$  Centered Kleene algebras  HA  Heyting algebras  NA  Nelson algebras  $$\mathsf{NA_{\mathrm{c}}}$$  Centered Nelson algebras  $$\mathsf{NL_{\mathrm{c}}}$$  Centered Nelson lattices  SH  Semi-Heyting algebras  PSN  Pre-semi-Nelson algebras  SN  Semi-Nelson algebras  $$\mathsf{SN_{c}}$$  Centered semi-Nelson algebras  KSH  Centered Kleene algebras endowed with a particular binary operation  Category  Objects  BDL  Bounded distributive lattices  $$\mathsf{KA_{\mathrm{c}}}$$  Centered Kleene algebras  HA  Heyting algebras  NA  Nelson algebras  $$\mathsf{NA_{\mathrm{c}}}$$  Centered Nelson algebras  $$\mathsf{NL_{\mathrm{c}}}$$  Centered Nelson lattices  SH  Semi-Heyting algebras  PSN  Pre-semi-Nelson algebras  SN  Semi-Nelson algebras  $$\mathsf{SN_{c}}$$  Centered semi-Nelson algebras  KSH  Centered Kleene algebras endowed with a particular binary operation  The categories $$\mathsf{NL_{\mathrm{c}}}$$ and KSH will be defined in Section 4. The results studied in the present paper are motivated by ideas coming from different varieties of algebras, as Heyting algebras and Nelson algebras, and by the equivalence between the category of Heyting algebras and the category of centered Nelson algebras (see Theorem 1.2). Our main goal is to extend the above mentioned equivalence by considering the category of semi-Heyting algebras and the category of centered semi-Nelson algebras. We think that the properties studied here can be of interest for future work about the understanding of the categories of semi-Heyting algebras and centered semi-Nelson algebras, respectively. The paper is organized as follows. In Section 2 we study some properties concerning centered semi-Nelson algebras. In Section 3 we generalize the equivalence between the categories HA and $$\mathsf{NA_{\mathrm{c}}}$$ (see Theorem 1.2) in the framework of SH and $$\mathsf{SN_{c}}$$. Finally, in Section 4 we study the relationship between $$\mathsf{SN_{c}}$$ and the category KSH (see Definition 4.4) introduced in [16], which is also equivalent to the category SH. 2 Basic results Let T ∈ PSN. We denote by ≤ to the partial order associated to the underlying lattice of T. In this section we give some basic properties about (centered) semi-Nelson algebras we shall use in the next section in order to show that there exists an equivalence between SH and $$\mathsf{SN_{c}}$$. Since the results given here are very technical, we recommend to the reader don’t read the proofs of them in a first lecture of the present paper. Lemma 2.1 [12] Let T ∈ PSN and x, y, z ∈ T. Then (a) $$1 \to _{N} x = x$$, (b) $$x \to _{N} x = 1$$, (c) if x ≤ y then $$x \to _{N} y = 1$$, (d) x ≤ y if and only if $$x \to _{N} y = 1$$ and $$\sim y \to _{N} \sim x = 1$$, (e) If $$x \to _{N} y = y \to _{N} z = 1$$ then $$x \to _{N} z = 1$$, (f) $$(x \wedge y) \to _{N} y = 1$$, (g) if $$x\to _{N} y=1$$ and $$y\to _{N} z=1$$ then $$x\to _{N} z=1$$. Lemma 2.2 [13] Let T ∈ SN and x, y, z ∈ T. Then (a) $$(x \wedge \sim x) \to _{N} y = 1$$, (b) $$(x \to _{N} y) \to _{N} ((x \to _{N} z) \to _{N} (x \to _{N} (y \wedge z))) = 1$$, (c) $$(x \to _{N} z) \to _{N} ((y \to _{N} z) \to _{N} ((x \vee y) \to _{N} z)) = 1$$, (d) $$x \to _{N} y = x \to _{N} (x\land y)$$. We will use the previous lemmas in order to show the following result. Lemma 2.3 Let T ∈ SN and x, y, z ∈ T. Then (a) If $$x \to _{N} y = 1$$ and $$x \to _{N} z = 1$$ then $$x \to _{N} (y \wedge z) = 1$$; (b) If $$x \to _{N} z = 1$$ and $$y \to _{N} z = 1$$ then $$(x \vee y) \to _{N} z = 1$$; (c) If $$x \to _{N} y = 1$$ then $$(x \vee z) \to _{N} (y \vee z) = 1$$. Proof. (a) By (b) of Lemma 2.2 we have that   $$ (x \to_{N} y) \to_{N} ((x \to_{N} z) \to_{N} (x \to_{N} (y \wedge z))) = 1. $$Hence, by (a) of Lemma 2.1 we obtain that $$x \to _{N} (y \wedge z) = 1$$. (b) It follows from (c) of Lemma 2.2 that   $$ (x \to_{N} z) \to_{N} ((y \to_{N} z) \to_{N} ((x \vee y) \to_{N} z)) = 1. $$Thus, by (a) of Lemma 2.1 we deduce the equality $$(x \vee y) \to _{N} z = 1$$. (c) By hypothesis we have that $$x \to _{N} y = 1$$. It follows from (c) of Lemma 2.1 the equality $$y \to _{N} (y \vee z) = 1$$. Then, using (g) of Lemma 2.1 we obtain that $$x \to _{N} (y \vee z) = 1$$. Observe that $$z \to _{N} (y \vee z) = 1$$, which follows again from (c) of Lemma 2.1. In view of (c) of Lemma 2.2 it holds that   $$ (x \to_{N} (y \vee z)) \to_{N} ((z \to_{N} (y \vee z)) \to_{N} ((x \vee z) \to_{N} (y \vee z))) = 1. $$Therefore, by (a) of Lemma 2.1 we conclude that $$(x \vee z) \to _{N} (y \vee z) = 1$$. The following lemma presents some useful properties of centered semi-Nelson algebras. Lemma 2.4 Let $$T \in \mathsf{SN_{c}}$$ and x, y, z ∈ T. Then (a) $$\mathrm{c} \to _{N} x = 1$$, (b) $$(x \wedge \mathrm{c}) \to _{N} y = 1$$, (c) $$(\sim (x \to y) \vee \mathrm{c}) \to _{N} ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c})) = 1$$, (d) $$((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c})) \to _{N} (\sim (x \to y) \vee \mathrm{c}) = 1$$, (e) $$[\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \to _{N} [\sim (\sim (x \to y) \vee \mathrm{c})] = 1$$, (f) $$[\sim (\sim (x \to y) \vee \mathrm{c})] \to _{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] = 1$$, (g) ∼ (x → y) ∨c = (x ∨c) ∧ (∼ y ∨c), (h) $$[\sim ((x \to y) \vee \mathrm{c})] \to _{N} [\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] = 1$$, (i) $$[\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \to _{N} [\sim ((x \to y) \vee \mathrm{c})] = 1$$, (j) $$(x \to y) \to _{N} ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) = 1$$, (k) $$((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) \to _{N} (x \to y) = 1$$, (l) (x ∨c) → (y ∨c) = (x → y) ∨c. Proof. Let x, y ∈ T. (a) Note that $$\mathrm{c} \to _{N} x = (\mathrm{c} \wedge \mathrm{c}) \to _{N} x = (\mathrm{c} \wedge \sim \mathrm{c}) \to _{N} x = 1$$, which follows from (a) of Lemma 2.2. (b) By item (a) we have that $$\mathrm{c} \to _ N y =1$$ and it follows from (f) of Lemma 2.1 that $$(x \wedge \mathrm{c}) \to _{N} \mathrm{c} = 1 $$. Thus, it follows from (g) of Lemma 2.1 that $$(x \wedge \mathrm{c}) \to _{N} y = 1$$. (c) The fact that $$\sim (x \to y) \to _{N} (x \wedge \sim y) = 1$$ follows from (SN10). By (f) of Lemma 2.1 we obtain that $$(x \wedge \sim y) \to _{N} x = 1$$. Then, applying (e) of Lemma 2.1 we obtain that $$\sim (x \to y) \to _{N} x = 1$$. Thus, taking into account (c) of Lemma 2.3 we have that   \begin{align} (\sim (x \to y) \vee \mathrm{c}) \to_{N} (x \vee \mathrm{c}) = 1. \end{align} (2)Similarly we can show that   \begin{align} (\sim (x \to y) \vee \mathrm{c}) \to_{N} (\sim y \vee \mathrm{c}) = 1. \end{align} (3)By (2), (3) and (a) of Lemma 2.3 it is possible to verify that   $$ (\sim (x \to y) \vee \mathrm{c}) \to_{N} ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c})). $$ (d) By (SN11) we have that $$(x \wedge \sim y) \to _{N} (\sim (x \to y)) = 1$$. Thus, by (c) of Lemma 2.3 we deduce the equality   $$ ((x \wedge \sim y) \vee \mathrm{c}) \to_{N} ((\sim (x \to y)) \vee \mathrm{c}) = 1. $$Consequently,   $$ ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c})) \to_{N} (\sim(x \to y) \vee \mathrm{c}) = ((x \wedge \sim y) \vee \mathrm{c}) \to_{N} (\sim(x \to y) \vee \mathrm{c}) = 1. $$ (e) Let us notice that (f) of Lemma 2.1 implies the equality   \begin{align} (\sim x \wedge \sim \mathrm{c}) \to_{N} (\sim \mathrm{c}) = 1. \end{align} (4)Hence, using (a) we have that   \begin{align} \mathrm{c} \to_{N} (\sim \mathrm{c}) = 1 \end{align} (5)and   \begin{align} \mathrm{c} \to_{N} (x \to y) = 1. \end{align} (6)Hence, (5), (6) and (a) of Lemma 2.3 can be used to verify that   $$ \mathrm{c} \to_{N} ((\sim \mathrm{c}) \wedge (x \to y)) = 1. $$Since c =∼c then   \begin{align} (\sim \mathrm{c}) \to_{N} ((\sim \mathrm{c}) \wedge (x \to y)) = 1. \end{align} (7)It follows from (4), (7) and (g) of Lemma 2.1 that   \begin{align} (\sim x \wedge \sim \mathrm{c}) \to_{N} ((\sim \mathrm{c}) \wedge (x \to y)) = 1. \end{align} (8)Similarly we can show the equality   \begin{align} (y \wedge \sim \mathrm{c}) \to_{N} ((\sim \mathrm{c}) \wedge (x \to y)) = 1. \end{align} (9)By (b) of Lemma 2.3 combined with (8) and (9) we deduce that   \begin{align} ((\sim x \wedge \sim \mathrm{c}) \vee (y \wedge \sim \mathrm{c})) \to_{N} ((\sim \mathrm{c}) \wedge (x \to y)) = 1. \end{align} (10)In consequence, it follows from (10) that   \begin{align*} & [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \to_{N} [\sim(\sim(x \to y) \vee \mathrm{c})]\\ &\quad= [[\sim ((x \vee \mathrm{c})] \vee [\sim (\sim y \vee \mathrm{c})]] \to_{N} [\sim(\sim(x \to y) \vee \mathrm{c})] \\ &\quad= ((\sim x \wedge \sim \mathrm{c}) \vee (\sim \sim y \wedge \sim \mathrm{c})) \to_{N} [\sim(\sim(x \to y) \vee \mathrm{c})] \\ &\quad= ((\sim x \wedge \sim \mathrm{c}) \vee (y \wedge \sim \mathrm{c})) \to_{N} [\sim(\sim(x \to y) \vee \mathrm{c})] \\ &\quad= ((\sim x \wedge \sim \mathrm{c}) \vee (y \wedge \sim \mathrm{c})) \to_{N} [(\sim \sim(x \to y) \wedge (\sim \mathrm{c}))] \\ &\quad= ((\sim x \wedge \sim \mathrm{c}) \vee (y \wedge \sim \mathrm{c})) \to_{N} [((x \to y) \wedge (\sim \mathrm{c}))] \\ &\quad= 1. \end{align*} (f) Item (b) implies the equality   \begin{align} ((x \to y) \wedge \mathrm{c}) \to_{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] = 1. \end{align} (11)Hence, it follows from (11) that   \begin{align*} & [\sim(\sim(x \to y) \vee \mathrm{c})] \to_{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \\ &\quad= [(\sim\sim(x \to y)) \wedge (\sim \mathrm{c})] \to_{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \\ &\quad= [((x \to y)) \wedge (\sim \mathrm{c})] \to_{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \\ &\quad= [((x \to y)) \wedge \mathrm{c}] \to_{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \\ &\quad= 1. \end{align*} (g) Using items (c), (d), (e), (f) and (d) of Lemma 2.1, the desired identity holds. (h) It is consequence from (b) as follows:   \begin{align*} & [\sim ((x \to y) \vee \mathrm{c})] \to_{N} [\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \\ &\quad= [(\sim (x \to y)) \wedge (\sim \mathrm{c})] \to_{N} [\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \\ &\quad= [(\sim (x \to y)) \wedge \mathrm{c}] \to_{N} [\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \\ &\quad= 1. \end{align*} (i) By (SN10) we obtain that   \begin{align} [\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \to_{N} [(x \vee \mathrm{c}) \wedge (\sim (y \vee \mathrm{c}))] = 1. \end{align} (12)Taking into account (b) we have that   \begin{align*} & [(x \vee \mathrm{c}) \wedge (\sim (y \vee \mathrm{c}))] \to_{N} [\sim ((x \to y) \vee \mathrm{c})] \\ &\quad= [(x \vee \mathrm{c}) \wedge (\sim y) \wedge (\sim \mathrm{c})] \to_{N} [\sim ((x \to y) \vee \mathrm{c})] \\ &\quad= [(x \vee \mathrm{c}) \wedge (\sim y) \wedge \mathrm{c}] \to_{N} [\sim ((x \to y) \vee \mathrm{c})] \\ &\quad= 1. \end{align*}Then,   \begin{align} [(x \vee \mathrm{c}) \wedge (\sim (y \vee \mathrm{c}))] \to_{N} [\sim ((x \to y) \vee \mathrm{c})] = 1. \end{align} (13)To finish off the proof of this item we can apply (g) of Lemma 2.1 to equations (12) and (13). (j) Using item (a) and (b) of Lemma 2.1 we can verify that $$\mathrm{c} \to _{N} x = 1$$ and $$x \to _{N} x = 1,$$ respectively. By (b) of Lemma 2.3 we obtain that   \begin{align} (x \vee \mathrm{c}) \to_{N} x = 1. \end{align} (14)Also, it follows from (c) of Lemma 2.1 that   \begin{align} x \to_{N} (x \vee \mathrm{c}) = 1. \end{align} (15)Similarly we can show that   \begin{align} (y \vee \mathrm{c}) \to_{N} y = 1 \ \ \ \mbox{ and } \ \ \ y \to_{N} (y \vee \mathrm{c}) = 1. \end{align} (16)Then,   \begin{align*} & (x \to y) \to ((x \vee \mathrm{c}) \to y) \\ &\quad= 1 \to_{N} [(x \to y) \to ((x \vee \mathrm{c}) \to y)] \\ &\qquad\ \, \mbox{by (a) of}\ \textrm{Lemma}\ {2.1} \\ &\quad= ((x \vee \mathrm{c}) \to_{N} x) \to_{N} [(x \to y) \to ((x \vee \mathrm{c}) \to y)]&\! \mbox{by} \ {(14)}\quad \\ &\quad= 1 \to_{N} [((x \vee \mathrm{c}) \to_{N} x) \to_{N} [(x \to y) \to ((x \vee \mathrm{c}) \to y)]] \\ &\qquad\ \, \mbox{by (a) of}\ \textrm{Lemma}\ {2.1} \\ &\quad= (x \to_{N} (x \vee \mathrm{c})) \to_{N} [((x \vee \mathrm{c}) \to_{N} x) \to_{N} [(x \to y) \to ((x \vee \mathrm{c}) \to y)]]& \mbox{by} \ {(15)}\quad \\ &\quad = 1 &\mbox{by (SN8)}. \end{align*}Then,   \begin{align} (x \to y) \to ((x \vee \mathrm{c}) \to y) = 1. \end{align} (17)On the other hand,   \begin{align*} & ((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) \\ &\quad= 1 \to_{N} [((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \\ &\qquad\ \, \mbox{by (a) of}\ \textrm{Lemma}\ {2.1} \\ &\quad= ((y \vee \mathrm{c}) \to_{N} y) \to_{N} [((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] & \mbox{by}\ {(16)}\quad \\ &\quad= 1 \to_{N} [((y \vee \mathrm{c}) \to_{N} y) \to_{N} [((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))]] \\ &\qquad\ \, \mbox{by (a) of}\ \textrm{Lemma}\ {2.1} \\ &\quad= (y \to_{N} (y \vee \mathrm{c}))\\ &\qquad\to_{N} [((y \vee \mathrm{c}) \to_{N} y) \to_{N} [((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))]] & \mbox{by}\ {(16)}\quad \\ &\quad= 1 & \mbox{by (SN9)}. \end{align*}Thus, we have the equality   \begin{align} ((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) = 1. \end{align} (18)Finally, taking into account (g) of Lemma 2.1 in conditions (17) and (18) we conclude that (x → y) → ((x ∨c) → (y ∨c)) = 1. (k) The proof of this item is similar to the one used in (j). (l) It follows from item (j) that $$(x \to y) \to _{N} ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) = 1$$. Also, $$\mathrm{c} \to _{N} ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) = 1$$ follows from (a). Hence, by (b) of Lemma 2.3 we deduce that   \begin{align} ((x \to y) \vee \mathrm{c}) \to_{N} ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) = 1. \end{align} (19)Besides, from item (k) we know that $$((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) \to _{N} (x \to y) = 1$$. Also, by (c) of Lemma 2.1 we have that $$(x \to y) \to _{N} ((x \to y) \vee \mathrm{c})$$. Thus, by (e) of Lemma 2.1 the equality   \begin{align} ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) \to_{N} ((x \to y) \vee \mathrm{c}) = 1 \end{align} (20)is satisfied. Therefore, applying (d) of Lemma 2.1 in (19), (20), (h) and (i) we can verify that (x ∨c) → (y ∨c) = (x → y) ∨c. In what follows we will use previous results in order to prove the following lemma. Lemma 2.5 Let $$T\in \mathsf{SN_{c}}$$ and x, y ∈ T. Then $$\mathrm{c} \leq x \rightarrow (y\vee \mathrm{c})$$. Proof. By (b) of Lemma 2.4 we have that   \begin{align*} (x \wedge \sim (y \vee \mathrm{c})) \to_{N} (\sim \mathrm{c}) &= (x \wedge (\sim y) \wedge (\sim \mathrm{c})) \to_{N} (\sim \mathrm{c}) & & \\ & = (x \wedge (\sim y) \wedge \mathrm{c}) \to_{N} (\sim \mathrm{c}) \\ & = 1. \end{align*} Hence, $$(x \wedge \sim (y \vee \mathrm{c})) \to _{N} (\sim \mathrm{c}) = 1$$. Besides, it follows from (SN10) that $$[\sim (x \to (y\vee \mathrm{c}))] \to _{N} (x \wedge \sim (y \vee \mathrm{c})) = 1$$. Then, using (g) of Lemma 2.1 we obtain that   \begin{align} [\sim (x \to (y\vee \mathrm{c}))] \to_{N} (\sim \mathrm{c}) = 1. \end{align} (21)Notice that   \begin{align} \mathrm{c} \to_{N} (x \to (y\vee \mathrm{c})) = 1 \end{align} (22)in view of (a) of Lemma 2.4. Therefore, taking into account (21), (22) and using (d) of Lemma 2.1 we conclude that c ≤ x → (y ∨c). 3 Categorical equivalence between SH and SNc In this section we will prove that there is a categorical equivalence between SH and $$\mathsf{SN_{c}}$$. We will use the same notations and constructions given in Section 1 about the functors $$\mathrm{K}:\mathsf{HA} \rightarrow \mathsf{NA_{\mathrm{c}}}$$, $$\mathrm{C}:\mathsf{NA_{\mathrm{c}}}\rightarrow \mathsf{HA}$$ and the isomorphisms $$\alpha _{H}$$ (for H ∈ HA) and $$\beta _{T}$$ (for $$T\in \mathsf{NA_{\mathrm{c}}}$$). Let H ∈ SH. In [12, Theorem 4.1] it was proved that (K(H), ∧, ∨, ⇒, ∼, 1) ∈ SN. Thus, if H ∈ SH then $$\mathrm{K}(H) \in \mathsf{SN_{c}}$$. It is immediate that if f is a morphism in SH, then K(f) is a morphism in $$\mathsf{SN_{c}}$$. Hence, we obtain the following result. Proposition 3.1 There exists a functor K from SH to $$\mathsf{SN_{c}}$$. Let T ∈ SN (or $$T\in \mathsf{SN_{c}}$$). The binary relation ≡ on T given by   $$ x \equiv y\;\textrm{if and only if}\;x\rightarrow y = 1\;\textrm{and}\; y\rightarrow x = 1 $$is an equivalence relation on T compatible with the operations ∧, ∨ and $$\rightarrow $$, as it is shown in [12, Lemma 3.1]. If x ∈ T we write $$[\![x]\!] $$ for the equivalence class associated to ≡. As usual, we write $$T/_{\equiv }$$ for the set of equivalence classes. We denote by sH(T) the algebra $$(T/_{\equiv },\cap ,\cup ,\rightsquigarrow ,[\![0]\!],[\![1]\!])$$, where the operations are defined by: • $$[\![x]\!]\ \cap\ [\![y]\!] = [\![x \wedge y]\!]$$, • $$[\![x]\!]\ \cup\ [\![y]\!] = [\![x \vee y]\!]$$, • $$[\![x]\!]\rightsquigarrow [\![x]\!] = [\![x \rightarrow y]\!]$$. It follows from [12, Theorem 3.4] that sH(T) ∈ SH. The following lemma will be used in the next proposition. Lemma 3.2 a) Let $$T\in \mathsf{SN_{c}}$$. Then C(T) is closed by the operations $$\wedge , \vee , \rightarrow $$ and the constants c and 1. Hence, C(T) can be seen as an algebra of type (2, 2, 2, 0, 0). b) Let $$T, U\in \mathsf{SN_{c}}$$ such that T ≅ U, i.e. T and U are isomorphic algebras. Then C(T) ≅ C(U). In particular, C(T) ∈ SH if and only if C(U) ∈ SH. Proof. It is immediate that c, 1 ∈ C(T). It is also immediate that C(T) is closed by ∧ and ∨. The fact that C(T) is closed by $$\rightarrow $$ follows from Lemma 2.5. Then we have proved a). In order to prove b), let $$T,U\in \mathsf{SN_{c}}$$ and $$f:T\rightarrow U$$ an isomorphism. Straightforward computations show that the map $$\hat{f}: \mathrm{C}(T) \rightarrow \mathrm{C}(U)$$ given by $$\hat{f}(x) = f(x)$$ is an isomorphism. Thus, C(T) ≅ C(U). The fact that C(T) ∈ SH if and only if C(U) ∈ SH follows from that SH is a variety. Proposition 3.3 Let $$T\in \mathsf{SN_{c}}$$. Then C(T) ∈ SH. Proof. Consider the map $$h: T \rightarrow \mathrm{K}(\mathbf{sH}(T))$$ given by $$h(x) = ([\![x]\!], [\![\sim x]\!])$$. We will write $$\rightarrow $$ for the implication of T and ⇒ for the implication of K(sH(T)). The function h is an injective morphism in SN, see [12, Corollary 5.2]. Straightforward computations show that h preserves the bottom and the center, so T ≅ h(T) in $$\mathsf{SN_{c}}$$. Since $$\mathsf{SN_{c}}$$ is a variety, h(T) is a subalgebra of K(sH(T)) and $$\mathrm{K}(\mathbf{sH}(T)) \in \mathsf{SN_{c}}$$ then $$h(T) \in \mathsf{SN_{c}}$$. Hence, by Lemma 3.2 we have that C(T) ∈ SH if and only if C(h(T)) ∈ SH. By this reason, in what follows we will prove that C(h(T)) ∈ SH. Our aim is to show that C(h(T)) ∈ SH. In order to prove it we will use the fact that sH(T) ∈ SH. Let x, y, z ∈ T such that h(x), h(y), h(z) ∈ C(h(T)). Then $$[\![c]\!]\le [\![x]\!], [\![c]\!]\le[\![y]\!], [\![c]\!]\le[\![z]\!], [\![\sim x]\!]\le[\![c]\!], [\![\sim y]\!]\le[\![c]\!]$$ and $$[\![\sim z]\!]\le[\![c]\!]$$. Taking into account that $$[\![\sim y]\!]\le[\![c]\!]\le[\![x]\!]$$ and $$[\![x \wedge (x\rightarrow y)]\!] = [\![x \wedge y]\!]$$, the condition (SH2) in sH(T), we obtain that   \begin{align*} h(x) \cap (h(x) \Rightarrow h(y)) & = ([\![x]\!],[\![\sim x]\!]) \cap (([\![x]\!],[\![\sim x]\!]) \Rightarrow ([\![y]\!], [\![\sim y]\!])) \\ & = ([\![x]\!], [\![\sim x]\!]) \cap ([\![x\rightarrow y]\!], [\![x]\!] \cap [\![\sim y]\!]) \\ & = ([\![x]\!], [\![\sim x]\!]) \cap ([\![x\rightarrow y]\!], [\![\sim y]\!]) \\ & = ([\![x \wedge (x\rightarrow y)]\!], [\![\sim x \vee \sim y]\!])\\ & = ([\![x \wedge y]\!], [\![\sim x \vee \sim y]\!])\\ & = ([\![x]\!] \cap [\![y]\!], [\![\sim x]\!] \cup [\![\sim y]\!])\\ & = h(x) \cap h(y). \end{align*} Then C(h(T)) satisfies (SH2). In order to prove (SH3), first note that since $$[\![\sim z]\!]\le[\![c]\!]\le[\![y]\!]$$ then   \begin{align*} h(x) \cap (h(y) \Rightarrow h(z)) & = ([\![x]\!],[\![\sim x]\!]) \cap (([\![y]\!],[\![\sim y]\!]) \Rightarrow ([\![z]\!], [\![\sim z]\!])) \\ & = ([\![x]\!], [\![\sim x]\!]) \cap ([\![y\rightarrow z]\!], [\![y]\!] \cap [\![\sim z]\!])\\ & = ([\![x]\!], [\![\sim x]\!]) \cap ([\![y\rightarrow z]\!], [\![\sim z]\!]) \\ & = ([\![x\wedge (y\rightarrow z)]\!], [\![\sim x \vee \sim z]\!]). \end{align*} Then   \begin{align} h(x) \cap (h(y) \Rightarrow h(z)) = ([\![x\wedge (y\rightarrow z)]\!], [\![\sim x \vee \sim z]\!]). \end{align} (23) Besides, straightforward computations shows that   \begin{align} h(x) \cap ((h(x)\cap h(y)) \Rightarrow (h(x) \cap h(z)) = ([\![x\wedge((x\wedge y) \rightarrow (x\wedge z))]\!], [\![\sim x \vee \sim z]\!]). \end{align} (24) Using (SH3) in sH(T) we have that $$[\![x\wedge (y\rightarrow z)]\!] = [\![x\wedge ((x\wedge y) \rightarrow (x\wedge z))]\!]$$. Thus, it follows from (23) and (24) that   $$ h(x) \cap (h(y) \Rightarrow h(z)) = h(x) \cap ((h(x)\cap h(y)) \Rightarrow (h(x) \cap h(z)), $$which is the condition (SH3) in C(h(T)). Finally,   \begin{align*} h(x) \Rightarrow h(x) & = ([\![x\rightarrow x]\!],[\![x\wedge \sim x]\!]) \\ & = ([\![1]\!], [\![0]\!]), \end{align*}i.e. the condition (SH4) is also satisfied in C(h(T)). Therefore, C(T) ∈ SH. It is immediate that if f is a morphism in SH, then C(f) is a morphism in $$\mathsf{SN_{c}}$$. Therefore, we conclude the following result. Proposition 3.4 There exists a functor C from $$\mathsf{SN_{c}}$$ to SH. Lemma 3.5 Let $$T \in \mathsf{SN_{c}}$$ and x, y ∈ T. Then $$x\vee \mathrm{c} \leq (y\vee c) \rightarrow ((x\vee c) \wedge (y\vee c))$$. Proof. Let $$T \in \mathsf{SN_{c}}$$ and x, y ∈ T. In particular, we have that x ∨c, y ∨c ∈ C(T). Besides, by Proposition 3.3 we have that C(T) ∈ SH. Hence, it follows from (SH3) and (SH4) that $$x\vee \mathrm{c} \leq (y\vee c) \rightarrow ((x\vee c) \wedge (y\vee c))$$. Remark 3.6 Let $$T \in \mathsf{SN_{c}}$$ and x, y ∈ T. Throughout the rest of this section we will use the equalities $$\sim (x\rightarrow y)\vee \mathrm{c} = (x\vee \mathrm{c})\wedge (\sim y \vee \mathrm{c})$$ and $$(x\rightarrow y) \vee \mathrm{c} = (x\vee \mathrm{c}) \rightarrow (y\vee \mathrm{c})$$, which appears in items (g) and (l) of Lemma 2.4, respectively. The following lemma will be used later. Lemma 3.7 Let $$T\in \mathsf{SN_{c}}$$. Then T satisfies (CK). Proof. In this proof we will use Lemma 3.5 and Remark 3.6. Let x, y ∈ T such that x ≥ c, y ≥ c and x ∧ y = c. Let $$z = (y\rightarrow \sim y) \wedge x$$. In particular, ∼ x ≤ c and ∼ y ≤ c. We will prove that x = z ∨ c and y =∼ z ∨ c. The equality z ∨ c = x can be proved as follows:   \begin{align*} z\vee \mathrm{c} & = ((y \rightarrow \sim y)\wedge x) \vee (\mathrm{c} \wedge x)\\ & = x \wedge ((y\rightarrow \sim y) \vee \mathrm{c})\\ & = x \wedge ((y\vee \mathrm{c})\rightarrow (\sim y \vee \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{align*} Finally we have that   \begin{align*} z\wedge \mathrm{c} & = ((y \rightarrow \sim y)\wedge \mathrm{c}) \wedge x \\ & = (\sim (\sim (y\rightarrow \sim y) \vee \mathrm{c})) \wedge x \\ & = (\sim (y\vee \mathrm{c})) \wedge x \\ & =\, \sim y \wedge \mathrm{c} \wedge x \\ & =\, \sim y \wedge \mathrm{c} \\ & =\, \sim y, \end{align*}so ∼ z ∨ c = y. Let $$T\in \mathsf{SN_{c}}$$. We will see that $$\beta _{T}$$ is an isomorphism in $$\mathsf{SN_{c}}$$. Proposition 3.8 Let $$T\in \mathsf{SN_{c}}$$. Then $$\beta _{T}$$ is an isomorphism in $$\mathsf{SN_{c}}$$. Proof. We know that $$\beta _{T}$$ is an injective morphism in $$\mathsf{KA_{\mathrm{c}}}$$. The fact that $$\beta _{T}$$ preserves the implication is a direct consequence of Remark 3.6. Thus, $$\beta _{T}$$ is an injective morphism in $$\mathsf{SN_{c}}$$. Finally, since the condition (CK) is equivalent to the surjectivity of $$\beta _{T}$$, then it follows from Lemma 3.7 that $$\beta _{T}$$ is a surjective map. Therefore, $$\beta _{T}$$ is an isomorphism in $$\mathsf{SN_{c}}$$. Taking into account the previous results of this section, the fact that if H ∈ SH then $$\alpha _{H}$$ is an isomorphism in SH, and the categorical equivalence between $$\mathsf{KA_{\mathrm{c}}}$$ and BDL, we obtain the main theorem of this paper. Theorem 3.9 The functors K and C establish a categorical equivalence between SH and $$\mathsf{SN_{c}}$$ with natural isomorphisms $$\alpha $$ and $$\beta $$. Theorem 3.9 allow us to give other presentation of the categorical equivalence between SH and $$\mathsf{SN_{c}}$$, as we show in what follows. Let $$T\in \mathsf{SN_{c}}$$. Then sH(T) ∈ SH. Moreover, if $$g:T\rightarrow U \in \mathsf{SN_{c}}$$ then straightforward computations show that the map $$\mathbf{sH}(g):\mathbf{sH}(T) \rightarrow \mathbf{sH}(U)$$ given by $$\mathbf{sH}(g)([\![x]\!])=[\![g(x)]\!]$$ is a morphism in SH. Hence, we have a functor   $$ \mathbf{sH}: \mathsf{SN_{c}} \rightarrow \mathsf{SH}. $$ Let $$T\in \mathsf{SN_{c}}$$ and x ∈ T. We will see that   \begin{align} [\![x \vee \mathrm{c}]\!] = [\![x]\!]. \end{align} (25)Note that $$[\![x \vee \mathrm{c}]\!] = [\![x]\!]