# Justification announcements in discrete time. Part I: Completeness results

Justification announcements in discrete time. Part I: Completeness results Abstract We axiomatize stit logic of justification announcements (JA-STIT) interpreted over models with discrete time structure. The resulting Hilbert-style axiomatic system ΣD is strictly stronger than the system Σ presented in the study by Olkhovikov (2018, J. Logic Comput., to appear) as axiomatization of JA-STIT over the general class of justification stit models. We then prove a rather general strong completeness theorem for ΣD. 1 Introduction The so-called stit logic of justification announcements (JA-STIT, for short) was introduced in [6] as an explicit fragment of the richer logic of E-notions defined in [8]. The underlying idea was to interpret the proving activity as an activity that results in presenting (or, as it were, demonstrating) proofs to the community, thus making them publicly available within this community. In this way, the JA-STIT model of proving activity involves the following three main components: (i) the agents presenting the proofs to the community, (ii) the proofs to be presented and (iii) an interface for the interaction of agents with proofs. JA-STIT borrows its picture of (i) from stit logic by N. Belnap et al. [4] and its picture of (i) from epistemic justification logic by S. Artemov et al. [2]. This logic, therefore, retains the full set of expressive means of the two abovementioned logics, which places it in the landscape of justification stit (or jstit, for short) logics explored from the conceptual viewpoint in [7] and [8]. As for (iii), the third main component above, the underlying intuitions here are an idealized and abstracted version of a (not uncommon) situation when a group of agents tries to produce a proof working on a common whiteboard. This whiteboard then is their main medium for making different possible proofs epistemically transparent to themselves and their colleagues. If in this setting we abstract away (i) from the other available media (like private notes, private messages, etc.), (ii) from the natural limitations of the actual whiteboard (like limited space and necessity to erase old proofs) and (iii) from the natural limitations of the agents’ communication capacities (like bad handwriting on the part of presenting agents or short sightedness on the part of spectators) we get the following picture for (iii): A common pool of presented proofs is supposed to exist; This pool is the unique medium connecting the realm of abstract proofs to the world in which agents act; This pool can hold an unlimited number of proofs; Once a proof is in the pool, it stays there forever; Once a proof is in the pool, it is immediately understood and recognized by every agent in the community. The state of this pool, or of the community whiteboard, as we will sometimes call it, is described in JA-STIT by modality Et, where t is an arbitrary proof polynomial of justification logic. The informal interpretation of Et is that the proof t is presented to the community, or that t is on the whiteboard. This reading also explains the choice of E as notation for this modality since it serves as a sort of existence predicate for the pool of proofs publicly announced within the community. The axiomatization of JA-STIT w.r.t. the full class of its intended models was given in [6]. At the same time, Proposition 1 of [6] showed that, rather surprisingly, this axiomatization is sensitive to the temporal structure of the underlying models, even though none of the modalities available within JA-STIT can look outside a given moment, and the standard temporal modalities seem to be undefinable. Nevertheless, it turned out that once the class of underlying models is restricted to the models based on discrete time, the axiomatization is no longer complete. This situation raises the natural question of whether it is possible to axiomatize the set of JA-STIT validities over the class of models with discrete temporal substructure. This first part of the present paper answers this question by providing the respective axiomatization and showing it to be strongly complete. The layout of the rest of the paper is as follows. In Section 2 we define the language and the semantics of the logic at hand. We then define the notion of a stit frame and consider some natural subclasses of stit frames, which will be relevant to the main part of the paper and will appear in the results presented in Section 5. We briefly discuss set-theoretical relations between these classes. Our axiomatization itself is then presented in Section 3. We show this system to be sound w.r.t. the classes introduced in Section 2, both in its minimal form and modulo the so-called constant specifications inherited by JA-STIT from justification logic. We also define an alternative axiomatization of this same system in Lemma 6. Section 4 then contains the bulk of technical work necessary for the completeness of the presented axiomatization. It provides a stepwise construction and adequacy check for all the numerous components of the canonical model and ends with a proof of a truth lemma. Section 5 gives a concise proof of the completeness result for most of the subclasses of frames defined in Section 2 and draws some standard corollaries. Section 6 then formulates some conclusions and contains an announcement of the contents of Part II. The final section is devoted to acknowledgments. In what follows we will be assuming, due to space limitations, a basic acquaintance with both stit logic and justification logic. We recommend to peruse [5, Ch. 2] for a quick introduction into the basics of stit logic, and [1] for the same w.r.t. justification logic. 2 Basic definitions and notation We fix some preliminaries. First, we choose a finite non-empty set Ag disjoint from all the other sets to be defined below. Individual agents from this set will be denoted by letters i and j. Then we fix countably infinite sets PV ar of proof variables (denoted by x, y, z) and PConst of proof constants (denoted by c, d). When needed, subscripts and superscripts will be used with the above notations or any other notations to be introduced in this paper. Set Pol of proof polynomials is then defined by the following BNF:1 $$t := x \mid c \mid s + t \mid s \cdot t \mid !t,$$ with x ∈ PV ar, c ∈ PConst and s, t ranging over elements of Pol. In the above definition + stands for the sum of proofs, ⋅ denotes application of its left argument to the right one and ! denotes the so-called proof checker, so that !t checks the correctness of proof t. In order to define the set FormAg of formulas we fix a countably infinite set V ar of propositional variables to be denoted by letters p, q. Formulas themselves will be denoted by letters A, B, C, D, and the definition of FormAg is supplied by the following BNF: $$A := p \mid A \wedge B \mid \neg A \mid [j]A \mid \Box \ A \mid t{:} A \mid K\ A \mid Et,$$ with p ∈ V ar, j ∈ Ag and t ∈ Pol. It is clear from the above definition of FormAg that we are considering a version of modal propositional language. As for the informal interpretations of modalities, $$\big [j\big ]A$$ is the so-called cstit action modality and $$\Box$$ is the historical necessity modality, both modalities are borrowed from stit logic. The next two modalities, K A and t:A, come from epistemic justification logic and the latter is interpreted as ‘t proves A’, whereas the former is the strong epistemic modality ‘A is known’. We assume ◊ as notation for the dual modality of $$\Box$$ and ⟨K⟩ for the dual of K. As usual, ω will denote the set of natural numbers including 0, ordered in the natural way. For the language at hand, we assume the following semantics. A justification stit (jstit) model for Ag is a structure $$\mathcal{M} = \big\langle Tree, \unlhd, Choice, Act, R, R_{e}, \mathcal{E}, V\big\rangle$$ such that Tree is a non-empty set. Elements of Tree are called moments. $$\unlhd$$ is a partial order on Tree for which a temporal interpretation is assumed. We will also freely use notations like $$\unrhd$$, $$\lhd$$ and $$\rhd$$ to denote the inverse relation and the irreflexive companions.2 Hist is a set of maximal chains in Tree w.r.t. $$\unlhd$$. Since Hist is completely determined by Tree and $$\unlhd$$, it is not included into the structure of model as a separate component. Elements of Hist are called histories. The set of histories containing a given moment m will be denoted Hm. The following set $$MH(\mathcal{M}) = \big\{ \big(m,h\big)\mid m \in Tree,\, h \in H_{m} \big\},$$ called the set of moment-history pairs, will be used to evaluate formulas of the above language. Choice is a function mapping Tree × Ag into $$2^{2^{Hist}}$$ in such a way that for any given j ∈ Ag and m ∈ Tree we have as $$Choice\big (m,j\big )$$ (to be denoted as $$Choic{e^{m}_{j}}$$ below) a partition of Hm. Intuitively, the sets of the form $$Choic{e^{m}_{j}}$$ may be thought of as the choice cells of agent j at moment m, the idea being that j cannot distinguish by her or his activity at m between histories that belong to one and the same choice cell. The histories in a choice cell from $$Choic{e^{m}_{j}}$$ are choice-equivalent for j at m. For a given h ∈ Hm we will denote by $$Choic{e^{m}_{j}}\big (h\big )$$ the element of partition $$Choic{e^{m}_{j}}$$ containing h. Act is a function mapping $$MH(\mathcal{M})$$ into 2Pol. R and Re are two pre-orders on Tree giving two versions of epistemic accessibility relation. They are assumed to be connected by inclusion R ⊆ Re. $$\mathcal{E}$$ is a function mapping Tree × Pol into $$2^{Form^{Ag}}$$ called admissible evidence function. V is an evaluation function, mapping the set V ar into $$2^{MH\left (\mathcal{M}\right )}$$. However, not all structures of the above described type are admitted as jstit models. A number of additional restrictions need to be satisfied. In order to facilitate the exposition of these restrictions, we introduce a couple of useful notations first. For a given m ∈ Tree and any given h, g ∈ Hm we stipulate that $$Act_{m} := \bigcap_{h \in H_{m}}\big(Act\left(m,h\right),$$ and $$h \approx_{m} g \Leftrightarrow h,g \in H_{m} \,\&\,\left(h = g \vee \left(\exists m^{\prime} \rhd m\right)\left(h, g \in H_{m^{\prime}}\right)\right).$$ Whenever we have h ≈mg, we say that h and g are undivided at m. We then demand satisfaction of the following constraints by the jstit models: Historical connection: $$\left(\forall m,m_{1} \in Tree\right)\left(\exists m_{2} \in Tree\right)\left(m_{2} \unlhd m \wedge m_{2} \unlhd m_{1}\right).$$ No backward branching: $$\left(\forall m,m_{1},m_{2} \in Tree\right)\left(\left(m_{1} \unlhd m \wedge m_{2} \unlhd m\right) \Rightarrow \left(m_{1} \unlhd m_{2} \vee m_{2} \unlhd m_{1}\right)\right).$$ No choice between undivided histories: $$\left(\forall m \in Tree\right)\left(\forall h,h^{\prime} \in H_{m}\right)\left(h \approx_{m} h^{\prime} \Rightarrow Choic{e^{m}_{j}}\left(h\right) = Choic{e^{m}_{j}}\left(h^{\prime}\right)\right)$$ for every j ∈ Ag. Independence of agents: $$\left(\forall m\in Tree\right)\left(\forall f:Ag \rightarrow 2^{H_{m}}\right)\left(\left(\forall j \in Ag\right)\left(f\big(j\big) \in Choic{e^{m}_{j}}\right) \Rightarrow \bigcap_{j \in Ag}f\big(j\big) \neq \emptyset\right).$$ Monotonicity of evidence: $$\left(\forall t \in Pol\right)\left(\forall m,m^{\prime} \in Tree\right)\left(R_{e}\left(m,m^{\prime}\right) \Rightarrow \mathcal{E}\left(m,t\right) \subseteq \mathcal{E}\left(m^{\prime},t\right)\right).$$ Evidence closure properties. For arbitrary m ∈ Tree, s, t ∈ Pol and A, B ∈ FormAg it is assumed that (a) $$A \rightarrow B \in \mathcal{E}(m,s) \wedge A \in \mathcal{E}(m,t) \Rightarrow B \in \mathcal{E}(m,s\cdot t)$$; (b) $$\mathcal{E}(m,s) \cup \mathcal{E}(m,t) \subseteq \mathcal{E}(m,s + t)$$; (c) $$A \in \mathcal{E}(m,t) \Rightarrow t{:} A \in \mathcal{E}(m,!\!t)$$. Expansion of presented proofs: $$\left(\forall m,m^{\prime} \in Tree\right)\left(m^{\prime} \lhd m \Rightarrow \forall h \in H_{m} \left(Act\left(m^{\prime},h\right) \subseteq Act\left(m,h\right)\right)\right).$$ No new proofs guaranteed: $$\left(\forall m \in Tree\right)\left(Act_{m} \subseteq \bigcup_{m^{\prime} \lhd m, h \in H_{m}}\left(Act\left(m^{\prime},h\right)\right)\right).$$ Presenting a new proof makes histories divide: $$\left(\forall m \in Tree\right)\left(\forall h,h^{\prime} \in H_{m}\right)\left(h \approx_{m} h^{\prime} \Rightarrow \left(Act\left(m,h\right) = Act\left(m,h^{\prime}\right)\right)\right).$$ Future always matters: $$\unlhd \subseteq R.$$ Presented proofs are epistemically transparent: $$\left(\forall m,m^{\prime} \in Tree\right)\left(R_{e}\left(m,m^{\prime}\right) \Rightarrow \left(Act_{m} \subseteq Act_{m^{\prime}}\right)\right).$$ We offer some intuitive explanation for the abovedefined notion of jstit model. Due to space limitations, we only explain the intuitions behind jstit models very briefly, and we urge the reader to consult [7, Section 3] for a more comprehensive explanations, whenever needed. The components like Tree, $$\unlhd$$, Choice and V are inherited from stit logic, whereas R, Re and $$\mathcal{E}$$ come from epistemic justification logic. The only new component is Act which represents the abovementioned common pool of proofs demonstrated to the community or the state of the community whiteboard at any given moment under a given history. When interpreting Act, we invoke the classical stit distinction between dynamic (agentive) and static (moment determinate) entities, assuming that the presence of a given proof polynomial t on the community whiteboard only becomes an accomplished fact at m when t is present in $$Act\left (m,h\right )$$ for every h ∈ Hm. On the other hand, if t is in Act(m, h) only for some h ∈ Hm this means that t is rather in a dynamic state of being presented, rather than being present, to the community. The numbered list of semantical constraints above then just builds on these intuitions. Constraints 1–4 are borrowed from stit logic, constraints 5 and 6 are inherited from justification logic. Constraint 7 just says that nothing gets erased from the whiteboard, constraint 8 says a new proof cannot spring into existence as a static (i.e. moment determinate) feature of the environment out of nothing, but rather has to come as a result (or a by-product) of a previous activity. Constraint 9 is just a corollary to constraint 3 in the richer environment of jstit models, constraint 10 says that the possible future of the given moment is always epistemically relevant in this moment and constraint 11 says that the community knows everything that has firmly made its way onto the whiteboard. For the members of FormAg, we will assume the following inductively defined satisfaction relation. For every jstit model $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e}, \mathcal{E}, V\big \rangle$$ and for every $$\left (m,h\right ) \in MH\left (\mathcal{M}\right )$$ we stipulate that \begin{align*} &\mathcal{M}, m, h \models p \Leftrightarrow \big(m,h\big) \in V\big(p\big);\\ &\mathcal{M}, m, h \models \big[j\big]\,A \Leftrightarrow \left(\forall h^{\prime} \in Choic{e^{m}_{j}}\big(h\big)\right)\left(\mathcal{M}, m, h^{\prime} \models A\right);\\ &\mathcal{M}, m, h \models \Box\, A \Leftrightarrow \left(\forall h^{\prime} \in H_{m}\right)\left(\mathcal{M}, m, h^{\prime} \models A\right);\\ &\mathcal{M}, m, h \models K\ A \Leftrightarrow \forall m^{\prime}\forall h^{\prime}\left(R\left(m,m^{\prime}\right) \& h^{\prime} \in H_{m^{\prime}} \Rightarrow \mathcal{M}, m^{\prime}, h^{\prime} \models A\right);\\ &\mathcal{M}, m, h \models t{:} A \Leftrightarrow A \in \mathcal{E}\left(m,t\right) \& \forall m^{\prime}\forall h^{\prime}\left(R_{e}\left(m,m^{\prime}\right)\! \& h^{\prime} \in H_{m^{\prime}} \Rightarrow \mathcal{M}, m^{\prime}, h^{\prime} \models A\right);\\ &\mathcal{M}, m, h \models Et \Leftrightarrow t \in Act\big(m,h\big). \end{align*} In the above clauses we assume that p ∈ Var; we also assume standard clauses for Boolean connectives. In addition to jstit models, we will need a notion of stit frame. If $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e},$$$$\mathcal{E}, V\big \rangle$$ is a jstit model for Ag, then $$C = \big \langle Tree, \unlhd , Choice\big \rangle$$ is a stit frame for Ag.3 In this case, $$\mathcal{M}$$ is said to be based on C. Given a class $$\mathcal{C}$$ of stit frames, we will denote the class of jstit models based on the frames from $$\mathcal{C}$$ by $$Mod(\mathcal{C})$$. We also observe that notations like Hist(C) and MH(C) still make perfect sense when C is a stit frame. The following lemmas collect some very basic technical facts which will be used in what follows: Lemma 1 Let $$C = \big \langle Tree, \unlhd , Choice \big \rangle$$ be a stit frame. Then (1) $$\big (\forall m \in Tree\big )\big (\forall h \in H_{m}\big )\big (\big \{ m_{1} \in Tree \mid m_{1} \unlhd m \big \} \subseteq h\big )$$; (2) $$\big (\forall m \in Tree\big )\big (\forall h,g \in H_{m}\big )\big (h \neq g \Rightarrow \big (\exists m^{\prime} \rhd m\big )\big (h \in H_{m^{\prime}}\big )\big )$$; (3) $$\big (\forall m, m^{\prime} \in Tree\big )\big (m \unlhd m^{\prime} \Rightarrow H_{m^{\prime}} \subseteq H_{m}\big )$$; (4) ≈m is an equivalence relation on Hm for every m ∈ Tree. Proof. (Part 1). We clearly have m ∈ h. Consider an arbitrary $$m_{1} \lhd m$$. Then h ∪ {m1} must be an $$\unlhd$$-chain. Indeed, if m′ ∈ h then either $$m \unlhd m^{\prime}$$ or $$m^{\prime} \lhd m$$. In the former case we get $$m_{1} \unlhd m^{\prime}$$ by transitivity of $$\unlhd$$, in the latter case we get $$m_{1} \unlhd m^{\prime} \vee m^{\prime} \unlhd m_{1}$$ by the absence of backward branching. But since h is a maximal chain, this means that we must have m1 ∈ h. (Part 2). To obtain a contradiction, assume that h, g ∈ Hm are different, but we have $$\big(\forall m^{\prime} \rhd m\big)\big(m^{\prime} \notin h\big)$$ (1) Given that every two elements of h must be $$\unlhd$$-comparable, this means that $$h \subseteq \big \{ m_{1} \in Tree \mid m_{1} \unlhd m \big \}$$ and, by Part 1, that $$h = \big \{ m_{1} \in Tree \mid m_{1} \unlhd m \big \}$$. Note that Part 1 also entails that $$g \supseteq \big \{ m_{1} \in Tree \mid m_{1} \unlhd m \big \}$$, so that in this case we must have $$g \supseteq h$$. We can have neither $$g \supset h$$, since this would violate the maximality of h, nor g = h, since this is in contradiction with our assumption. Therefore, (1) must be false. (Part 3). Immediately by the absence of backward branching. (Part 4). Reflexivity and symmetry are straightforward. We establish transitivity. Let m ∈ Tree and let h, h$$^{\prime}$$, h$$^{\prime\prime}$$ ∈ Hm are such that h ≈mh$$^{\prime}$$ and h$$^{\prime}$$ ≈mh$$^{\prime\prime}$$. If h = h$$^{\prime}$$ or h$$^{\prime}$$ = h$$^{\prime\prime}$$, then h ≈ h$$^{\prime\prime}$$ follows trivially. Therefore, assume that both $$h \neq h^{\prime}$$ and $$h^{\prime} \neq h^{\prime\prime}$$. Then choose $$m^{\prime}, m^{\prime\prime} \rhd m$$ such that m$$^{\prime}$$ ∈ h ∩ h$$^{\prime}$$ and m$$^{\prime\prime}$$ ∈ h$$^{\prime}$$ ∩ h$$^{\prime\prime}$$. Then both m$$^{\prime}$$ and m$$^{\prime\prime}$$ are in h$$^{\prime}$$ and hence must be $$\unlhd$$-comparable. Setting m0 as the minimal of the two moments, we get that, by Part 3, $$h, h^{\prime}, h^{\prime\prime} \in H_{m_{0}}$$ and $$m_{0} \rhd m$$, whence h ≈mh$$^{\prime\prime}$$. Lemma 2 Let $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e}, \mathcal{E}, V \big \rangle$$ be a jstit model. Then $$\big(\forall m,m^{\prime} \in Tree\big)\big(\forall h \in H_{m^{\prime}}\big)\big(\forall t \in Pol\big)\big(m \lhd m^{\prime} \& t\in Act\big(m,h\big) \Rightarrow t \in Act_{m^{\prime}}\big).$$ Proof. Let m, m$$^{\prime}$$ ∈ Tree, h ∈ Hm′ and t ∈ Pol be such that $$m \lhd m^{\prime} \& t\in Act\big (m,h\big )$$. Choose an arbitrary g ∈ Hm′. By Lemma 1.3, we know that g ∈ Hm and that also h ∈ Hm. Therefore, h ≈mg whence by the presenting a new proof makes histories divide constraint we get that $$Act\big (m,h\big ) = Act\big (m,g\big )$$. Therefore, $$t \in Act\big (m,g\big )$$ and, by the expansion of presented proofs, $$t \in Act\big (m^{\prime},g\big )$$. Since g ∈ Hm′ was chosen arbitrarily, this means that t ∈ Actm′. An important subclass of jstit models is made up of what we will call unirelational jstit models. These are the models satisfying the additional constraint that Re ⊆ R. It is known (see [2, Comment 6.5]) that switching from the full class of models to the unirelational models in the context of pure epistemic justification logic still leaves us with a class of models w.r.t. which the logic is complete and we have shown in [6] that the same holds for JA-STIT. In this paper, we will eventually show that this situation does not change if the class of intended models for JA-STIT is restricted to any of its natural subclasses that we define within the present section.4 Since the class of unirelational jstit models will be thus important to us in what follows, we observe that switching to unirelational models allows to somewhat simplify the whole setting. First, one can simply throw away Re from the model structure and redefine the model as a tuple of the form $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, \mathcal{E}, V\big \rangle$$, as long as one also updates the semantical constraints, replacing Re everywhere with R. The same simplification can be introduced to the definition of jstit frame. Second, one can somewhat simplify the satisfaction clause for t: A replacing it with: $$\mathcal{M}, m, h \models t{:} A \Leftrightarrow A \in \mathcal{E}(m,t) \& \mathcal{M}, m, h \models KA.$$ Moreover, whenever $$\mathcal{C}$$ is a class of stit frames we will denote by $$Mod^{\downarrow }(\mathcal{C})$$ the class of unirelational jstit models based on frames from $$\mathcal{C}$$. Before we move on, we need to introduce the notation for an immediate $$\lhd$$-successor of a given moment as it will play an important part in the frame restrictions to be introduced below. So whenever $$C = \big \langle Tree, \unlhd , Choice\big \rangle$$ is a stit frame and m, m$$^{\prime}$$ ∈ Tree, we set that $$Next\left(m,m^{\prime}\right) \Leftrightarrow \left(m \lhd m^{\prime} \& \left(\forall m^{\prime\prime} \lhd m^{\prime}\right)\left(m^{\prime\prime} \unlhd m\right)\right).