$$ if and only if $$x\rightarrow (x\vee \mathrm{c}) = 1$$ and $$(x\vee \mathrm{c})\rightarrow x = 1$$. It can be proved that these equations hold in K(H) for every H ∈ SH. By Proposition 3.3 we have that C(T) ∈ SH, so the equations hold in K(C(T)). Thus, by Proposition 3.8 the equations hold in T, which was our aim. Now consider the injective morphism $$h:T\rightarrow \mathrm{K}(\mathbf{sH}(T))$$ given in the proof of Proposition 3.3. Lemma 3.10 Let $$T\in \mathsf{SN_{c}}$$. The map h is an isomorphism in $$\mathsf{SN_{c}}$$. Proof. First we will prove that the morphism $$\rho :\mathrm{C}(T) \rightarrow \mathbf{sH}(T)$$ given by $$\rho (x) = [\![x]\!]$$ is an isomorphism in SH. It follows from (25) that $$\rho $$ is a surjective map. In order to prove the injectivity, let x, y ≥ c such that $$[\![x ]\!] = [\![y]\!]$$, i.e. $$x\rightarrow y = 1$$ and $$y\rightarrow x = 1$$. By Proposition 3.3 we have that C(T) ∈ SH, so $$x = x\wedge (x\rightarrow y) = x\wedge y$$ and $$y = y\wedge (y\rightarrow x) = y\wedge x$$. Thus, x ≤ y and y ≤ x, so x = y. Thus, $$\rho $$ is an injective map. Hence, it is immediate that the map $$\mathrm{K}(\rho ): \mathrm{K}(\mathrm{C}(T))\rightarrow \mathrm{K}(\mathbf{sH}(T))$$ is an isomorphism in $$\mathsf{SN_{c}}$$. By Proposition 3.8 we also have that $$\beta _{T}:T\rightarrow \mathrm{K}(\mathrm{C}(T))$$ is an isomorphism in $$\mathsf{SN_{c}}$$. Then the map $$\mathrm{K}(\rho )\circ \beta _{T}: T\rightarrow \mathrm{K}(\mathbf{sH}(T))$$ is an isomorphism in $$\mathsf{SN_{c}}$$. Finally, it follows from (25) that $$h = \mathrm{K}(\rho )\circ \beta _{T}$$. Therefore, h is an isomorphism in $$\mathsf{SN_{c}}$$. Let H ∈ SH. The map $$i:H \rightarrow \mathbf{sH}(\mathrm{K}(H))$$ given by $$i(a) = [\![(a, a\rightarrow 0)]\!]$$ is an isomorphism in SH, as it was proved in [12, Theorem 5.3]. Therefore, straightforward computations show that the functors K and sH establish a categorical equivalence between SH and $$\mathsf{SN_{c}}$$ with natural isomorphisms i and h. 4 Connection with existing literature 4.1 Categorical equivalence between $$\mathsf{SN_{c}}$$ and KSH In [16] it was proved that there exists a categorical equivalence between SH and an algebraic category denoted by KSH (see Definition 4.4). The original motivation to consider this algebraic category comes from a different definition of the binary operation given in (1) of Section 1 on (K(H), ∧, ∨, ∼, c, 0, 1), where H ∈ SH. Combining the categorical equivalence between SH and KSH with Theorem 3.9, we obtain that there exists a categorical equivalence between $$\mathsf{SN_{c}}$$ and KSH. In this section we do a more detailed study about the connection between the categories $$\mathsf{SN_{c}}$$ and KSH. We assume the reader is familiar with commutative residuated lattices [15]. An involutive residuated lattice is a bounded, integral and commutative residuated lattice $$(T,\wedge , \vee , \ast ,\rightarrow , 0, 1)$$ such that for every x ∈ T it holds that ¬¬x = x, where $$\neg x: = x\rightarrow 0$$ and 0 is the first element of T [8]. 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 [8] 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 [22]. Remark 4.1 a) Let (T, ∧, ∨, ⇒, ∼, 0, 1) be a Nelson algebra. We define on T the binary operations * and $$\rightarrow $$ by   \begin{align*} x*y & :={\sim} (x \Rightarrow{\sim} y) \vee{\sim} (y \Rightarrow{\sim} x), \\ x \rightarrow y & := (x \Rightarrow y) \wedge ({\sim} y\Rightarrow{\sim} x). \end{align*}Then, [8, Theorem 3.1] says that $$(T,\wedge , \vee , \rightarrow ,*, 0,1)$$ is a Nelson lattice. Moreover, $${\sim } x = \neg x = x\rightarrow 0$$. b) Let $$(T,\wedge ,\vee ,*,\rightarrow ,0,1)$$ be a Nelson lattice. We define on T a binary operation ⇒ and a unary operation ∼ by   \begin{align*} x \Rightarrow y & := x^{2} \rightarrow y, \\{\sim} x & := \neg x, \end{align*}where $$x^{2}: = x*x$$. In particular, $$x \Rightarrow y= (\sim (x \rightarrow \sim x))\rightarrow y$$. Then, [8, Theorem 3.11] says that the algebra (T, ∧, ∨, ⇒, ∼, 0, 1) is a Nelson algebra. c) Notice that in [8, 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 (see [8]), we obtain 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 in items a) and b). The results from [8] about the connections between Nelson algebras and Nelson lattices mentioned in Remark 4.1 are based on results from Spinks and Veroff [22]. More precisely, 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 $$(T,\vee,\wedge ,*,\rightarrow ,\mathrm{c},0,1)$$, where the reduct $$(T,\vee ,\wedge ,*,\rightarrow,0,1)$$ is a Nelson lattice and c is an element of T such that ¬c = c. It follows from Remark 4.1 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 4.2 Let $$(H,\wedge , \vee ,\rightarrow ,0,1) \in \mathsf{HA}$$. We know that $$(\mathrm{K}(H),\wedge ,\vee ,\Rightarrow ,{\sim },\mathrm{c},0,1) \in \mathsf{NA_{\mathrm{c}}}$$, where ⇒ is the operation given in (1) of Section 1. Hence, it follows from Remark 4.1 that $$(\mathrm{K}(H),\wedge,\vee,*,\rightarrow,\mathrm{c},0,1) \in \mathsf{NL_{\mathrm{c}}}$$, where for (a, b) and (d, e) in K(H) the operations * and $$\rightarrow $$ take the form   \begin{align} (a,b) * (d,e) = (a\wedge d, (a\rightarrow e)\wedge (d\rightarrow b)),\!\!\!\! \end{align} (26)  \begin{align} (a,b) \rightarrow (d,e) = ((a\rightarrow d)\wedge (e\rightarrow b), a \wedge e). \end{align} (27)We write $$\rightarrow $$ both for the implication in H as for the implication in K(H) as Nelson lattice. It was proved in [16] that K defines a functor from HA to $$\mathsf{NL_{\mathrm{c}}}$$, where K is defined using the same construction given in Section 1 but changing the binary operation ⇒ given in (1.1) of Section 1 by the binary operation given in (27) of Remark 4.2. Also it was proved in [16] that C is a functor from KSH to SH, where C is defined as in Section 1. Moreover, we have that the maps $$\alpha _{H}$$ for H ∈ SH and $$\beta _{T}$$ for T ∈ KSH are isomorphisms. The following result is [16, Proposition 7]. Proposition 4.3 The functors K and C establish a categorical equivalence between HA and $$\mathsf{NL_{\mathrm{c}}}$$ with natural isomorphisms $$\alpha $$ and $$\beta $$. In what follows we recall the definition of the category KSH given in [16]. Definition 4.4 We write KSH for the algebraic category whose objects are algebras $$(T,\wedge ,\vee , \rightarrow ,{\sim },\mathrm{c},0,1)$$ of type (2, 2, 2, 1, 0, 0, 0) such that (T, ∧, ∨, ∼, c, 0, 1) is a centered Kleene algebra and $$\rightarrow $$ is a binary operation on T which satisfies the following conditions for every x, y ∈ T: (1) $$\mathrm{c} \leq x \rightarrow (y\vee \mathrm{c})$$, (2) $$x \rightarrow x = 1$$, (3) $$(x\rightarrow y)\wedge \mathrm{c} = ({\sim } x \wedge \mathrm{c}) \vee (y\wedge \mathrm{c})$$, (4) $$(x\rightarrow{\sim } y) \vee \mathrm{c} = ((x\vee \mathrm{c}) \rightarrow ({\sim } y \vee \mathrm{c})) \wedge ((y\vee \mathrm{c}) \rightarrow ({\sim } x \vee \mathrm{c}))$$, (5) $$x\wedge ((x\vee \mathrm{c})\rightarrow (y\vee \mathrm{c})) = x\wedge (y\vee \mathrm{c})$$, (6) $$x \wedge ((y\vee \mathrm{c}) \rightarrow (z\vee \mathrm{c})) = x \wedge (((x\vee \mathrm{c}) \wedge (y\vee \mathrm{c}))\rightarrow ((x\vee \mathrm{c})\wedge (z\vee \mathrm{c})))$$. By considering the objects of KSH as algebras $$(T,\wedge ,\vee , \rightarrow ,*,{\sim },\mathrm{c},0,1)$$, where * is defined as in (26) of Remark 4.2, we have that $$\mathsf{NL_{\mathrm{c}}}$$ is a full subcategory of KSH. Proposition 4.3 can be generalized in order to prove a categorical equivalence between SH and KSH, as it is shown in [16, Theorem 51]. Theorem 4.5 The functors K and C establish a categorical equivalence between SH and KSH with natural isomorphisms $$\alpha $$ and $$\beta $$. The following result follows from Theorems 3.9 and 4.5. Theorem 4.6 There exists a categorical equivalence between $$\mathsf{SN_{c}}$$ and KSH. Notice we are using the same notation K to refer us to a functor from SH to $$\mathsf{SN_{c}}$$ and also for a functor from SH to KSH (similarly with the notation C). We believe it is clear which is the corresponding functor considered in each case. We know that the varieties $$\mathsf{NA_{\mathrm{c}}}$$ and $$\mathsf{NL_{\mathrm{c}}}$$ are term equivalent. We will prove that the varieties $$\mathsf{SN_{c}}$$ and KSH are not term equivalent by using the construction given in Remark 4.1. Consider an algebra $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1)\in \mathsf{SN_{c}}$$ and define a binary operation $$\rightarrow _{\mathsf{KSH}}$$ by   $$ x\rightarrow_{\mathsf{KSH}} y = (x\rightarrow y)\wedge (\sim y \rightarrow \sim x). $$ Remark 4.7 Let H ∈ SH and $$\rightarrow $$ the implication of H. Let a, b, d, e ∈ H such that (a, b), (d, e) ∈ K(H). Then   \begin{align*} (a,b)\rightarrow_{\mathsf{KSH}}(d,e) & = ((a,b)\Rightarrow (d,e))\cap ((e,d)\Rightarrow (b,a))\\ & = (a\rightarrow d,a\wedge e)\cap ((e\rightarrow b),a\wedge e)\\ & = ((a\rightarrow d)\wedge(e\rightarrow b),a\wedge e). \end{align*} Therefore, the implication of the algebra K(H) ∈ KSH (see Remark 4.2) is exactly the binary operation $$\rightarrow _{\mathsf{KSH}}$$ defined in K(H). In the following proposition we will see that if we consider an algebra in $$\mathsf{SN_{c}}$$ then we can define an algebra in KSH. Proposition 4.8 Let $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1) \in \mathsf{SN_{c}}$$. Then, $$(T,\wedge ,\vee ,\rightarrow _{\mathsf{KSH}},\sim ,\mathrm{c},0,1)\in \mathsf{KSH}$$. Proof. We will write $$\mathrm{K}^{\mathsf{SN_{c}}}$$ for the functor K from SH to $$\mathsf{SN_{c}}$$ and $$\mathrm{K}^{\mathsf{KSH}}$$ for the functor K from SH to KSH. In this proof we will use theorems 3.9 and 4.5. Let $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1) \in \mathsf{SN_{c}}$$. Hence, $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1) \cong \mathrm{K}^{\mathsf{SN_{c}}}(\mathrm{C}(T))$$, where the isomorphism from $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1)$$ to $$\mathrm{K}^{\mathsf{SN_{c}}}(\mathrm{C}(T))$$ is given by $$\beta _{T}$$. Since $$\rightarrow _{\mathsf{KSH}}$$ can be written in terms of $$\rightarrow $$, ∧ and ∼, then $$\beta _{T}$$ preserves the operation $$\rightarrow _{\mathsf{KSH}}$$, i.e., for every x, y ∈ T it holds that $$\beta _{T}(x\!\rightarrow _{\mathsf{KSH}}\! y) \!=\! \beta _{T}(x) \!\rightarrow _{\mathsf{KSH}}\! \beta _{T} (y)$$. Thus, it follows from Remark 4.7 that $$(T,\wedge ,\vee,\!\rightarrow _{\mathsf{KSH}},\sim ,\mathrm{c},0,1) \cong \mathrm{K}^{\mathsf{KSH}}(\mathrm{C}(T))$$. However, $$\mathrm{K}^{\mathsf{KSH}}(\mathrm{C}(T)) \in \mathsf{KSH}$$, so $$(T,\wedge ,\vee ,\rightarrow _{\mathsf{KSH}},\sim ,\mathrm{c},0,1) \in \mathsf{KSH}$$ because KSH is a variety. Let $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1)\in \mathsf{KSH}$$. Define a binary operation $$\hat{\rightarrow }$$ by   $$ x \hat{\rightarrow} y: = (\sim (x\rightarrow \sim x))\rightarrow y. $$The definition of $$\hat{\rightarrow }$$ is motivated by item b) of Remark 4.1. Then, it naturally arises the following question: does it hold that $$(T,\wedge ,\hat{\rightarrow },\sim ,\mathrm{c},0,1)\in \mathsf{SN_{c}}$$? The answer is negative, as we show in what follows. Let $$(H,\wedge ,\vee ,\rightarrow ,0,1) \in \mathsf{SH}$$ and also write $$\rightarrow $$ for the implication in K(H) ∈ KSH. Let (a, b), (d, e) ∈ K(H). Then   \begin{align*} (a,b) \hat{\rightarrow} (d,e) &= (\sim((a,b)\rightarrow \sim (a,b)))\rightarrow (d,e) \\ & = (\sim(a\rightarrow b, a))\rightarrow (d,e) \\ & = (a,a\rightarrow b)\rightarrow (d,e) \\ & = ((a\rightarrow d) \wedge (e\rightarrow (a\rightarrow b)),a\wedge e). \end{align*} Note that $$(a,b) \hat{\rightarrow } (d,e) = (a\rightarrow d,a\wedge e)$$ if and only if $$a\rightarrow d \leq e\rightarrow (a\rightarrow b)$$. In Heyting algebras the condition $$a\rightarrow d \leq e\rightarrow (a\rightarrow b)$$ is true whenever a ∧ b = d ∧ e = 0. However, in semi-Heyting algebras this condition is not necessarily true. For instance, consider the semi-Heyting algebra given by with a = d = 0 and b = e = 1. In this case a ∧ b = d ∧ e = 0, $$a\rightarrow d = 1$$ and $$e\rightarrow (a\rightarrow b) = 0$$, so $$a\rightarrow d \nleq e\rightarrow (a\rightarrow b)$$. Hence, we have that there exists $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0, 1) \in \mathsf{KSH}$$ such that the algebra $$(T,\wedge ,\vee , \hat{\rightarrow },\sim ,\mathrm{c},0, 1)\notin \mathsf{SN_{c}}$$. Therefore, the varieties $$\mathsf{SN_{c}}$$ and KSH are not term equivalent by using the construction given in Remark 4.1. 4.2 Congruences Finally we study the connection between the congruences of $$\mathsf{SN_{c}}$$ and KSH. For $$T_{\mathsf{SN_{c}}} = (T, \wedge, \vee,\rightarrow, \sim,\mathrm{c},0,1) \in \mathsf{SN_{c}}$$ consider the algebra $$T_{\mathsf{KSH}} = (T,\wedge ,\vee ,\rightarrow _{\mathsf{KSH}},\sim ,\mathrm{c},0,1) \in \mathsf{KSH}$$. We start with the following preliminary lemma. Lemma 4.9 If $$T_{\mathsf{SN_{c}}} \in \mathsf{SN_{c}}$$ then $$\mathrm{C}(T_{\mathsf{SN_{c}}}) = \mathrm{C}(T_{\mathsf{KSH}})$$. Proof. Let x, y ≥ c. We will prove that $$x\rightarrow y = x \rightarrow _{\mathsf{KSH}} y$$. Since $$x\rightarrow _{\mathsf{KSH}} y = (x\rightarrow y) \wedge (\sim y \rightarrow \sim x)$$, it is enough to prove that $$\sim y \rightarrow \sim x = 1$$. In order to show it we use Remark 3.6 as follows:   \begin{align*} (\sim y \rightarrow \sim x) \vee \mathrm{c} &= (\sim y \vee \mathrm{c})\rightarrow (\sim x \vee \mathrm{c}) \\ & = c\rightarrow c \\ & = 1 \\ & = 1\vee \mathrm{c} \end{align*}and   \begin{align*} (\sim y \rightarrow \sim x) \wedge \mathrm{c} &=\, \sim(\sim (\sim y \rightarrow \sim x) \vee \mathrm{c}) \\ & =\, \sim((\sim y \vee \mathrm{c})\wedge(x\vee \mathrm{c}))\\ & =\, \sim \mathrm{c} \\ & = \mathrm{c} \\ & = 1\wedge \mathrm{c}. \end{align*} Hence, taking into account the distributivity of the underlying lattice of $$T_{\mathsf{SN_{c}}}$$ we deduce the equality $$\sim y \rightarrow \sim x = 1$$, which was our aim. If T is an algebra we write Con(T) for the lattice of congruences of T. Proposition 4.10 Let $$T_{\mathsf{SN_{c}}} \in \mathsf{SN_{c}}$$. Then $$\mathrm{Con}(T_{\mathsf{SN_{c}}}) = \mathrm{Con}(T_{\mathsf{KSH}})$$. Proof. It follows from Proposition 4.8 that $$\mathrm{Con}(T_{\mathsf{SN_{c}}})\subseteq \mathrm{Con}(T_{\mathsf{KSH}})$$. Conversely, let $$\theta \in \mathrm{Con}(T_{\mathsf{KSH}})$$ and x, y, z, w ∈ T such that $$(x,y)\in \theta $$ and $$(z,w)\in \theta $$. First note that it follows from Remark 3.6 that   \begin{align} (x\vee \mathrm{c})\rightarrow (z\vee \mathrm{c}) = (x\rightarrow z) \vee \mathrm{c}, \end{align} (28)  \begin{align} (y\vee \mathrm{c})\rightarrow (w\vee \mathrm{c}) = (y\rightarrow w) \vee \mathrm{c}. \end{align} (29)Besides $$(x\vee \mathrm{c}, y\vee \mathrm{c})\in \theta $$ and $$(z\vee \mathrm{c},w\vee \mathrm{c})\in \theta $$, so   $$ ((x\vee \mathrm{c})\rightarrow_{\mathsf{KSH}}(z\vee \mathrm{c}), (y\vee \mathrm{c})\rightarrow_{\mathsf{KSH}}(w\vee \mathrm{c})) \in \theta. $$ It follows from Lemma 4.9 that $$(x\vee \mathrm{c})\rightarrow _{\mathsf{KSH}}(z\vee \mathrm{c}) = (x\vee \mathrm{c})\rightarrow (z\vee \mathrm{c})$$ and $$(y\vee \mathrm{c})\rightarrow _{\mathsf{KSH}}(w\vee \mathrm{c}) = (y\vee \mathrm{c})\rightarrow (w\vee \mathrm{c})$$. Hence, it follows from (28) and (29) that   \begin{align} ((x\rightarrow z) \vee \mathrm{c},(y\rightarrow w)\vee \mathrm{c})\in \theta. \end{align} (30)On the other hand, it follows from Remark 3.6 that   \begin{align} (x\rightarrow z)\wedge \mathrm{c} = (\sim x\wedge \mathrm{c})\vee (z\wedge \mathrm{c}), \end{align} (31)  \begin{align} (y\rightarrow w)\wedge \mathrm{c} = (\sim y \wedge \mathrm{c})\vee(w\wedge \mathrm{c}). \end{align} (32)Besides, since $$(x,y)\in \theta $$ and $$(z,w)\in \theta $$ then $$(\sim x\wedge \mathrm{c}, \sim y \wedge \mathrm{c}) \in \theta $$ and $$(z\wedge \mathrm{c}, w \wedge \mathrm{c}) \in \theta $$, so $$((\sim x \wedge \mathrm{c})\vee (z\wedge \mathrm{c}), (\sim y \wedge \mathrm{c})\vee (w\wedge \mathrm{c}))\in \theta $$. Thus, taking into account (31) and (32) we conclude that   \begin{align} ((x\rightarrow z)\wedge \mathrm{c}, (y\rightarrow w)\wedge \mathrm{c})\in \theta. \end{align} (33)Hence, it follows from (30), (33) and the distributivity of the underlying lattice of $$T_{\mathsf{SN_{c}}}$$ that $$(x\rightarrow z, y\rightarrow w) \in \theta $$, so $$\theta \in \mathrm{Con}(T_{\mathsf{SN_{c}}})$$. Then, $$\mathrm{Con}(T_{\mathsf{SN_{c}}}) = \mathrm{Con}(T_{\mathsf{KSH}})$$. Let H be a distributive lattice. Recall that a non empty subset F of H is said to be a filter the following two conditions are satisfied for each a, b ∈ H: 1) if a ≤ b and a ∈ F then b ∈ F, 2) a ∧ b ∈ F whenever a, b ∈ F. We write Fil(H) by the set of filters of H. Recall that in [20, Theorem 5.4] it was proved that if H ∈ SH then there is an order isomorphism between Con(H) and Fil(H). Corollary 4.11 Let $$T_{\mathsf{SN_{c}}} \in \mathsf{SN_{c}}$$. Then $$\mathrm{Con}(T_{\mathsf{SN_{c}}}) \cong \mathrm{Fil}(\mathrm{C}(T_{\mathsf{SN_{c}}}))$$. Proof. It follows from Proposition 4.10 that $$\mathrm{Con}(T_{\mathsf{SN_{c}}}) = \mathrm{Con}(T_{\mathsf{KSH}})$$. Besides, it follows from [16, Proposition 61] that $$\mathrm{Con}(T_{\mathsf{KSH}}) \cong \mathrm{Con}(\mathrm{C}(T_{\mathsf{KSH}}))$$. Also note that $$\mathrm{Con}(\mathrm{C}(T_{\mathsf{KSH}})) \cong \mathrm{Fil}(\mathrm{C}(T_{\mathsf{KSH}}))$$ because $$\mathrm{C}(T_{\mathsf{KSH}}) \in \mathsf{SH}$$. Finally, taking into account Lemma 4.9 we have that $$\mathrm{Fil}(\mathrm{C}(T_{\mathsf{KSH}})) = \mathrm{Fil}(\mathrm{C}(T_{\mathsf{SN_{c}}}))$$. Therefore, $$\mathrm{Con}(T_{\mathsf{SN_{c}}}) \cong \mathrm{Fil}(\mathrm{C}(T_{\mathsf{SN_{c}}}))$$. In what follows we will use the notation of the proof of Proposition 4.8. By Proposition 4.10 we have that if H ∈ SH then   $$ \mathrm{Con}(\mathrm{K}^{\mathsf{KSH}}(H)) = \mathrm{Con}(\mathrm{K}^{\mathsf{SN_{c}}}(H)). $$ Proposition 4.12 Let T ∈ KSH. Then $$\mathrm{Con}(T) \cong \mathrm{Con}(\mathrm{K}^{\mathsf{SN_{c}}}(\mathrm{C}(T)))$$. Acknowledgements The authors acknowledge many helpful comments from the anonymous referee, which considerably improved the presentation of this paper. Both authors want to thank the institutional support of Consejo Nacional de Investigaciones Científicas y Técnicas. References [1] M. Abad, J. M. Cornejo and J. P. Díaz Varela. The variety generated by semi-Heyting chains. Soft Computing , 15, 721– 728, 2010. Google Scholar CrossRef Search ADS   [2] M. Abad, J. M. Cornejo and J. P. Díaz Varela. The variety of semi-Heyting algebras satisfying the equation (0→1)*∨ (0→1)**≈ 1. Reports on Mathematical Logic , 46, 75– 90, 2011. [3] M. Abad, J. M. Cornejo and J. P. Díaz Varela. 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Mathematical Logic, Proceedings of the First Brazilian Vonference . A. I. Arruda, N. C. A. Da Costa, R. Chuaqui, eds. Lectures in Pure and Applied Mathematics, vol. 39. New York and Basel: Marcel Dekker, pp. 99– 117, 1978. [15] J. Hart, L. Raftery and C. Tsinakis. The structure of commutative residuated lattices. International Journal of Algebra and Compututation,  12, 509– 524, 2002. Google Scholar CrossRef Search ADS   [16] R. Jansana and H. J. San Martín. On Kalman’s functor for bounded hemi-implicative semilattices and hemi-implicative lattices. Logic Journal of the IGPL,  25, 348– 364, 2017. Google Scholar CrossRef Search ADS   [17] J. A. Kalman. Lattices with involution. Transactions of the American Math Society , 87, 485– 491, 1958. Google Scholar CrossRef Search ADS   [18] H. Rasiowa. An algebraic approach to non-classical logics. In Studies in Logic and the Foundations of Mathematics, 78. Nort-Holland and PNN 1974. [19] H. P. Sankappanavar. 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Magister dissertation in Mathematics, Universidad Nacional del Sur, Bahía Blanca, available at https://sites.google.com/site/viglizzo/viglizzo99nelson. © The Author(s) 2018. Published by Oxford University Press. All rights reserved. For permissions, please e-mail: journals.permission@oup.com. This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/about_us/legal/notices) For permissions, please e-mail: journals. permissions@oup.com http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Logic Journal of the IGPL Oxford University Press

A categorical equivalence between semi-Heyting algebras and centered semi-Nelson algebras

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Abstract

Abstract Motivated by a construction due to R. Cignoli that relates Heyting algebras and centered Nelson algebras, in this paper we prove that there exists an equivalence between the category of semi-Heyting algebras and the category of centered semi-Nelson algebras. 1 Introduction Inspired by results from [17] due to J. Kalman relating to lattices, R. Cignoli proved in [11, Theorem 2.4] that there exists an equivalence between the category of bounded distributive lattices and a particular full subcategory of centered Kleene algebras. Moreover, he also proved that there exists an equivalence between the category of Heyting algebras and the category of centered Nelson algebras [11, Theorem 3.14] (see also [10, 16]). In this paper we extend the previous result in order to show that there is an equivalence between the category of semi-Heyting algebras [20] and the category of semi-Nelson algebras [12] with center. In the process of our research on the topic of the present paper, we have found useful to place our problems in the following general context. We assume that the reader is familiar with bounded distributive lattices and Heyting algebras [5]. A De Morgan algebra is an algebra (H, ∧, ∨, ∼, 0, 1) of type (2, 2, 1, 0, 0) such that (H, ∧, ∨, 0, 1) is a bounded distributive lattice and ∼ fulfills the equations ∼∼x = x and ∼(x ∨ y) = ∼x ∧ ∼y. An operation ∼ which satisfies the previous two equations is called involution. A Kleene algebra is a De Morgan algebra in which the inequality x ∧ ∼x ≤ y ∨ ∼y holds. We say that an algebra (H, ∧, ∨, ∼, c, 0, 1) of type (2, 2, 1, 0, 0, 0) is a centered Kleene algebra if (H, ∧, ∨, ∼, 0, 1) is a Kleene algebra and c is such that c = ∼c (this element is called center). It is immediate to see that c is necessarily unique. We write 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 center, i.e. f(c) = c. For an object H ∈ BDL we define   $$ \mathrm{K}(H): =\{(a,b) \in H\times H: a\wedge b = 0\}. $$This set could be endowed with the operations and the distinguished elements defined by:   \begin{align*} (a,b)\vee (d,e) & := (a\vee d,b\wedge e)\\[-2pt] (a,b)\wedge (d,e)& := (a\wedge d,b\vee e)\\[-2pt] {\sim} (a,b)& := (b,a)\\[-2pt] 0 & := (0,1)\\[-2pt] 1 & := (1,0)\\[-2pt] \mathrm{c} & := (0,0). \end{align*} In particular, $$(\mathrm{K}(H),\wedge ,\vee , \sim ,\mathrm{c}, 0,1)\in \mathsf{KA_{\mathrm{c}}}$$. For a morphism $$f:H \rightarrow G \in \mathsf{BDL}$$, the map $$\mathrm{K}(\,f):\mathrm{K}(H) \rightarrow \mathrm{K}(G)$$, defined by K(f)(a, b) = (f(a), f(b)), is a morphism in $$\mathsf{KA_{\mathrm{c}}}$$. Moreover, K is a functor from 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, ∧, ∨, c, 1) and (C(T), ∧, ∨, c, 1) ∈ BDL. For a morphism $$g:T\rightarrow U \in \mathsf{KA_{\mathrm{c}}}$$, the map $$\mathrm{C}(g): \mathrm{C}(T) \rightarrow \mathrm{C}(U)$$, given by C(g)(x) = g(x), is a morphism in BDL. Moreover, C is a functor from $$\mathsf{KA_{\mathrm{c}}}$$ to BDL. Let H ∈ BDL. The map $$\alpha _{H}: H \rightarrow \mathrm{C}(\mathrm{K}(H))$$ given by $$\alpha _{H}(a) = (a,0)$$ is an isomorphism in BDL. If $$T \in \mathsf{KA_{\mathrm{c}}}$$, 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 it is a morphism in $$\mathsf{KA_{\mathrm{c}}}$$, but it is not necessarily surjective (see [16]). Let $$T\in \mathsf{KA_{\mathrm{c}}}$$. Consider the following algebraic condition:   \begin{align}(\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{align} (CK)This condition characterizes the surjectivity of $$\beta _{T}$$, that is, for every $$T\in \mathsf{KA_{\mathrm{c}}}$$ we have that T satisfies (CK) if and only if $$\beta _{T}$$ is a surjective map. The condition (CK) is not necessarily verified in every centered Kleene algebra, see for instance [10, Figure 1]. We write $$\mathsf{KA_{\mathrm{c}}^{CK}}$$ for the full subcategory of $$\mathsf{KA_{\mathrm{c}}}$$ whose objects satisfy (CK). The functor K can then be seen as a functor from BDL to $$\mathsf{KA_{\mathrm{c}}^{CK}}$$. The following result is [10, Theorem 2.7] (see also [11, Theorem 2.4]). Theorem 1.1 The functors K and C establish a categorical equivalence between BDL and $$\mathsf{KA_{\mathrm{c}}^{CK}}$$ with natural isomorphisms $$\alpha $$ and $$\beta $$. Let H ∈ BDL and a, b ∈ H. If the relative pseudocomplement of a with respect to b exists, then we denote it by $$a \rightarrow _{H} b$$. Recall that a Nelson algebra [11, 24] is a Kleene algebra such that for each pair x, y there exists the binary operation ⇒ given by $$x\Rightarrow y: = x \rightarrow _{H} ({\sim x} \vee y)$$ and for every x, y, z it holds that (x ∧ y) ⇒ z = x ⇒ (y ⇒ z). The binary operation ⇒ so defined is called the weak implication. Nelson algebras can be seen as algebras (H, ∧, ∨, ⇒, ∼, 0, 1) of type (2, 2, 2, 1, 0, 0). The class of Nelson algebras is a variety [6, 7, 18]. We say that an algebra (T, ∧, ∨, ⇒, ∼, c, 0, 1) is a