$$ We now list the defining conditions for some natural subclasses of stit frames: Definition 1 Let $$C = \big \langle Tree, \unlhd , Choice\big \rangle$$ be a stit frame. Then we say that C is based on discrete time iff every history in Hist(C) can be embedded into an initial (but maybe improper) segment of ω; C is a well-ordered frame iff every h ∈ Hist(C) is well ordered by $$\unlhd$$. C is a successor frame iff the following holds: $$\big(\forall m,m_{1} \in Tree\big)\big(m \lhd m_{1} \Rightarrow \big(\exists m_{2} \unlhd m_{1}\big)\big(Next\big(m, m_{2}\big)\big)$$ (succ) C is a colinear frame iff $$\big (\forall m \in Tree\big )\big (\forall h,g \in H_{m}\big )\big (h \approx _{m} g\big )$$. C is a mixed successor frame iff for all m, m1 ∈ Tree it is true that $$\big[m \lhd m_{1} \Rightarrow \big(\exists m_{2} \unlhd m_{1}\big)\big(Next\big(m, m_{2}\big)\big)\big] \vee \big[\big(\forall h,g \in H_{m}\big)\big(h \approx_{m} g\big)\big]$$ (mixsucc) Every property in the above definition induces the corresponding class of stit frames. We will denote the class of stit frames based on discrete time, the class of well-ordered stit frames, the class of successor stit frames, the class of colinear stit frames, and the class of mixed successor stit frames by $$\mathcal{C}_{discr}$$, $$\mathcal{C}_{wo}$$, $$\mathcal{C}_{succ}$$, $$\mathcal{C}_{col}$$ and $$\mathcal{C}_{mixsucc}$$, respectively. Further, whenever $$\mathcal{C}$$ is a class of stit frames we let $$\mathcal{C}^{Ag}$$ denote the class of all frames in $$\mathcal{C}$$ for a given community Ag. The following lemma sums up some quick facts about the defined classes: Lemma 3 Let Ag be a community of agents. Then $$\mathcal{C}^{Ag}_{discr} \subset \mathcal{C}^{Ag}_{wo} \subset \mathcal{C}^{Ag}_{succ}$$; $$\mathcal{C}^{Ag}_{succ} \cup \mathcal{C}^{Ag}_{col} \subset \mathcal{C}^{Ag}_{mixsucc}$$. Proof. Part 1 is straightforward. As for Part 2, it is clear that we must have $$\mathcal{C}^{Ag}_{succ} \cup \mathcal{C}^{Ag}_{col} \subseteq \mathcal{C}^{Ag}_{mixsucc}$$. To see that this inclusion is proper, consider for instance the stit frame $$C = \big \langle Tree, \unlhd , Choice\big \rangle$$ for Ag, where Tree = {1/2n∣n ∈ ω} ∪ {0, −1, a}, $$\unlhd = \leq \cup \{ (-1, a), (a,a) \}$$ (with ≤ being the natural order on rational numbers) and $$Choic{e^{m}_{j}} = H_{m}$$ for all m ∈ Tree and j ∈ Ag. It is easy to see that $$Hist(C) = \big \{ h_{1}, h_{2} \big \}$$, where h1 = (−1, a) and h2 = (−1, 0, …, 1/2n, …, 1/2, 1). Since h1 ≉−1h2, we get that $$C \notin \mathcal{C}^{Ag}_{col}$$. Further, we have $$0 \lhd 1$$, but there exists no minimal $$\lhd$$-successor of 0. Therefore, $$C \notin \mathcal{C}^{Ag}_{succ}$$ as well. On the other hand, the only moment in Tree for which (succ) is violated is 0, and we clearly have H0 = {h2}, therefore $$C \in \mathcal{C}^{Ag}_{mixsucc}$$. Note that we have a natural hierarchy of restrictions for the interval $$\big (\mathcal{C}_{discr},\mathcal{C}_{succ}\big )$$, whereas $$\mathcal{C}_{col}$$ uses a different principle and the construction of $$\mathcal{C}_{mixsucc}$$ looks artificial. The reason for our mentioning $$\mathcal{C}_{mixsucc}$$ is that due to the limitations of expressive power of JA-STIT language the stit frame definability result to be given in Part II of this paper has to use $$\mathcal{C}_{mixsucc}$$ rather than the much more natural $$\mathcal{C}_{succ}$$. Before we move on to the next section, it is also worth noting that JA-STIT lacks finite model property in a rather strong sense since some satisfiable formulas cannot be satisfied over finite models, or even over infinite models where all histories are finite. The argument for this is the same as in implicit jstit logic, for which this claim was proved in [6] using formula $$K\big (\Diamond p \wedge \Diamond \neg p\big )$$ as an example. Here we add that, as a consequence of Lemma 3, this example is still valid if the full set of jstit models is restricted to $$Mod\big (\mathcal{F}\big )$$ whenever $$\mathcal{C} \in \big \{ \mathcal{C}_{discr}, \mathcal{C}_{wo}, \mathcal{C}_{succ}, \mathcal{C}_{mixsucc} \big \}$$, since the model used in [6] for $$K\big (\Diamond p \wedge \Diamond \neg p\big )$$ extends a stit frame based on discrete time. 3 Axiomatic system and soundness From this moment on, Ag will refer to an arbitrary but fixed agent community. We start by defining a Hilbert-style axiomatic system ΣD. For this system, we fix the following set of axiomatic schemes: $$\text{A full set of axioms for classical propositional logic}$$ (A0) $$S5\text{ axioms for }\Box \text{ and }\big[j\big]\text{ for every }j \in Ag$$ (A1) $$\Box A \rightarrow \big[j\big]\,A \text{ for every }j \in Ag$$ (A2) $$\big(\Diamond\big[j_{1}\big]\,A_{1} \wedge\ldots \wedge \Diamond\big[j_{n}\big]\,A_{n}\big) \rightarrow \Diamond\big(\big[j_{1}\big]\,A_{1} \wedge\ldots \wedge\big[j_{n}\big]\,A_{n}\big)$$ (A3) $$(s{:}(A \rightarrow B) \rightarrow (t{:} A \rightarrow (s\cdot t){:} B)$$ (A4) $$t{:} A \rightarrow (!t{:}(t{:} A) \wedge K\ A)$$ (A5) $$(s{:} A \vee t{:} A) \rightarrow (s+t){:} A$$ (A6) $$S4\text{ axioms for }K$$ (A7) $$K\ A \rightarrow \Box K\Box \,A$$ (A8) $$\Box \, Et \rightarrow K\Box \,Et$$ (A9) \begin{align} &K\big(\neg\Box\, Et_{1} \vee\ldots\vee\neg\Box \, Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k}\big) \rightarrow\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\rightarrow\big(\neg Et_{1} \vee\ldots\vee\neg Et_{n}\vee Es_{1} \vee\ldots\vee Es_{k}\big) \end{align} (A10D) The assumption is that in (A3) j1, …, jn are pairwise different. The rules of inferences are then as follows: $$\text{From }A, A \rightarrow B \text{ infer } B;$$ (R1) $$\!\!\!\!\!\!\!\!\!\!\!\!\! \text{From }A\text{ infer }K\ A.$$ (R2) The axiomatic schemes (A0)–(A9), as well as the rules (R1)–(R2), are borrowed from the system Σ defined in [6] and retain their original names. The axiom (A10D) is a strengthening of the axiom (A10) from Σ. ΣD is a minimal system in which we make no assumptions as to the properties of proof constants. One standard way to extend a minimal system in the pure justification logic is to add a number of such assumptions. Normally, such extensions are called for to ensure that enough proofs for the axioms of the system are available. Although sets of such assumptions (also called constant specifications) are not valid in an arbitrary jstit model and thus require some narrowing of the class of intended models, the corresponding narrowing is rather straightforward and the completeness proof for the minimal system can be very easily accommodated to any such constant specification. In this way, an infinite number of systems extending the minimal system with a constant specification is taken on board in one sweeping move. In JA-STIT the situation is very similar, which allows us to give a somewhat more general form to our completeness proofs. More precisely, let us call a constant specification any set $$\mathcal{CS}$$ such that $$\mathcal{CS} \subseteq \{ c_{n}{:}\ldots c_{1}{:} A\mid c_{1},\ldots ,c_{n} \in PConst, A \text{ an instance of }$$ (A0)–(A10D)}; Whenever $$c_{n+1}{:} c_{n}{:}\ldots c_{1}{:} A \in \mathcal{CS}$$, then also $$c_{n}{:}\ldots c_{1}{:} A \in \mathcal{CS}$$. For a given constant specification, we can define the corresponding inference rule $$R_{\mathcal{CS}}$$ as follows: ($$R_{\mathcal{CS}}$$) $$\text{If }c_{n}{:}\ldots c_{1}{:} A \in \mathcal{CS},\text{ infer } c_{n}{:}\ldots c_{1}{:} A.)$$ We now define that $$\Sigma _{D}\left (\mathcal{CS}\right )$$ is just ΣD extended with the rule ($$R_{\mathcal{CS}}$$). Since ∅ is clearly one example of constant specification, we have that $$\Sigma _{D}\big (\emptyset \big ) = \Sigma _{D}$$ so that our initial axiomatic system is also in the class of systems of the form $$\Sigma _{D}(\mathcal{CS})$$. However, when $$\mathcal{CS} \neq \emptyset$$, the corresponding system $$\Sigma _{D}(\mathcal{CS})$$ will prove some formulas which are not valid even if we restrict our attention to jstit models based on any class of frames defined in Section 2. We therefore have to describe the restriction on jstit models which comes with a commitment to a given $$\mathcal{CS}$$. We then say that a jstit model $$\mathcal{M}$$ is $$\mathcal{CS}$$-normal iff it is true that $$\big(\forall c \in PConst\big)\big(\forall m \in Tree\big)\big(\big\{ A \mid c{:} A\in \mathcal{CS} \big\} \subseteq \mathcal{E}(m,c)\big),$$ where $$\mathcal{E}$$ is the $$\mathcal{M}$$s admissible evidence function. Again, it is easy to see that the class of ∅-normal jstit models is just the whole class of jstit models so that the representation $$\Sigma _{D}\big (\emptyset \big ) = \Sigma _{D}$$ does not place any additional restrictions on the class of intended models of ΣD. Whenever $$\mathcal{C}$$ is a class of stit frames, we will denote the class of $$\mathcal{CS}$$-normal (resp. unirelational) jstit models based on the frames from $$\mathcal{C}$$ by $$Mod_{\mathcal{CS}}\left (\mathcal{C}\right )$$ (resp. $$Mod^{\downarrow }_{\mathcal{CS}}(\mathcal{C})$$). We then define a proof in $$\Sigma _{D}\left (\mathcal{CS}\right )$$ as a finite sequence of formulas such that every formula in it is either an axiom or is obtained from earlier elements of the sequence by one of inference rules. A proof is a proof of its last formula. If an A ∈ FormAg is provable in $$\Sigma _{D}(\mathcal{CS})$$, we will write $$\vdash _{\mathcal{CS}} A$$. We say that Γ ⊆ FormAg is inconsistent in $$\Sigma _{D}(\mathcal{CS})$$ (or $$\mathcal{CS}$$-inconsistent) iff for some A1, …, An ∈ Γ we have $$\vdash _{\mathcal{CS}} (A_{1} \wedge \ldots \wedge A_{n}) \rightarrow \bot$$, and we say that Γ is consistent in $$\Sigma _{D}(\mathcal{CS})$$ (or $$\mathcal{CS}$$-consistent) iff it is not inconsistent in $$\Sigma _{D}(\mathcal{CS})$$. Γ is maxiconsistent in $$\Sigma _{D}(\mathcal{CS})$$ (or $$\mathcal{CS}$$-maxiconsistent) iff it is consistent in $$\Sigma _{D}(\mathcal{CS})$$ and no subset of FormAg, which is consistent in $$\Sigma _{D}(\mathcal{CS})$$, properly extends Γ. With this definition of inconsistency, we can do many familiar things, e.g. extend consistent sets with new formulas and eventually make them maxiconsistent. More precisely, the following lemma holds: Lemma 4 Let $$\mathcal{CS}$$ be a constant specification, let Γ ⊆ FormAg be consistent in $$\Sigma _{D}\left (\mathcal{CS}\right )$$ and let A, B ∈ FormAg. Then There exists a Δ ⊆ FormAg such that Δ is $$\mathcal{CS}$$-maxiconsistent and Γ ⊆ Δ. If Γ is $$\mathcal{CS}$$-maxiconsistent, then exactly one element of {A, ¬A} is in Γ. If Γ is $$\mathcal{CS}$$-maxiconsistent, then A ∨ B ∈ Γ iff (A ∈ Γ or B ∈ Γ). If Γ is $$\mathcal{CS}$$-maxiconsistent and A, (A → B) ∈ Γ, then B ∈ Γ. If Γ is $$\mathcal{CS}$$-maxiconsistent, then A ∧ B ∈ Γ iff (A ∈ Γ and B ∈ Γ). Proof. (Part 1) Just as in the standard case, we enumerate the elements of FormAg as A1, …, An, … and form the sequence of sets Γ1, …, Γn, …, such that Γ1 := Γ and for every natural i ≥ 1: $$\Gamma_{i + 1} := \begin{cases} \Gamma_{i}, & \text{ if }\Gamma_{i} \cup \big\{ A_ i \big\}\text{ is }\mathcal{CS}\text{-inconsistent;} \\ \Gamma_{i} \cup \big\{ A_ i \big\}, & \text{ otherwise.} \\ \end{cases}$$ We now define $$\Delta := \bigcup _{i \geq 1}\Gamma _{i}$$. Of course, we have Γ ⊆ Δ, and moreover, Δ is $$\mathcal{CS}$$-maxiconsistent. To see this, note that for every i ≥ 1 the set Γi is $$\mathcal{CS}$$-consistent by construction. Now, if Δ is $$\mathcal{CS}$$-inconsistent, then there must be a $$\Sigma _{D}\left (\mathcal{CS}\right )$$-valid implication from a finite conjunction of formulas in Δ to ⊥. These formulas must be mentioned in our numeration of FormAg so that the valid implication in question can presented as $$\vdash _{\mathcal{CS}} \left (A_{i_{1}} \wedge \ldots \wedge A_{i_{n}}\right ) \rightarrow \bot$$ for appropriate natural i1, …, in. Since all of $$A_{i_{1}}, \ldots , A_{i_{n}}$$ are in Δ, we must have, by the construction of Γ1, …, Γn, …, that $$A_{i_{1}}, \ldots , A_{i_{n}} \in \Gamma _{max\big (i_{1},\ldots , i_{n}\big )+1}$$. But then this latter set must be $$\mathcal{CS}$$-inconsistent which contradicts our construction. Further, if some $$\mathcal{CS}$$-consistent Ξ ⊆ FormAg is such that Δ ⊂ Ξ, then let An ∈ Ξ∖Δ. We must have then $$\Gamma _{n} \cup \big \{ A_{n} \big \}$$$$\mathcal{CS}$$-inconsistent, but we also have $$\Gamma _{n} \cup \big \{ A_{n} \big \} \subseteq \Xi$$, which implies $$\mathcal{CS}$$-inconsistency of Ξ, in contradiction to our assumptions. Therefore, Δ is not only $$\mathcal{CS}$$-consistent, but also $$\mathcal{CS}$$-maxiconsistent. (Part 2) We cannot have both A and ¬A in Γ, since we have, of course, $$\vdash _{\mathcal{CS}} (A \wedge \neg A) \rightarrow \bot$$. If, on the other hand, neither A nor ¬A is in Γ, then both Γ ∪ {A} and Γ ∪ {¬A} must be $$\mathcal{CS}$$-inconsistent, so that for some B1, …, Bn ∈ Γ we will have $$\vdash_{\mathcal{CS}} \big(B_{1}\wedge \ldots\wedge B_{n} \wedge A\big) \rightarrow \bot,$$ whereas for some C1, …, Ck ∈ Γ we will have $$\vdash_{\mathcal{CS}} \big(C_{1}\wedge \ldots\wedge C_{k} \wedge \neg A\big) \rightarrow \bot,$$ whence we get, using (A0) and (R1), $$\vdash_{\mathcal{CS}} \big(C_{1}\wedge \ldots\wedge C_{k}\big) \rightarrow A,$$ and further $$\vdash_{\mathcal{CS}} \big(B_{1}\wedge \ldots\wedge B_{n} \wedge C_{1}\wedge \ldots\wedge C_{k}\big) \rightarrow \bot,$$ so that Γ turns out to be $$\mathcal{CS}$$-inconsistent, contrary to our assumptions. (Part 3) Assume (A ∨ B) ∈ Γ. If neither A nor B are in Γ, then, by Part 2, both ¬A and ¬B are in Γ. Using (A0) and (R1) we get that $$\vdash_{\mathcal{CS}} \big((A \vee B) \wedge \neg A \wedge \neg B\big) \rightarrow \bot,$$ showing that Γ is $$\mathcal{CS}$$-inconsistent, contrary to our assumptions. In the other direction, if say A ∈ Γ and (A ∨ B) ∉ Γ, then, by Part 2, we must have ¬(A ∨ B) ∈ Γ. Using (A0) and (R1) we get that $$\vdash_{\mathcal{CS}} \big(\neg(A \vee B) \wedge A\big) \rightarrow \bot,$$ showing again that Γ is $$\mathcal{CS}$$-inconsistent, contrary to our assumptions. The case when B ∈ Γ is similar. Parts 4 and 5 are similar to Part 3. Our goal is now to obtain, for any given constant specification $$\mathcal{CS}$$, a strong completeness theorem for $$\Sigma _{D}(\mathcal{CS})$$, and we start by establishing some soundness claims: Theorem 1 Let $$\mathcal{CS}$$ be an arbitrary constant specification. If A ∈ FormAg is such that $$\vdash _{\mathcal{CS}} A$$, then A is valid over the class $$Mod_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{mixsucc}\right )$$. Proof. Given the above notion of proof, it is sufficient to show that every instance of (A0)–(A10D) is valid over the class $$Mod_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{mixsucc}\right )$$, and every application of rules (R1), (R2) and ($$R_{\mathcal{CS}}$$) to formulas which are valid over $$Mod_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{mixsucc}\right )$$ yields a formula which is valid over the same class. First, note that if $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e}, \mathcal{E}, V\big \rangle$$ is a $$\mathcal{CS}$$-normal jstit model based on a jstit frame from $$\mathcal{C}^{Ag}_{mixsucc}$$, then $$\big \langle Tree, \unlhd , Choice, V\big \rangle$$ is a model of stit logic. Therefore, axioms (A0)–(A3), which were copy pasted from the standard axiomatization of dstit logic5 must be valid. Second, note that if $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e}, \mathcal{E}, V\big \rangle$$ is a $$\mathcal{CS}$$-normal jstit model, then $$\mathcal{M} = \big \langle Tree, R, R_{e}, \mathcal{E}, V\big \rangle$$ is what it is called in [2, Section 6] an AF-model with the form of constant specification given by $$\mathcal{CS}$$.6 This means that also all of the (A4)–(A7) must be valid, whereas (R1), (R2) and ($$R_{\mathcal{CS}}$$) must preserve validity, given that all these parts of our axiomatic system were borrowed from the standard axiomatization of the epistemic justification logic. The validity of other parts of $$\Sigma _{D}\left (\mathcal{CS}\right )$$ will be motivated below in some detail. In what follows, $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e},\mathcal{E}, V\big \rangle$$ will always stand for an arbitrary jstit model in $$Mod_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{mixsucc}\right )$$, and $$\big (m,h\big )$$ for an arbitrary element of $$MH\left (\mathcal{M}\right )$$. As for (A8), assume for reductio that $$\mathcal{M}, m,h \models KA \wedge \Diamond \big \langle K\big \rangle \Diamond \neg A$$. Then $$\mathcal{M}, m,h \models KA$$ and also $$\mathcal{M}, m,h^{\prime} \models \langle K\rangle \Diamond \neg A$$ for some h′∈ Hm. Therefore, there must be an m′∈ Tree and a g ∈ Hm′ such that both $$R\big (m,m^{\prime}\big )$$ and $$\mathcal{M}, m^{\prime},g \models \Diamond \neg A$$, whence we can choose a g$$^{\prime}$$ ∈ Hm′ such that $$\mathcal{M}, m^{\prime},g^{\prime} \not \models A$$. The latter is in obvious contradiction with $$\mathcal{M}, m,h \models KA$$. We consider next (A9). If $$\Box Et$$ is true at $$\big (m,h\big )$$ in $$\mathcal{M}$$, then, by definition, t ∈ Actm. Now, if m$$^{\prime}$$ ∈ Tree is such that $$R\big (m,m^{\prime}\big )$$, then, by R ⊆ Re we will have $$R_{e}\big (m,m^{\prime}\big )$$, and, by the epistemic transparency of presented proofs constraint, we must have t ∈ Actm′ so that for every g ∈ Hm′ we will have $$\mathcal{M},m^{\prime},g \models \Box Et$$. Therefore, we must have $$\mathcal{M},m,h \models K\Box Et$$ as well. It only remains to show the validity of (A10D), and we again use reductio ad absurdum. Assume that we have both $$\mathcal{M}, m, h \models K\big(\neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k}\big)$$ (2) and $$\mathcal{M}, m, h \models Et_{1} \wedge\ldots\wedge Et_{n}\wedge\neg Es_{1} \wedge\ldots\wedge\neg Es_{k}.$$ (3) Whence, by (2) and the validity of (A7), we know that also $$\mathcal{M}, m, h \models \neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k}.$$ (4) By validity of (A1), it follows from (3) that $$\mathcal{M}, m, h \models \neg\Box Es_{1} \wedge\ldots\wedge\neg\Box Es_{k}.$$ (5) Whence, by (4) and (5) we know that also $$\mathcal{M}, m, h \models \neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}.$$ (6) Therefore, we can choose a natural u such that 1 ≤ u ≤ n and $$\mathcal{M}, m, h \models \neg\Box Et_{u}.$$ The latter, in turn, means that for some h′∈ Hm we have that $$\mathcal{M}, m, h^{\prime} \models \neg Et_{u}.$$ (7) Comparison between (3) and (7) shows that $$Act\big (m,h\big ) \neq Act\big (m,h^{\prime}\big )$$, whence by the presenting new proof makes histories divide constraint we get that h ≉mh$$^{\prime}$$. Hence, we know that the second disjunct of condition (mixsucc) fails for m, so that we must have $$\big(\forall m_{1} \in Tree\big)\left(m \lhd m_{1} \Rightarrow \big(\exists m_{2} \unlhd m_{1}\big)\big(Next\big(m, m_{2}\big)\big)\right).$$ (8) Further, it follows from h ≉mh$$^{\prime}$$ that $$h \neq h^{\prime}$$. Therefore, by Lemma 1.2, we can choose in Tree some $$m^{\prime} \rhd m$$ such that m$$^{\prime}$$ ∈ h. By (8), we must be able to choose an m2 ∈ Tree such that both $$m_{2} \unlhd m^{\prime}$$ and $$Next\big (m, m_{2}\big )$$. So we consider such an m2. Given that m$$^{\prime}$$ ∈ h, we know, by Lemma 1.3, that $$h \in H_{m_{2}}$$. Therefore, it follows from (3) and Lemma 2 that $$t_{1},\ldots , t_{n} \in Act_{m_{2}}$$ or, equivalently, $$\mathcal{M}, m_{2}, h \models \Box Et_{1} \wedge\ldots \wedge \Box Et_{n}.$$ (9) Furthermore, by the future always matters constraint we know that $$R\big (m,m_{2}\big )$$, whence it follows, by (2), that $$\mathcal{M}, m_{2}, h \models \neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k}.$$ (10) Finally, choose an arbitrary r between 1 and k. If $$s_{r} \in Act_{m_{2}}$$, then, by the no new proofs guaranteed constraint, there must be some $$g \in H_{m_{2}}$$ and some $$m_{0} \lhd m_{2}$$ such that sr ∈ Act(m0, g). Then, by Lemma 1.3, g ∈ Hm, hence h ≈mg. Therefore, by the presenting a new proof makes histories divide constraint, $$Act\big (m, g\big ) = Act\big (m,h\big )$$. By $$Next\big (m, m_{2}\big )$$ we must have $$m_{0} \unlhd m$$, therefore, by the expansion of presented proofs, $$s_{r} \in Act\big (m,g\big )$$, whence also $$s_{r} \in Act\big (m,h\big )$$. But this plainly contradicts (3). Since 1 ≤ r ≤ k was chosen arbitrarily, this means that all of s1, …, sk are outside $$Act_{m_{2}}$$ so that we have $$\mathcal{M}, m_{2}, h \models \neg\Box Es_{1} \wedge\ldots \wedge \neg\Box Es_{k}.$$ (11) Taken together, (9)–(11) give us a plain contradiction. We are now prepared to formulate our completeness result: Theorem 2 Let Γ ⊆ FormAg and let $$\mathcal{C}$$ be a class of stit frames such that $$\mathcal{C}^{Ag}_{discr} \subseteq \mathcal{C} \subseteq \mathcal{C}^{Ag}_{mixsucc}$$. Then Γ is $$\mathcal{CS}$$-consistent iff it is satisfiable in $$Mod_{\mathcal{CS}}\left (\mathcal{C}\right )$$ iff it is satisfiable in $$Mod^{\downarrow }_{\mathcal{CS}}(\mathcal{C})$$. One part of the completeness results we have, of course, right away, as a consequence of Theorem 1: Corollary 1 Let Γ ⊆ FormAg and let $$\mathcal{C}$$ be a class of stit frames such that $$\mathcal{C}^{Ag}_{discr} \subseteq \mathcal{C} \subseteq \mathcal{C}^{Ag}_{mixsucc}$$. If Γ ⊆ FormAg is satisfiable in $$Mod_{\mathcal{CS}}\left (\mathcal{C}\right )$$ (thus also when it is satisfiable in $$Mod^{\downarrow }_{\mathcal{CS}}\left (\mathcal{C}\right )$$), then Γ is $$\mathcal{CS}$$-consistent. Proof. Let Γ ⊆ FormAg be satisfiable in $$Mod_{\mathcal{CS}}(\mathcal{C})$$ so that for some $$\mathcal{M} \in Mod_{\mathcal{CS}}(\mathcal{C})$$ and some $$\big (m,h\big ) \in MH\left (\mathcal{M}\right )$$ we have $$\mathcal{M}, m, h \models \Gamma$$. Then we must have $$\mathcal{M} \in Mod_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{mixsucc}\right )$$. If Γ were $$\mathcal{CS}$$-inconsistent this would mean that for some A1, …, An ∈ Γ we would have $$\vdash _{\mathcal{CS}} \big (A_{1} \wedge \ldots \wedge A_{n}\big ) \rightarrow \bot$$. By Theorem 1, this would mean that $$\mathcal{M}, m, h \models \big(A_{1} \wedge\ldots \wedge A_{n}\big) \rightarrow \bot,$$ whence clearly $$\mathcal{M}, m, h \models \bot$$, which is impossible. Therefore, Γ must be $$\mathcal{CS}$$-consistent. Before we move further, we prove some claims about provability in $$\Sigma _{D}\left (\mathcal{CS}\right )$$ to be used below: Lemma 5 Let $$\mathcal{CS}$$ be a constant specification. Then all of the following theorems and derived rules are provable in $$\Sigma _{D}\left (\mathcal{CS}\right )$$ for every A ∈ FormAg, t ∈ Pol, x, y ∈ PVar and j ∈ Ag: $$K\ A \rightarrow \Box A$$ (T0) ($$\text{Nec}\Box$$) $$\text{From }A\text{ infer }\Box A$$ $$\text{From }A\text{ infer }\big[j\big]A$$ (Nec[j]) $$t{:} A \rightarrow Kt{:} A$$ (T1) $$t{:} A \rightarrow \Box t{:} A$$ (T2) $$K\ A \rightarrow \Box K\ A$$ (T3) \begin{align} &\text{From }K\ A \rightarrow \big(\neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k}\big) \nonumber\\ &\qquad\qquad\text{ infer }K\ A \rightarrow \big(\neg Et_{1} \vee\ldots\vee\neg Et_{n}\vee Es_{1} \vee\ldots\vee Es_{k}\big). \end{align} (ρD) Proof. (T0). We use the transitivity of implication w.r.t. the following set of formulas: \begin{align*} &K\ A \rightarrow \Box K\Box A &&\text{(by A8)}\\ &\Box K\Box A \rightarrow K\Box A &&\text{(by A1)}\\ &K\Box A \rightarrow \Box A &&\text{(by A7)} \end{align*} (Nec$$\Box$$). From A we infer KA by (R2) and then use (T0) and (R1) to get $$\Box A$$. (Nec[j]). From A we infer $$\Box A$$ by (Nec$$\Box$$) and then apply (A2) and (R1). We pause to note that by (Nec$$\Box$$) and (Nec[j]) we know that every modality in the set $$\{\Box \} \cup \big \{ [j] \mid j \in Ag \big \}$$ is an S5-modality. (T1). We have both $$t{:} A \rightarrow !