centered Nelson algebra if the reduct (T, ∧, ∨, ⇒, ∼, 0, 1) is a Nelson algebra and c satisfies ∼c = c. It is a known fact that centered Nelson algebras satisfy the condition (CK) (see [10]). M. Fidel [14] and D. Vakarelov [23] proved independently that if $$(H,\wedge ,\vee ,\rightarrow ,0,1)$$ is a Heyting algebra then (K(H), ∧, ∨, ⇒, ∼, c, 0, 1) is a centered Nelson algebra, where ⇒ is defined as follows, for pairs (a, b) and (d, e) in K(H):   \begin{align} (a,b)\Rightarrow(d,e):= (a\rightarrow d, a\wedge e). \end{align} (1) Let HA be the category of Heyting algebras and $$\mathsf{NA_{\mathrm{c}}}$$ the category of centered Nelson algebras. The following result is a reformulation of [11, Theorem 3.14] (see also [9]). Theorem 1.2 The functors K and C establish a categorical equivalence between HA and $$\mathsf{NA_{\mathrm{c}}}$$ with natural isomorphisms $$\alpha $$ and $$\beta $$. In what follows we recall the definition of semi-Heyting algebras, which were introduced by H. P. Sankappanavar in [20] as an abstraction of Heyting algebras. Semi-Heyting algebras share with Heyting algebras the following properties: they are pseudocomplemented, distributive lattices and their congruences are determined by the lattice filters. The relationship between the variety of semi-Heyting algebras and the varieties of Heyting algebras (and its expansions) have been studied lately in [1, 4, 19]. Definition 1.3 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: (SH1) (H, ∧, ∨, 0, 1) is a bounded lattice, (SH2) $$a\wedge (a\rightarrow b) = a \wedge b$$, (SH3) $$a\wedge (b\rightarrow d) = a \wedge [(a\wedge b) \rightarrow (a\wedge d)]$$, (SH4) $$a\rightarrow a = 1$$. We write SH for the category of semi-Heyting algebras. It was proved in [20] that the underlying lattice of a semi-Heyting algebra is necessarily distributive. In what follows we recall the definition of pre-semi-Nelson algebra and semi-Nelson algebra respectively. These definitions were introduced in [12]. Definition 1.4 An algebra (T, ∧, ∨, →, ∼, 1) of type (2, 2, 2, 1, 0) is a pre-semi-Nelson algebra if for every x, y, z ∈ H the following conditions are satisfied: (SN1) x ∧ (x ∨ y) = x, (SN2) x ∧ (y ∨ z) = (z ∧ x) ∨ (y ∧ x), (SN3) ∼∼ x = x, (SN4) ∼ (x ∧ y) =∼ x∨ ∼ y, (SN5) x∧ ∼ x = (x∧ ∼ x) ∧ (y∨ ∼ y), (SN6) $$x \wedge (x \to _{N} y) = x \wedge (\sim x \vee y)$$, (SN7) $$x \to _{N} (y \to _{N} z) = (x \wedge y) \to _{N} z$$, (SN8) $$(x \to _{N} y) \to _{N} [(y \to _{N} x) \to _{N} [(x \to z) \to _{N} (y \to z)]] = 1$$, (SN9) $$(x \to _{N} y) \to _{N} [(y \to _{N} x) \to _{N} [(z \to x) \to _{N} (z \to y)]] = 1$$, where $$x \to _{N} y$$ stands for the term x → (x ∧ y). Definition 1.5 A pre-semi-Nelson algebra (T, ∧, ∨, →, ∼, 1) is a semi-Nelson algebra if it also verifies the following conditions for every x, y ∈ H: (SN10) $$(\sim (x \to y)) \to _{N} (x \wedge \sim y) = 1$$, (SN11) $$(x \wedge \sim y) \to _{N} (\sim (x \to y)) = 1$$. We write PSN for the category of pre-semi-Nelson algebras. Notice that the conditions (SN1) and (SN2) are those given by Sholander in [21] which define distributive lattices. In addition, the conditions (SN3), (SN4) and (SN5) define Kleene algebras. Thus, every pre-semi-Nelson algebra is in particular a Kleene algebra. Semi-Nelson algebras were introduced by J. M. Cornejo and I. Viglizzo in [12] as a generalization of Nelson algebras. Moreover, in [12, Corollary 5.2] it was proved that if T is a semi-Nelson algebra then there exists a particular semi-Heyting algebra H such that T is isomorphic to a subalgebra of the semi-Nelson algebra (K(H), ∧, ∨, ⇒, ∼, 1). We write SN for the category of semi-Nelson algebras and $$\mathsf{SN_{c}}$$ for the category of centered semi-Nelson algebras. More precisely, the objects of $$\mathsf{SN_{c}}$$ are defined as algebras $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1)$$ of type (2, 2, 2, 1, 0, 0, 0), where $$(T,\wedge ,\vee ,\rightarrow ,\sim ,1)$$ is a semi-Nelson algebra, 0 =∼ 1 and c satisfies that c =∼ c. The morphisms of $$\mathsf{SN_{c}}$$ are the algebra homomorphisms. 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 SH and $$\mathsf{SN_{c}}$$. More precisely, it arises the natural question if it is possible to prove that there exists an equivalence between SH and $$\mathsf{SN_{c}}$$, making the following diagram commute: where $$i_{1}$$ is the inclusion functor from HA to SH and $$i_{2}$$ is the inclusion functor from $$\mathsf{NA_{\mathrm{c}}}$$ to $$\mathsf{SN_{c}}$$. One might think, since that there exists an equivalence between Heyting algebras and centered Nelson algebras, that the equivalence between semi-Heyting algebras and centered semi-Nelson algebras is a routine exercise to prove. However, the natural construction contains some subtleties and is not entirely straightforward because we need some additional results given in sections 2 and 3 in order to show that the condition (CK) is satisfied in every centered semi-Nelson algebra. We give the following table with the categories we consider in this paper. Category  Objects  BDL  Bounded distributive lattices  $$\mathsf{KA_{\mathrm{c}}}$$  Centered Kleene algebras  HA  Heyting algebras  NA  Nelson algebras  $$\mathsf{NA_{\mathrm{c}}}$$  Centered Nelson algebras  $$\mathsf{NL_{\mathrm{c}}}$$  Centered Nelson lattices  SH  Semi-Heyting algebras  PSN  Pre-semi-Nelson algebras  SN  Semi-Nelson algebras  $$\mathsf{SN_{c}}$$  Centered semi-Nelson algebras  KSH  Centered Kleene algebras endowed with a particular binary operation  Category  Objects  BDL  Bounded distributive lattices  $$\mathsf{KA_{\mathrm{c}}}$$  Centered Kleene algebras  HA  Heyting algebras  NA  Nelson algebras  $$\mathsf{NA_{\mathrm{c}}}$$  Centered Nelson algebras  $$\mathsf{NL_{\mathrm{c}}}$$  Centered Nelson lattices  SH  Semi-Heyting algebras  PSN  Pre-semi-Nelson algebras  SN  Semi-Nelson algebras  $$\mathsf{SN_{c}}$$  Centered semi-Nelson algebras  KSH  Centered Kleene algebras endowed with a particular binary operation  Category  Objects  BDL  Bounded distributive lattices  $$\mathsf{KA_{\mathrm{c}}}$$  Centered Kleene algebras  HA  Heyting algebras  NA  Nelson algebras  $$\mathsf{NA_{\mathrm{c}}}$$  Centered Nelson algebras  $$\mathsf{NL_{\mathrm{c}}}$$  Centered Nelson lattices  SH  Semi-Heyting algebras  PSN  Pre-semi-Nelson algebras  SN  Semi-Nelson algebras  $$\mathsf{SN_{c}}$$  Centered semi-Nelson algebras  KSH  Centered Kleene algebras endowed with a particular binary operation  Category  Objects  BDL  Bounded distributive lattices  $$\mathsf{KA_{\mathrm{c}}}$$  Centered Kleene algebras  HA  Heyting algebras  NA  Nelson algebras  $$\mathsf{NA_{\mathrm{c}}}$$  Centered Nelson algebras  $$\mathsf{NL_{\mathrm{c}}}$$  Centered Nelson lattices  SH  Semi-Heyting algebras  PSN  Pre-semi-Nelson algebras  SN  Semi-Nelson algebras  $$\mathsf{SN_{c}}$$  Centered semi-Nelson algebras  KSH  Centered Kleene algebras endowed with a particular binary operation  The categories $$\mathsf{NL_{\mathrm{c}}}$$ and KSH will be defined in Section 4. The results studied in the present paper are motivated by ideas coming from different varieties of algebras, as Heyting algebras and Nelson algebras, and by the equivalence between the category of Heyting algebras and the category of centered Nelson algebras (see Theorem 1.2). Our main goal is to extend the above mentioned equivalence by considering the category of semi-Heyting algebras and the category of centered semi-Nelson algebras. We think that the properties studied here can be of interest for future work about the understanding of the categories of semi-Heyting algebras and centered semi-Nelson algebras, respectively. The paper is organized as follows. In Section 2 we study some properties concerning centered semi-Nelson algebras. In Section 3 we generalize the equivalence between the categories HA and $$\mathsf{NA_{\mathrm{c}}}$$ (see Theorem 1.2) in the framework of SH and $$\mathsf{SN_{c}}$$. Finally, in Section 4 we study the relationship between $$\mathsf{SN_{c}}$$ and the category KSH (see Definition 4.4) introduced in [16], which is also equivalent to the category SH. 2 Basic results Let T ∈ PSN. We denote by ≤ to the partial order associated to the underlying lattice of T. In this section we give some basic properties about (centered) semi-Nelson algebras we shall use in the next section in order to show that there exists an equivalence between SH and $$\mathsf{SN_{c}}$$. Since the results given here are very technical, we recommend to the reader don’t read the proofs of them in a first lecture of the present paper. Lemma 2.1 [12] Let T ∈ PSN and x, y, z ∈ T. Then (a) $$1 \to _{N} x = x$$, (b) $$x \to _{N} x = 1$$, (c) if x ≤ y then $$x \to _{N} y = 1$$, (d) x ≤ y if and only if $$x \to _{N} y = 1$$ and $$\sim y \to _{N} \sim x = 1$$, (e) If $$x \to _{N} y = y \to _{N} z = 1$$ then $$x \to _{N} z = 1$$, (f) $$(x \wedge y) \to _{N} y = 1$$, (g) if $$x\to _{N} y=1$$ and $$y\to _{N} z=1$$ then $$x\to _{N} z=1$$. Lemma 2.2 [13] Let T ∈ SN and x, y, z ∈ T. Then (a) $$(x \wedge \sim x) \to _{N} y = 1$$, (b) $$(x \to _{N} y) \to _{N} ((x \to _{N} z) \to _{N} (x \to _{N} (y \wedge z))) = 1$$, (c) $$(x \to _{N} z) \to _{N} ((y \to _{N} z) \to _{N} ((x \vee y) \to _{N} z)) = 1$$, (d) $$x \to _{N} y = x \to _{N} (x\land y)$$. We will use the previous lemmas in order to show the following result. Lemma 2.3 Let T ∈ SN and x, y, z ∈ T. Then (a) If $$x \to _{N} y = 1$$ and $$x \to _{N} z = 1$$ then $$x \to _{N} (y \wedge z) = 1$$; (b) If $$x \to _{N} z = 1$$ and $$y \to _{N} z = 1$$ then $$(x \vee y) \to _{N} z = 1$$; (c) If $$x \to _{N} y = 1$$ then $$(x \vee z) \to _{N} (y \vee z) = 1$$. Proof. (a) By (b) of Lemma 2.2 we have that   $$ (x \to_{N} y) \to_{N} ((x \to_{N} z) \to_{N} (x \to_{N} (y \wedge z))) = 1. $$Hence, by (a) of Lemma 2.1 we obtain that $$x \to _{N} (y \wedge z) = 1$$. (b) It follows from (c) of Lemma 2.2 that   $$ (x \to_{N} z) \to_{N} ((y \to_{N} z) \to_{N} ((x \vee y) \to_{N} z)) = 1. $$Thus, by (a) of Lemma 2.1 we deduce the equality $$(x \vee y) \to _{N} z = 1$$. (c) By hypothesis we have that $$x \to _{N} y = 1$$. It follows from (c) of Lemma 2.1 the equality $$y \to _{N} (y \vee z) = 1$$. Then, using (g) of Lemma 2.1 we obtain that $$x \to _{N} (y \vee z) = 1$$. Observe that $$z \to _{N} (y \vee z) = 1$$, which follows again from (c) of Lemma 2.1. In view of (c) of Lemma 2.2 it holds that   $$ (x \to_{N} (y \vee z)) \to_{N} ((z \to_{N} (y \vee z)) \to_{N} ((x \vee z) \to_{N} (y \vee z))) = 1. $$Therefore, by (a) of Lemma 2.1 we conclude that $$(x \vee z) \to _{N} (y \vee z) = 1$$. The following lemma presents some useful properties of centered semi-Nelson algebras. Lemma 2.4 Let $$T \in \mathsf{SN_{c}}$$ and x, y, z ∈ T. Then (a) $$\mathrm{c} \to _{N} x = 1$$, (b) $$(x \wedge \mathrm{c}) \to _{N} y = 1$$, (c) $$(\sim (x \to y) \vee \mathrm{c}) \to _{N} ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c})) = 1$$, (d) $$((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c})) \to _{N} (\sim (x \to y) \vee \mathrm{c}) = 1$$, (e) $$[\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \to _{N} [\sim (\sim (x \to y) \vee \mathrm{c})] = 1$$, (f) $$[\sim (\sim (x \to y) \vee \mathrm{c})] \to _{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] = 1$$, (g) ∼ (x → y) ∨c = (x ∨c) ∧ (∼ y ∨c), (h) $$[\sim ((x \to y) \vee \mathrm{c})] \to _{N} [\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] = 1$$, (i) $$[\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \to _{N} [\sim ((x \to y) \vee \mathrm{c})] = 1$$, (j) $$(x \to y) \to _{N} ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) = 1$$, (k) $$((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) \to _{N} (x \to y) = 1$$, (l) (x ∨c) → (y ∨c) = (x → y) ∨c. Proof. Let x, y ∈ T. (a) Note that $$\mathrm{c} \to _{N} x = (\mathrm{c} \wedge \mathrm{c}) \to _{N} x = (\mathrm{c} \wedge \sim \mathrm{c}) \to _{N} x = 1$$, which follows from (a) of Lemma 2.2. (b) By item (a) we have that $$\mathrm{c} \to _ N y =1$$ and it follows from (f) of Lemma 2.1 that $$(x \wedge \mathrm{c}) \to _{N} \mathrm{c} = 1 $$. Thus, it follows from (g) of Lemma 2.1 that $$(x \wedge \mathrm{c}) \to _{N} y = 1$$. (c) The fact that $$\sim (x \to y) \to _{N} (x \wedge \sim y) = 1$$ follows from (SN10). By (f) of Lemma 2.1 we obtain that $$(x \wedge \sim y) \to _{N} x = 1$$. Then, applying (e) of Lemma 2.1 we obtain that $$\sim (x \to y) \to _{N} x = 1$$. Thus, taking into account (c) of Lemma 2.3 we have that   \begin{align} (\sim (x \to y) \vee \mathrm{c}) \to_{N} (x \vee \mathrm{c}) = 1. \end{align} (2)Similarly we can show that   \begin{align} (\sim (x \to y) \vee \mathrm{c}) \to_{N} (\sim y \vee \mathrm{c}) = 1. \end{align} (3)By (2), (3) and (a) of Lemma 2.3 it is possible to verify that   $$ (\sim (x \to y) \vee \mathrm{c}) \to_{N} ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c})). $$ (d) By (SN11) we have that $$(x \wedge \sim y) \to _{N} (\sim (x \to y)) = 1$$. Thus, by (c) of Lemma 2.3 we deduce the equality   $$ ((x \wedge \sim y) \vee \mathrm{c}) \to_{N} ((\sim (x \to y)) \vee \mathrm{c}) = 1. $$Consequently,   $$ ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c})) \to_{N} (\sim(x \to y) \vee \mathrm{c}) = ((x \wedge \sim y) \vee \mathrm{c}) \to_{N} (\sim(x \to y) \vee \mathrm{c}) = 1. $$ (e) Let us notice that (f) of Lemma 2.1 implies the equality   \begin{align} (\sim x \wedge \sim \mathrm{c}) \to_{N} (\sim \mathrm{c}) = 1. \end{align} (4)Hence, using (a) we have that   \begin{align} \mathrm{c} \to_{N} (\sim \mathrm{c}) = 1 \end{align} (5)and   \begin{align} \mathrm{c} \to_{N} (x \to y) = 1. \end{align} (6)Hence, (5), (6) and (a) of Lemma 2.3 can be used to verify that   $$ \mathrm{c} \to_{N} ((\sim \mathrm{c}) \wedge (x \to y)) = 1. $$Since c =∼c then   \begin{align} (\sim \mathrm{c}) \to_{N} ((\sim \mathrm{c}) \wedge (x \to y)) = 1. \end{align} (7)It follows from (4), (7) and (g) of Lemma 2.1 that   \begin{align} (\sim x \wedge \sim \mathrm{c}) \to_{N} ((\sim \mathrm{c}) \wedge (x \to y)) = 1. \end{align} (8)Similarly we can show the equality   \begin{align} (y \wedge \sim \mathrm{c}) \to_{N} ((\sim \mathrm{c}) \wedge (x \to y)) = 1. \end{align} (9)By (b) of Lemma 2.3 combined with (8) and (9) we deduce that   \begin{align} ((\sim x \wedge \sim \mathrm{c}) \vee (y \wedge \sim \mathrm{c})) \to_{N} ((\sim \mathrm{c}) \wedge (x \to y)) = 1. \end{align} (10)In consequence, it follows from (10) that   \begin{align*} & [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \to_{N} [\sim(\sim(x \to y) \vee \mathrm{c})]\\ &\quad= [[\sim ((x \vee \mathrm{c})] \vee [\sim (\sim y \vee \mathrm{c})]] \to_{N} [\sim(\sim(x \to y) \vee \mathrm{c})] \\ &\quad= ((\sim x \wedge \sim \mathrm{c}) \vee (\sim \sim y \wedge \sim \mathrm{c})) \to_{N} [\sim(\sim(x \to y) \vee \mathrm{c})] \\ &\quad= ((\sim x \wedge \sim \mathrm{c}) \vee (y \wedge \sim \mathrm{c})) \to_{N} [\sim(\sim(x \to y) \vee \mathrm{c})] \\ &\quad= ((\sim x \wedge \sim \mathrm{c}) \vee (y \wedge \sim \mathrm{c})) \to_{N} [(\sim \sim(x \to y) \wedge (\sim \mathrm{c}))] \\ &\quad= ((\sim x \wedge \sim \mathrm{c}) \vee (y \wedge \sim \mathrm{c})) \to_{N} [((x \to y) \wedge (\sim \mathrm{c}))] \\ &\quad= 1. \end{align*} (f) Item (b) implies the equality   \begin{align} ((x \to y) \wedge \mathrm{c}) \to_{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] = 1. \end{align} (11)Hence, it follows from (11) that   \begin{align*} & [\sim(\sim(x \to y) \vee \mathrm{c})] \to_{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \\ &\quad= [(\sim\sim(x \to y)) \wedge (\sim \mathrm{c})] \to_{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \\ &\quad= [((x \to y)) \wedge (\sim \mathrm{c})] \to_{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \\ &\quad= [((x \to y)) \wedge \mathrm{c}] \to_{N} [\sim ((x \vee \mathrm{c}) \wedge (\sim y \vee \mathrm{c}))] \\ &\quad= 1. \end{align*} (g) Using items (c), (d), (e), (f) and (d) of Lemma 2.1, the desired identity holds. (h) It is consequence from (b) as follows:   \begin{align*} & [\sim ((x \to y) \vee \mathrm{c})] \to_{N} [\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \\ &\quad= [(\sim (x \to y)) \wedge (\sim \mathrm{c})] \to_{N} [\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \\ &\quad= [(\sim (x \to y)) \wedge \mathrm{c}] \to_{N} [\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \\ &\quad= 1. \end{align*} (i) By (SN10) we obtain that   \begin{align} [\sim ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \to_{N} [(x \vee \mathrm{c}) \wedge (\sim (y \vee \mathrm{c}))] = 1. \end{align} (12)Taking into account (b) we have that   \begin{align*} & [(x \vee \mathrm{c}) \wedge (\sim (y \vee \mathrm{c}))] \to_{N} [\sim ((x \to y) \vee \mathrm{c})] \\ &\quad= [(x \vee \mathrm{c}) \wedge (\sim y) \wedge (\sim \mathrm{c})] \to_{N} [\sim ((x \to y) \vee \mathrm{c})] \\ &\quad= [(x \vee \mathrm{c}) \wedge (\sim y) \wedge \mathrm{c}] \to_{N} [\sim ((x \to y) \vee \mathrm{c})] \\ &\quad= 1. \end{align*}Then,   \begin{align} [(x \vee \mathrm{c}) \wedge (\sim (y \vee \mathrm{c}))] \to_{N} [\sim ((x \to y) \vee \mathrm{c})] = 1. \end{align} (13)To finish off the proof of this item we can apply (g) of Lemma 2.1 to equations (12) and (13). (j) Using item (a) and (b) of Lemma 2.1 we can verify that $$\mathrm{c} \to _{N} x = 1$$ and $$x \to _{N} x = 1,$$ respectively. By (b) of Lemma 2.3 we obtain that   \begin{align} (x \vee \mathrm{c}) \to_{N} x = 1. \end{align} (14)Also, it follows from (c) of Lemma 2.1 that   \begin{align} x \to_{N} (x \vee \mathrm{c}) = 1. \end{align} (15)Similarly we can show that   \begin{align} (y \vee \mathrm{c}) \to_{N} y = 1 \ \ \ \mbox{ and } \ \ \ y \to_{N} (y \vee \mathrm{c}) = 1. \end{align} (16)Then,   \begin{align*} & (x \to y) \to ((x \vee \mathrm{c}) \to y) \\ &\quad= 1 \to_{N} [(x \to y) \to ((x \vee \mathrm{c}) \to y)] \\ &\qquad\ \, \mbox{by (a) of}\ \textrm{Lemma}\ {2.1} \\ &\quad= ((x \vee \mathrm{c}) \to_{N} x) \to_{N} [(x \to y) \to ((x \vee \mathrm{c}) \to y)]&\! \mbox{by} \ {(14)}\quad \\ &\quad= 1 \to_{N} [((x \vee \mathrm{c}) \to_{N} x) \to_{N} [(x \to y) \to ((x \vee \mathrm{c}) \to y)]] \\ &\qquad\ \, \mbox{by (a) of}\ \textrm{Lemma}\ {2.1} \\ &\quad= (x \to_{N} (x \vee \mathrm{c})) \to_{N} [((x \vee \mathrm{c}) \to_{N} x) \to_{N} [(x \to y) \to ((x \vee \mathrm{c}) \to y)]]& \mbox{by} \ {(15)}\quad \\ &\quad = 1 &\mbox{by (SN8)}. \end{align*}Then,   \begin{align} (x \to y) \to ((x \vee \mathrm{c}) \to y) = 1. \end{align} (17)On the other hand,   \begin{align*} & ((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) \\ &\quad= 1 \to_{N} [((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] \\ &\qquad\ \, \mbox{by (a) of}\ \textrm{Lemma}\ {2.1} \\ &\quad= ((y \vee \mathrm{c}) \to_{N} y) \to_{N} [((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))] & \mbox{by}\ {(16)}\quad \\ &\quad= 1 \to_{N} [((y \vee \mathrm{c}) \to_{N} y) \to_{N} [((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))]] \\ &\qquad\ \, \mbox{by (a) of}\ \textrm{Lemma}\ {2.1} \\ &\quad= (y \to_{N} (y \vee \mathrm{c}))\\ &\qquad\to_{N} [((y \vee \mathrm{c}) \to_{N} y) \to_{N} [((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c}))]] & \mbox{by}\ {(16)}\quad \\ &\quad= 1 & \mbox{by (SN9)}. \end{align*}Thus, we have the equality   \begin{align} ((x \vee \mathrm{c}) \to y) \to ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) = 1. \end{align} (18)Finally, taking into account (g) of Lemma 2.1 in conditions (17) and (18) we conclude that (x → y) → ((x ∨c) → (y ∨c)) = 1. (k) The proof of this item is similar to the one used in (j). (l) It follows from item (j) that $$(x \to y) \to _{N} ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) = 1$$. Also, $$\mathrm{c} \to _{N} ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) = 1$$ follows from (a). Hence, by (b) of Lemma 2.3 we deduce that   \begin{align} ((x \to y) \vee \mathrm{c}) \to_{N} ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) = 1. \end{align} (19)Besides, from item (k) we know that $$((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) \to _{N} (x \to y) = 1$$. Also, by (c) of Lemma 2.1 we have that $$(x \to y) \to _{N} ((x \to y) \vee \mathrm{c})$$. Thus, by (e) of Lemma 2.1 the equality   \begin{align} ((x \vee \mathrm{c}) \to (y \vee \mathrm{c})) \to_{N} ((x \to y) \vee \mathrm{c}) = 1 \end{align} (20)is satisfied. Therefore, applying (d) of Lemma 2.1 in (19), (20), (h) and (i) we can verify that (x ∨c) → (y ∨c) = (x → y) ∨c. In what follows we will use previous results in order to prove the following lemma. Lemma 2.5 Let $$T\in \mathsf{SN_{c}}$$ and x, y ∈ T. Then $$\mathrm{c} \leq x \rightarrow (y\vee \mathrm{c})$$. Proof. By (b) of Lemma 2.4 we have that   \begin{align*} (x \wedge \sim (y \vee \mathrm{c})) \to_{N} (\sim \mathrm{c}) &= (x \wedge (\sim y) \wedge (\sim \mathrm{c})) \to_{N} (\sim \mathrm{c}) & & \\ & = (x \wedge (\sim y) \wedge \mathrm{c}) \to_{N} (\sim \mathrm{c}) \\ & = 1. \end{align*} Hence, $$(x \wedge \sim (y \vee \mathrm{c})) \to _{N} (\sim \mathrm{c}) = 1$$. Besides, it follows from (SN10) that $$[\sim (x \to (y\vee \mathrm{c}))] \to _{N} (x \wedge \sim (y \vee \mathrm{c})) = 1$$. Then, using (g) of Lemma 2.1 we obtain that   \begin{align} [\sim (x \to (y\vee \mathrm{c}))] \to_{N} (\sim \mathrm{c}) = 1. \end{align} (21)Notice that   \begin{align} \mathrm{c} \to_{N} (x \to (y\vee \mathrm{c})) = 1 \end{align} (22)in view of (a) of Lemma 2.4. Therefore, taking into account (21), (22) and using (d) of Lemma 2.1 we conclude that c ≤ x → (y ∨c). 3 Categorical equivalence between SH and SNc In this section we will prove that there is a categorical equivalence between SH and $$\mathsf{SN_{c}}$$. We will use the same notations and constructions given in Section 1 about the functors $$\mathrm{K}:\mathsf{HA} \rightarrow \mathsf{NA_{\mathrm{c}}}$$, $$\mathrm{C}:\mathsf{NA_{\mathrm{c}}}\rightarrow \mathsf{HA}$$ and the isomorphisms $$\alpha _{H}$$ (for H ∈ HA) and $$\beta _{T}$$ (for $$T\in \mathsf{NA_{\mathrm{c}}}$$). Let H ∈ SH. In [12, Theorem 4.1] it was proved that (K(H), ∧, ∨, ⇒, ∼, 1) ∈ SN. Thus, if H ∈ SH then $$\mathrm{K}(H) \in \mathsf{SN_{c}}$$. It is immediate that if f is a morphism in SH, then K(f) is a morphism in $$\mathsf{SN_{c}}$$. Hence, we obtain the following result. Proposition 3.1 There exists a functor K from SH to $$\mathsf{SN_{c}}$$. Let T ∈ SN (or $$T\in \mathsf{SN_{c}}$$). The binary relation ≡ on T given by   $$ x \equiv y\;\textrm{if and only if}\;x\rightarrow y = 1\;\textrm{and}\; y\rightarrow x = 1 $$is an equivalence relation on T compatible with the operations ∧, ∨ and $$\rightarrow $$, as it is shown in [12, Lemma 3.1]. If x ∈ T we write $$[\![x]\!] $$ for the equivalence class associated to ≡. As usual, we write $$T/_{\equiv }$$ for the set of equivalence classes. We denote by sH(T) the algebra $$(T/_{\equiv },\cap ,\cup ,\rightsquigarrow ,[\![0]\!],[\![1]\!])$$, where the operations are defined by: • $$[\![x]\!]\ \cap\ [\![y]\!] = [\![x \wedge y]\!]$$, • $$[\![x]\!]\ \cup\ [\![y]\!] = [\![x \vee y]\!]$$, • $$[\![x]\!]\rightsquigarrow [\![x]\!] = [\![x \rightarrow y]\!]$$. It follows from [12, Theorem 3.4] that sH(T) ∈ SH. The following lemma will be used in the next proposition. Lemma 3.2 a) Let $$T\in \mathsf{SN_{c}}$$. Then C(T) is closed by the operations $$\wedge , \vee , \rightarrow $$ and the constants c and 1. Hence, C(T) can be seen as an algebra of type (2, 2, 2, 0, 0). b) Let $$T, U\in \mathsf{SN_{c}}$$ such that T ≅ U, i.e. T and U are isomorphic algebras. Then C(T) ≅ C(U). In particular, C(T) ∈ SH if and only if C(U) ∈ SH. Proof. It is immediate that c, 1 ∈ C(T). It is also immediate that C(T) is closed by ∧ and ∨. The fact that C(T) is closed by $$\rightarrow $$ follows from Lemma 2.5. Then we have proved a). In order to prove b), let $$T,U\in \mathsf{SN_{c}}$$ and $$f:T\rightarrow U$$ an isomorphism. Straightforward computations show that the map $$\hat{f}: \mathrm{C}(T) \rightarrow \mathrm{C}(U)$$ given by $$\hat{f}(x) = f(x)$$ is an isomorphism. Thus, C(T) ≅ C(U). The fact that C(T) ∈ SH if and only if C(U) ∈ SH follows from that SH is a variety. Proposition 3.3 Let $$T\in \mathsf{SN_{c}}$$. Then C(T) ∈ SH. Proof. Consider the map $$h: T \rightarrow \mathrm{K}(\mathbf{sH}(T))$$ given by $$h(x) = ([\![x]\!], [\![\sim x]\!])$$. We will write $$\rightarrow $$ for the implication of T and ⇒ for the implication of K(sH(T)). The function h is an injective morphism in SN, see [12, Corollary 5.2]. Straightforward computations show that h preserves the bottom and the center, so T ≅ h(T) in $$\mathsf{SN_{c}}$$. Since $$\mathsf{SN_{c}}$$ is a variety, h(T) is a subalgebra of K(sH(T)) and $$\mathrm{K}(\mathbf{sH}(T)) \in \mathsf{SN_{c}}$$ then $$h(T) \in \mathsf{SN_{c}}$$. Hence, by Lemma 3.2 we have that C(T) ∈ SH if and only if C(h(T)) ∈ SH. By this reason, in what follows we will prove that C(h(T)) ∈ SH. Our aim is to show that C(h(T)) ∈ SH. In order to prove it we will use the fact that sH(T) ∈ SH. Let x, y, z ∈ T such that h(x), h(y), h(z) ∈ C(h(T)). Then $$[\![c]\!]\le [\![x]\!], [\![c]\!]\le[\![y]\!], [\![c]\!]\le[\![z]\!], [\![\sim x]\!]\le[\![c]\!], [\![\sim y]\!]\le[\![c]\!]$$ and $$[\![\sim z]\!]\le[\![c]\!]$$. Taking into account that $$[\![\sim y]\!]\le[\![c]\!]