\!t{:}\big (t{:} A\big )$$ and $$!\!t{:}\big (t{:} A\big ) \rightarrow Kt{:} A$$ by (A5) so that we get (T1) by transitivity of implication. (T2). By (T1) and (T0). (T3). By K A → K K A (a part of (A7)) and (T0). (ρD). For a given sequence t1, …, tn, s1, …, sk ∈ Pol, we introduce the following abbreviations: \begin{align*} &C := \neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k},\\ &D := \neg Et_{1} \vee\ldots\vee\neg Et_{n}\vee Es_{1} \vee\ldots\vee Es_{k}. \end{align*} Now, we construct the following inference: $$K \ A \rightarrow C \qquad\qquad(\text{premise})$$ (12) $$K(K\ A \rightarrow C) \qquad\qquad(\text{by}\ (12)\ \text{and}\ (R2))$$ (13) $$K\ A \rightarrow K\ C \qquad\qquad(\text{by}\ (13)\ \text{and}\ K\ \text{is}\ \textrm{S}4)$$ (14) $$K\ C \rightarrow D \qquad\qquad(\text{by}\ (A10_D))$$ (15) $$K \ A \rightarrow D \qquad\qquad(\text{by } (14)\ \text{and}\ (15))$$ (16) The next lemma shows that (ρD) can be also used in an alternative axiomatization of $$\Sigma _{D}\left (\mathcal{CS}\right )$$: Lemma 6 Let $$\Sigma ^{\prime}_{D}(\mathcal{CS})$$ be the axiomatic system obtained from $$\Sigma _{D}(\mathcal{CS})$$ by replacing (A10D) with (ρD). Then, for every A ∈ FormAg, it is true that $$\vdash _{\mathcal{CS}} A$$ iff A is provable in $$\Sigma ^{\prime}_{D}(\mathcal{CS})$$. Proof. Assuming the conditions of the lemma, it is easy to see that whenever A is provable in $$\Sigma ^{\prime}_{D}\left (\mathcal{CS}\right )$$, it is also provable in $$\Sigma _{D}\left (\mathcal{CS}\right )$$, since (ρD) is derivable, and all the other axioms and inference rules of $$\Sigma ^{\prime}_{D}\left (\mathcal{CS}\right )$$ are also present in $$\Sigma _{D}\left (\mathcal{CS}\right )$$. In the other direction, it is sufficient to show that every instance of (A10D) is provable in $$\Sigma ^{\prime}_{D}\left (\mathcal{CS}\right )$$. Using abbreviations C, D in the same sense as in the derivation of (ρD), we construct the following proof: $$K\ C \rightarrow C \qquad\qquad(\text{by}\ (A7))$$ (17) $$K\ C \rightarrow D \qquad\qquad(\text{by}\ (17)\, \text{and}\ (\rho_D))$$ (18) However, note that in $$\Sigma ^{\prime}_{D}(\mathcal{CS})$$ the rule ($$R_{\mathcal{CS}}$$) still gets applied to the instances of (A10D) even though this scheme is no longer an axiom. Thus the alternative form of axiomatization is also somewhat less neat. 4 The canonical model The main aim of the present section is to prove the inverse of Corollary 1. The method used is a variant of the canonical model technique, but, due to the complexity of the case, we do not define our model in one sweeping definition. Rather, we proceed piecewise, defining elements of the model one by one, and checking the relevant constraints as soon as we have got enough parts of the model in place. The last subsection proves the truth lemma for the defined model. Throughout this section, we fix some arbitrary constant specification $$\mathcal{CS}$$ to serve as a parameter in the canonical model construction. The canonical model $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ to be built in this section will be a member of $$Mod^{\downarrow }_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{discr}\right )$$. The ultimate building blocks of $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ we will call elements. Before going on with the definition of $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$, we define what these elements are and explore some of their properties. Definition 2 An element is a sequence of the form $$\big (\Gamma _{1},\ldots ,\Gamma _{n}\big )$$ for some natural n ≥ 1 such that For every i ≤ n, Γi is $$\mathcal{CS}$$-maxiconsistent; For every i < n, for all A ∈ FormAg, if KA ∈ Γi, then KA ∈ Γi+1; For every i < n, for all t ∈ Pol, if Et ∈ Γi, then $$\Box Et \in \Gamma _{i + 1}$$; For every i < n, for all t ∈ Pol, if ¬Et ∈ Γi, then $$\neg \Box Et \in \Gamma _{i + 1}$$. We prove the following lemma: Lemma 7 Whenever $$\big (\Gamma _{1},\ldots ,\Gamma _{n}\big )$$ is an element, then, for some Γn+1 ⊆ FormAg, the sequence $$\big (\Gamma _{1},\ldots ,\Gamma _{n + 1}\big )$$ is also an element. Proof. Assume $$\big (\Gamma _{1},\ldots ,\Gamma _{n}\big )$$ is an element, and consider the following set: $$\Delta := \big\{ K \, A \mid K \, A \in \Gamma_{n} \big\} \cup \big\{ \Box Et \mid Et \in \Gamma_{n} \big\} \cup \big\{ \neg\Box Et \mid \neg Et \in \Gamma_{n} \big\}.$$ We show that Δ is $$\mathcal{CS}$$-consistent. Of course, the set $$\big \{ K\, A \mid K\, A \in \Gamma _{n} \big \}$$ is $$\mathcal{CS}$$-consistent since it is a subset of Γn and the latter is assumed to be $$\mathcal{CS}$$-consistent. Further, if Δ is $$\mathcal{CS}$$-inconsistent, then, wlog, for some KB1, …, KBk, Et1, …, Etl, ¬Es1, …, ¬Esr ∈ Γn we will have $$\vdash_{\mathcal{CS}}\big(K \, B_{1}\wedge\ldots \wedge K\, B_{k}\big) \rightarrow \big(\neg\Box Et_{1} \vee\ldots \vee \neg\Box Et_{l}\vee \Box Es_{1}\vee\ldots\vee \Box Es_{r}\big),$$ whence, by (A7): $$\vdash_{\mathcal{CS}} K\big(B_{1}\wedge\ldots \wedge B_{k}\big) \rightarrow \big(\neg\Box Et_{1} \vee\ldots \vee \neg\Box Et_{l}\vee \Box Es_{1}\vee\ldots\vee \Box Es_{r}\big),$$ and further, by (ρD): $$\vdash_{\mathcal{CS}} K\big(B_{1}\wedge\ldots \wedge B_{k}\big) \rightarrow \big(\neg Et_{1} \vee\ldots \vee \neg Et_{l}\vee Es_{1}\vee\ldots\vee Es_{r}\big).$$ The latter formula shows that Γn is $$\mathcal{CS}$$-inconsistent which contradicts the assumption that $$\big (\Gamma _{1},\ldots ,\Gamma _{n}\big )$$ is an element. Therefore, Δ must be $$\mathcal{CS}$$-consistent and, by Lemma 4.1, it is also extendable to a maxiconsistent Γn+1. By the choice of Δ, this means that $$\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Gamma _{n + 1}\big )$$ must be an element. The structure of elements will be important in what follows. If $$\xi = \big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ is an element, then its initial segment is any element τ of the form $$\big (\Gamma _{1},\ldots , \Gamma _{k}\big )$$ with k ≤ n. If, moreover, k < n, then τ is a proper initial segment of ξ, and if k = n − 1, then τ is the greatest proper initial segment of ξ. Moreover, we define n to be the length of ξ. Thus, any element of length 1 has no proper initial segments. Furthermore, we define that Γn is the end element of ξ and write $$\Gamma _{n} = end\big (\xi \big )$$. We now define the canonical model using elements as our building blocks. We start by defining the following relation ≡ between elements of equal length: \begin{align*} \big(\Gamma_{1},\ldots, \Gamma_{n}, \Gamma\big) \equiv \big(\Delta_{1},\ldots, \Delta_{n}, \Delta\big) \Leftrightarrow &\big(\Gamma_{1} = \Delta_{1} \& \ldots \& \Gamma_{n} = \Delta_{n} \&\\ &\& \big(\forall A \in Form^{Ag}\big)\big(\Box A \in \Gamma \Rightarrow A \in \Delta\big). \end{align*} It is routine to check that ≡ is an equivalence relation given that $$\Box$$ is an S5-modality. We will denote the equivalence class of element $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ generated by ≡ by $$\big [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\big ]_{\equiv }$$. Since all the elements inside a given ≡-equivalence class are of the same length, we may extend the notion of length to these classes setting that the length of $$\big [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\big ]_{\equiv }$$ also equals n. The next lemma will be repeatedly used in what follows: Lemmma 8 Let $$\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Gamma \big )$$ be an element, let Δ ⊆ FormAg be $$\mathcal{CS}$$-maxiconsistent and let $$\big\{ \Box A \mid \Box A \in \Gamma \big\} \subseteq \Delta.$$ Then $$\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Delta \big )$$ is also an element and, moreover, $$\big(\Gamma_{1},\ldots,\Gamma_{n}, \Gamma\big) \equiv \big(\Gamma_{1},\ldots,\Gamma_{n}, \Delta\big).$$ Proof. We first show that $$\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Delta \big )$$ is an element. Indeed, if KA ∈ Γn, then KA ∈ Γ by definition of an element. But then $$\Box KA \in \Gamma$$ by (T3) and $$\mathcal{CS}$$-maxiconsistency of Γ, whence $$\Box KA \in \Delta$$. By (A1) and $$\mathcal{CS}$$-maxiconsistency of Δ we get then KA ∈ Δ. Similarly, if Et ∈ Γn, then $$\Box Et \in \Gamma$$ by definition of an element. But then $$\Box Et \in \Delta$$. Finally, if ¬Et ∈ Γn, then $$\neg \Box Et \in \Gamma$$ by definition of an element and, further, $$\Box \neg \Box Et \in \Gamma$$ by (A1) and $$\mathcal{CS}$$-maxiconsistency of Γ. But then we must have $$\Box \neg \Box Et \in \Delta$$. By (A1) and $$\mathcal{CS}$$-maxiconsistency of Δ we get then $$\neg \Box Et \in \Delta$$. Given the inclusion $$\big \{ \Box A \mid \Box A \in \Gamma \big \} \subseteq \Delta$$, the other part of the lemma is straightforward. We now proceed to definitions of components for the canonical model. 4.1 Tree, $$\unlhd$$, and Hist The first two components of the canonical model $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ are as follows: Tree is the set of ≡-equivalence classes of elements plus † and ‡ as additional moments; We set that both † $$\lhd m$$ and $$m \ntriangleleft$$ † for every m ∈ Tree ∖{†}. We further set that ‡ is only $$\lhd$$-comparable to †, and for any two ≡-equivalence classes of elements m and m$$^{\prime}$$, we have that $$m \lhd m^{\prime}$$ iff there is an element ξ ∈ m such that ξ is a proper initial segment of every element τ ∈ m$$^{\prime}$$. The relation $$\unlhd$$ is then defined as the reflexive companion to $$\lhd$$. Before we move on to the choice- and justifications-related components, let us pause to check that the restraints imposed by our semantics on Tree and $$\unlhd$$ are satisfied: Lemma 9 The relation $$\unlhd$$, as defined above, is a partial order on Tree, which satisfies both historical connection and no backward branching constraints. Proof. Transitivity and reflexivity of $$\unlhd$$ are obvious. As for anti-symmetry, assume that we have both $$m \lhd m^{\prime}$$ and $$m^{\prime} \lhd m$$. Then m and m$$^{\prime}$$ must be equivalence classes of elements. Let ξ ∈ m be a proper intial segment of every element in m$$^{\prime}$$ and let τ ∈ m$$^{\prime}$$ be a proper initial segment of every element in m. It follows that ξ is a proper initial segment of τ and also τ is a proper initial segment of ξ, a contradiction. Historical connection is satisfied since † is the $$\unlhd$$-least element of Tree. Let us prove the absence of backward branching. Assume that we have both $$m \unlhd m^{\prime\prime}$$ and $$m^{\prime} \unlhd m^{\prime\prime}$$ but neither $$m \unlhd m^{\prime}$$ nor $$m^{\prime} \unlhd m$$ holds. This means that all the three moments are pairwise different and none of them is either † or ‡, otherwise our assumptions about them would be immediately falsified. Therefore, all the three moments are some equivalence classes of elements and we also have $$m \neq m^{\prime}$$, $$m \lhd m^{\prime\prime}$$ and $$m^{\prime} \lhd m^{\prime\prime}$$. So let ξ ∈ m and τ ∈ m$$^{\prime}$$ be such that both ξ and τ are proper initial segments of every element in $$m^{\prime\prime}$$. We first show that $$\xi \neq \tau$$. Indeed, since ξ ∈ m and τ ∈ m$$^{\prime}$$, we must have both m = [ξ]≡ and m$$^{\prime}$$ = [τ]≡. Therefore, from ξ = τ it would immediately follow that [ξ]≡ = [τ]≡, hence m = m$$^{\prime}$$. The latter is in contradiction with our assumption that $$m \neq m^{\prime}$$. Therefore either ξ must be a proper initial segment of τ or τ must be a proper initial segment of ξ. Assume, wlog, that ξ is a proper initial segment of τ. Then ξ is included into the greatest proper intitial segment of τ. Let τ′ be any element in m$$^{\prime}$$. It follows from the definition of ≡ that all the elements within m$$^{\prime}$$ share the same greatest proper initial segment, therefore ξ must be a proper initial segment of τ$$^{\prime}$$ as well. It follows that $$m \lhd m^{\prime}$$, contrary to our assumptions. Before we move on, let us have a quick look into the structure of histories induced by Tree and $$\unlhd$$. Lemma 10 Let h be a history induced by Tree and $$\unlhd$$ as defined in this subsection. If ‡ ∉ h, then for every n ≥ 1, h contains exactly one equivalence class of elements of the length n. Proof. A history cannot contain more than one class of elements of the same length. Indeed, if m, m$$^{\prime}$$ are both equivalence classes of elements of the length n then no element in m can be a proper initial segment of any element in m′ or vice versa. Therefore, whenever $$m \neq m^{\prime}$$, we will also have $$\left(m \ntrianglelefteq m^{\prime}\right) \& \left( m^{\prime} \ntrianglelefteq m\right),$$ which means that no chain of elements can contain both m and m$$^{\prime}$$. We still have to show that every history contains at least one equivalence class of elements of the length n for every n ≥ 1. Assume the contrary. Then let h be a history induced by Tree and $$\unlhd$$ and let k ≥ 1 be the least number such that the given h does not contain any equivalence classes of elements of the length k. We have then two cases to consider. Case 1. k = 1. Since h is a maximal $$\unlhd$$-chain and † is the $$\unlhd$$-least element of Tree, we know that † ∈ h. However, h cannot be {†}, given that we would have then h ⊂ {†, ‡}. Since {†, ‡} is clearly a $$\unlhd$$-chain, this would contradict the maximality of h. Therefore, we can choose an m ∈ h ∖ {†}. Given that ‡ ∉ h, m must be an equivalence class of elements, say $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{r}\big )\big ]_{\equiv }$$ for some appropriate Γ1, …, Γr ⊆ FormAg. Then we clearly have $$\dagger \lhd \big[\big(\Gamma_{1}\big)\big]_{\equiv} \unlhd \big[\big(\Gamma_{1},\ldots, \Gamma_{r}\big)\big]_{\equiv}.$$ (19) Now, let m$$^{\prime}$$ be an arbitrary moment in h. Then either $$m \unlhd m^{\prime}$$ or $$m^{\prime} \lhd m$$. In the former case we will have $$\big [\big (\Gamma _{1}\big )\big ]_{\equiv } \unlhd m^{\prime}$$ by (19) and transitivity, in the latter case we will have $$\big [\big (\Gamma _{1}\big )\big ]_{\equiv } \unlhd m^{\prime} \vee m^{\prime} \unlhd \big [\big (\Gamma _{1}\big )\big ]_{\equiv }$$ by (19) and the absence of backward branching. Therefore $$h \cup \big \{ [(\Gamma _{1})]_{\equiv } \big \}$$ is a $$\unlhd$$-chain, and by maximality of histories we get that $$\big [\big (\Gamma _{1}\big )\big ]_{\equiv } \in h$$. Since the length of $$\big [\big (\Gamma _{1}\big )\big ]_{\equiv }$$ equals 1, this contradicts our assumption. Case 2. k > 1, say k = l + 1. Then h must contain at least one moment m of the length l, say $$\big [\big (\Gamma _{1},\ldots , \Gamma _{l}\big )\big ]_{\equiv } = m \in h$$. We have to distinguish between two subcases: Case 2.1. m is the $$\unlhd$$-greatest moment in h. Then, by Lemma 7, choose an appropriate Γ ⊆ FormAg such that $$\big (\Gamma _{1},\ldots , \Gamma _{l}, \Gamma \big )$$ is an element. We obviously have $$m \lhd m^{\prime}$$ for $$m^{\prime} = \big [\big (\Gamma _{1},\ldots , \Gamma _{l}, \Gamma \big )\big ]_{\equiv }$$, and since m is $$\unlhd$$-greatest in h, this means that $$h \subset h \cup \big \{ m^{\prime} \big \}$$ and that $$h \cup \big \{ m^{\prime} \big \}$$ is a $$\unlhd$$-chain, which contradicts the maximality of h. Case 2.2. m is not the $$\unlhd$$-greatest moment in h. Then choose an m$$^{\prime}$$ ∈ h, such that $$m \lhd m^{\prime}$$. By definition of $$\unlhd$$, m$$^{\prime}$$ must be an equivalence class of elements, and some element in m must be a proper segment of every element in m$$^{\prime}$$. This means that the length of m$$^{\prime}$$ must be strictly greater than l. By our assumption, m$$^{\prime}$$ cannot have the length l + 1, therefore, the length of m$$^{\prime}$$ must also exceed l + 1. But then m$$^{\prime}$$ must be of the form $$\big [\big (\Delta _{1},\ldots , \Delta _{l}, \Delta _{l+1},\ldots , \Delta _{r}\big )\big ]_{\equiv }$$ for some r > l + 1 and appropriate Δ1, …, Δn, Δn+1, …, Δr ⊆ FormAg. We obviously have then that $$m_{0} = \big[\big(\Delta_{1},\ldots, \Delta_{l}, \Delta_{l+1}\big)\big]_{\equiv} \lhd \big[\big(\Delta_{1},\ldots, \Delta_{l}, \Delta_{l+1},\ldots, \Delta_{r}\big)\big]_{\equiv} = m^{\prime}.$$ (20) Now, consider an arbitrary $$m^{\prime\prime}$$ ∈ h. Since h is a $$\unlhd$$-chain, we must have either $$m^{\prime\prime} \lhd m^{\prime}$$ or $$m^{\prime} \unlhd m^{\prime\prime}$$. In the latter case we will have $$m_{0} \lhd m^{\prime\prime}$$ by (20) and transitivity. In the former case we will have $$m_{0} \unlhd m^{\prime\prime} \vee m^{\prime\prime} \unlhd m_{0}$$ by (20) and the absence of backward branching. Therefore, h ∪ {m0} is a $$\unlhd$$-chain, and by maximality of h we must have m0 ∈ h. But the latter contradicts our assumption since the length of m0 equals l + 1. Lemma 11 Let h be a history induced by $$\big \langle Tree, \unlhd \big \rangle$$ as defined in this subsection. Then either h = (†, ‡) or h is of the form: $$\big(\dagger, [\xi_{1}]_{\equiv},\ldots, [\xi_{n}]_{\equiv},\ldots\big),$$ where for every n ≥ 1, ξn has the length n and is the greatest proper initial segment of ξn+1. Proof. We first show that every such h is indeed a history. This is clear for $$h = \big (\dagger , \ddagger \big )$$. Assume that h is of the form $$\big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots \big )$$ for the appropriate sequence of ξs. Then h is obviously a $$\unlhd$$-chain. Now, if m ∈ Tree ∖ h, then m is either ‡ or an equivalence class of elements. It is clear that h ∪ {‡} is not a $$\unlhd$$-chain, since ‡ is $$\unlhd$$-incomparable with [ξ1]≡. If m is an equivalence class of elements, then consider the length of m, say n. Since m ∉ h, we must have $$m \neq [\xi _{n}]_{\equiv }$$, therefore, h ∪ {m} will contain at least two different equivalence classes of moments of the same length n, hence by Lemma 10 cannot be a history. Therefore, every chain of the form $$\big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots \big )$$ is maximal and thus a history. In the other direction, let h be a history induced by Tree and $$\unlhd$$. Then ‡ is either in h or outside h. If ‡ ∈ h, then we must have h = (†, ‡) since † is the only other element in Tree which is $$\unlhd$$-comparable to ‡. On the other hand, if ‡ ∉ h, then, by Lemma 10, h must be of the form (†, m1, …, mn, …), where, for n ≥ 1, mn is an equivalence class of elements of length n († will be in h since it is the $$\unlhd$$-least). For a given n ≥ 1, consider the pair mn, mn+1. Since both of these moments are in h they must be $$\unlhd$$-comparable, but we cannot have $$m_{n+1} \unlhd m_{n}$$ since, by the length considerations, no element in mn+1 can be a proper initial segment of any element in mn. Therefore, we must have $$m_{n} \lhd m_{n+1}$$, which means that there must be an element in mn which is a proper segment of every element in mn+1. We now set this element as ξn. It is clear that under these settings, we get that $$h = \big(\dagger, [\xi_{1}]_{\equiv},\ldots, [\xi_{n}]_{\equiv},\ldots\big),$$ since we have ξn ∈ mn for every n ≥ 1. Further, every ξn is the greatest proper segment of its corresponding ξn+1, since ξn is chosen as a proper initial segment of every element in mn+1, ξn+1 ∈ mn+1, and the length of ξn+1 exceeds the length of ξn by one. Corollary 2 Let h be a history induced by Tree and $$\unlhd$$ as defined in this subsection. Then h can be embedded into an initial segment of ω with its natural order. Proof. If h is a history induced by the abovedefined Tree and $$\unlhd$$, then either h = (†, ‡) or $$h = \big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots \big )$$ for an appropriate sequence ξ1, …, ξn, … of elements. In the former case we set the embedding f as f(†) = 0, f(‡) = 1; in the latter case we set f(†) = 0 and $$f\big ([\xi _{n}]_{\equiv }\big ) = n$$ for every n ≥ 1 thus embedding h into ω which, of course, is its own (improper) initial segment. Whenever a history h is of the form $$\big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots \big )$$, we will call the sequence of elements $$\big (\xi _{1},\ldots , \xi _{n},\ldots \big )$$ a representation of h. One can show that such a representation is unique. Indeed, suppose that $$\big (\dagger , \xi _{1},\ldots , \xi _{n},\ldots \big )$$ and (†, ξ1′, …, ξn′, …) are two different representations for $$\big (\dagger , m_{1},\ldots , m_{n},\ldots \big )$$. Then let i ≥ 1 be the first natural number such that $$\xi _{i} \neq \xi ^{\prime}_{i}$$. Consider mi+1. We have ξi+1, ξi+1′∈ mi+1 so that ξi+1 ≡ ξi+1′. Since ξi and ξi′ are the greatest proper initial segments of ξi+1, ξi+1′, respectively, the greatest proper initial segments of ξi+1 and ξi+1′ are non-empty and, by ξi+1 ≡ ξi+1′, must coincide, which cannot be the case since $$\xi _{i} \neq \xi ^{\prime}_{i}$$. If $$h = \big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots \big )$$ is a history in the canonical model, and m ∈ h is an equivalence class of elements, we know that for some n ≥ 1 we must have m = [ξn]≡. Since the representations of histories are unique, we may unequivocally define the intersection of m and h (write m ⊓ h) as the element in the representation of h which induces m, so that, in our case, m ⊓ h = ξn. The following lemma will be extremely useful in what follows: Lemma 12 Let ξ be an element. Then there is at least one history $$h \in H_{[\xi ]_{\equiv }}$$ such that [ξ]≡ ⊓ h = ξ. Proof. Assume that $$\xi = \big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ for appropriate n ≥ 1, Γ1, …, Γn. Using Lemma 7, we choose the infinite sequence $$\Gamma_{n + 1},\ldots, \Gamma_{n + k},\ldots,$$ in such a way that the following: \begin{align*} \xi_{1} &:= \big(\Gamma_{1}\big);\\ \xi_{2} &:= \big(\Gamma_{1}, \Gamma_{2}\big);\\ &\ldots;\\ \xi_{n} &:= \xi = \big(\Gamma_{1},\ldots, \Gamma_{n}\big);\\ \xi_{n + 1} &:= \big(\Gamma_{1},\ldots, \Gamma_{n}, \Gamma_{n + 1}\big);\\ &\ldots;\\ \xi_{n + k} &:= \big(\Gamma_{1},\ldots, \Gamma_{n}, \Gamma_{n + 1}, \ldots, \Gamma_{n + k}\big);\\ &\ldots. \end{align*} is a sequence of elements in which every element is the greatest proper initial segment of the next one, so that, by Lemma 11, there must be a history h in $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ such that $$h = \big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots , [\xi _{n + k}]_{\equiv },\ldots \big )$$, and ξ = ξn = [ξn]≡⊓ h. 4.2 Choice We now define the choice structures of our canonical model. Let j ∈ Ag be arbitrary and let $$\big (m,h\big )$$ be a moment-history pair. Then $$Choic{e^{m}_{j}}(h) = H_{m}$$, if m ∈ {†, ‡}; $$Choic{e^{m}_{j}}(h) = \left \{ h^{\prime} \in H_{m} \mid \left (\forall A \in Form^{Ag}\right )\big (\big [j\big ]A \in end\left (m \sqcap h\big ) \Rightarrow A \in end\big (m \sqcap h^{\prime}\big )\right )\right \}$$, if m is an equivalence class of elements. Since for every j ∈ Ag, $$\big [j\big ]$$ is an S5-modality, Choice induces a partition on Hm for every given m ∈ Tree. We check that the choice function verifies the relevant semantic constraints: Lemma 13 The tuple $$\big \langle Tree, \unlhd , Choice\big \rangle$$, as defined above, verifies both independence of agents and no choice between undivided histories constraints. Proof. We first tackle the no choice between undivided histories constraint. Consider a moment m and two histories h, h$$^{\prime}$$ ∈ Hm such that h ≈mh$$^{\prime}$$. Since the agents’ choices can only be non-vacuous (that is to say, not equal to Hm), at moments represented by equivalence classes of elements, we may safely assume that m is such a class. We may also assume that $$h \neq h^{\prime}$$ since in this case the constraint is trivially satisfied. By h ≈mh$$^{\prime}$$, there must be a moment m$$^{\prime}$$ such that $$m \lhd m^{\prime}$$ and m$$^{\prime}$$ ∈ h ∩ h$$^{\prime}$$. Hence we know that also m$$^{\prime}$$ is some equivalence class of elements. Suppose the length of m is n and the length of m$$^{\prime}$$ is n$$^{\prime}$$. Then n < n$$^{\prime}$$, and m ⊓ h, m$$^{\prime}$$ ⊓ h are both in the representation of h so that m ⊓ h is the initial segment of length n of m$$^{\prime}$$ ⊓ h. Similarly, m ⊓ h$$^{\prime}$$ is the initial segment of length n of m$$^{\prime}$$ ⊓ h$$^{\prime}$$. But both m$$^{\prime}$$⊓ h and m$$^{\prime}$$⊓ h′ are, by definition, in m$$^{\prime}$$, therefore, they must share the greatest proper initial segment. Hence, their initial segments of length n must coincide as well, and we must have m ⊓ h = m ⊓ h$$^{\prime}$$, whence $$end\big (m \sqcap h\big ) = end\big (m \sqcap h^{\prime}\big )$$. Now, if j ∈ Ag and $$\big [j\big ]A \in end\big (m \sqcap h\big )$$, then, by (A1) and $$\mathcal{CS}$$-maxiconsistency of $$end\big (m \sqcap h\big )$$, we will have also $$A \in end\big (m \sqcap h\big ) = end\big (m \sqcap h^{\prime}\big )$$, and thus $$h^{\prime} \in Choic{e^{m}_{j}}\big (h\big )$$, so that $$Choic{e^{m}_{j}}\big (h\big ) = Choic{e^{m}_{j}}\big (h^{\prime}\big )$$ since Choice is a partition of Hm. Consider, next, the independence of agents. Let m ∈ Tree and let f be a function on Ag such that $$\forall j \in Ag\big (f\big (j\big ) \in Choic{e^{m}_{j}}\big )$$. We are going to show that in this case $$\bigcap _{j \in Ag}f\big (j\big ) \neq \emptyset$$. If m ∈ {†, ‡}, then this is obvious, since every agent will have a vacuous choice. We treat the case, when m is an equivalence class of elements. Assume that $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{n + 1}\big )\big ]_{\equiv }$$. By (A1) we know that there is a set Δ of formulas of the form $$\Box A$$ which is shared by all sets of the form end(ξ) with ξ ∈ m in the sense that if ξ ∈ m, then $$\Box A \in end\big (\xi \big )$$ iff $$\Box A \in \Delta$$. By the same axiom scheme and Lemma 12, we also know that for every j ∈ Ag there is a set Δj of formulas of the form $$\big [j\big ]A$$ which is shared by all sets of the form $$end\big (\xi \big )$$ such that $$\exists h\big (h \in f\big (j\big ) \& \xi = m \sqcap h\big )$$. More precisely, define: $$\Delta_{j} : = \bigcap_{h \in f\big(j\big)}\big\{\big[j\big]A \in end\big(\xi\big)\mid \xi = m \sqcap h \big\}.$$ We claim the following: Claim Let ξ ∈ m. Then $$\exists h\left(h \in f\big(j\big) \& \xi = m \sqcap h\right) \Leftrightarrow \left(\forall A \in Form^{Ag}\right)\big([j]A \in end\big(\xi\big) \Leftrightarrow \big[j\big]A \in \Delta_{j}\big).$$ Proof of the Claim. (⇒). If $$h \in f\big (j\big )$$ is such that ξ = m ⊓ h and $$\big [j\big ]A \in \Delta _{j}$$, then $$\big [j\big ]A \in end\big (\xi \big )$$ immediately by definition of Δj. In the other direction, if $$\big [j\big ]A \in end\big (\xi \big )$$, then choose an arbitrary $$g \in f\big (j\big )$$ and let τ = m ⊓ g. Since $$f\big (j\big ) \in Choic{e^{m}_{j}}$$ and $$h \in f\big (j\big )$$, this means that $$f\big (j\big ) = Choic{e^{m}_{j}}\big (h\big )$$ so that $$g \in Choic{e^{m}_{j}}\big (h\big )$$. Therefore, whenever $$\big [j\big ]A \in end(\xi ) = end\big (m \sqcap h\big )$$, we also have $$\big [j\big ]\big [j\big ]A \in end\big (m\ \sqcap \ h\big )$$ by (A1) and $$\mathcal{CS}$$-maxiconsistency of $$end\big (m\ \sqcap \ h\big )$$, whence further $$\big [j\big ]A \in end(\tau ) = end\big (m\ \sqcap \ g\big )$$ by the definition of Choice. Since g was chosen in $$f\big (j\big )$$ arbitrarily, this means that $$\big [j\big ]A \in \Delta _{j}$$. (⇐). Assume that ξ ∈ m is such that $$\left(\forall A \in Form^{Ag}\right)\left(\big[j\big]A \in end\big(\xi\big) \Leftrightarrow \big[j\big]A \in \Delta_{j}\right)$$ (21) Using Lemma 12, choose an h ∈ Hm such that m ⊓ h = ξ. Now, choose an arbitrary g ∈ f(j) and an arbitrary τ such that τ = m ⊓ g. By the (⇒)-part of the Claim, we know that $$\left(\forall A \in Form^{Ag}\right)([j]A \in end(\tau) \Leftrightarrow [j]A \in \Delta_{j}).$$ (22) It follows from (21) and (22) that $$\left(\forall A \in Form^{Ag}\right)\left(\big[j\big]A \in end(\tau) \Leftrightarrow \big[j\big]A \in end\big(\xi\big)\right),$$ (23) whence, by (A1) and $$\mathcal{CS}$$-maxiconsistency of end(ξ), end(τ), it follows that $$\left(\forall A \in Form^{Ag}\right)\left(\big[j\big]A \in end(\tau) \Rightarrow A \in end\big(\xi\big)\right).$$ (24) We now use the fact that ξ = m ⊓ h and τ = m ⊓ g to get that $$\left(\forall A \in Form^{Ag}\right)\left(\big[j\big]A \in end\big(m \sqcap g\big) \Rightarrow A \in end\big(m \sqcap h\big)\right)$$ (25) By the above definition of Choice, this means that $$h \in Choic{e^{m}_{j}}\big (g\big )$$. Moreover, since $$g \in f\big (j\big )$$, we also have $$f\big (j\big ) = Choic{e^{m}_{j}}\big (g\big )$$, so that $$h \in f\big (j\big )$$, as desired. The Claim has therefore been proven. We now consider the set $$\Delta \cup \bigcup \left \{ \Delta _{j}\mid j \in Ag \right \}$$ and show its consistency. Indeed, if this set is inconsistent, then, wlog, we would have a provable formula of the following form: $$\vdash_{\mathcal{CS}} \left(\Box A \wedge \bigwedge_{j \in Ag}\big[j\big]A_{j}\right) \rightarrow \bot,$$ (26) where $$\Box A \in \Delta$$, and for every j ∈ Ag we will have $$\big [j\big ]A_{j} \in \Delta _{j}$$. But then, choose for every j ∈ Ag an element ξj ∈ m such that $$\left(\forall A \in Form^{Ag}\right)\left(\big[j\big]A \in end\left(\xi_{j}\right) \Leftrightarrow \big[j\big]A \in \Delta_{j}\right).$$ This is possible, since we may simply choose an arbitrary $$h_{j} \in f\big (j\big )$$ and set ξj := m ⊓ hj. Then we will have $$\big [j\big ]A_{j} \in end\big (\xi _{j}\big )$$ for every j ∈ Ag. Next, consider Γn+1. Since $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{n + 1}\big )\big ]_{\equiv }$$ and $$\Box$$ is an S5-modality, we must have $$\left\{ \Diamond\big[j\big]A_{j}\mid j \in Ag \right\} \subseteq \Gamma_{n + 1},$$ whence, by Lemma 4.5: $$\bigwedge_{j \in Ag}\Diamond\big[j\big]A_{j} \in \Gamma_{n + 1},$$ and further, by (A3) and Lemma 4.4: $$\Diamond\bigwedge_{j \in Ag}\big[j\big]A_{j} \in \Gamma_{n + 1}.$$ Also, by definition of Δ and the fact that $$\big (\Gamma _{1},\ldots , \Gamma _{n + 1}\big ) \in m$$, we get successively $$\Box A \in \Gamma_{n + 1},$$ then, by Lemma 4.5 $$\Box A \wedge \Diamond\bigwedge_{j \in Ag}\big[j\big]A_{j} \in \Gamma_{n + 1},$$ and finally, by the fact that $$\Box$$ is an S5-modality: \begin{align} \Diamond\left(\Box A \wedge \bigwedge_{j \in Ag}\big[j\big]A_{j}\right) \in \Gamma_{n + 1}. \end{align} (27) From (26), together with (27), it follows by S5 reasoning for $$\Box$$ that ◊⊥ ∈ Γn+1, so that, again by S5 properties of $$\Box$$ and Lemma 4.4, it follows that ⊥ ∈ Γn+1, which is in contradiction with $$\mathcal{CS}$$-maxiconsistency of Γn+1. Hence $$\Delta \cup \bigcup \left \{ \Delta _{j}\mid j \in Ag \right \}$$ is $$\mathcal{CS}$$-consistent, and we can extend it to a $$\mathcal{CS}$$-maxiconsistent Ξ. By Lemma 8 and Δ ⊆ Ξ, we know that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Xi \big )$$ is an element and that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Xi \big ) \in m$$ so that $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{n}, \Xi \big )\big ]_{\equiv }$$. Using Lemma 12, we can choose a g ∈ Hm such that $$m \sqcap g = \big (\Gamma _{1},\ldots , \Gamma _{n}, \Xi \big )$$. We also know that for every j ∈ Ag, there is a history $$h_{j} \in f\big (j\big )$$ such that m ⊓ hj = ξj by the choice of ξj. Therefore, for every j ∈ Ag, $$Choic{e^{m}_{j}}\big (h_{j}\big ) = f\big (j\big )$$. Also, if $$\big [j\big ]A \in end\big (\xi _{j}\big ) = end\big (m \sqcap h_{j}\big )$$, then $$\big [j\big ]A \in \Delta _{j}$$, hence $$[j]A \in \Xi = end\big (m \sqcap g\big )$$, therefore, by (A1), $$A \in end\big (g \sqcap m\big )$$. Thus we get that $$g \in \bigcap _{j \in Ag}Choic{e^{m}_{j}}\big (h_{j}\big ) = \bigcap _{j \in Ag}f\big (j\big )$$ so that the independence of agents is verified. 4.3 R and $$\mathcal{E}$$ We now define the justifications-related components of our canonical model. We first define R as follows: $$R\big (\big [\big (\Gamma _{1},\ldots ,\Gamma _{n}\big )\big ]_{\equiv }, m^{\prime}\big )\Leftrightarrow \big (m^{\prime} \in Tree \setminus \{ \dagger ,\ddagger \}\big )\&\left (\forall \tau \in m^{\prime}\right )\left (\forall A \in Form^{Ag}\right )\big (KA \in \Gamma _{n} \Rightarrow KA \in end(\tau )\big )$$; R(†, m), for all m ∈ Tree; R(‡, m) ⇔ m = ‡. Since our canonical model is unirelational, we will assume that Re is the same as R and will use the simplified satisfaction clause for t:A. Now, for the definition of $$\mathcal{E}$$: For all t ∈ Pol: $$\mathcal{E}(\dagger , t) = \mathcal{E}(\ddagger , t) = \left \{ A \in Form^{Ag} \mid \vdash _{\mathcal{CS}} t{:} A \right \}$$; For all t ∈ Pol and m ∈ Tree ∖ {†, ‡}: $$\left(\forall A \in Form^{Ag}\right)\big(A \in \mathcal{E}(m, t) \Leftrightarrow \big(\forall \xi \in m\big)\big(t{:} A \in end\big(\xi\big)\big)\big).$$ We start by mentioning a straightforward corollary to the above definition: Lemma 14 For all m ∈ Tree and t ∈ Pol it is true that $$\left \{ A \in Form^{Ag} \mid \vdash _{\mathcal{CS}} t{:} A \right \} \subseteq \mathcal{E}(m,t)$$. Proof. This holds simply by the definition of $$\mathcal{E}$$ when m ∈ {†, ‡}. If m ∈ Tree ∖ {†, ‡}, then, for every ξ ∈ m, $$end\big (\xi \big )$$ is a $$\mathcal{CS}$$-maxiconsistent subset of FormAg and must contain every provable formula. Note that it follows from Lemma 14 that the abovedefined function $$\mathcal{E}$$ satisfies the $$\mathcal{CS}$$-normality condition on jstit models. Lemma 15 The relation R, as defined above, is a preorder on Tree and, together with $$\unlhd$$, verifies the future always matters constraint. Proof. Transitivity of R is immediate from its definition. We show that R is reflexive. If $$m = \big [\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Gamma \big )\big ]_{\equiv }$$ and KA ∈ Γ, then, by (T3) and maxiconsistency of Γ, we will also have that $$\Box KA \in \Gamma$$, whence, by definition of ≡, we know that KA ∈ end(τ) for every $$\tau \in \big [\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Gamma \big )\big ]_{\equiv }$$. Next, we look into future always matters constraint. Assume m ∈ Tree. If m = †, then it is connected to all the elements in Tree by both $$\unlhd$$ and R, and if m = ‡, then it is connected only to itself by both $$\unlhd$$ and R, so these moments cannot falsify the constraint. So let us assume that m is a class of equivalence generated by some element, say $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\big ]_{\equiv }$$. If m = m$$^{\prime}$$, then we are done by reflexivity of R. On the other hand, if $$m \lhd m^{\prime}$$ then we may assume, wlog, that $$m^{\prime} = \big [\big (\Gamma _{1},\ldots , \Gamma _{k}\big )\big ]_{\equiv }$$ for some k > n. But then take an arbitrary A ∈ FormAg. If KA ∈ Γn, then, since $$\big (\Gamma _{1},\ldots , \Gamma _{k}\big )$$ is an element, KA ∈ Γk. By (T3) and $$\mathcal{CS}$$-maxiconsistency of Γk, we will also have $$\Box KA \in \Gamma _{k}$$. Now, by definition of ≡, we get KA ∈ end(τ) for any given τ ∈ m′. It follows then that $$R\left (m, m^{\prime}\right )$$ as desired. We now check the semantical constraints for $$\mathcal{E}$$: Lemma 16 The function $$\mathcal{E}$$, as defined above, satisfies both monotonicity of evidence and evidence closure properties. Proof. We start with monotonicity of evidence. Assume $$R\left (m,m^{\prime}\right )$$ and t ∈ Pol. If $$m \in \left \{ \dagger , \ddagger \right \}$$ then, by Lemma 14, $$\mathcal{E}(m,t) = \left \{ A \in Form^{Ag} \mid \vdash _{\mathcal{CS}} t{:} A \right \} \subseteq \mathcal{E}\left (m^{\prime},t\right )$$ for any m′∈ Tree. Assume, further, that m is an equivalence class of elements. Let t ∈ Pol and A ∈ FormAg be such that $$A \in \mathcal{E}(m,t)$$. Then, for every ξ ∈ m, t:A ∈ end(ξ) and, by (T1), also Kt:A ∈ end(ξ). Therefore, by $$R\left (m,m^{\prime}\right )$$, we get that, for every τ ∈ m$$^{\prime}$$, Kt:A ∈ end(τ), so that by (A7) and maxiconsistency of every end(τ), also t:A ∈ end(τ). Therefore, $$A \in \mathcal{E}\left (m^{\prime},t\right )$$, as desired. We turn now to the closure conditions. We verify the first two conditions, and the third one can be verified in a similar way, restricting attention to t rather than considering both s and t. Let s, t ∈ Pol. We need to consider two cases: Case 1. m ∈ {†, ‡}. If $$A \in \mathcal{E}(m,s)$$, then $$\vdash _{\mathcal{CS}} s{:} A$$. Therefore, by (A6), we must also have $$\vdash _{\mathcal{CS}} (s + t){:}\ A$$ so that $$A \in \mathcal{E}(m,s + t)$$. Similarly, if $$A \in \mathcal{E}(m,t)$$, then also $$A \in \mathcal{E}(m,s + t)$$ and the closure constraint (b) is verified. If, on the other hand, it is true that for some A, B ∈ FormAg we have both $$A \rightarrow B \in \mathcal{E}(m,s)$$ and $$A \in \mathcal{E}(m,t)$$, then, again, this means that both $$\vdash _{\mathcal{CS}} s{:} A \rightarrow B$$ and $$\vdash _{\mathcal{CS}} t{:} A$$. By (A4), it follows that $$\vdash _{\mathcal{CS}} s\cdot t{:} B$$ and, therefore, also $$B \in \mathcal{E}(m,s\cdot t)$$, so that the closure condition (a) is also verified. Case 2. m ∈ Tree ∖{†, ‡}. If A ∈ FormAg and $$A \in \mathcal{E}(m,s)$$, then, for every ξ ∈ m, $$s{:} A \in end\big (\xi \big )$$, and, by (A6) and $$\mathcal{CS}$$-maxiconsistency of every $$end\big (\xi \big )$$, we get that s + t:A ∈ end(ξ). Therefore, $$A \in \mathcal{E}(m,s + t)$$. Similarly, if $$A \in \mathcal{E}(m,t)$$, then $$A \in \mathcal{E}(m,s + t)$$ as well, and closure condition (b) is verified. On the other hand, if A, B ∈ FormAg and we have both $$A \rightarrow B \in \mathcal{E}(m,s)$$ and $$A \in \mathcal{E}(m,t)$$, then, for every ξ ∈ m, we have $$t{:} A, s{:}\big (A \rightarrow B\big ) \in end\big (\xi \big )$$. By (A4) and $$\mathcal{CS}$$-maxiconsistency of every $$end\big (\xi \big )$$, we get that $$s \cdot t{:} B \in end\big (\xi \big )$$, thus $$B \in \mathcal{E}(m,s \cdot t)$$, and closure condition (a) is verified. 4.4 Act and V It only remains to define Act and V for our canonical model, and we define them as follows: $$\big (m,h\big ) \in V\big (p\big ) \Leftrightarrow p \in end\big (m \sqcap h\big )$$, for all p ∈ V ar; $$Act\left (\dagger , \big (\dagger ,\ddagger \big )\right ) = Act\left (\ddagger , \big (\dagger ,\ddagger \big )\right ) = \emptyset$$; $$Act\left (\dagger , \big (\dagger ,[\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots , \big )\right ) = \left \{ t \in Pol\mid \Box Et \in end\big (\xi _{1}\big ) \right \}$$; $$Act\big (m,h\big ) = \left \{ t \in Pol \mid Et \in end\big (m \sqcap h\big ) \right \}$$, if m ∈ Tree ∖ {†, ‡}. We first establish a technical claim: Lemma 17 If m ∈ Tree ∖ {†, ‡} and t ∈ Pol, then $$t \in Act_{m} \Leftrightarrow \Box Et \in \bigcap_{\xi \in m}end\big(\xi\big).$$ Proof. (⇒) Assume that t ∈ Actm. Since m ∈ Tree ∖ {†, ‡}, then, for appropriate Γ1, …, Γn, Γ ⊆ FormAg and n ≥ 0 we will have $$m = \left [\big (\Gamma _{1},\ldots ,\Gamma _{n},\Gamma \big )\right ]_{\equiv }$$. Note that, by Lemma 12, we know that for every ξ ∈ m we can choose a history h ∈ Hm such that ξ = m ⊓ h. It follows, then, by our assumption, that $$Et \in end\big (\xi \big )$$ for all ξ ∈ m. But this means that the set $$\Xi = \big\{ \Box B \mid \Box B \in \Gamma \big\} \cup \big\{ \neg Et \big\}$$ must be $$\mathcal{CS}$$-inconsistent. Suppose otherwise. Then we can extend Ξ to a $$\mathcal{CS}$$-maxiconsistent Δ. By Lemma 8 and Ξ ⊆ Δ we get that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Delta \big )$$ is an element and that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Delta \big ) \equiv \big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )$$, so that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Delta \big ) \in m$$, but also that $$\neg Et \in \Delta = end\left (\big (\Gamma _{1},\ldots , \Gamma _{n}, \Delta \big )\right )$$ which contradicts the earlier established fact that $$Et \in end\big (\xi \big )$$ for all ξ ∈ m. The obtained contradiction shows that Ξ must be $$\mathcal{CS}$$-inconsistent, and this further means that there are some $$\Box B_{1},\ldots ,\Box B_{n} \in \Gamma _{1}$$ for which we have $$\vdash_{\mathcal{CS}} \big(\Box B_{1}\wedge\ldots \wedge \Box B_{n}\big) \rightarrow Et,$$ which, by S5 reasoning for $$\Box$$, would further mean that $$\vdash_{\mathcal{CS}} \big(\Box B_{1}\wedge\ldots \wedge \Box B_{n}\big) \rightarrow \Box Et,$$ so that, by $$\mathcal{CS}$$-maxiconsistency of Γ we know that $$\Box Et \in \Gamma$$, whence, again by $$\mathcal{CS}$$-maxiconsistency of Γ and (A1), we get that $$\Box \Box Et \in \Gamma = end\big (\big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )\big )$$. Now, if ξ ∈ m is arbitrary, then we must have $$\xi \equiv \big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )$$, whence by definition of ≡ we obtain that $$\Box Et \in end\big (\xi \big )$$ and, further, $$\Box Et \in \bigcap _{\xi \in m}end\big (\xi \big )$$. (⇐) If $$\Box Et \in \bigcap _{\xi \in m}end\big (\xi \big )$$, then, for every h ∈ Hm, m ⊓ h ∈ m, so that, by (A1), we must have $$Et \in end\big (m \sqcap h\big )$$, and, by definition of Act, $$t \in Act\big (m,h\big )$$. We now check that the remaining semantic constraints on jstit models are satisfied: Lemma 18 The canonical model, as defined above, satisfies the constraints as to the expansion of presented proofs, no new proofs guaranteed, presenting new proofs makes histories divide and epistemic transparency of presented proofs. Proof. We consider the expansion of presented proofs first. Let $$m^{\prime} \lhd m$$ and let h ∈ Hm. Then $$m^{\prime} \neq \ddagger$$, since ‡ has no $$\lhd$$-successors. If m$$^{\prime}$$ = † and m = ‡, then h must be (†, ‡) and we have $$Act\!\left (\dagger , \big (\dagger ,\ddagger \big )\right ) = Act\!\left (\ddagger , \big (\dagger ,\ddagger \big )\right ) = \emptyset$$, so that the expansion of presented proofs holds. If m$$^{\prime}$$ = † and m is an equivalence class of elements, then h is of the form $$\big (\dagger ,[\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots , \big )$$, and, for some natural n, m ⊓ h = ξn. Also, we must have $$\xi _{n} = \big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ for appropriate Γ1, …, Γn, whereas ξ1 must be the initial segment of ξn of length 1, so that we have $$\xi _{1} = \big (\Gamma _{1}\big )$$. Let t ∈ Pol be arbitrary. If $$t \in Act\big (\dagger , h\big )$$, then $$\Box Et \in \Gamma _{1}$$, therefore, by (A1) and $$\mathcal{CS}$$-maxiconsistency of Γ1, Et ∈ Γ1. Hence we must have $$\Box Et \in \Gamma _{n}$$, since $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ is an element. Again, using (A1) plus $$\mathcal{CS}$$-maxiconsistency of Γn, we infer that Et ∈ Γn, or, equivalently $$Et \in end\big (\xi _{n}\big ) = end\big (m \sqcap h\big )$$, which means that $$t \in Act\big (m,h\big )$$. Finally, if m$$^{\prime}$$ is an equivalence class of elements, then m is also an equivalence class of elements. In this case, m ⊓ h must be of the form $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ for the respective Γ1, …, Γn ⊆ FormAg. But then, for some k < n, m$$^{\prime}$$⊓ h must be of the form $$\big (\Gamma _{1},\ldots , \Gamma _{k}\big )$$. Let t ∈ Pol be arbitrary. If $$t \in Act\big (m^{\prime},h\big )$$, then $$Et \in end\big (m^{\prime} \sqcap h\big ) = \Gamma _{k}$$. Since $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ is an element, we must have then $$\Box Et \in \Gamma _{n}$$. Further, using (A1) plus maxiconsistency of Γn, we infer that Et ∈ Γn or, equivalently $$Et \in end\big (\xi _{n}\big ) = end\big (m \sqcap h\big )$$, which means that $$t \in Act\big (m,h\big )$$. We consider next the no new proofs guaranteed constraint. Let m ∈ Tree. If m ∈ {†, ‡}, then Actm = ∅ and the constraint is trivially satisfied. Assume that m ∈ Tree ∖ {†, ‡}. We need to distinguish then between two cases: Case 1. The length of m equals 1. Then m is of the form $$\left [\big (\Gamma \big )\right ]_{\equiv }$$ for the respective Γ ⊆ FormAg. Let t ∈ Pol be such that t ∈ Actm and choose, by Lemma 12, an h ∈ Hm such that $$\big (\Gamma \big ) = m \sqcap h$$. By Lemma 17, we know that $$\Box Et \in \Gamma = end\big (m \sqcap h\big )$$, therefore, by definition of Act, we have $$t \in Act(\dagger , h) \subseteq \bigcup _{m^{\prime} \lhd m, h \in H_{m}}\left (Act\big (m^{\prime},h\big )\right )$$, as desired. Case 2. The length of m is greater than 1. Then m must be of the form $$\left [\big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )\right ]_{\equiv }$$ for the respective Γ1, …, Γn, Γ ⊆ FormAg and n > 0. Furthermore, $$m^{\prime} = \left [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\right ]_{\equiv }$$ is the immediate $$\lhd$$-predecessor of m. Let t ∈ Pol be such that t ∈ Actm and choose, by Lemma 12, an h ∈ Hm such that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big ) = m \sqcap h$$. By Lemma 17, we know that $$\Box Et \in \Gamma = end\big (m \sqcap h\big )$$. Since $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )$$ is an element, Γn must be $$\mathcal{CS}$$-maxiconsistent; therefore, by Lemma 4.2, we either have Et ∈ Γn, or ¬Et ∈ Γn. In the latter case we would have $$\neg \Box Et \in \Gamma$$, which, given that also $$\Box Et \in \Gamma$$, is in contradiction with $$\mathcal{CS}$$-maxiconsistency of Γ. Therefore, we must have Et ∈Γn. Moreover, since $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )$$ is in the representation of h, it is clear that its greatest proper initial segment, $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$, must also be in the representation of h. Therefore, we get that $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big ) = m^{\prime} \sqcap h$$ so that $$\Gamma _{n} = end\big (m^{\prime} \sqcap h\big )$$. It follows, by definition of Act, that $$t \in Act\big (m^{\prime},h\big )$$, and, given $$m^{\prime} \lhd m$$, also that $$t \in \bigcup _{m^{\prime} \lhd m, h \in H_{m}}\big (Act\big (m^{\prime},h\big )\big )$$, as desired. We turn next to the presenting a new proof makes histories divide constraint. Consider an m ∈ Tree and arbitrary h, h$$^{\prime}$$ ∈ Hm′ such that h ≈mh$$^{\prime}$$. We may assume that $$h \neq h^{\prime}$$ since otherwise the constraint is trivially satisfied. Therefore, there must be an $$m^{\prime} \rhd m$$ such that m′∈ h ∩ h$$^{\prime}$$. Then $$m \neq \ddagger$$, since ‡ has no $$\lhd$$-successors. If m = † and m$$^{\prime}$$ = ‡, then we must have h = h$$^{\prime}$$ = (†, ‡) and the constraint is verified trivially. If m = † and $$m^{\prime} \neq \ddagger$$, then m$$^{\prime}$$ must be an equivalence class of elements, say, $$m^{\prime} = \left [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\right ]_{\equiv }$$. Since m$$^{\prime}$$ ∈ h ∩ h$$^{\prime}$$, then for some ξ, τ ∈ m$$^{\prime}$$ we must have ξ = m$$^{\prime}$$⊓ h and τ = m$$^{\prime}$$⊓ h$$^{\prime}$$, respectively. By definition of ≡ we know that for some maxiconsistent Δ, Ξ ⊆ FormAg we will have both $$\xi = \big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Delta \big )$$ and $$\tau = \big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Xi \big )$$, and we will also have ξ ≡ τ. We now have two cases to consider: Case 1. n > 1. Then $$\big (\Gamma _{1}\big )$$ is a proper initial segment of both ξ and τ which means that, by $$\left [\big (\Gamma _{1}\big )\right ]_{\equiv } \lhd m^{\prime}$$ and no backward branching, we must have $$\left [\big (\Gamma _{1}\big )\right ]_{\equiv } \in h \cap h^{\prime}$$. Since $$\big (\Gamma _{1},\ldots , \Gamma _{n - 1}\big )$$ is in the representations of both h and h$$^{\prime}$$, we know that also $$\big (\Gamma _{1}\big )$$ must be in both representations so that $$\big(\Gamma_{1}) = \big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h = \big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h^{\prime}.$$ It follows immediately that $$\Gamma_{1} = end\left(\big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h\right) = end\left(\big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h^{\prime}\right),$$ and, further, that $$Act\big(\dagger, h\big)\!=\! \left\{ t \in Pol \mid \Box Et\! \in\! end\left(\big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h\right) \right\}\! =\! \left\{ t\! \in\! Pol \mid \Box Et\! \in\! end\big(\big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h^{\prime}\big) \right\}\! =\! Act\big(\dagger, h^{\prime}\big),$$ and we are done. Case 2. n = 1. Then we must have (Δ) = ξ ≡ τ = (Ξ) so that, by definition of ≡ and S5 properties of $$\Box$$ we get, among other things, that $$\Box Et \in \Delta = end\big(m^{\prime} \sqcap h\big) \Leftrightarrow \Box Et \in \Xi =end\big(m^{\prime} \sqcap h^{\prime}\big)$$ for every t ∈ Pol. Whence it clearly follows that $$Act\big(\dagger, h\big) = \left\{ t \in Pol \mid \Box Et \in end\big(m^{\prime} \sqcap h\big) \right\} = \left\{ t \in Pol \mid \Box Et \in end\big(m^{\prime} \sqcap h^{\prime}\big) \right\} = Act\big(\dagger, h^{\prime}\big),$$ and we are done. The only case that is left now is when m is an equivalence class of elements. In this case m$$^{\prime}$$ is an equivalence class of elements either, so that, say, $$m^{\prime} = \big [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\big ]_{\equiv }$$. Since m$$^{\prime}$$ ∈ h ∩ h$$^{\prime}$$, then for some ξ, τ ∈ m$$^{\prime}$$ we must have ξ = m$$^{\prime}$$⊓ h and τ = m$$^{\prime}$$⊓ h$$^{\prime}$$, respectively. By definition of ≡ we know that for some $$\mathcal{CS}$$-maxiconsistent Δ, Ξ ⊆ FormAg we will have both $$\xi = \big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Delta \big )$$ and $$\tau = \big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Xi \big )$$, and, moreover, we will have ξ ≡ τ. Since $$m \lhd m^{\prime}$$ and m ∈ h ∩ h$$^{\prime}$$, then m ⊓ h must be a proper initial segment of $$\big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Delta \big )$$, whereas m ⊓ h$$^{\prime}$$ must be a proper initial segment of $$\big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Xi \big )$$ and both segments must be the elements in m so that they must be of the same length. It follows then, that for some k < n we will have $$m \sqcap h = \big(\Gamma_{1},\ldots, \Gamma_{k}\big) = m \sqcap h^{\prime},$$ whence $$end\big(m \sqcap h\big) = \Gamma_{k} = end\big(m \sqcap h^{\prime}\big).$$ The latter equation entails that $$Act\big(m, h\big) = \left\{ t \in Pol \mid Et \in end\big(m \sqcap h\big) \right\} = \left\{ t \in Pol \mid Et \in end\big(m \sqcap h^{\prime}\big) \right\} = Act\big(m,h^{\prime}\big),$$ as desired. It remains to check the epistemic transparency of presented proofs constraint. Assume that m, m′∈ Tree are such that $$R\big (m,m^{\prime}\big )$$. If we have m ∈ {†, ‡}, then, by definition, we must have Actm = ∅, and the constraint is verified in a trivial way. If, on the other hand, m ∈ Tree ∖ {†, ‡}, then $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\big ]_{\equiv }$$ for appropriate Γ1, …, Γn ⊆ FormAg and n > 0. By R(m, m$$^{\prime}$$), we must also have m$$^{\prime}$$ ∈ Tree ∖ {†, ‡}. Now, if for some t ∈ Pol we have t ∈ Actm, then, by Lemma 17, we will also have $$\Box Et \in \Gamma _{n}$$, whence, by (A9) and $$\mathcal{CS}$$-maxiconsistency of Γn, that $$K\Box Et \in \Gamma _{n}$$. The latter means, by definition of R, that $$K\Box Et \in \bigcap _{\tau \in m^{\prime}}end(\tau )$$. $$\mathcal{CS}$$-maxiconsistency of end(τ) for every τ ∈ m$$^{\prime}$$, together with (A7), yields then $$\Box Et \in \bigcap _{\tau \in m^{\prime}}end(\tau )$$ so that, by Lemma 17 again, it follows that t ∈ Actm′ and thus we are done. 4.5 The truth lemma It follows from Lemmas 9, 13, 15, 16 and 18 that our abovedefined canonical model $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ is a unirelational jstit model for Ag. By Corollary 2 we know that $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ is based on discrete time and by Lemma 14 we know that $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ is $$\mathcal{CS}$$-normal. Summing everything up, we get that $$\mathcal{M}^{Ag}_{\mathcal{CS}} \in Mod^{\downarrow }_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{discr}\right )$$. Now we need to supply a truth lemma: Lemma 19 Let A ∈ FormAg, let m ∈ Tree ∖ {†, ‡}, and let h ∈ Hm. Then $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models A \Leftrightarrow A \in end\big(m \sqcap h\big).$$ Proof. As is usual, we prove the lemma by induction on the construction of A. The basis of induction with A = p ∈ V ar we have by definition of V, whereas Boolean cases for the induction step are trivial. We treat the modal cases: Case 1. $$A = \Box B$$. If $$\Box B \in end\big (m \sqcap h\big )$$, then note that for every h′∈ Hm we must have m ⊓ h′ ∈ m so that m ⊓ h′ ≡ m ⊓ h. By definition of ≡ and the fact that m ∈ Tree ∖ {†, ‡}, we must have then $$B \in end\big (m \sqcap h^{\prime}\big )$$ for all h′∈ Hm and thus, by induction hypothesis, we obtain that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models \Box B$$. If, on the other hand, $$\Box B \notin end(m \sqcap h)$$, we must have then $$m \sqcap h = \big (\Gamma _{1},\ldots ,\Gamma _{n},\Gamma \big )$$ for some appropriate n ≥ 0 and Γ1, …, Γn, Γ so that Γ = end(m ⊓ h). Then the set $$\Xi = \left\{ \Box C \mid \Box C \in \Gamma \right\} \cup \left\{ \neg B \right\}$$ must be $$\mathcal{CS}$$-consistent, since otherwise we would have $$\vdash_{\mathcal{CS}} \big(\Box C_{1}\wedge\ldots\wedge\Box C_{n}\big) \rightarrow B$$ for some $$\Box C_{1},\ldots ,\Box C_{n} \in \Gamma$$, whence, since $$\Box$$ is an S5-modality, we would get $$\vdash_{\mathcal{CS}} \big(\Box C_{1}\wedge\ldots\wedge\Box C_{n}\big) \rightarrow \Box B,$$ which would mean that $$\Box B \in \Gamma$$, contrary to our assumption. Therefore, Ξ is $$\mathcal{CS}$$-consistent and we can extend Ξ to a $$\mathcal{CS}$$-maxiconsistent Δ ∈ FormAg. Of course, in this case B ∉ Δ. By Lemma 8 and Ξ ⊆ Δ, we know that $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big )$$ is actually an element and that $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Gamma \big ) \equiv \big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big )$$. By Lemma 12, for some h′ ∈ Hm we will have $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big ) = m \sqcap h^{\prime}$$ and, therefore, $$\Delta = end\big (m \sqcap h^{\prime}\big )$$. Since B ∉ Δ, it follows, by induction hypothesis, that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h^{\prime} \not \models B$$, hence $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \not \models \Box B$$ as desired. Case 2. $$A = \big [j\big ]B$$ for some j ∈ Ag. Then, if $$\big [j\big ]B \in end(m \sqcap h)$$, by definition of Choice and the fact that m ∈ Tree ∖ {†, ‡} we must have $$Choic{e^{m}_{j}}\big(h\big) = \left\{ h^{\prime} \in H_{m} \mid \left(\forall C \in Form^{Ag}\right)\left(\big[j\big]C \in end\big(m \sqcap h\big) \Rightarrow C \in end\big(m \sqcap h^{\prime}\big)\right)\right\}.$$ Therefore, if $$h^{\prime} \in Choic{e^{m}_{j}}\big (h\big )$$, then we must have $$B \in end\big (m \sqcap h^{\prime}\big )$$, and, by induction hypothesis, $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h^{\prime} \models B$$, so that we get $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models \big [j\big ]B$$. On the other hand, if $$\big [j\big ]B \notin end\big (m \sqcap h\big )$$, then we must have $$m \sqcap h = \big (\Gamma _{1},\ldots ,\Gamma _{n},\Gamma \big )$$ for some appropriate n ≥ 0 and Γ1, …, Γn, Γ so that $$\Gamma = end\big (m \sqcap h\big )$$. Then the set $$\Xi = \left\{ \big[j\big]C \mid \big[j\big]C \in \Gamma \right\} \cup \left\{ \neg B \right\}$$ must be $$\mathcal{CS}$$-consistent, since otherwise we would have $$\vdash_{\mathcal{CS}} \big(\big[j\big]C_{1}\wedge\ldots\wedge\big[j\big]C_{n}\big) \rightarrow B$$ for some $$\big [j\big ]C_{1},\ldots ,\big [j\big ]C_{n} \in \Gamma$$, whence, since $$\big [j\big ]$$ is an S5-modality, we would get $$\vdash_{\mathcal{CS}} \left(\big[j\big]C_{1}\wedge\ldots\wedge\big[j\big]C_{n}\right) \rightarrow \big[j\big]B,$$ which would mean that $$\big [j\big ]B \in \Gamma$$, contrary to our assumption. Therefore, Ξ is $$\mathcal{CS}$$-consistent and we can extend Ξ to a $$\mathcal{CS}$$-maxiconsistent Δ ⊆ FormAg. Of course, in this case B ∉ Δ. Furthermore, note that if D ∈ FormAg is such that $$\Box D \in \Gamma$$, then, by (A1), (A2) and $$\mathcal{CS}$$-maxiconsistency of Γ, we know that $$\big [j\big ]\Box D \in \Gamma$$, so that also $$\big [j\big ]\Box D \in \Delta$$, and hence, by (A1) and $$\mathcal{CS}$$-maxiconsistency of Δ, $$\Box D \in \Delta$$. We have thus shown that $$\left \{ \Box D \mid \Box D \in \Gamma \right \} \subseteq \Delta$$. Again, by Lemma 8 it follows that $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big )$$ is an element and that $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Gamma \big ) \equiv \big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big )$$. By Lemma 12, for some h′∈ Hm we will have $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big ) = m \sqcap h^{\prime}$$ and, therefore, $$\Delta = end\big (m \sqcap h^{\prime}\big )$$. Also, since Δ contains all the $$\big [j\big ]$$-modalized formulas from Γ, we know that for any such h′ we will have $$h^{\prime} \in Choic{e^{m}_{j}}\big (h\big )$$. Since B ∉ Δ, it follows, by induction hypothesis, that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,\,h^{\prime} \not \models B$$, hence $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \not \models \big [j\big ]B$$ as desired. Case 3. A = KB. Assume that $$KB \in end\big (m \sqcap h\big )$$. We clearly have then $$m = \big [\big (m \sqcap h\big )\big ]_{\equiv }$$. Hence, by definition of R and the fact that m ∈ Tree ∖ {†, ‡} we must have for every m′∈ Tree: $$R\big(m,m^{\prime}\big) \Rightarrow \big(\forall \tau \in m^{\prime}\big)\big(\forall C \in Form^{Ag}\big)\big(KC \in end\big(m \sqcap h\big) \Rightarrow KC \in end(\tau)\big).$$ Therefore, if $$R\big (m,m^{\prime}\big )$$ and h′∈ Hm′ is arbitrary, then, of course, (m′⊓ h′) ∈ m′ so that $$KB \in end\big (m^{\prime} \sqcap h^{\prime}\big )$$, and, further, $$B \in end\big (m^{\prime} \sqcap h^{\prime}\big )$$ by S4 reasoning for K. Therefore, by induction hypothesis, we get that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m^{\prime},h^{\prime} \models B$$, whence $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models KB$$. On the other hand, if $$KB \notin end\big (m \sqcap h\big )$$, then consider the set $$\Xi = \big\{ KC \mid KC \in end\big(m \sqcap h\big) \big\} \cup \big\{ \neg\Box B \big\}.$$ This set must be $$\mathcal{CS}$$-consistent, since otherwise we would have $$\vdash_{\mathcal{CS}} \big(KC_{1}\wedge\ldots\wedge KC_{n}\big) \rightarrow \Box B$$ for some KC1, …, KCn ∈Γ, whence, since K is an S4-modality, we would get $$\vdash_{\mathcal{CS}} \big(KC_{1}\wedge\ldots\wedge KC_{n}\big) \rightarrow K\Box B,$$ which would mean that $$K\Box B \in end\big (m \sqcap h\big )$$, hence, by (A1), (A7) and $$\mathcal{CS}$$-maxiconsistency of $$end\big (m \sqcap h\big )$$, that $$KB \in end\big (m \sqcap h\big )$$, contrary to our assumption. Therefore, Ξ is $$\mathcal{CS}$$-consistent and we can extend Ξ to a $$\mathcal{CS}$$-maxiconsistent Δ ⊆ FormAg. Of course, in this case $$\Box B \notin \Delta$$. We will have then that $$\big (\Delta \big )$$ is an element. So we set $$m^{\prime} = \big [\big (\Delta \big )\big ]_{\equiv }$$. Now choose an arbitrary (Δ′) ∈ m′. We will have, by the choice of m′, that $$\big (\Delta ^{\prime}\big ) \equiv \big (\Delta \big )$$. Then every boxed formula from Δ will be in Δ′. In particular, whenever $$KC \in end\big (m \sqcap h\big )$$ for a given C ∈ FormAg, it follows that KC ∈ Δ, whence $$\Box KC \in \Delta$$ and thus $$KC \in \Delta ^{\prime} = end\big (\big (\Delta ^{\prime}\big )\big )$$, by (T3) and maxiconsistency of Δ. Since $$\big (\Delta ^{\prime}\big )$$ was chosen as an arbitrary element of m′, this means that whenever KC ∈ end(m ⊓ h), it is also the case that $$KC \in \bigcap _{\tau \in m^{\prime}}end(\tau )$$ so that we must have R(m, m′). On the other hand, since $$\Box B \notin \Delta$$, then, by Case 1, there must be a τ ∈ m′ such that B ∉ end(τ). But then, by Lemma 12, we can choose an h′ ∈ Hm′ in such a way that τ = m′ ⊓ h′, and we get that $$B \notin end\big (m^{\prime} \sqcap h^{\prime}\big )$$. Therefore, by induction hypothesis, we get $$\mathcal{M}^{Ag}_{\mathcal{CS}},m^{\prime},h^{\prime} \not \models B$$. In view of the fact that also $$R\big (m,m^{\prime}\big )$$, this means that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \not \models KB$$ as desired. Case 4. A = t:B for some t ∈ Pol. If $$t{:} B \in end\big (m\ \sqcap\ h\big )$$, then, by $$\mathcal{CS}$$-maxiconsistency of $$end\big (m\ \sqcap\ h\big )$$ and (T2), we must have $$\Box t{:} B \in end\big (m \sqcap h\big )$$. Now if ξ ∈ m, then we must have, of course, ξ ≡ m ⊓ h, whence $$t{:} B \in end\big (\xi \big )$$. Therefore, we must have $$B \in \mathcal{E}(m,t)$$. Also, by maxiconsistency of $$end\big (m\ \sqcap \ h\big )$$ and (A5), we will have $$KB \in end\big (m\ \sqcap \ h\big )$$. Therefore, by Case 3, we will have that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models KB$$ and further, by $$B \in \mathcal{E}(m,t)$$, that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models t{:} B$$. On the other hand, if $$t{:} B \notin end\big (m \sqcap h\big )$$, then since clearly m ⊓ h ∈ m, we must have $$B \notin \mathcal{E}(m,t)$$, whence $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \not \models t{:} B$$. Case 5. A = Et for some t ∈ Pol. Then, given that m ∈ Tree ∖ {†, ‡}, we have, simply by definition of Act, that $$Et \in end\big(m \sqcap h\big) \Leftrightarrow t \in Act\big(m,h\big) \Leftrightarrow \mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models Et.$$ This finishes the list of the modal induction cases at hand, and thus the proof of our truth lemma is complete. 5 The main result We are now in a position to prove Theorem 2. One direction of the theorem was proved as Corollary 1. In the other direction, assume that Γ ⊆ FormAg is $$\mathcal{CS}$$-consistent. Then, by Lemma 4.1, Γ can be extended to a $$\mathcal{CS}$$-maxiconsistent Δ. Now, consider $$\mathcal{M}^{Ag}_{\mathcal{CS}} = \big \langle Tree, \unlhd , Choice, Act, R, \mathcal{E}, V\big \rangle$$, the canonical model defined in Section 4. Since $$\mathcal{M}^{Ag}_{\mathcal{CS}} \in Mod^{\downarrow }_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{discr}\right )$$, we know that $$\mathcal{M}^{Ag}_{\mathcal{CS}} \in Mod^{\downarrow }_{\mathcal{CS}}\left (\mathcal{C}\right ) \subseteq Mod_{\mathcal{CS}}\left (\mathcal{C}\right )$$. The structure $$\big (\Delta \big )$$ is an element, therefore $$\big [\big (\Delta \big )\big ]_{\equiv } \in Tree$$. By Lemma 12, there is a history $$h \in H_{[(\Delta )]_{\equiv }}$$ such that $$\big (\Delta \big ) = \big [\big (\Delta \big )\big ]_{\equiv } \sqcap h$$. For this h, we will also have $$\Delta = end\big (\big [\big (\Delta \big )\big ]_{\equiv } \sqcap h\big )$$. By Lemma 19, we therefore get that $$\mathcal{M}^{Ag}_{\mathcal{CS}}, \big[\big(\Delta\big)\big]_{\equiv}, h \models \Delta \supseteq \Gamma,$$ and thus Γ is shown to be satisfiable in a jstit model based on frame from $$\mathcal{C}$$. We observe that it follows from the above proof that the canonical model $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ defined in Section 4 is $$\mathcal{CS}$$-universal in the sense that it satisfies every $$\mathcal{CS}$$-consistent subset of FormAg. As standard corollaries to Theorem 2, we get the respective weak completeness and compactness results. 6 Conclusions We have shown that the Hilbert-style axiomatic system ΣD is strongly complete w.r.t. the set of JA-STIT formulas valid over jstit models based on discrete times and that this set does not change as long as one adds further models to this class, provided that these models are based on mixed successor stit frames. We have also shown that this result is stable modulo constant specifications that might be added to ΣD. The fact that claiming the full set of ΣD-theorems for the class of models based on a given stit frame C does not ensure that the C will be based on discrete time lights up some essential limitations of expressive powers of JA-STIT and calls for further investigation. Therefore, in the second part of the paper we will look in the reverse direction, and instead of finding an axiomatic system for a given class of frames we will identify classes of frames defined by ΣD. We will present two main results, one for the stit frame as defined above and another one for jstit frames, which extend stit frames with the epistemic accessibility relations R and Re. The presence of these relations happens to be a game changer in that the form of frame definability result for jstit frames is very different from the respective theorem for stit frames. Funding The author would like to acknowledge financial support from the Deutsche Forschungsgemeinschaft, DFG, project WA 936/11-1. Acknowledgements The author would like to thank the anonymous reviewer of this paper, who made a lot of useful comments which led to some corrections and other improvements in the presentation. Footnotes 1  Backus-Naur Form. 2  A more common notation ≤ is not convenient for us since we also widely use ≤ in this paper to denote the natural order relation between elements of ω. 3  In other words, our stit frames are just a propositional version of what [4] calls BT+AC structures. A more inclusive notion of jstit frame will be considered in some detail in Part II of this paper. 4  One exception is the class of colinear frames only mentioned in this paper for technical reasons, although our axiomatization is sound also w.r.t. this class. 5  See, e.g. [4, Ch. 17], although ΣD uses a simpler format closer to that given in [3, Section 2.3]. 6  The format for the variable assignment V is slightly different, but this is of no consequence for the present setting. References [1] S. Artemov and M. Fitting . Justification Logic , E. N. Zalta ed. , winter 2016 edn. The Stanford Encyclopedia of Philosophy . Metaphysics Research Lab, Stanford University, 2016 . [2] S. N. Artemov and E. Nogina . Introducing justification into epistemic logic . Journal of Logic and Computation , 15 , 1059 -- 1073 , 2005 . Google Scholar CrossRef Search ADS [3] P. Balbiani , A. Herzig and N. Troquard . Alternative axiomatics and complexity of deliberative stit theories . Journal of Philosophical Logic , 37 , 387 -- 406 , 2008 . Google Scholar CrossRef Search ADS [4] N. Belnap , M. Perloff , and M. Xu . Facing the Future: Agents and Choices in Our Indeterminist World . Oxford University Press , 2001 . [5] J. Horty . Agency and Deontic Logic . Oxford University Press , USA , 2001 . Google Scholar CrossRef Search ADS [6] G. Olkhovikov . Stit logic of justification announcements: a completeness result . Journal of Logic and Computation , (to appear) , 2018 . Google Scholar CrossRef Search ADS [7] G. Olkhovikov and H. Wansing . Inference as doxastic agency. Part I: the basics of justification stit logic. Studia Logica, 2018 . Online first : https://doi.org/10.1007/s11225-017-9779-z. [8] G. Olkhovikov and H. Wansing . Inference as doxastic agency. Part II: ramifications and refinements . Australasian Journal of Philosophy , 14 , 408 -- 438 , 2017 . © The Author(s) 2018. Published by Oxford University Press. All rights reserved. For Permissions, please email: journals.permissions@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/open_access/funder_policies/chorus/standard_publication_model) http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Logic Journal of the IGPL Oxford University Press

# Justification announcements in discrete time. Part I: Completeness results

, Volume Advance Article (3) – Feb 27, 2018
31 pages

/lp/ou_press/justification-announcements-in-discrete-time-part-i-completeness-ueD0Z52EX1
Publisher
Oxford University Press
ISSN
1367-0751
eISSN
1368-9894
D.O.I.