\le[\![x]\!]$$ and $$[\![x \wedge (x\rightarrow y)]\!] = [\![x \wedge y]\!]$$, the condition (SH2) in sH(T), we obtain that   \begin{align*} h(x) \cap (h(x) \Rightarrow h(y)) & = ([\![x]\!],[\![\sim x]\!]) \cap (([\![x]\!],[\![\sim x]\!]) \Rightarrow ([\![y]\!], [\![\sim y]\!])) \\ & = ([\![x]\!], [\![\sim x]\!]) \cap ([\![x\rightarrow y]\!], [\![x]\!] \cap [\![\sim y]\!]) \\ & = ([\![x]\!], [\![\sim x]\!]) \cap ([\![x\rightarrow y]\!], [\![\sim y]\!]) \\ & = ([\![x \wedge (x\rightarrow y)]\!], [\![\sim x \vee \sim y]\!])\\ & = ([\![x \wedge y]\!], [\![\sim x \vee \sim y]\!])\\ & = ([\![x]\!] \cap [\![y]\!], [\![\sim x]\!] \cup [\![\sim y]\!])\\ & = h(x) \cap h(y). \end{align*} Then C(h(T)) satisfies (SH2). In order to prove (SH3), first note that since $$[\![\sim z]\!]\le[\![c]\!]\le[\![y]\!]$$ then   \begin{align*} h(x) \cap (h(y) \Rightarrow h(z)) & = ([\![x]\!],[\![\sim x]\!]) \cap (([\![y]\!],[\![\sim y]\!]) \Rightarrow ([\![z]\!], [\![\sim z]\!])) \\ & = ([\![x]\!], [\![\sim x]\!]) \cap ([\![y\rightarrow z]\!], [\![y]\!] \cap [\![\sim z]\!])\\ & = ([\![x]\!], [\![\sim x]\!]) \cap ([\![y\rightarrow z]\!], [\![\sim z]\!]) \\ & = ([\![x\wedge (y\rightarrow z)]\!], [\![\sim x \vee \sim z]\!]). \end{align*} Then   \begin{align} h(x) \cap (h(y) \Rightarrow h(z)) = ([\![x\wedge (y\rightarrow z)]\!], [\![\sim x \vee \sim z]\!]). \end{align} (23) Besides, straightforward computations shows that   \begin{align} h(x) \cap ((h(x)\cap h(y)) \Rightarrow (h(x) \cap h(z)) = ([\![x\wedge((x\wedge y) \rightarrow (x\wedge z))]\!], [\![\sim x \vee \sim z]\!]). \end{align} (24) Using (SH3) in sH(T) we have that $$[\![x\wedge (y\rightarrow z)]\!] = [\![x\wedge ((x\wedge y) \rightarrow (x\wedge z))]\!]$$. Thus, it follows from (23) and (24) that   $$ h(x) \cap (h(y) \Rightarrow h(z)) = h(x) \cap ((h(x)\cap h(y)) \Rightarrow (h(x) \cap h(z)), $$which is the condition (SH3) in C(h(T)). Finally,   \begin{align*} h(x) \Rightarrow h(x) & = ([\![x\rightarrow x]\!],[\![x\wedge \sim x]\!]) \\ & = ([\![1]\!], [\![0]\!]), \end{align*}i.e. the condition (SH4) is also satisfied in C(h(T)). Therefore, C(T) ∈ SH. It is immediate that if f is a morphism in SH, then C(f) is a morphism in $$\mathsf{SN_{c}}$$. Therefore, we conclude the following result. Proposition 3.4 There exists a functor C from $$\mathsf{SN_{c}}$$ to SH. Lemma 3.5 Let $$T \in \mathsf{SN_{c}}$$ and x, y ∈ T. Then $$x\vee \mathrm{c} \leq (y\vee c) \rightarrow ((x\vee c) \wedge (y\vee c))$$. Proof. Let $$T \in \mathsf{SN_{c}}$$ and x, y ∈ T. In particular, we have that x ∨c, y ∨c ∈ C(T). Besides, by Proposition 3.3 we have that C(T) ∈ SH. Hence, it follows from (SH3) and (SH4) that $$x\vee \mathrm{c} \leq (y\vee c) \rightarrow ((x\vee c) \wedge (y\vee c))$$. Remark 3.6 Let $$T \in \mathsf{SN_{c}}$$ and x, y ∈ T. Throughout the rest of this section we will use the equalities $$\sim (x\rightarrow y)\vee \mathrm{c} = (x\vee \mathrm{c})\wedge (\sim y \vee \mathrm{c})$$ and $$(x\rightarrow y) \vee \mathrm{c} = (x\vee \mathrm{c}) \rightarrow (y\vee \mathrm{c})$$, which appears in items (g) and (l) of Lemma 2.4, respectively. The following lemma will be used later. Lemma 3.7 Let $$T\in \mathsf{SN_{c}}$$. Then T satisfies (CK). Proof. In this proof we will use Lemma 3.5 and Remark 3.6. Let x, y ∈ T such that x ≥ c, y ≥ c and x ∧ y = c. Let $$z = (y\rightarrow \sim y) \wedge x$$. In particular, ∼ x ≤ c and ∼ y ≤ c. We will prove that x = z ∨ c and y =∼ z ∨ c. The equality z ∨ c = x can be proved as follows:   \begin{align*} z\vee \mathrm{c} & = ((y \rightarrow \sim y)\wedge x) \vee (\mathrm{c} \wedge x)\\ & = x \wedge ((y\rightarrow \sim y) \vee \mathrm{c})\\ & = x \wedge ((y\vee \mathrm{c})\rightarrow (\sim y \vee \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{align*} Finally we have that   \begin{align*} z\wedge \mathrm{c} & = ((y \rightarrow \sim y)\wedge \mathrm{c}) \wedge x \\ & = (\sim (\sim (y\rightarrow \sim y) \vee \mathrm{c})) \wedge x \\ & = (\sim (y\vee \mathrm{c})) \wedge x \\ & =\, \sim y \wedge \mathrm{c} \wedge x \\ & =\, \sim y \wedge \mathrm{c} \\ & =\, \sim y, \end{align*}so ∼ z ∨ c = y. Let $$T\in \mathsf{SN_{c}}$$. We will see that $$\beta _{T}$$ is an isomorphism in $$\mathsf{SN_{c}}$$. Proposition 3.8 Let $$T\in \mathsf{SN_{c}}$$. Then $$\beta _{T}$$ is an isomorphism in $$\mathsf{SN_{c}}$$. Proof. We know that $$\beta _{T}$$ is an injective morphism in $$\mathsf{KA_{\mathrm{c}}}$$. The fact that $$\beta _{T}$$ preserves the implication is a direct consequence of Remark 3.6. Thus, $$\beta _{T}$$ is an injective morphism in $$\mathsf{SN_{c}}$$. Finally, since the condition (CK) is equivalent to the surjectivity of $$\beta _{T}$$, then it follows from Lemma 3.7 that $$\beta _{T}$$ is a surjective map. Therefore, $$\beta _{T}$$ is an isomorphism in $$\mathsf{SN_{c}}$$. Taking into account the previous results of this section, the fact that if H ∈ SH then $$\alpha _{H}$$ is an isomorphism in SH, and the categorical equivalence between $$\mathsf{KA_{\mathrm{c}}}$$ and BDL, we obtain the main theorem of this paper. Theorem 3.9 The functors K and C establish a categorical equivalence between SH and $$\mathsf{SN_{c}}$$ with natural isomorphisms $$\alpha $$ and $$\beta $$. Theorem 3.9 allow us to give other presentation of the categorical equivalence between SH and $$\mathsf{SN_{c}}$$, as we show in what follows. Let $$T\in \mathsf{SN_{c}}$$. Then sH(T) ∈ SH. Moreover, if $$g:T\rightarrow U \in \mathsf{SN_{c}}$$ then straightforward computations show that the map $$\mathbf{sH}(g):\mathbf{sH}(T) \rightarrow \mathbf{sH}(U)$$ given by $$\mathbf{sH}(g)([\![x]\!])=[\![g(x)]\!]$$ is a morphism in SH. Hence, we have a functor   $$ \mathbf{sH}: \mathsf{SN_{c}} \rightarrow \mathsf{SH}. $$ Let $$T\in \mathsf{SN_{c}}$$ and x ∈ T. We will see that   \begin{align} [\![x \vee \mathrm{c}]\!] = [\![x]\!]. \end{align} (25)Note that $$[\![x \vee \mathrm{c}]\!] = [\![x]\!]$$ if and only if $$x\rightarrow (x\vee \mathrm{c}) = 1$$ and $$(x\vee \mathrm{c})\rightarrow x = 1$$. It can be proved that these equations hold in K(H) for every H ∈ SH. By Proposition 3.3 we have that C(T) ∈ SH, so the equations hold in K(C(T)). Thus, by Proposition 3.8 the equations hold in T, which was our aim. Now consider the injective morphism $$h:T\rightarrow \mathrm{K}(\mathbf{sH}(T))$$ given in the proof of Proposition 3.3. Lemma 3.10 Let $$T\in \mathsf{SN_{c}}$$. The map h is an isomorphism in $$\mathsf{SN_{c}}$$. Proof. First we will prove that the morphism $$\rho :\mathrm{C}(T) \rightarrow \mathbf{sH}(T)$$ given by $$\rho (x) = [\![x]\!]$$ is an isomorphism in SH. It follows from (25) that $$\rho $$ is a surjective map. In order to prove the injectivity, let x, y ≥ c such that $$[\![x ]\!] = [\![y]\!]$$, i.e. $$x\rightarrow y = 1$$ and $$y\rightarrow x = 1$$. By Proposition 3.3 we have that C(T) ∈ SH, so $$x = x\wedge (x\rightarrow y) = x\wedge y$$ and $$y = y\wedge (y\rightarrow x) = y\wedge x$$. Thus, x ≤ y and y ≤ x, so x = y. Thus, $$\rho $$ is an injective map. Hence, it is immediate that the map $$\mathrm{K}(\rho ): \mathrm{K}(\mathrm{C}(T))\rightarrow \mathrm{K}(\mathbf{sH}(T))$$ is an isomorphism in $$\mathsf{SN_{c}}$$. By Proposition 3.8 we also have that $$\beta _{T}:T\rightarrow \mathrm{K}(\mathrm{C}(T))$$ is an isomorphism in $$\mathsf{SN_{c}}$$. Then the map $$\mathrm{K}(\rho )\circ \beta _{T}: T\rightarrow \mathrm{K}(\mathbf{sH}(T))$$ is an isomorphism in $$\mathsf{SN_{c}}$$. Finally, it follows from (25) that $$h = \mathrm{K}(\rho )\circ \beta _{T}$$. Therefore, h is an isomorphism in $$\mathsf{SN_{c}}$$. Let H ∈ SH. The map $$i:H \rightarrow \mathbf{sH}(\mathrm{K}(H))$$ given by $$i(a) = [\![(a, a\rightarrow 0)]\!]$$ is an isomorphism in SH, as it was proved in [12, Theorem 5.3]. Therefore, straightforward computations show that the functors K and sH establish a categorical equivalence between SH and $$\mathsf{SN_{c}}$$ with natural isomorphisms i and h. 4 Connection with existing literature 4.1 Categorical equivalence between $$\mathsf{SN_{c}}$$ and KSH In [16] it was proved that there exists a categorical equivalence between SH and an algebraic category denoted by KSH (see Definition 4.4). The original motivation to consider this algebraic category comes from a different definition of the binary operation given in (1) of Section 1 on (K(H), ∧, ∨, ∼, c, 0, 1), where H ∈ SH. Combining the categorical equivalence between SH and KSH with Theorem 3.9, we obtain that there exists a categorical equivalence between $$\mathsf{SN_{c}}$$ and KSH. In this section we do a more detailed study about the connection between the categories $$\mathsf{SN_{c}}$$ and KSH. We assume the reader is familiar with commutative residuated lattices [15]. An involutive residuated lattice is a bounded, integral and commutative residuated lattice $$(T,\wedge , \vee , \ast ,\rightarrow , 0, 1)$$ such that for every x ∈ T it holds that ¬¬x = x, where $$\neg x: = x\rightarrow 0$$ and 0 is the first element of T [8]. 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 [8] 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 [22]. Remark 4.1 a) Let (T, ∧, ∨, ⇒, ∼, 0, 1) be a Nelson algebra. We define on T the binary operations * and $$\rightarrow $$ by   \begin{align*} x*y & :={\sim} (x \Rightarrow{\sim} y) \vee{\sim} (y \Rightarrow{\sim} x), \\ x \rightarrow y & := (x \Rightarrow y) \wedge ({\sim} y\Rightarrow{\sim} x). \end{align*}Then, [8, Theorem 3.1] says that $$(T,\wedge , \vee , \rightarrow ,*, 0,1)$$ is a Nelson lattice. Moreover, $${\sim } x = \neg x = x\rightarrow 0$$. b) Let $$(T,\wedge ,\vee ,*,\rightarrow ,0,1)$$ be a Nelson lattice. We define on T a binary operation ⇒ and a unary operation ∼ by   \begin{align*} x \Rightarrow y & := x^{2} \rightarrow y, \\{\sim} x & := \neg x, \end{align*}where $$x^{2}: = x*x$$. In particular, $$x \Rightarrow y= (\sim (x \rightarrow \sim x))\rightarrow y$$. Then, [8, Theorem 3.11] says that the algebra (T, ∧, ∨, ⇒, ∼, 0, 1) is a Nelson algebra. c) Notice that in [8, 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 (see [8]), we obtain 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 in items a) and b). The results from [8] about the connections between Nelson algebras and Nelson lattices mentioned in Remark 4.1 are based on results from Spinks and Veroff [22]. More precisely, 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 $$(T,\vee,\wedge ,*,\rightarrow ,\mathrm{c},0,1)$$, where the reduct $$(T,\vee ,\wedge ,*,\rightarrow,0,1)$$ is a Nelson lattice and c is an element of T such that ¬c = c. It follows from Remark 4.1 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 4.2 Let $$(H,\wedge , \vee ,\rightarrow ,0,1) \in \mathsf{HA}$$. We know that $$(\mathrm{K}(H),\wedge ,\vee ,\Rightarrow ,{\sim },\mathrm{c},0,1) \in \mathsf{NA_{\mathrm{c}}}$$, where ⇒ is the operation given in (1) of Section 1. Hence, it follows from Remark 4.1 that $$(\mathrm{K}(H),\wedge,\vee,*,\rightarrow,\mathrm{c},0,1) \in \mathsf{NL_{\mathrm{c}}}$$, where for (a, b) and (d, e) in K(H) the operations * and $$\rightarrow $$ take the form   \begin{align} (a,b) * (d,e) = (a\wedge d, (a\rightarrow e)\wedge (d\rightarrow b)),\!\!\!