10.1093/jigpal/jzy002
Publisher site
See Article on Publisher Site

### Abstract

Abstract We axiomatize stit logic of justification announcements (JA-STIT) interpreted over models with discrete time structure. The resulting Hilbert-style axiomatic system ΣD is strictly stronger than the system Σ presented in the study by Olkhovikov (2018, J. Logic Comput., to appear) as axiomatization of JA-STIT over the general class of justification stit models. We then prove a rather general strong completeness theorem for ΣD. 1 Introduction The so-called stit logic of justification announcements (JA-STIT, for short) was introduced in [6] as an explicit fragment of the richer logic of E-notions defined in [8]. The underlying idea was to interpret the proving activity as an activity that results in presenting (or, as it were, demonstrating) proofs to the community, thus making them publicly available within this community. In this way, the JA-STIT model of proving activity involves the following three main components: (i) the agents presenting the proofs to the community, (ii) the proofs to be presented and (iii) an interface for the interaction of agents with proofs. JA-STIT borrows its picture of (i) from stit logic by N. Belnap et al. [4] and its picture of (i) from epistemic justification logic by S. Artemov et al. [2]. This logic, therefore, retains the full set of expressive means of the two abovementioned logics, which places it in the landscape of justification stit (or jstit, for short) logics explored from the conceptual viewpoint in [7] and [8]. As for (iii), the third main component above, the underlying intuitions here are an idealized and abstracted version of a (not uncommon) situation when a group of agents tries to produce a proof working on a common whiteboard. This whiteboard then is their main medium for making different possible proofs epistemically transparent to themselves and their colleagues. If in this setting we abstract away (i) from the other available media (like private notes, private messages, etc.), (ii) from the natural limitations of the actual whiteboard (like limited space and necessity to erase old proofs) and (iii) from the natural limitations of the agents’ communication capacities (like bad handwriting on the part of presenting agents or short sightedness on the part of spectators) we get the following picture for (iii): A common pool of presented proofs is supposed to exist; This pool is the unique medium connecting the realm of abstract proofs to the world in which agents act; This pool can hold an unlimited number of proofs; Once a proof is in the pool, it stays there forever; Once a proof is in the pool, it is immediately understood and recognized by every agent in the community. The state of this pool, or of the community whiteboard, as we will sometimes call it, is described in JA-STIT by modality Et, where t is an arbitrary proof polynomial of justification logic. The informal interpretation of Et is that the proof t is presented to the community, or that t is on the whiteboard. This reading also explains the choice of E as notation for this modality since it serves as a sort of existence predicate for the pool of proofs publicly announced within the community. The axiomatization of JA-STIT w.r.t. the full class of its intended models was given in [6]. At the same time, Proposition 1 of [6] showed that, rather surprisingly, this axiomatization is sensitive to the temporal structure of the underlying models, even though none of the modalities available within JA-STIT can look outside a given moment, and the standard temporal modalities seem to be undefinable. Nevertheless, it turned out that once the class of underlying models is restricted to the models based on discrete time, the axiomatization is no longer complete. This situation raises the natural question of whether it is possible to axiomatize the set of JA-STIT validities over the class of models with discrete temporal substructure. This first part of the present paper answers this question by providing the respective axiomatization and showing it to be strongly complete. The layout of the rest of the paper is as follows. In Section 2 we define the language and the semantics of the logic at hand. We then define the notion of a stit frame and consider some natural subclasses of stit frames, which will be relevant to the main part of the paper and will appear in the results presented in Section 5. We briefly discuss set-theoretical relations between these classes. Our axiomatization itself is then presented in Section 3. We show this system to be sound w.r.t. the classes introduced in Section 2, both in its minimal form and modulo the so-called constant specifications inherited by JA-STIT from justification logic. We also define an alternative axiomatization of this same system in Lemma 6. Section 4 then contains the bulk of technical work necessary for the completeness of the presented axiomatization. It provides a stepwise construction and adequacy check for all the numerous components of the canonical model and ends with a proof of a truth lemma. Section 5 gives a concise proof of the completeness result for most of the subclasses of frames defined in Section 2 and draws some standard corollaries. Section 6 then formulates some conclusions and contains an announcement of the contents of Part II. The final section is devoted to acknowledgments. In what follows we will be assuming, due to space limitations, a basic acquaintance with both stit logic and justification logic. We recommend to peruse [5, Ch. 2] for a quick introduction into the basics of stit logic, and [1] for the same w.r.t. justification logic. 2 Basic definitions and notation We fix some preliminaries. First, we choose a finite non-empty set Ag disjoint from all the other sets to be defined below. Individual agents from this set will be denoted by letters i and j. Then we fix countably infinite sets PV ar of proof variables (denoted by x, y, z) and PConst of proof constants (denoted by c, d). When needed, subscripts and superscripts will be used with the above notations or any other notations to be introduced in this paper. Set Pol of proof polynomials is then defined by the following BNF:1 $$t := x \mid c \mid s + t \mid s \cdot t \mid !t,$$ with x ∈ PV ar, c ∈ PConst and s, t ranging over elements of Pol. In the above definition + stands for the sum of proofs, ⋅ denotes application of its left argument to the right one and ! denotes the so-called proof checker, so that !t checks the correctness of proof t. In order to define the set FormAg of formulas we fix a countably infinite set V ar of propositional variables to be denoted by letters p, q. Formulas themselves will be denoted by letters A, B, C, D, and the definition of FormAg is supplied by the following BNF: $$A := p \mid A \wedge B \mid \neg A \mid [j]A \mid \Box \ A \mid t{:} A \mid K\ A \mid Et,$$ with p ∈ V ar, j ∈ Ag and t ∈ Pol. It is clear from the above definition of FormAg that we are considering a version of modal propositional language. As for the informal interpretations of modalities, $$\big [j\big ]A$$ is the so-called cstit action modality and $$\Box$$ is the historical necessity modality, both modalities are borrowed from stit logic. The next two modalities, K A and t:A, come from epistemic justification logic and the latter is interpreted as ‘t proves A’, whereas the former is the strong epistemic modality ‘A is known’. We assume ◊ as notation for the dual modality of $$\Box$$ and ⟨K⟩ for the dual of K. As usual, ω will denote the set of natural numbers including 0, ordered in the natural way. For the language at hand, we assume the following semantics. A justification stit (jstit) model for Ag is a structure $$\mathcal{M} = \big\langle Tree, \unlhd, Choice, Act, R, R_{e}, \mathcal{E}, V\big\rangle$$ such that Tree is a non-empty set. Elements of Tree are called moments. $$\unlhd$$ is a partial order on Tree for which a temporal interpretation is assumed. We will also freely use notations like $$\unrhd$$, $$\lhd$$ and $$\rhd$$ to denote the inverse relation and the irreflexive companions.2 Hist is a set of maximal chains in Tree w.r.t. $$\unlhd$$. Since Hist is completely determined by Tree and $$\unlhd$$, it is not included into the structure of model as a separate component. Elements of Hist are called histories. The set of histories containing a given moment m will be denoted Hm. The following set $$MH(\mathcal{M}) = \big\{ \big(m,h\big)\mid m \in Tree,\, h \in H_{m} \big\},$$ called the set of moment-history pairs, will be used to evaluate formulas of the above language. Choice is a function mapping Tree × Ag into $$2^{2^{Hist}}$$ in such a way that for any given j ∈ Ag and m ∈ Tree we have as $$Choice\big (m,j\big )$$ (to be denoted as $$Choic{e^{m}_{j}}$$ below) a partition of Hm. Intuitively, the sets of the form $$Choic{e^{m}_{j}}$$ may be thought of as the choice cells of agent j at moment m, the idea being that j cannot distinguish by her or his activity at m between histories that belong to one and the same choice cell. The histories in a choice cell from $$Choic{e^{m}_{j}}$$ are choice-equivalent for j at m. For a given h ∈ Hm we will denote by $$Choic{e^{m}_{j}}\big (h\big )$$ the element of partition $$Choic{e^{m}_{j}}$$ containing h. Act is a function mapping $$MH(\mathcal{M})$$ into 2Pol. R and Re are two pre-orders on Tree giving two versions of epistemic accessibility relation. They are assumed to be connected by inclusion R ⊆ Re. $$\mathcal{E}$$ is a function mapping Tree × Pol into $$2^{Form^{Ag}}$$ called admissible evidence function. V is an evaluation function, mapping the set V ar into $$2^{MH\left (\mathcal{M}\right )}$$. However, not all structures of the above described type are admitted as jstit models. A number of additional restrictions need to be satisfied. In order to facilitate the exposition of these restrictions, we introduce a couple of useful notations first. For a given m ∈ Tree and any given h, g ∈ Hm we stipulate that $$Act_{m} := \bigcap_{h \in H_{m}}\big(Act\left(m,h\right),$$ and $$h \approx_{m} g \Leftrightarrow h,g \in H_{m} \,\&\,\left(h = g \vee \left(\exists m^{\prime} \rhd m\right)\left(h, g \in H_{m^{\prime}}\right)\right).$$ Whenever we have h ≈mg, we say that h and g are undivided at m. We then demand satisfaction of the following constraints by the jstit models: Historical connection: $$\left(\forall m,m_{1} \in Tree\right)\left(\exists m_{2} \in Tree\right)\left(m_{2} \unlhd m \wedge m_{2} \unlhd m_{1}\right).$$ No backward branching: $$\left(\forall m,m_{1},m_{2} \in Tree\right)\left(\left(m_{1} \unlhd m \wedge m_{2} \unlhd m\right) \Rightarrow \left(m_{1} \unlhd m_{2} \vee m_{2} \unlhd m_{1}\right)\right).$$ No choice between undivided histories: $$\left(\forall m \in Tree\right)\left(\forall h,h^{\prime} \in H_{m}\right)\left(h \approx_{m} h^{\prime} \Rightarrow Choic{e^{m}_{j}}\left(h\right) = Choic{e^{m}_{j}}\left(h^{\prime}\right)\right)$$ for every j ∈ Ag. Independence of agents: $$\left(\forall m\in Tree\right)\left(\forall f:Ag \rightarrow 2^{H_{m}}\right)\left(\left(\forall j \in Ag\right)\left(f\big(j\big) \in Choic{e^{m}_{j}}\right) \Rightarrow \bigcap_{j \in Ag}f\big(j\big) \neq \emptyset\right).$$ Monotonicity of evidence: $$\left(\forall t \in Pol\right)\left(\forall m,m^{\prime} \in Tree\right)\left(R_{e}\left(m,m^{\prime}\right) \Rightarrow \mathcal{E}\left(m,t\right) \subseteq \mathcal{E}\left(m^{\prime},t\right)\right).$$ Evidence closure properties. For arbitrary m ∈ Tree, s, t ∈ Pol and A, B ∈ FormAg it is assumed that (a) $$A \rightarrow B \in \mathcal{E}(m,s) \wedge A \in \mathcal{E}(m,t) \Rightarrow B \in \mathcal{E}(m,s\cdot t)$$; (b) $$\mathcal{E}(m,s) \cup \mathcal{E}(m,t) \subseteq \mathcal{E}(m,s + t)$$; (c) $$A \in \mathcal{E}(m,t) \Rightarrow t{:} A \in \mathcal{E}(m,!\!t)$$. Expansion of presented proofs: $$\left(\forall m,m^{\prime} \in Tree\right)\left(m^{\prime} \lhd m \Rightarrow \forall h \in H_{m} \left(Act\left(m^{\prime},h\right) \subseteq Act\left(m,h\right)\right)\right).$$ No new proofs guaranteed: $$\left(\forall m \in Tree\right)\left(Act_{m} \subseteq \bigcup_{m^{\prime} \lhd m, h \in H_{m}}\left(Act\left(m^{\prime},h\right)\right)\right).$$ Presenting a new proof makes histories divide: $$\left(\forall m \in Tree\right)\left(\forall h,h^{\prime} \in H_{m}\right)\left(h \approx_{m} h^{\prime} \Rightarrow \left(Act\left(m,h\right) = Act\left(m,h^{\prime}\right)\right)\right).$$ Future always matters: $$\unlhd \subseteq R.$$ Presented proofs are epistemically transparent: $$\left(\forall m,m^{\prime} \in Tree\right)\left(R_{e}\left(m,m^{\prime}\right) \Rightarrow \left(Act_{m} \subseteq Act_{m^{\prime}}\right)\right).$$ We offer some intuitive explanation for the abovedefined notion of jstit model. Due to space limitations, we only explain the intuitions behind jstit models very briefly, and we urge the reader to consult [7, Section 3] for a more comprehensive explanations, whenever needed. The components like Tree, $$\unlhd$$, Choice and V are inherited from stit logic, whereas R, Re and $$\mathcal{E}$$ come from epistemic justification logic. The only new component is Act which represents the abovementioned common pool of proofs demonstrated to the community or the state of the community whiteboard at any given moment under a given history. When interpreting Act, we invoke the classical stit distinction between dynamic (agentive) and static (moment determinate) entities, assuming that the presence of a given proof polynomial t on the community whiteboard only becomes an accomplished fact at m when t is present in $$Act\left (m,h\right )$$ for every h ∈ Hm. On the other hand, if t is in Act(m, h) only for some h ∈ Hm this means that t is rather in a dynamic state of being presented, rather than being present, to the community. The numbered list of semantical constraints above then just builds on these intuitions. Constraints 1–4 are borrowed from stit logic, constraints 5 and 6 are inherited from justification logic. Constraint 7 just says that nothing gets erased from the whiteboard, constraint 8 says a new proof cannot spring into existence as a static (i.e. moment determinate) feature of the environment out of nothing, but rather has to come as a result (or a by-product) of a previous activity. Constraint 9 is just a corollary to constraint 3 in the richer environment of jstit models, constraint 10 says that the possible future of the given moment is always epistemically relevant in this moment and constraint 11 says that the community knows everything that has firmly made its way onto the whiteboard. For the members of FormAg, we will assume the following inductively defined satisfaction relation. For every jstit model $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e}, \mathcal{E}, V\big \rangle$$ and for every $$\left (m,h\right ) \in MH\left (\mathcal{M}\right )$$ we stipulate that \begin{align*} &\mathcal{M}, m, h \models p \Leftrightarrow \big(m,h\big) \in V\big(p\big);\\ &\mathcal{M}, m, h \models \big[j\big]\,A \Leftrightarrow \left(\forall h^{\prime} \in Choic{e^{m}_{j}}\big(h\big)\right)\left(\mathcal{M}, m, h^{\prime} \models A\right);\\ &\mathcal{M}, m, h \models \Box\, A \Leftrightarrow \left(\forall h^{\prime} \in H_{m}\right)\left(\mathcal{M}, m, h^{\prime} \models A\right);\\ &\mathcal{M}, m, h \models K\ A \Leftrightarrow \forall m^{\prime}\forall h^{\prime}\left(R\left(m,m^{\prime}\right) \& h^{\prime} \in H_{m^{\prime}} \Rightarrow \mathcal{M}, m^{\prime}, h^{\prime} \models A\right);\\ &\mathcal{M}, m, h \models t{:} A \Leftrightarrow A \in \mathcal{E}\left(m,t\right) \& \forall m^{\prime}\forall h^{\prime}\left(R_{e}\left(m,m^{\prime}\right)\! \& h^{\prime} \in H_{m^{\prime}} \Rightarrow \mathcal{M}, m^{\prime}, h^{\prime} \models A\right);\\ &\mathcal{M}, m, h \models Et \Leftrightarrow t \in Act\big(m,h\big). \end{align*} In the above clauses we assume that p ∈ Var; we also assume standard clauses for Boolean connectives. In addition to jstit models, we will need a notion of stit frame. If $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e},$$$$\mathcal{E}, V\big \rangle$$ is a jstit model for Ag, then $$C = \big \langle Tree, \unlhd , Choice\big \rangle$$ is a stit frame for Ag.3 In this case, $$\mathcal{M}$$ is said to be based on C. Given a class $$\mathcal{C}$$ of stit frames, we will denote the class of jstit models based on the frames from $$\mathcal{C}$$ by $$Mod(\mathcal{C})$$. We also observe that notations like Hist(C) and MH(C) still make perfect sense when C is a stit frame. The following lemmas collect some very basic technical facts which will be used in what follows: Lemma 1 Let $$C = \big \langle Tree, \unlhd , Choice \big \rangle$$ be a stit frame. Then (1) $$\big (\forall m \in Tree\big )\big (\forall h \in H_{m}\big )\big (\big \{ m_{1} \in Tree \mid m_{1} \unlhd m \big \} \subseteq h\big )$$; (2) $$\big (\forall m \in Tree\big )\big (\forall h,g \in H_{m}\big )\big (h \neq g \Rightarrow \big (\exists m^{\prime} \rhd m\big )\big (h \in H_{m^{\prime}}\big )\big )$$; (3) $$\big (\forall m, m^{\prime} \in Tree\big )\big (m \unlhd m^{\prime} \Rightarrow H_{m^{\prime}} \subseteq H_{m}\big )$$; (4) ≈m is an equivalence relation on Hm for every m ∈ Tree. Proof. (Part 1). We clearly have m ∈ h. Consider an arbitrary $$m_{1} \lhd m$$. Then h ∪ {m1} must be an $$\unlhd$$-chain. Indeed, if m′ ∈ h then either $$m \unlhd m^{\prime}$$ or $$m^{\prime} \lhd m$$. In the former case we get $$m_{1} \unlhd m^{\prime}$$ by transitivity of $$\unlhd$$, in the latter case we get $$m_{1} \unlhd m^{\prime} \vee m^{\prime} \unlhd m_{1}$$ by the absence of backward branching. But since h is a maximal chain, this means that we must have m1 ∈ h. (Part 2). To obtain a contradiction, assume that h, g ∈ Hm are different, but we have $$\big(\forall m^{\prime} \rhd m\big)\big(m^{\prime} \notin h\big)$$ (1) Given that every two elements of h must be $$\unlhd$$-comparable, this means that $$h \subseteq \big \{ m_{1} \in Tree \mid m_{1} \unlhd m \big \}$$ and, by Part 1, that $$h = \big \{ m_{1} \in Tree \mid m_{1} \unlhd m \big \}$$. Note that Part 1 also entails that $$g \supseteq \big \{ m_{1} \in Tree \mid m_{1} \unlhd m \big \}$$, so that in this case we must have $$g \supseteq h$$. We can have neither $$g \supset h$$, since this would violate the maximality of h, nor g = h, since this is in contradiction with our assumption. Therefore, (1) must be false. (Part 3). Immediately by the absence of backward branching. (Part 4). Reflexivity and symmetry are straightforward. We establish transitivity. Let m ∈ Tree and let h, h$$^{\prime}$$, h$$^{\prime\prime}$$ ∈ Hm are such that h ≈mh$$^{\prime}$$ and h$$^{\prime}$$ ≈mh$$^{\prime\prime}$$. If h = h$$^{\prime}$$ or h$$^{\prime}$$ = h$$^{\prime\prime}$$, then h ≈ h$$^{\prime\prime}$$ follows trivially. Therefore, assume that both $$h \neq h^{\prime}$$ and $$h^{\prime} \neq h^{\prime\prime}$$. Then choose $$m^{\prime}, m^{\prime\prime} \rhd m$$ such that m$$^{\prime}$$ ∈ h ∩ h$$^{\prime}$$ and m$$^{\prime\prime}$$ ∈ h$$^{\prime}$$ ∩ h$$^{\prime\prime}$$. Then both m$$^{\prime}$$ and m$$^{\prime\prime}$$ are in h$$^{\prime}$$ and hence must be $$\unlhd$$-comparable. Setting m0 as the minimal of the two moments, we get that, by Part 3, $$h, h^{\prime}, h^{\prime\prime} \in H_{m_{0}}$$ and $$m_{0} \rhd m$$, whence h ≈mh$$^{\prime\prime}$$. Lemma 2 Let $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e}, \mathcal{E}, V \big \rangle$$ be a jstit model. Then $$\big(\forall m,m^{\prime} \in Tree\big)\big(\forall h \in H_{m^{\prime}}\big)\big(\forall t \in Pol\big)\big(m \lhd m^{\prime} \& t\in Act\big(m,h\big) \Rightarrow t \in Act_{m^{\prime}}\big).$$ Proof. Let m, m$$^{\prime}$$ ∈ Tree, h ∈ Hm′ and t ∈ Pol be such that $$m \lhd m^{\prime} \& t\in Act\big (m,h\big )$$. Choose an arbitrary g ∈ Hm′. By Lemma 1.3, we know that g ∈ Hm and that also h ∈ Hm. Therefore, h ≈mg whence by the presenting a new proof makes histories divide constraint we get that $$Act\big (m,h\big ) = Act\big (m,g\big )$$. Therefore, $$t \in Act\big (m,g\big )$$ and, by the expansion of presented proofs, $$t \in Act\big (m^{\prime},g\big )$$. Since g ∈ Hm′ was chosen arbitrarily, this means that t ∈ Actm′. An important subclass of jstit models is made up of what we will call unirelational jstit models. These are the models satisfying the additional constraint that Re ⊆ R. It is known (see [2, Comment 6.5]) that switching from the full class of models to the unirelational models in the context of pure epistemic justification logic still leaves us with a class of models w.r.t. which the logic is complete and we have shown in [6] that the same holds for JA-STIT. In this paper, we will eventually show that this situation does not change if the class of intended models for JA-STIT is restricted to any of its natural subclasses that we define within the present section.4 Since the class of unirelational jstit models will be thus important to us in what follows, we observe that switching to unirelational models allows to somewhat simplify the whole setting. First, one can simply throw away Re from the model structure and redefine the model as a tuple of the form $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, \mathcal{E}, V\big \rangle$$, as long as one also updates the semantical constraints, replacing Re everywhere with R. The same simplification can be introduced to the definition of jstit frame. Second, one can somewhat simplify the satisfaction clause for t: A replacing it with: $$\mathcal{M}, m, h \models t{:} A \Leftrightarrow A \in \mathcal{E}(m,t) \& \mathcal{M}, m, h \models KA.$$ Moreover, whenever $$\mathcal{C}$$ is a class of stit frames we will denote by $$Mod^{\downarrow }(\mathcal{C})$$ the class of unirelational jstit models based on frames from $$\mathcal{C}$$. Before we move on, we need to introduce the notation for an immediate $$\lhd$$-successor of a given moment as it will play an important part in the frame restrictions to be introduced below. So whenever $$C = \big \langle Tree, \unlhd , Choice\big \rangle$$ is a stit frame and m, m$$^{\prime}$$ ∈ Tree, we set that $$Next\left(m,m^{\prime}\right) \Leftrightarrow \left(m \lhd m^{\prime} \& \left(\forall m^{\prime\prime} \lhd m^{\prime}\right)\left(m^{\prime\prime} \unlhd m\right)\right).$$ We now list the defining conditions for some natural subclasses of stit frames: Definition 1 Let $$C = \big \langle Tree, \unlhd , Choice\big \rangle$$ be a stit frame. Then we say that C is based on discrete time iff every history in Hist(C) can be embedded into an initial (but maybe improper) segment of ω; C is a well-ordered frame iff every h ∈ Hist(C) is well ordered by $$\unlhd$$. C is a successor frame iff the following holds: $$\big(\forall m,m_{1} \in Tree\big)\big(m \lhd m_{1} \Rightarrow \big(\exists m_{2} \unlhd m_{1}\big)\big(Next\big(m, m_{2}\big)\big)$$ (succ) C is a colinear frame iff $$\big (\forall m \in Tree\big )\big (\forall h,g \in H_{m}\big )\big (h \approx _{m} g\big )$$. C is a mixed successor frame iff for all m, m1 ∈ Tree it is true that $$\big[m \lhd m_{1} \Rightarrow \big(\exists m_{2} \unlhd m_{1}\big)\big(Next\big(m, m_{2}\big)\big)\big] \vee \big[\big(\forall h,g \in H_{m}\big)\big(h \approx_{m} g\big)\big]$$ (mixsucc) Every property in the above definition induces the corresponding class of stit frames. We will denote the class of stit frames based on discrete time, the class of well-ordered stit frames, the class of successor stit frames, the class of colinear stit frames, and the class of mixed successor stit frames by $$\mathcal{C}_{discr}$$, $$\mathcal{C}_{wo}$$, $$\mathcal{C}_{succ}$$, $$\mathcal{C}_{col}$$ and $$\mathcal{C}_{mixsucc}$$, respectively. Further, whenever $$\mathcal{C}$$ is a class of stit frames we let $$\mathcal{C}^{Ag}$$ denote the class of all frames in $$\mathcal{C}$$ for a given community Ag. The following lemma sums up some quick facts about the defined classes: Lemma 3 Let Ag be a community of agents. Then $$\mathcal{C}^{Ag}_{discr} \subset \mathcal{C}^{Ag}_{wo} \subset \mathcal{C}^{Ag}_{succ}$$; $$\mathcal{C}^{Ag}_{succ} \cup \mathcal{C}^{Ag}_{col} \subset \mathcal{C}^{Ag}_{mixsucc}$$. Proof. Part 1 is straightforward. As for Part 2, it is clear that we must have $$\mathcal{C}^{Ag}_{succ} \cup \mathcal{C}^{Ag}_{col} \subseteq \mathcal{C}^{Ag}_{mixsucc}$$. To see that this inclusion is proper, consider for instance the stit frame $$C = \big \langle Tree, \unlhd , Choice\big \rangle$$ for Ag, where Tree = {1/2n∣n ∈ ω} ∪ {0, −1, a}, $$\unlhd = \leq \cup \{ (-1, a), (a,a) \}$$ (with ≤ being the natural order on rational numbers) and $$Choic{e^{m}_{j}} = H_{m}$$ for all m ∈ Tree and j ∈ Ag. It is easy to see that $$Hist(C) = \big \{ h_{1}, h_{2} \big \}$$, where h1 = (−1, a) and h2 = (−1, 0, …, 1/2n, …, 1/2, 1). Since h1 ≉−1h2, we get that $$C \notin \mathcal{C}^{Ag}_{col}$$. Further, we have $$0 \lhd 1$$, but there exists no minimal $$\lhd$$-successor of 0. Therefore, $$C \notin \mathcal{C}^{Ag}_{succ}$$ as well. On the other hand, the only moment in Tree for which (succ) is violated is 0, and we clearly have H0 = {h2}, therefore $$C \in \mathcal{C}^{Ag}_{mixsucc}$$. Note that we have a natural hierarchy of restrictions for the interval $$\big (\mathcal{C}_{discr},\mathcal{C}_{succ}\big )$$, whereas $$\mathcal{C}_{col}$$ uses a different principle and the construction of $$\mathcal{C}_{mixsucc}$$ looks artificial. The reason for our mentioning $$\mathcal{C}_{mixsucc}$$ is that due to the limitations of expressive power of JA-STIT language the stit frame definability result to be given in Part II of this paper has to use $$\mathcal{C}_{mixsucc}$$ rather than the much more natural $$\mathcal{C}_{succ}$$. Before we move on to the next section, it is also worth noting that JA-STIT lacks finite model property in a rather strong sense since some satisfiable formulas cannot be satisfied over finite models, or even over infinite models where all histories are finite. The argument for this is the same as in implicit jstit logic, for which this claim was proved in [6] using formula $$K\big (\Diamond p \wedge \Diamond \neg p\big )$$ as an example. Here we add that, as a consequence of Lemma 3, this example is still valid if the full set of jstit models is restricted to $$Mod\big (\mathcal{F}\big )$$ whenever $$\mathcal{C} \in \big \{ \mathcal{C}_{discr}, \mathcal{C}_{wo}, \mathcal{C}_{succ}, \mathcal{C}_{mixsucc} \big \}$$, since the model used in [6] for $$K\big (\Diamond p \wedge \Diamond \neg p\big )$$ extends a stit frame based on discrete time. 3 Axiomatic system and soundness From this moment on, Ag will refer to an arbitrary but fixed agent community. We start by defining a Hilbert-style axiomatic system ΣD. For this system, we fix the following set of axiomatic schemes: $$\text{A full set of axioms for classical propositional logic}$$ (A0) $$S5\text{ axioms for }\Box \text{ and }\big[j\big]\text{ for every }j \in Ag$$ (A1) $$\Box A \rightarrow \big[j\big]\,A \text{ for every }j \in Ag$$ (A2) $$\big(\Diamond\big[j_{1}\big]\,A_{1} \wedge\ldots \wedge \Diamond\big[j_{n}\big]\,A_{n}\big) \rightarrow \Diamond\big(\big[j_{1}\big]\,A_{1} \wedge\ldots \wedge\big[j_{n}\big]\,A_{n}\big)$$ (A3) $$(s{:}(A \rightarrow B) \rightarrow (t{:} A \rightarrow (s\cdot t){:} B)$$ (A4) $$t{:} A \rightarrow (!t{:}(t{:} A) \wedge K\ A)$$ (A5) $$(s{:} A \vee t{:} A) \rightarrow (s+t){:} A$$ (A6) $$S4\text{ axioms for }K$$ (A7) $$K\ A \rightarrow \Box K\Box \,A$$ (A8) $$\Box \, Et \rightarrow K\Box \,Et$$ (A9) \begin{align} &K\big(\neg\Box\, Et_{1} \vee\ldots\vee\neg\Box \, Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k}\big) \rightarrow\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\rightarrow\big(\neg Et_{1} \vee\ldots\vee\neg Et_{n}\vee Es_{1} \vee\ldots\vee Es_{k}\big) \end{align} (A10D) The assumption is that in (A3) j1, …, jn are pairwise different. The rules of inferences are then as follows: $$\text{From }A, A \rightarrow B \text{ infer } B;$$ (R1) $$\!\!\!\!\!\!\!\!\!\!\!\!\! \text{From }A\text{ infer }K\ A.$$ (R2) The axiomatic schemes (A0)–(A9), as well as the rules (R1)–(R2), are borrowed from the system Σ defined in [6] and retain their original names. The axiom (A10D) is a strengthening of the axiom (A10) from Σ. ΣD is a minimal system in which we make no assumptions as to the properties of proof constants. One standard way to extend a minimal system in the pure justification logic is to add a number of such assumptions. Normally, such extensions are called for to ensure that enough proofs for the axioms of the system are available. Although sets of such assumptions (also called constant specifications) are not valid in an arbitrary jstit model and thus require some narrowing of the class of intended models, the corresponding narrowing is rather straightforward and the completeness proof for the minimal system can be very easily accommodated to any such constant specification. In this way, an infinite number of systems extending the minimal system with a constant specification is taken on board in one sweeping move. In JA-STIT the situation is very similar, which allows us to give a somewhat more general form to our completeness proofs. More precisely, let us call a constant specification any set $$\mathcal{CS}$$ such that $$\mathcal{CS} \subseteq \{ c_{n}{:}\ldots c_{1}{:} A\mid c_{1},\ldots ,c_{n} \in PConst, A \text{ an instance of }$$ (A0)–(A10D)}; Whenever $$c_{n+1}{:} c_{n}{:}\ldots c_{1}{:} A \in \mathcal{CS}$$, then also $$c_{n}{:}\ldots c_{1}{:} A \in \mathcal{CS}$$. For a given constant specification, we can define the corresponding inference rule $$R_{\mathcal{CS}}$$ as follows: ($$R_{\mathcal{CS}}$$) $$\text{If }c_{n}{:}\ldots c_{1}{:} A \in \mathcal{CS},\text{ infer } c_{n}{:}\ldots c_{1}{:} A.)$$ We now define that $$\Sigma _{D}\left (\mathcal{CS}\right )$$ is just ΣD extended with the rule ($$R_{\mathcal{CS}}$$). Since ∅ is clearly one example of constant specification, we have that $$\Sigma _{D}\big (\emptyset \big ) = \Sigma _{D}$$ so that our initial axiomatic system is also in the class of systems of the form $$\Sigma _{D}(\mathcal{CS})$$. However, when $$\mathcal{CS} \neq \emptyset$$, the corresponding system $$\Sigma _{D}(\mathcal{CS})$$ will prove some formulas which are not valid even if we restrict our attention to jstit models based on any class of frames defined in Section 2. We therefore have to describe the restriction on jstit models which comes with a commitment to a given $$\mathcal{CS}$$. We then say that a jstit model $$\mathcal{M}$$ is $$\mathcal{CS}$$-normal iff it is true that $$\big(\forall c \in PConst\big)\big(\forall m \in Tree\big)\big(\big\{ A \mid c{:} A\in \mathcal{CS} \big\} \subseteq \mathcal{E}(m,c)\big),$$ where $$\mathcal{E}$$ is the $$\mathcal{M}$$s admissible evidence function. Again, it is easy to see that the class of ∅-normal jstit models is just the whole class of jstit models so that the representation $$\Sigma _{D}\big (\emptyset \big ) = \Sigma _{D}$$ does not place any additional restrictions on the class of intended models of ΣD. Whenever $$\mathcal{C}$$ is a class of stit frames, we will denote the class of $$\mathcal{CS}$$-normal (resp. unirelational) jstit models based on the frames from $$\mathcal{C}$$ by $$Mod_{\mathcal{CS}}\left (\mathcal{C}\right )$$ (resp. $$Mod^{\downarrow }_{\mathcal{CS}}(\mathcal{C})$$). We then define a proof in $$\Sigma _{D}\left (\mathcal{CS}\right )$$ as a finite sequence of formulas such that every formula in it is either an axiom or is obtained from earlier elements of the sequence by one of inference rules. A proof is a proof of its last formula. If an A ∈ FormAg is provable in $$\Sigma _{D}(\mathcal{CS})$$, we will write $$\vdash _{\mathcal{CS}} A$$. We say that Γ ⊆ FormAg is inconsistent in $$\Sigma _{D}(\mathcal{CS})$$ (or $$\mathcal{CS}$$-inconsistent) iff for some A1, …, An ∈ Γ we have $$\vdash _{\mathcal{CS}} (A_{1} \wedge \ldots \wedge A_{n}) \rightarrow \bot$$, and we say that Γ is consistent in $$\Sigma _{D}(\mathcal{CS})$$ (or $$\mathcal{CS}$$-consistent) iff it is not inconsistent in $$\Sigma _{D}(\mathcal{CS})$$. Γ is maxiconsistent in $$\Sigma _{D}(\mathcal{CS})$$ (or $$\mathcal{CS}$$-maxiconsistent) iff it is consistent in $$\Sigma _{D}(\mathcal{CS})$$ and no subset of FormAg, which is consistent in $$\Sigma _{D}(\mathcal{CS})$$, properly extends Γ. With this definition of inconsistency, we can do many familiar things, e.g. extend consistent sets with new formulas and eventually make them maxiconsistent. More precisely, the following lemma holds: Lemma 4 Let $$\mathcal{CS}$$ be a constant specification, let Γ ⊆ FormAg be consistent in $$\Sigma _{D}\left (\mathcal{CS}\right )$$ and let A, B ∈ FormAg. Then There exists a Δ ⊆ FormAg such that Δ is $$\mathcal{CS}$$-maxiconsistent and Γ ⊆ Δ. If Γ is $$\mathcal{CS}$$-maxiconsistent, then exactly one element of {A, ¬A} is in Γ. If Γ is $$\mathcal{CS}$$-maxiconsistent, then A ∨ B ∈ Γ iff (A ∈ Γ or B ∈ Γ). If Γ is $$\mathcal{CS}$$-maxiconsistent and A, (A → B) ∈ Γ, then B ∈ Γ. If Γ is $$\mathcal{CS}$$-maxiconsistent, then A ∧ B ∈ Γ iff (A ∈ Γ and B ∈ Γ). Proof. (Part 1) Just as in the standard case, we enumerate the elements of FormAg as A1, …, An, … and form the sequence of sets Γ1, …, Γn, …, such that Γ1 := Γ and for every natural i ≥ 1: $$\Gamma_{i + 1} := \begin{cases} \Gamma_{i}, & \text{ if }\Gamma_{i} \cup \big\{ A_ i \big\}\text{ is }\mathcal{CS}\text{-inconsistent;} \\ \Gamma_{i} \cup \big\{ A_ i \big\}, & \text{ otherwise.} \\ \end{cases}$$ We now define $$\Delta := \bigcup _{i \geq 1}\Gamma _{i}$$. Of course, we have Γ ⊆ Δ, and moreover, Δ is $$\mathcal{CS}$$-maxiconsistent. To see this, note that for every i ≥ 1 the set Γi is $$\mathcal{CS}$$-consistent by construction. Now, if Δ is $$\mathcal{CS}$$-inconsistent, then there must be a $$\Sigma _{D}\left (\mathcal{CS}\right )$$-valid implication from a finite conjunction of formulas in Δ to ⊥. These formulas must be mentioned in our numeration of FormAg so that the valid implication in question can presented as $$\vdash _{\mathcal{CS}} \left (A_{i_{1}} \wedge \ldots \wedge A_{i_{n}}\right ) \rightarrow \bot$$ for appropriate natural i1, …, in. Since all of $$A_{i_{1}}, \ldots , A_{i_{n}}$$ are in Δ, we must have, by the construction of Γ1, …, Γn, …, that $$A_{i_{1}}, \ldots , A_{i_{n}} \in \Gamma _{max\big (i_{1},\ldots , i_{n}\big )+1}$$. But then this latter set must be $$\mathcal{CS}$$-inconsistent which contradicts our construction. Further, if some $$\mathcal{CS}$$-consistent Ξ ⊆ FormAg is such that Δ ⊂ Ξ, then let An ∈ Ξ∖Δ. We must have then $$\Gamma _{n} \cup \big \{ A_{n} \big \}$$$$\mathcal{CS}$$-inconsistent, but we also have $$\Gamma _{n} \cup \big \{ A_{n} \big \} \subseteq \Xi$$, which implies $$\mathcal{CS}$$-inconsistency of Ξ, in contradiction to our assumptions. Therefore, Δ is not only $$\mathcal{CS}$$-consistent, but also $$\mathcal{CS}$$-maxiconsistent. (Part 2) We cannot have both A and ¬A in Γ, since we have, of course, $$\vdash _{\mathcal{CS}} (A \wedge \neg A) \rightarrow \bot$$. If, on the other hand, neither A nor ¬A is in Γ, then both Γ ∪ {A} and Γ ∪ {¬A} must be $$\mathcal{CS}$$-inconsistent, so that for some B1, …, Bn ∈ Γ we will have $$\vdash_{\mathcal{CS}} \big(B_{1}\wedge \ldots\wedge B_{n} \wedge A\big) \rightarrow \bot,$$ whereas for some C1, …, Ck ∈ Γ we will have $$\vdash_{\mathcal{CS}} \big(C_{1}\wedge \ldots\wedge C_{k} \wedge \neg A\big) \rightarrow \bot,$$ whence we get, using (A0) and (R1), $$\vdash_{\mathcal{CS}} \big(C_{1}\wedge \ldots\wedge C_{k}\big) \rightarrow A,$$ and further $$\vdash_{\mathcal{CS}} \big(B_{1}\wedge \ldots\wedge B_{n} \wedge C_{1}\wedge \ldots\wedge C_{k}\big) \rightarrow \bot,$$ so that Γ turns out to be $$\mathcal{CS}$$-inconsistent, contrary to our assumptions. (Part 3) Assume (A ∨ B) ∈ Γ. If neither A nor B are in Γ, then, by Part 2, both ¬A and ¬B are in Γ. Using (A0) and (R1) we get that $$\vdash_{\mathcal{CS}} \big((A \vee B) \wedge \neg A \wedge \neg B\big) \rightarrow \bot,$$ showing that Γ is $$\mathcal{CS}$$-inconsistent, contrary to our assumptions. In the other direction, if say A ∈ Γ and (A ∨ B) ∉ Γ, then, by Part 2, we must have ¬(A ∨ B) ∈ Γ. Using (A0) and (R1) we get that $$\vdash_{\mathcal{CS}} \big(\neg(A \vee B) \wedge A\big) \rightarrow \bot,$$ showing again that Γ is $$\mathcal{CS}$$-inconsistent, contrary to our assumptions. The case when B ∈ Γ is similar. Parts 4 and 5 are similar to Part 3. Our goal is now to obtain, for any given constant specification $$\mathcal{CS}$$, a strong completeness theorem for $$\Sigma _{D}(\mathcal{CS})$$, and we start by establishing some soundness claims: Theorem 1 Let $$\mathcal{CS}$$ be an arbitrary constant specification. If A ∈ FormAg is such that $$\vdash _{\mathcal{CS}} A$$, then A is valid over the class $$Mod_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{mixsucc}\right )$$. Proof. Given the above notion of proof, it is sufficient to show that every instance of (A0)–(A10D) is valid over the class $$Mod_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{mixsucc}\right )$$, and every application of rules (R1), (R2) and ($$R_{\mathcal{CS}}$$) to formulas which are valid over $$Mod_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{mixsucc}\right )$$ yields a formula which is valid over the same class. First, note that if $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e}, \mathcal{E}, V\big \rangle$$ is a $$\mathcal{CS}$$-normal jstit model based on a jstit frame from $$\mathcal{C}^{Ag}_{mixsucc}$$, then $$\big \langle Tree, \unlhd , Choice, V\big \rangle$$ is a model of stit logic. Therefore, axioms (A0)–(A3), which were copy pasted from the standard axiomatization of dstit logic5 must be valid. Second, note that if $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e}, \mathcal{E}, V\big \rangle$$ is a $$\mathcal{CS}$$-normal jstit model, then $$\mathcal{M} = \big \langle Tree, R, R_{e}, \mathcal{E}, V\big \rangle$$ is what it is called in [2, Section 6] an AF-model with the form of constant specification given by $$\mathcal{CS}$$.6 This means that also all of the (A4)–(A7) must be valid, whereas (R1), (R2) and ($$R_{\mathcal{CS}}$$) must preserve validity, given that all these parts of our axiomatic system were borrowed from the standard axiomatization of the epistemic justification logic. The validity of other parts of $$\Sigma _{D}\left (\mathcal{CS}\right )$$ will be motivated below in some detail. In what follows, $$\mathcal{M} = \big \langle Tree, \unlhd , Choice, Act, R, R_{e},\mathcal{E}, V\big \rangle$$ will always stand for an arbitrary jstit model in $$Mod_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{mixsucc}\right )$$, and $$\big (m,h\big )$$ for an arbitrary element of $$MH\left (\mathcal{M}\right )$$. As for (A8), assume for reductio that $$\mathcal{M}, m,h \models KA \wedge \Diamond \big \langle K\big \rangle \Diamond \neg A$$. Then $$\mathcal{M}, m,h \models KA$$ and also $$\mathcal{M}, m,h^{\prime} \models \langle K\rangle \Diamond \neg A$$ for some h′∈ Hm. Therefore, there must be an m′∈ Tree and a g ∈ Hm′ such that both $$R\big (m,m^{\prime}\big )$$ and $$\mathcal{M}, m^{\prime},g \models \Diamond \neg A$$, whence we can choose a g$$^{\prime}$$ ∈ Hm′ such that $$\mathcal{M}, m^{\prime},g^{\prime} \not \models A$$. The latter is in obvious contradiction with $$\mathcal{M}, m,h \models KA$$. We consider next (A9). If $$\Box Et$$ is true at $$\big (m,h\big )$$ in $$\mathcal{M}$$, then, by definition, t ∈ Actm. Now, if m$$^{\prime}$$ ∈ Tree is such that $$R\big (m,m^{\prime}\big )$$, then, by R ⊆ Re we will have $$R_{e}\big (m,m^{\prime}\big )$$, and, by the epistemic transparency of presented proofs constraint, we must have t ∈ Actm′ so that for every g ∈ Hm′ we will have $$\mathcal{M},m^{\prime},g \models \Box Et$$. Therefore, we must have $$\mathcal{M},m,h \models K\Box Et$$ as well. It only remains to show the validity of (A10D), and we again use reductio ad absurdum. Assume that we have both $$\mathcal{M}, m, h \models K\big(\neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k}\big)$$ (2) and $$\mathcal{M}, m, h \models Et_{1} \wedge\ldots\wedge Et_{n}\wedge\neg Es_{1} \wedge\ldots\wedge\neg Es_{k}.$$ (3) Whence, by (2) and the validity of (A7), we know that also $$\mathcal{M}, m, h \models \neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k}.$$ (4) By validity of (A1), it follows from (3) that $$\mathcal{M}, m, h \models \neg\Box Es_{1} \wedge\ldots\wedge\neg\Box Es_{k}.$$ (5) Whence, by (4) and (5) we know that also $$\mathcal{M}, m, h \models \neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}.$$ (6) Therefore, we can choose a natural u such that 1 ≤ u ≤ n and $$\mathcal{M}, m, h \models \neg\Box Et_{u}.$$ The latter, in turn, means that for some h′∈ Hm we have that $$\mathcal{M}, m, h^{\prime} \models \neg Et_{u}.$$ (7) Comparison between (3) and (7) shows that $$Act\big (m,h\big ) \neq Act\big (m,h^{\prime}\big )$$, whence by the presenting new proof makes histories divide constraint we get that h ≉mh$$^{\prime}$$. Hence, we know that the second disjunct of condition (mixsucc) fails for m, so that we must have $$\big(\forall m_{1} \in Tree\big)\left(m \lhd m_{1} \Rightarrow \big(\exists m_{2} \unlhd m_{1}\big)\big(Next\big(m, m_{2}\big)\big)\right).$$ (8) Further, it follows from h ≉mh$$^{\prime}$$ that $$h \neq h^{\prime}$$. Therefore, by Lemma 1.2, we can choose in Tree some $$m^{\prime} \rhd m$$ such that m$$^{\prime}$$ ∈ h. By (8), we must be able to choose an m2 ∈ Tree such that both $$m_{2} \unlhd m^{\prime}$$ and $$Next\big (m, m_{2}\big )$$. So we consider such an m2. Given that m$$^{\prime}$$ ∈ h, we know, by Lemma 1.3, that $$h \in H_{m_{2}}$$. Therefore, it follows from (3) and Lemma 2 that $$t_{1},\ldots , t_{n} \in Act_{m_{2}}$$ or, equivalently, $$\mathcal{M}, m_{2}, h \models \Box Et_{1} \wedge\ldots \wedge \Box Et_{n}.$$ (9) Furthermore, by the future always matters constraint we know that $$R\big (m,m_{2}\big )$$, whence it follows, by (2), that $$\mathcal{M}, m_{2}, h \models \neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k}.$$ (10) Finally, choose an arbitrary r between 1 and k. If $$s_{r} \in Act_{m_{2}}$$, then, by the no new proofs guaranteed constraint, there must be some $$g \in H_{m_{2}}$$ and some $$m_{0} \lhd m_{2}$$ such that sr ∈ Act(m0, g). Then, by Lemma 1.3, g ∈ Hm, hence h ≈mg. Therefore, by the presenting a new proof makes histories divide constraint, $$Act\big (m, g\big ) = Act\big (m,h\big )$$. By $$Next\big (m, m_{2}\big )$$ we must have $$m_{0} \unlhd m$$, therefore, by the expansion of presented proofs, $$s_{r} \in Act\big (m,g\big )$$, whence also $$s_{r} \in Act\big (m,h\big )$$. But this plainly contradicts (3). Since 1 ≤ r ≤ k was chosen arbitrarily, this means that all of s1, …, sk are outside $$Act_{m_{2}}$$ so that we have $$\mathcal{M}, m_{2}, h \models \neg\Box Es_{1} \wedge\ldots \wedge \neg\Box Es_{k}.$$ (11) Taken together, (9)–(11) give us a plain contradiction. We are now prepared to formulate our completeness result: Theorem 2 Let Γ ⊆ FormAg and let $$\mathcal{C}$$ be a class of stit frames such that $$\mathcal{C}^{Ag}_{discr} \subseteq \mathcal{C} \subseteq \mathcal{C}^{Ag}_{mixsucc}$$. Then Γ is $$\mathcal{CS}$$-consistent iff it is satisfiable in $$Mod_{\mathcal{CS}}\left (\mathcal{C}\right )$$ iff it is satisfiable in $$Mod^{\downarrow }_{\mathcal{CS}}(\mathcal{C})$$. One part of the completeness results we have, of course, right away, as a consequence of Theorem 1: Corollary 1 Let Γ ⊆ FormAg and let $$\mathcal{C}$$ be a class of stit frames such that $$\mathcal{C}^{Ag}_{discr} \subseteq \mathcal{C} \subseteq \mathcal{C}^{Ag}_{mixsucc}$$. If Γ ⊆ FormAg is satisfiable in $$Mod_{\mathcal{CS}}\left (\mathcal{C}\right )$$ (thus also when it is satisfiable in $$Mod^{\downarrow }_{\mathcal{CS}}\left (\mathcal{C}\right )$$), then Γ is $$\mathcal{CS}$$-consistent. Proof. Let Γ ⊆ FormAg be satisfiable in $$Mod_{\mathcal{CS}}(\mathcal{C})$$ so that for some $$\mathcal{M} \in Mod_{\mathcal{CS}}(\mathcal{C})$$ and some $$\big (m,h\big ) \in MH\left (\mathcal{M}\right )$$ we have $$\mathcal{M}, m, h \models \Gamma$$. Then we must have $$\mathcal{M} \in Mod_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{mixsucc}\right )$$. If Γ were $$\mathcal{CS}$$-inconsistent this would mean that for some A1, …, An ∈ Γ we would have $$\vdash _{\mathcal{CS}} \big (A_{1} \wedge \ldots \wedge A_{n}\big ) \rightarrow \bot$$. By Theorem 1, this would mean that $$\mathcal{M}, m, h \models \big(A_{1} \wedge\ldots \wedge A_{n}\big) \rightarrow \bot,$$ whence clearly $$\mathcal{M}, m, h \models \bot$$, which is impossible. Therefore, Γ must be $$\mathcal{CS}$$-consistent. Before we move further, we prove some claims about provability in $$\Sigma _{D}\left (\mathcal{CS}\right )$$ to be used below: Lemma 5 Let $$\mathcal{CS}$$ be a constant specification. Then all of the following theorems and derived rules are provable in $$\Sigma _{D}\left (\mathcal{CS}\right )$$ for every A ∈ FormAg, t ∈ Pol, x, y ∈ PVar and j ∈ Ag: $$K\ A \rightarrow \Box A$$ (T0) ($$\text{Nec}\Box$$) $$\text{From }A\text{ infer }\Box A$$ $$\text{From }A\text{ infer }\big[j\big]A$$ (Nec[j]) $$t{:} A \rightarrow Kt{:} A$$ (T1) $$t{:} A \rightarrow \Box t{:} A$$ (T2) $$K\ A \rightarrow \Box K\ A$$ (T3) \begin{align} &\text{From }K\ A \rightarrow \big(\neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k}\big) \nonumber\\ &\qquad\qquad\text{ infer }K\ A \rightarrow \big(\neg Et_{1} \vee\ldots\vee\neg Et_{n}\vee Es_{1} \vee\ldots\vee Es_{k}\big). \end{align} (ρD) Proof. (T0). We use the transitivity of implication w.r.t. the following set of formulas: \begin{align*} &K\ A \rightarrow \Box K\Box A &&\text{(by A8)}\\ &\Box K\Box A \rightarrow K\Box A &&\text{(by A1)}\\ &K\Box A \rightarrow \Box A &&\text{(by A7)} \end{align*} (Nec$$\Box$$). From A we infer KA by (R2) and then use (T0) and (R1) to get $$\Box A$$. (Nec[j]). From A we infer $$\Box A$$ by (Nec$$\Box$$) and then apply (A2) and (R1). We pause to note that by (Nec$$\Box$$) and (Nec[j]) we know that every modality in the set $$\{\Box \} \cup \big \{ [j] \mid j \in Ag \big \}$$ is an S5-modality. (T1). We have both $$t{:} A \rightarrow !\!t{:}\big (t{:} A\big )$$ and $$!\!t{:}\big (t{:} A\big ) \rightarrow Kt{:} A$$ by (A5) so that we get (T1) by transitivity of implication. (T2). By (T1) and (T0). (T3). By K A → K K A (a part of (A7)) and (T0). (ρD). For a given sequence t1, …, tn, s1, …, sk ∈ Pol, we introduce the following abbreviations: \begin{align*} &C := \neg\Box Et_{1} \vee\ldots\vee\neg\Box Et_{n}\vee\Box Es_{1} \vee\ldots\vee\Box Es_{k},\\ &D := \neg Et_{1} \vee\ldots\vee\neg Et_{n}\vee Es_{1} \vee\ldots\vee Es_{k}. \end{align*} Now, we construct the following inference: $$K \ A \rightarrow C \qquad\qquad(\text{premise})$$ (12) $$K(K\ A \rightarrow C) \qquad\qquad(\text{by}\ (12)\ \text{and}\ (R2))$$ (13) $$K\ A \rightarrow K\ C \qquad\qquad(\text{by}\ (13)\ \text{and}\ K\ \text{is}\ \textrm{S}4)$$ (14) $$K\ C \rightarrow D \qquad\qquad(\text{by}\ (A10_D))$$ (15) $$K \ A \rightarrow D \qquad\qquad(\text{by } (14)\ \text{and}\ (15))$$ (16) The next lemma shows that (ρD) can be also used in an alternative axiomatization of $$\Sigma _{D}\left (\mathcal{CS}\right )$$: Lemma 6 Let $$\Sigma ^{\prime}_{D}(\mathcal{CS})$$ be the axiomatic system obtained from $$\Sigma _{D}(\mathcal{CS})$$ by replacing (A10D) with (ρD). Then, for every A ∈ FormAg, it is true that $$\vdash _{\mathcal{CS}} A$$ iff A is provable in $$\Sigma ^{\prime}_{D}(\mathcal{CS})$$. Proof. Assuming the conditions of the lemma, it is easy to see that whenever A is provable in $$\Sigma ^{\prime}_{D}\left (\mathcal{CS}\right )$$, it is also provable in $$\Sigma _{D}\left (\mathcal{CS}\right )$$, since (ρD) is derivable, and all the other axioms and inference rules of $$\Sigma ^{\prime}_{D}\left (\mathcal{CS}\right )$$ are also present in $$\Sigma _{D}\left (\mathcal{CS}\right )$$. In the other direction, it is sufficient to show that every instance of (A10D) is provable in $$\Sigma ^{\prime}_{D}\left (\mathcal{CS}\right )$$. Using abbreviations C, D in the same sense as in the derivation of (ρD), we construct the following proof: $$K\ C \rightarrow C \qquad\qquad(\text{by}\ (A7))$$ (17) $$K\ C \rightarrow D \qquad\qquad(\text{by}\ (17)\, \text{and}\ (\rho_D))$$ (18) However, note that in $$\Sigma ^{\prime}_{D}(\mathcal{CS})$$ the rule ($$R_{\mathcal{CS}}$$) still gets applied to the instances of (A10D) even though this scheme is no longer an axiom. Thus the alternative form of axiomatization is also somewhat less neat. 4 The canonical model The main aim of the present section is to prove the inverse of Corollary 1. The method used is a variant of the canonical model technique, but, due to the complexity of the case, we do not define our model in one sweeping definition. Rather, we proceed piecewise, defining elements of the model one by one, and checking the relevant constraints as soon as we have got enough parts of the model in place. The last subsection proves the truth lemma for the defined model. Throughout this section, we fix some arbitrary constant specification $$\mathcal{CS}$$ to serve as a parameter in the canonical model construction. The canonical model $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ to be built in this section will be a member of $$Mod^{\downarrow }_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{discr}\right )$$. The ultimate building blocks of $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ we will call elements. Before going on with the definition of $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$, we define what these elements are and explore some of their properties. Definition 2 An element is a sequence of the form $$\big (\Gamma _{1},\ldots ,\Gamma _{n}\big )$$ for some natural n ≥ 1 such that For every i ≤ n, Γi is $$\mathcal{CS}$$-maxiconsistent; For every i < n, for all A ∈ FormAg, if KA ∈ Γi, then KA ∈ Γi+1; For every i < n, for all t ∈ Pol, if Et ∈ Γi, then $$\Box Et \in \Gamma _{i + 1}$$; For every i < n, for all t ∈ Pol, if ¬Et ∈ Γi, then $$\neg \Box Et \in \Gamma _{i + 1}$$. We prove the following lemma: Lemma 7 Whenever $$\big (\Gamma _{1},\ldots ,\Gamma _{n}\big )$$ is an element, then, for some Γn+1 ⊆ FormAg, the sequence $$\big (\Gamma _{1},\ldots ,\Gamma _{n + 1}\big )$$ is also an element. Proof. Assume $$\big (\Gamma _{1},\ldots ,\Gamma _{n}\big )$$ is an element, and consider the following set: $$\Delta := \big\{ K \, A \mid K \, A \in \Gamma_{n} \big\} \cup \big\{ \Box Et \mid Et \in \Gamma_{n} \big\} \cup \big\{ \neg\Box Et \mid \neg Et \in \Gamma_{n} \big\}.