\! \end{align} (26)  \begin{align} (a,b) \rightarrow (d,e) = ((a\rightarrow d)\wedge (e\rightarrow b), a \wedge e). \end{align} (27)We write $$\rightarrow $$ both for the implication in H as for the implication in K(H) as Nelson lattice. It was proved in [16] that K defines a functor from HA to $$\mathsf{NL_{\mathrm{c}}}$$, where K is defined using the same construction given in Section 1 but changing the binary operation ⇒ given in (1.1) of Section 1 by the binary operation given in (27) of Remark 4.2. Also it was proved in [16] that C is a functor from KSH to SH, where C is defined as in Section 1. Moreover, we have that the maps $$\alpha _{H}$$ for H ∈ SH and $$\beta _{T}$$ for T ∈ KSH are isomorphisms. The following result is [16, Proposition 7]. Proposition 4.3 The functors K and C establish a categorical equivalence between HA and $$\mathsf{NL_{\mathrm{c}}}$$ with natural isomorphisms $$\alpha $$ and $$\beta $$. In what follows we recall the definition of the category KSH given in [16]. Definition 4.4 We write KSH for the algebraic category whose objects are algebras $$(T,\wedge ,\vee , \rightarrow ,{\sim },\mathrm{c},0,1)$$ of type (2, 2, 2, 1, 0, 0, 0) such that (T, ∧, ∨, ∼, c, 0, 1) is a centered Kleene algebra and $$\rightarrow $$ is a binary operation on T which satisfies the following conditions for every x, y ∈ T: (1) $$\mathrm{c} \leq x \rightarrow (y\vee \mathrm{c})$$, (2) $$x \rightarrow x = 1$$, (3) $$(x\rightarrow y)\wedge \mathrm{c} = ({\sim } x \wedge \mathrm{c}) \vee (y\wedge \mathrm{c})$$, (4) $$(x\rightarrow{\sim } y) \vee \mathrm{c} = ((x\vee \mathrm{c}) \rightarrow ({\sim } y \vee \mathrm{c})) \wedge ((y\vee \mathrm{c}) \rightarrow ({\sim } x \vee \mathrm{c}))$$, (5) $$x\wedge ((x\vee \mathrm{c})\rightarrow (y\vee \mathrm{c})) = x\wedge (y\vee \mathrm{c})$$, (6) $$x \wedge ((y\vee \mathrm{c}) \rightarrow (z\vee \mathrm{c})) = x \wedge (((x\vee \mathrm{c}) \wedge (y\vee \mathrm{c}))\rightarrow ((x\vee \mathrm{c})\wedge (z\vee \mathrm{c})))$$. By considering the objects of KSH as algebras $$(T,\wedge ,\vee , \rightarrow ,*,{\sim },\mathrm{c},0,1)$$, where * is defined as in (26) of Remark 4.2, we have that $$\mathsf{NL_{\mathrm{c}}}$$ is a full subcategory of KSH. Proposition 4.3 can be generalized in order to prove a categorical equivalence between SH and KSH, as it is shown in [16, Theorem 51]. Theorem 4.5 The functors K and C establish a categorical equivalence between SH and KSH with natural isomorphisms $$\alpha $$ and $$\beta $$. The following result follows from Theorems 3.9 and 4.5. Theorem 4.6 There exists a categorical equivalence between $$\mathsf{SN_{c}}$$ and KSH. Notice we are using the same notation K to refer us to a functor from SH to $$\mathsf{SN_{c}}$$ and also for a functor from SH to KSH (similarly with the notation C). We believe it is clear which is the corresponding functor considered in each case. We know that the varieties $$\mathsf{NA_{\mathrm{c}}}$$ and $$\mathsf{NL_{\mathrm{c}}}$$ are term equivalent. We will prove that the varieties $$\mathsf{SN_{c}}$$ and KSH are not term equivalent by using the construction given in Remark 4.1. Consider an algebra $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1)\in \mathsf{SN_{c}}$$ and define a binary operation $$\rightarrow _{\mathsf{KSH}}$$ by   $$ x\rightarrow_{\mathsf{KSH}} y = (x\rightarrow y)\wedge (\sim y \rightarrow \sim x). $$ Remark 4.7 Let H ∈ SH and $$\rightarrow $$ the implication of H. Let a, b, d, e ∈ H such that (a, b), (d, e) ∈ K(H). Then   \begin{align*} (a,b)\rightarrow_{\mathsf{KSH}}(d,e) & = ((a,b)\Rightarrow (d,e))\cap ((e,d)\Rightarrow (b,a))\\ & = (a\rightarrow d,a\wedge e)\cap ((e\rightarrow b),a\wedge e)\\ & = ((a\rightarrow d)\wedge(e\rightarrow b),a\wedge e). \end{align*} Therefore, the implication of the algebra K(H) ∈ KSH (see Remark 4.2) is exactly the binary operation $$\rightarrow _{\mathsf{KSH}}$$ defined in K(H). In the following proposition we will see that if we consider an algebra in $$\mathsf{SN_{c}}$$ then we can define an algebra in KSH. Proposition 4.8 Let $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1) \in \mathsf{SN_{c}}$$. Then, $$(T,\wedge ,\vee ,\rightarrow _{\mathsf{KSH}},\sim ,\mathrm{c},0,1)\in \mathsf{KSH}$$. Proof. We will write $$\mathrm{K}^{\mathsf{SN_{c}}}$$ for the functor K from SH to $$\mathsf{SN_{c}}$$ and $$\mathrm{K}^{\mathsf{KSH}}$$ for the functor K from SH to KSH. In this proof we will use theorems 3.9 and 4.5. Let $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1) \in \mathsf{SN_{c}}$$. Hence, $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1) \cong \mathrm{K}^{\mathsf{SN_{c}}}(\mathrm{C}(T))$$, where the isomorphism from $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1)$$ to $$\mathrm{K}^{\mathsf{SN_{c}}}(\mathrm{C}(T))$$ is given by $$\beta _{T}$$. Since $$\rightarrow _{\mathsf{KSH}}$$ can be written in terms of $$\rightarrow $$, ∧ and ∼, then $$\beta _{T}$$ preserves the operation $$\rightarrow _{\mathsf{KSH}}$$, i.e., for every x, y ∈ T it holds that $$\beta _{T}(x\!\rightarrow _{\mathsf{KSH}}\! y) \!=\! \beta _{T}(x) \!\rightarrow _{\mathsf{KSH}}\! \beta _{T} (y)$$. Thus, it follows from Remark 4.7 that $$(T,\wedge ,\vee,\!\rightarrow _{\mathsf{KSH}},\sim ,\mathrm{c},0,1) \cong \mathrm{K}^{\mathsf{KSH}}(\mathrm{C}(T))$$. However, $$\mathrm{K}^{\mathsf{KSH}}(\mathrm{C}(T)) \in \mathsf{KSH}$$, so $$(T,\wedge ,\vee ,\rightarrow _{\mathsf{KSH}},\sim ,\mathrm{c},0,1) \in \mathsf{KSH}$$ because KSH is a variety. Let $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0,1)\in \mathsf{KSH}$$. Define a binary operation $$\hat{\rightarrow }$$ by   $$ x \hat{\rightarrow} y: = (\sim (x\rightarrow \sim x))\rightarrow y. $$The definition of $$\hat{\rightarrow }$$ is motivated by item b) of Remark 4.1. Then, it naturally arises the following question: does it hold that $$(T,\wedge ,\hat{\rightarrow },\sim ,\mathrm{c},0,1)\in \mathsf{SN_{c}}$$? The answer is negative, as we show in what follows. Let $$(H,\wedge ,\vee ,\rightarrow ,0,1) \in \mathsf{SH}$$ and also write $$\rightarrow $$ for the implication in K(H) ∈ KSH. Let (a, b), (d, e) ∈ K(H). Then   \begin{align*} (a,b) \hat{\rightarrow} (d,e) &= (\sim((a,b)\rightarrow \sim (a,b)))\rightarrow (d,e) \\ & = (\sim(a\rightarrow b, a))\rightarrow (d,e) \\ & = (a,a\rightarrow b)\rightarrow (d,e) \\ & = ((a\rightarrow d) \wedge (e\rightarrow (a\rightarrow b)),a\wedge e). \end{align*} Note that $$(a,b) \hat{\rightarrow } (d,e) = (a\rightarrow d,a\wedge e)$$ if and only if $$a\rightarrow d \leq e\rightarrow (a\rightarrow b)$$. In Heyting algebras the condition $$a\rightarrow d \leq e\rightarrow (a\rightarrow b)$$ is true whenever a ∧ b = d ∧ e = 0. However, in semi-Heyting algebras this condition is not necessarily true. For instance, consider the semi-Heyting algebra given by with a = d = 0 and b = e = 1. In this case a ∧ b = d ∧ e = 0, $$a\rightarrow d = 1$$ and $$e\rightarrow (a\rightarrow b) = 0$$, so $$a\rightarrow d \nleq e\rightarrow (a\rightarrow b)$$. Hence, we have that there exists $$(T,\wedge ,\vee ,\rightarrow ,\sim ,\mathrm{c},0, 1) \in \mathsf{KSH}$$ such that the algebra $$(T,\wedge ,\vee , \hat{\rightarrow },\sim ,\mathrm{c},0, 1)\notin \mathsf{SN_{c}}$$. Therefore, the varieties $$\mathsf{SN_{c}}$$ and KSH are not term equivalent by using the construction given in Remark 4.1. 4.2 Congruences Finally we study the connection between the congruences of $$\mathsf{SN_{c}}$$ and KSH. For $$T_{\mathsf{SN_{c}}} = (T, \wedge, \vee,\rightarrow, \sim,\mathrm{c},0,1) \in \mathsf{SN_{c}}$$ consider the algebra $$T_{\mathsf{KSH}} = (T,\wedge ,\vee ,\rightarrow _{\mathsf{KSH}},\sim ,\mathrm{c},0,1) \in \mathsf{KSH}$$. We start with the following preliminary lemma. Lemma 4.9 If $$T_{\mathsf{SN_{c}}} \in \mathsf{SN_{c}}$$ then $$\mathrm{C}(T_{\mathsf{SN_{c}}}) = \mathrm{C}(T_{\mathsf{KSH}})$$. Proof. Let x, y ≥ c. We will prove that $$x\rightarrow y = x \rightarrow _{\mathsf{KSH}} y$$. Since $$x\rightarrow _{\mathsf{KSH}} y = (x\rightarrow y) \wedge (\sim y \rightarrow \sim x)$$, it is enough to prove that $$\sim y \rightarrow \sim x = 1$$. In order to show it we use Remark 3.6 as follows:   \begin{align*} (\sim y \rightarrow \sim x) \vee \mathrm{c} &= (\sim y \vee \mathrm{c})\rightarrow (\sim x \vee \mathrm{c}) \\ & = c\rightarrow c \\ & = 1 \\ & = 1\vee \mathrm{c} \end{align*}and   \begin{align*} (\sim y \rightarrow \sim x) \wedge \mathrm{c} &=\, \sim(\sim (\sim y \rightarrow \sim x) \vee \mathrm{c}) \\ & =\, \sim((\sim y \vee \mathrm{c})\wedge(x\vee \mathrm{c}))\\ & =\, \sim \mathrm{c} \\ & = \mathrm{c} \\ & = 1\wedge \mathrm{c}. \end{align*} Hence, taking into account the distributivity of the underlying lattice of $$T_{\mathsf{SN_{c}}}$$ we deduce the equality $$\sim y \rightarrow \sim x = 1$$, which was our aim. If T is an algebra we write Con(T) for the lattice of congruences of T. Proposition 4.10 Let $$T_{\mathsf{SN_{c}}} \in \mathsf{SN_{c}}$$. Then $$\mathrm{Con}(T_{\mathsf{SN_{c}}}) = \mathrm{Con}(T_{\mathsf{KSH}})$$. Proof. It follows from Proposition 4.8 that $$\mathrm{Con}(T_{\mathsf{SN_{c}}})\subseteq \mathrm{Con}(T_{\mathsf{KSH}})$$. Conversely, let $$\theta \in \mathrm{Con}(T_{\mathsf{KSH}})$$ and x, y, z, w ∈ T such that $$(x,y)\in \theta $$ and $$(z,w)\in \theta $$. First note that it follows from Remark 3.6 that   \begin{align} (x\vee \mathrm{c})\rightarrow (z\vee \mathrm{c}) = (x\rightarrow z) \vee \mathrm{c}, \end{align} (28)  \begin{align} (y\vee \mathrm{c})\rightarrow (w\vee \mathrm{c}) = (y\rightarrow w) \vee \mathrm{c}. \end{align} (29)Besides $$(x\vee \mathrm{c}, y\vee \mathrm{c})\in \theta $$ and $$(z\vee \mathrm{c},w\vee \mathrm{c})\in \theta $$, so   $$ ((x\vee \mathrm{c})\rightarrow_{\mathsf{KSH}}(z\vee \mathrm{c}), (y\vee \mathrm{c})\rightarrow_{\mathsf{KSH}}(w\vee \mathrm{c})) \in \theta. $$ It follows from Lemma 4.9 that $$(x\vee \mathrm{c})\rightarrow _{\mathsf{KSH}}(z\vee \mathrm{c}) = (x\vee \mathrm{c})\rightarrow (z\vee \mathrm{c})$$ and $$(y\vee \mathrm{c})\rightarrow _{\mathsf{KSH}}(w\vee \mathrm{c}) = (y\vee \mathrm{c})\rightarrow (w\vee \mathrm{c})$$. Hence, it follows from (28) and (29) that   \begin{align} ((x\rightarrow z) \vee \mathrm{c},(y\rightarrow w)\vee \mathrm{c})\in \theta. \end{align} (30)On the other hand, it follows from Remark 3.6 that   \begin{align} (x\rightarrow z)\wedge \mathrm{c} = (\sim x\wedge \mathrm{c})\vee (z\wedge \mathrm{c}), \end{align} (31)  \begin{align} (y\rightarrow w)\wedge \mathrm{c} = (\sim y \wedge \mathrm{c})\vee(w\wedge \mathrm{c}). \end{align} (32)Besides, since $$(x,y)\in \theta $$ and $$(z,w)\in \theta $$ then $$(\sim x\wedge \mathrm{c}, \sim y \wedge \mathrm{c}) \in \theta $$ and $$(z\wedge \mathrm{c}, w \wedge \mathrm{c}) \in \theta $$, so $$((\sim x \wedge \mathrm{c})\vee (z\wedge \mathrm{c}), (\sim y \wedge \mathrm{c})\vee (w\wedge \mathrm{c}))\in \theta $$. Thus, taking into account (31) and (32) we conclude that   \begin{align} ((x\rightarrow z)\wedge \mathrm{c}, (y\rightarrow w)\wedge \mathrm{c})\in \theta. \end{align} (33)Hence, it follows from (30), (33) and the distributivity of the underlying lattice of $$T_{\mathsf{SN_{c}}}$$ that $$(x\rightarrow z, y\rightarrow w) \in \theta $$, so $$\theta \in \mathrm{Con}(T_{\mathsf{SN_{c}}})$$. Then, $$\mathrm{Con}(T_{\mathsf{SN_{c}}}) = \mathrm{Con}(T_{\mathsf{KSH}})$$. Let H be a distributive lattice. Recall that a non empty subset F of H is said to be a filter the following two conditions are satisfied for each a, b ∈ H: 1) if a ≤ b and a ∈ F then b ∈ F, 2) a ∧ b ∈ F whenever a, b ∈ F. We write Fil(H) by the set of filters of H. Recall that in [20, Theorem 5.4] it was proved that if H ∈ SH then there is an order isomorphism between Con(H) and Fil(H). Corollary 4.11 Let $$T_{\mathsf{SN_{c}}} \in \mathsf{SN_{c}}$$. Then $$\mathrm{Con}(T_{\mathsf{SN_{c}}}) \cong \mathrm{Fil}(\mathrm{C}(T_{\mathsf{SN_{c}}}))$$. Proof. It follows from Proposition 4.10 that $$\mathrm{Con}(T_{\mathsf{SN_{c}}}) = \mathrm{Con}(T_{\mathsf{KSH}})$$. Besides, it follows from [16, Proposition 61] that $$\mathrm{Con}(T_{\mathsf{KSH}}) \cong \mathrm{Con}(\mathrm{C}(T_{\mathsf{KSH}}))$$. Also note that $$\mathrm{Con}(\mathrm{C}(T_{\mathsf{KSH}})) \cong \mathrm{Fil}(\mathrm{C}(T_{\mathsf{KSH}}))$$ because $$\mathrm{C}(T_{\mathsf{KSH}}) \in \mathsf{SH}$$. Finally, taking into account Lemma 4.9 we have that $$\mathrm{Fil}(\mathrm{C}(T_{\mathsf{KSH}})) = \mathrm{Fil}(\mathrm{C}(T_{\mathsf{SN_{c}}}))$$. Therefore, $$\mathrm{Con}(T_{\mathsf{SN_{c}}}) \cong \mathrm{Fil}(\mathrm{C}(T_{\mathsf{SN_{c}}}))$$. In what follows we will use the notation of the proof of Proposition 4.8. By Proposition 4.10 we have that if H ∈ SH then   $$ \mathrm{Con}(\mathrm{K}^{\mathsf{KSH}}(H)) = \mathrm{Con}(\mathrm{K}^{\mathsf{SN_{c}}}(H)). $$ Proposition 4.12 Let T ∈ KSH. Then $$\mathrm{Con}(T) \cong \mathrm{Con}(\mathrm{K}^{\mathsf{SN_{c}}}(\mathrm{C}(T)))$$. Acknowledgements The authors acknowledge many helpful comments from the anonymous referee, which considerably improved the presentation of this paper. Both authors want to thank the institutional support of Consejo Nacional de Investigaciones Científicas y Técnicas. References [1] M. Abad, J. M. Cornejo and J. P. Díaz Varela. The variety generated by semi-Heyting chains. Soft Computing , 15, 721– 728, 2010. Google Scholar CrossRef Search ADS   [2] M. Abad, J. M. Cornejo and J. P. Díaz Varela. The variety of semi-Heyting algebras satisfying the equation (0→1)*∨ (0→1)**≈ 1. Reports on Mathematical Logic , 46, 75– 90, 2011. [3] M. Abad, J. M. Cornejo and J. P. Díaz Varela. 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Magister dissertation in Mathematics, Universidad Nacional del Sur, Bahía Blanca, available at https://sites.google.com/site/viglizzo/viglizzo99nelson. © The Author(s) 2018. Published by Oxford University Press. All rights reserved. For permissions, please e-mail: journals.permission@oup.com. This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/about_us/legal/notices) For permissions, please e-mail: journals. permissions@oup.com

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