$$ We show that Δ is $$\mathcal{CS}$$-consistent. Of course, the set $$\big \{ K\, A \mid K\, A \in \Gamma _{n} \big \}$$ is $$\mathcal{CS}$$-consistent since it is a subset of Γn and the latter is assumed to be $$\mathcal{CS}$$-consistent. Further, if Δ is $$\mathcal{CS}$$-inconsistent, then, wlog, for some KB1, …, KBk, Et1, …, Etl, ¬Es1, …, ¬Esr ∈ Γn we will have $$\vdash_{\mathcal{CS}}\big(K \, B_{1}\wedge\ldots \wedge K\, B_{k}\big) \rightarrow \big(\neg\Box Et_{1} \vee\ldots \vee \neg\Box Et_{l}\vee \Box Es_{1}\vee\ldots\vee \Box Es_{r}\big),$$ whence, by (A7): $$\vdash_{\mathcal{CS}} K\big(B_{1}\wedge\ldots \wedge B_{k}\big) \rightarrow \big(\neg\Box Et_{1} \vee\ldots \vee \neg\Box Et_{l}\vee \Box Es_{1}\vee\ldots\vee \Box Es_{r}\big),$$ and further, by (ρD): $$\vdash_{\mathcal{CS}} K\big(B_{1}\wedge\ldots \wedge B_{k}\big) \rightarrow \big(\neg Et_{1} \vee\ldots \vee \neg Et_{l}\vee Es_{1}\vee\ldots\vee Es_{r}\big).$$ The latter formula shows that Γn is $$\mathcal{CS}$$-inconsistent which contradicts the assumption that $$\big (\Gamma _{1},\ldots ,\Gamma _{n}\big )$$ is an element. Therefore, Δ must be $$\mathcal{CS}$$-consistent and, by Lemma 4.1, it is also extendable to a maxiconsistent Γn+1. By the choice of Δ, this means that $$\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Gamma _{n + 1}\big )$$ must be an element. The structure of elements will be important in what follows. If $$\xi = \big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ is an element, then its initial segment is any element τ of the form $$\big (\Gamma _{1},\ldots , \Gamma _{k}\big )$$ with k ≤ n. If, moreover, k < n, then τ is a proper initial segment of ξ, and if k = n − 1, then τ is the greatest proper initial segment of ξ. Moreover, we define n to be the length of ξ. Thus, any element of length 1 has no proper initial segments. Furthermore, we define that Γn is the end element of ξ and write $$\Gamma _{n} = end\big (\xi \big )$$. We now define the canonical model using elements as our building blocks. We start by defining the following relation ≡ between elements of equal length: \begin{align*} \big(\Gamma_{1},\ldots, \Gamma_{n}, \Gamma\big) \equiv \big(\Delta_{1},\ldots, \Delta_{n}, \Delta\big) \Leftrightarrow &\big(\Gamma_{1} = \Delta_{1} \& \ldots \& \Gamma_{n} = \Delta_{n} \&\\ &\& \big(\forall A \in Form^{Ag}\big)\big(\Box A \in \Gamma \Rightarrow A \in \Delta\big). \end{align*} It is routine to check that ≡ is an equivalence relation given that $$\Box$$ is an S5-modality. We will denote the equivalence class of element $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ generated by ≡ by $$\big [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\big ]_{\equiv }$$. Since all the elements inside a given ≡-equivalence class are of the same length, we may extend the notion of length to these classes setting that the length of $$\big [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\big ]_{\equiv }$$ also equals n. The next lemma will be repeatedly used in what follows: Lemmma 8 Let $$\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Gamma \big )$$ be an element, let Δ ⊆ FormAg be $$\mathcal{CS}$$-maxiconsistent and let $$\big\{ \Box A \mid \Box A \in \Gamma \big\} \subseteq \Delta.$$ Then $$\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Delta \big )$$ is also an element and, moreover, $$\big(\Gamma_{1},\ldots,\Gamma_{n}, \Gamma\big) \equiv \big(\Gamma_{1},\ldots,\Gamma_{n}, \Delta\big).$$ Proof. We first show that $$\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Delta \big )$$ is an element. Indeed, if KA ∈ Γn, then KA ∈ Γ by definition of an element. But then $$\Box KA \in \Gamma$$ by (T3) and $$\mathcal{CS}$$-maxiconsistency of Γ, whence $$\Box KA \in \Delta$$. By (A1) and $$\mathcal{CS}$$-maxiconsistency of Δ we get then KA ∈ Δ. Similarly, if Et ∈ Γn, then $$\Box Et \in \Gamma$$ by definition of an element. But then $$\Box Et \in \Delta$$. Finally, if ¬Et ∈ Γn, then $$\neg \Box Et \in \Gamma$$ by definition of an element and, further, $$\Box \neg \Box Et \in \Gamma$$ by (A1) and $$\mathcal{CS}$$-maxiconsistency of Γ. But then we must have $$\Box \neg \Box Et \in \Delta$$. By (A1) and $$\mathcal{CS}$$-maxiconsistency of Δ we get then $$\neg \Box Et \in \Delta$$. Given the inclusion $$\big \{ \Box A \mid \Box A \in \Gamma \big \} \subseteq \Delta$$, the other part of the lemma is straightforward. We now proceed to definitions of components for the canonical model. 4.1 Tree, $$\unlhd$$, and Hist The first two components of the canonical model $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ are as follows: Tree is the set of ≡-equivalence classes of elements plus † and ‡ as additional moments; We set that both † $$\lhd m$$ and $$m \ntriangleleft$$ † for every m ∈ Tree ∖{†}. We further set that ‡ is only $$\lhd$$-comparable to †, and for any two ≡-equivalence classes of elements m and m$$^{\prime}$$, we have that $$m \lhd m^{\prime}$$ iff there is an element ξ ∈ m such that ξ is a proper initial segment of every element τ ∈ m$$^{\prime}$$. The relation $$\unlhd$$ is then defined as the reflexive companion to $$\lhd$$. Before we move on to the choice- and justifications-related components, let us pause to check that the restraints imposed by our semantics on Tree and $$\unlhd$$ are satisfied: Lemma 9 The relation $$\unlhd$$, as defined above, is a partial order on Tree, which satisfies both historical connection and no backward branching constraints. Proof. Transitivity and reflexivity of $$\unlhd$$ are obvious. As for anti-symmetry, assume that we have both $$m \lhd m^{\prime}$$ and $$m^{\prime} \lhd m$$. Then m and m$$^{\prime}$$ must be equivalence classes of elements. Let ξ ∈ m be a proper intial segment of every element in m$$^{\prime}$$ and let τ ∈ m$$^{\prime}$$ be a proper initial segment of every element in m. It follows that ξ is a proper initial segment of τ and also τ is a proper initial segment of ξ, a contradiction. Historical connection is satisfied since † is the $$\unlhd$$-least element of Tree. Let us prove the absence of backward branching. Assume that we have both $$m \unlhd m^{\prime\prime}$$ and $$m^{\prime} \unlhd m^{\prime\prime}$$ but neither $$m \unlhd m^{\prime}$$ nor $$m^{\prime} \unlhd m$$ holds. This means that all the three moments are pairwise different and none of them is either † or ‡, otherwise our assumptions about them would be immediately falsified. Therefore, all the three moments are some equivalence classes of elements and we also have $$m \neq m^{\prime}$$, $$m \lhd m^{\prime\prime}$$ and $$m^{\prime} \lhd m^{\prime\prime}$$. So let ξ ∈ m and τ ∈ m$$^{\prime}$$ be such that both ξ and τ are proper initial segments of every element in $$m^{\prime\prime}$$. We first show that $$\xi \neq \tau$$. Indeed, since ξ ∈ m and τ ∈ m$$^{\prime}$$, we must have both m = [ξ]≡ and m$$^{\prime}$$ = [τ]≡. Therefore, from ξ = τ it would immediately follow that [ξ]≡ = [τ]≡, hence m = m$$^{\prime}$$. The latter is in contradiction with our assumption that $$m \neq m^{\prime}$$. Therefore either ξ must be a proper initial segment of τ or τ must be a proper initial segment of ξ. Assume, wlog, that ξ is a proper initial segment of τ. Then ξ is included into the greatest proper intitial segment of τ. Let τ′ be any element in m$$^{\prime}$$. It follows from the definition of ≡ that all the elements within m$$^{\prime}$$ share the same greatest proper initial segment, therefore ξ must be a proper initial segment of τ$$^{\prime}$$ as well. It follows that $$m \lhd m^{\prime}$$, contrary to our assumptions. Before we move on, let us have a quick look into the structure of histories induced by Tree and $$\unlhd$$. Lemma 10 Let h be a history induced by Tree and $$\unlhd$$ as defined in this subsection. If ‡ ∉ h, then for every n ≥ 1, h contains exactly one equivalence class of elements of the length n. Proof. A history cannot contain more than one class of elements of the same length. Indeed, if m, m$$^{\prime}$$ are both equivalence classes of elements of the length n then no element in m can be a proper initial segment of any element in m′ or vice versa. Therefore, whenever $$m \neq m^{\prime}$$, we will also have $$\left(m \ntrianglelefteq m^{\prime}\right) \& \left( m^{\prime} \ntrianglelefteq m\right),$$ which means that no chain of elements can contain both m and m$$^{\prime}$$. We still have to show that every history contains at least one equivalence class of elements of the length n for every n ≥ 1. Assume the contrary. Then let h be a history induced by Tree and $$\unlhd$$ and let k ≥ 1 be the least number such that the given h does not contain any equivalence classes of elements of the length k. We have then two cases to consider. Case 1. k = 1. Since h is a maximal $$\unlhd$$-chain and † is the $$\unlhd$$-least element of Tree, we know that † ∈ h. However, h cannot be {†}, given that we would have then h ⊂ {†, ‡}. Since {†, ‡} is clearly a $$\unlhd$$-chain, this would contradict the maximality of h. Therefore, we can choose an m ∈ h ∖ {†}. Given that ‡ ∉ h, m must be an equivalence class of elements, say $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{r}\big )\big ]_{\equiv }$$ for some appropriate Γ1, …, Γr ⊆ FormAg. Then we clearly have $$\dagger \lhd \big[\big(\Gamma_{1}\big)\big]_{\equiv} \unlhd \big[\big(\Gamma_{1},\ldots, \Gamma_{r}\big)\big]_{\equiv}.$$ (19) Now, let m$$^{\prime}$$ be an arbitrary moment in h. Then either $$m \unlhd m^{\prime}$$ or $$m^{\prime} \lhd m$$. In the former case we will have $$\big [\big (\Gamma _{1}\big )\big ]_{\equiv } \unlhd m^{\prime}$$ by (19) and transitivity, in the latter case we will have $$\big [\big (\Gamma _{1}\big )\big ]_{\equiv } \unlhd m^{\prime} \vee m^{\prime} \unlhd \big [\big (\Gamma _{1}\big )\big ]_{\equiv }$$ by (19) and the absence of backward branching. Therefore $$h \cup \big \{ [(\Gamma _{1})]_{\equiv } \big \}$$ is a $$\unlhd$$-chain, and by maximality of histories we get that $$\big [\big (\Gamma _{1}\big )\big ]_{\equiv } \in h$$. Since the length of $$\big [\big (\Gamma _{1}\big )\big ]_{\equiv }$$ equals 1, this contradicts our assumption. Case 2. k > 1, say k = l + 1. Then h must contain at least one moment m of the length l, say $$\big [\big (\Gamma _{1},\ldots , \Gamma _{l}\big )\big ]_{\equiv } = m \in h$$. We have to distinguish between two subcases: Case 2.1. m is the $$\unlhd$$-greatest moment in h. Then, by Lemma 7, choose an appropriate Γ ⊆ FormAg such that $$\big (\Gamma _{1},\ldots , \Gamma _{l}, \Gamma \big )$$ is an element. We obviously have $$m \lhd m^{\prime}$$ for $$m^{\prime} = \big [\big (\Gamma _{1},\ldots , \Gamma _{l}, \Gamma \big )\big ]_{\equiv }$$, and since m is $$\unlhd$$-greatest in h, this means that $$h \subset h \cup \big \{ m^{\prime} \big \}$$ and that $$h \cup \big \{ m^{\prime} \big \}$$ is a $$\unlhd$$-chain, which contradicts the maximality of h. Case 2.2. m is not the $$\unlhd$$-greatest moment in h. Then choose an m$$^{\prime}$$ ∈ h, such that $$m \lhd m^{\prime}$$. By definition of $$\unlhd$$, m$$^{\prime}$$ must be an equivalence class of elements, and some element in m must be a proper segment of every element in m$$^{\prime}$$. This means that the length of m$$^{\prime}$$ must be strictly greater than l. By our assumption, m$$^{\prime}$$ cannot have the length l + 1, therefore, the length of m$$^{\prime}$$ must also exceed l + 1. But then m$$^{\prime}$$ must be of the form $$\big [\big (\Delta _{1},\ldots , \Delta _{l}, \Delta _{l+1},\ldots , \Delta _{r}\big )\big ]_{\equiv }$$ for some r > l + 1 and appropriate Δ1, …, Δn, Δn+1, …, Δr ⊆ FormAg. We obviously have then that $$m_{0} = \big[\big(\Delta_{1},\ldots, \Delta_{l}, \Delta_{l+1}\big)\big]_{\equiv} \lhd \big[\big(\Delta_{1},\ldots, \Delta_{l}, \Delta_{l+1},\ldots, \Delta_{r}\big)\big]_{\equiv} = m^{\prime}.$$ (20) Now, consider an arbitrary $$m^{\prime\prime}$$ ∈ h. Since h is a $$\unlhd$$-chain, we must have either $$m^{\prime\prime} \lhd m^{\prime}$$ or $$m^{\prime} \unlhd m^{\prime\prime}$$. In the latter case we will have $$m_{0} \lhd m^{\prime\prime}$$ by (20) and transitivity. In the former case we will have $$m_{0} \unlhd m^{\prime\prime} \vee m^{\prime\prime} \unlhd m_{0}$$ by (20) and the absence of backward branching. Therefore, h ∪ {m0} is a $$\unlhd$$-chain, and by maximality of h we must have m0 ∈ h. But the latter contradicts our assumption since the length of m0 equals l + 1. Lemma 11 Let h be a history induced by $$\big \langle Tree, \unlhd \big \rangle$$ as defined in this subsection. Then either h = (†, ‡) or h is of the form: $$\big(\dagger, [\xi_{1}]_{\equiv},\ldots, [\xi_{n}]_{\equiv},\ldots\big),$$ where for every n ≥ 1, ξn has the length n and is the greatest proper initial segment of ξn+1. Proof. We first show that every such h is indeed a history. This is clear for $$h = \big (\dagger , \ddagger \big )$$. Assume that h is of the form $$\big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots \big )$$ for the appropriate sequence of ξs. Then h is obviously a $$\unlhd$$-chain. Now, if m ∈ Tree ∖ h, then m is either ‡ or an equivalence class of elements. It is clear that h ∪ {‡} is not a $$\unlhd$$-chain, since ‡ is $$\unlhd$$-incomparable with [ξ1]≡. If m is an equivalence class of elements, then consider the length of m, say n. Since m ∉ h, we must have $$m \neq [\xi _{n}]_{\equiv }$$, therefore, h ∪ {m} will contain at least two different equivalence classes of moments of the same length n, hence by Lemma 10 cannot be a history. Therefore, every chain of the form $$\big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots \big )$$ is maximal and thus a history. In the other direction, let h be a history induced by Tree and $$\unlhd$$. Then ‡ is either in h or outside h. If ‡ ∈ h, then we must have h = (†, ‡) since † is the only other element in Tree which is $$\unlhd$$-comparable to ‡. On the other hand, if ‡ ∉ h, then, by Lemma 10, h must be of the form (†, m1, …, mn, …), where, for n ≥ 1, mn is an equivalence class of elements of length n († will be in h since it is the $$\unlhd$$-least). For a given n ≥ 1, consider the pair mn, mn+1. Since both of these moments are in h they must be $$\unlhd$$-comparable, but we cannot have $$m_{n+1} \unlhd m_{n}$$ since, by the length considerations, no element in mn+1 can be a proper initial segment of any element in mn. Therefore, we must have $$m_{n} \lhd m_{n+1}$$, which means that there must be an element in mn which is a proper segment of every element in mn+1. We now set this element as ξn. It is clear that under these settings, we get that $$h = \big(\dagger, [\xi_{1}]_{\equiv},\ldots, [\xi_{n}]_{\equiv},\ldots\big),$$ since we have ξn ∈ mn for every n ≥ 1. Further, every ξn is the greatest proper segment of its corresponding ξn+1, since ξn is chosen as a proper initial segment of every element in mn+1, ξn+1 ∈ mn+1, and the length of ξn+1 exceeds the length of ξn by one. Corollary 2 Let h be a history induced by Tree and $$\unlhd$$ as defined in this subsection. Then h can be embedded into an initial segment of ω with its natural order. Proof. If h is a history induced by the abovedefined Tree and $$\unlhd$$, then either h = (†, ‡) or $$h = \big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots \big )$$ for an appropriate sequence ξ1, …, ξn, … of elements. In the former case we set the embedding f as f(†) = 0, f(‡) = 1; in the latter case we set f(†) = 0 and $$f\big ([\xi _{n}]_{\equiv }\big ) = n$$ for every n ≥ 1 thus embedding h into ω which, of course, is its own (improper) initial segment. Whenever a history h is of the form $$\big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots \big )$$, we will call the sequence of elements $$\big (\xi _{1},\ldots , \xi _{n},\ldots \big )$$ a representation of h. One can show that such a representation is unique. Indeed, suppose that $$\big (\dagger , \xi _{1},\ldots , \xi _{n},\ldots \big )$$ and (†, ξ1′, …, ξn′, …) are two different representations for $$\big (\dagger , m_{1},\ldots , m_{n},\ldots \big )$$. Then let i ≥ 1 be the first natural number such that $$\xi _{i} \neq \xi ^{\prime}_{i}$$. Consider mi+1. We have ξi+1, ξi+1′∈ mi+1 so that ξi+1 ≡ ξi+1′. Since ξi and ξi′ are the greatest proper initial segments of ξi+1, ξi+1′, respectively, the greatest proper initial segments of ξi+1 and ξi+1′ are non-empty and, by ξi+1 ≡ ξi+1′, must coincide, which cannot be the case since $$\xi _{i} \neq \xi ^{\prime}_{i}$$. If $$h = \big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots \big )$$ is a history in the canonical model, and m ∈ h is an equivalence class of elements, we know that for some n ≥ 1 we must have m = [ξn]≡. Since the representations of histories are unique, we may unequivocally define the intersection of m and h (write m ⊓ h) as the element in the representation of h which induces m, so that, in our case, m ⊓ h = ξn. The following lemma will be extremely useful in what follows: Lemma 12 Let ξ be an element. Then there is at least one history $$h \in H_{[\xi ]_{\equiv }}$$ such that [ξ]≡ ⊓ h = ξ. Proof. Assume that $$\xi = \big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ for appropriate n ≥ 1, Γ1, …, Γn. Using Lemma 7, we choose the infinite sequence $$\Gamma_{n + 1},\ldots, \Gamma_{n + k},\ldots,$$ in such a way that the following: \begin{align*} \xi_{1} &:= \big(\Gamma_{1}\big);\\ \xi_{2} &:= \big(\Gamma_{1}, \Gamma_{2}\big);\\ &\ldots;\\ \xi_{n} &:= \xi = \big(\Gamma_{1},\ldots, \Gamma_{n}\big);\\ \xi_{n + 1} &:= \big(\Gamma_{1},\ldots, \Gamma_{n}, \Gamma_{n + 1}\big);\\ &\ldots;\\ \xi_{n + k} &:= \big(\Gamma_{1},\ldots, \Gamma_{n}, \Gamma_{n + 1}, \ldots, \Gamma_{n + k}\big);\\ &\ldots. \end{align*} is a sequence of elements in which every element is the greatest proper initial segment of the next one, so that, by Lemma 11, there must be a history h in $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ such that $$h = \big (\dagger , [\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots , [\xi _{n + k}]_{\equiv },\ldots \big )$$, and ξ = ξn = [ξn]≡⊓ h. 4.2 Choice We now define the choice structures of our canonical model. Let j ∈ Ag be arbitrary and let $$\big (m,h\big )$$ be a moment-history pair. Then $$Choic{e^{m}_{j}}(h) = H_{m}$$, if m ∈ {†, ‡}; $$Choic{e^{m}_{j}}(h) = \left \{ h^{\prime} \in H_{m} \mid \left (\forall A \in Form^{Ag}\right )\big (\big [j\big ]A \in end\left (m \sqcap h\big ) \Rightarrow A \in end\big (m \sqcap h^{\prime}\big )\right )\right \}$$, if m is an equivalence class of elements. Since for every j ∈ Ag, $$\big [j\big ]$$ is an S5-modality, Choice induces a partition on Hm for every given m ∈ Tree. We check that the choice function verifies the relevant semantic constraints: Lemma 13 The tuple $$\big \langle Tree, \unlhd , Choice\big \rangle$$, as defined above, verifies both independence of agents and no choice between undivided histories constraints. Proof. We first tackle the no choice between undivided histories constraint. Consider a moment m and two histories h, h$$^{\prime}$$ ∈ Hm such that h ≈mh$$^{\prime}$$. Since the agents’ choices can only be non-vacuous (that is to say, not equal to Hm), at moments represented by equivalence classes of elements, we may safely assume that m is such a class. We may also assume that $$h \neq h^{\prime}$$ since in this case the constraint is trivially satisfied. By h ≈mh$$^{\prime}$$, there must be a moment m$$^{\prime}$$ such that $$m \lhd m^{\prime}$$ and m$$^{\prime}$$ ∈ h ∩ h$$^{\prime}$$. Hence we know that also m$$^{\prime}$$ is some equivalence class of elements. Suppose the length of m is n and the length of m$$^{\prime}$$ is n$$^{\prime}$$. Then n < n$$^{\prime}$$, and m ⊓ h, m$$^{\prime}$$ ⊓ h are both in the representation of h so that m ⊓ h is the initial segment of length n of m$$^{\prime}$$ ⊓ h. Similarly, m ⊓ h$$^{\prime}$$ is the initial segment of length n of m$$^{\prime}$$ ⊓ h$$^{\prime}$$. But both m$$^{\prime}$$⊓ h and m$$^{\prime}$$⊓ h′ are, by definition, in m$$^{\prime}$$, therefore, they must share the greatest proper initial segment. Hence, their initial segments of length n must coincide as well, and we must have m ⊓ h = m ⊓ h$$^{\prime}$$, whence $$end\big (m \sqcap h\big ) = end\big (m \sqcap h^{\prime}\big )$$. Now, if j ∈ Ag and $$\big [j\big ]A \in end\big (m \sqcap h\big )$$, then, by (A1) and $$\mathcal{CS}$$-maxiconsistency of $$end\big (m \sqcap h\big )$$, we will have also $$A \in end\big (m \sqcap h\big ) = end\big (m \sqcap h^{\prime}\big )$$, and thus $$h^{\prime} \in Choic{e^{m}_{j}}\big (h\big )$$, so that $$Choic{e^{m}_{j}}\big (h\big ) = Choic{e^{m}_{j}}\big (h^{\prime}\big )$$ since Choice is a partition of Hm. Consider, next, the independence of agents. Let m ∈ Tree and let f be a function on Ag such that $$\forall j \in Ag\big (f\big (j\big ) \in Choic{e^{m}_{j}}\big )$$. We are going to show that in this case $$\bigcap _{j \in Ag}f\big (j\big ) \neq \emptyset$$. If m ∈ {†, ‡}, then this is obvious, since every agent will have a vacuous choice. We treat the case, when m is an equivalence class of elements. Assume that $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{n + 1}\big )\big ]_{\equiv }$$. By (A1) we know that there is a set Δ of formulas of the form $$\Box A$$ which is shared by all sets of the form end(ξ) with ξ ∈ m in the sense that if ξ ∈ m, then $$\Box A \in end\big (\xi \big )$$ iff $$\Box A \in \Delta$$. By the same axiom scheme and Lemma 12, we also know that for every j ∈ Ag there is a set Δj of formulas of the form $$\big [j\big ]A$$ which is shared by all sets of the form $$end\big (\xi \big )$$ such that $$\exists h\big (h \in f\big (j\big ) \& \xi = m \sqcap h\big )$$. More precisely, define: $$\Delta_{j} : = \bigcap_{h \in f\big(j\big)}\big\{\big[j\big]A \in end\big(\xi\big)\mid \xi = m \sqcap h \big\}.$$ We claim the following: Claim Let ξ ∈ m. Then $$\exists h\left(h \in f\big(j\big) \& \xi = m \sqcap h\right) \Leftrightarrow \left(\forall A \in Form^{Ag}\right)\big([j]A \in end\big(\xi\big) \Leftrightarrow \big[j\big]A \in \Delta_{j}\big).$$ Proof of the Claim. (⇒). If $$h \in f\big (j\big )$$ is such that ξ = m ⊓ h and $$\big [j\big ]A \in \Delta _{j}$$, then $$\big [j\big ]A \in end\big (\xi \big )$$ immediately by definition of Δj. In the other direction, if $$\big [j\big ]A \in end\big (\xi \big )$$, then choose an arbitrary $$g \in f\big (j\big )$$ and let τ = m ⊓ g. Since $$f\big (j\big ) \in Choic{e^{m}_{j}}$$ and $$h \in f\big (j\big )$$, this means that $$f\big (j\big ) = Choic{e^{m}_{j}}\big (h\big )$$ so that $$g \in Choic{e^{m}_{j}}\big (h\big )$$. Therefore, whenever $$\big [j\big ]A \in end(\xi ) = end\big (m \sqcap h\big )$$, we also have $$\big [j\big ]\big [j\big ]A \in end\big (m\ \sqcap \ h\big )$$ by (A1) and $$\mathcal{CS}$$-maxiconsistency of $$end\big (m\ \sqcap \ h\big )$$, whence further $$\big [j\big ]A \in end(\tau ) = end\big (m\ \sqcap \ g\big )$$ by the definition of Choice. Since g was chosen in $$f\big (j\big )$$ arbitrarily, this means that $$\big [j\big ]A \in \Delta _{j}$$. (⇐). Assume that ξ ∈ m is such that $$\left(\forall A \in Form^{Ag}\right)\left(\big[j\big]A \in end\big(\xi\big) \Leftrightarrow \big[j\big]A \in \Delta_{j}\right)$$ (21) Using Lemma 12, choose an h ∈ Hm such that m ⊓ h = ξ. Now, choose an arbitrary g ∈ f(j) and an arbitrary τ such that τ = m ⊓ g. By the (⇒)-part of the Claim, we know that $$\left(\forall A \in Form^{Ag}\right)([j]A \in end(\tau) \Leftrightarrow [j]A \in \Delta_{j}).$$ (22) It follows from (21) and (22) that $$\left(\forall A \in Form^{Ag}\right)\left(\big[j\big]A \in end(\tau) \Leftrightarrow \big[j\big]A \in end\big(\xi\big)\right),$$ (23) whence, by (A1) and $$\mathcal{CS}$$-maxiconsistency of end(ξ), end(τ), it follows that $$\left(\forall A \in Form^{Ag}\right)\left(\big[j\big]A \in end(\tau) \Rightarrow A \in end\big(\xi\big)\right).$$ (24) We now use the fact that ξ = m ⊓ h and τ = m ⊓ g to get that $$\left(\forall A \in Form^{Ag}\right)\left(\big[j\big]A \in end\big(m \sqcap g\big) \Rightarrow A \in end\big(m \sqcap h\big)\right)$$ (25) By the above definition of Choice, this means that $$h \in Choic{e^{m}_{j}}\big (g\big )$$. Moreover, since $$g \in f\big (j\big )$$, we also have $$f\big (j\big ) = Choic{e^{m}_{j}}\big (g\big )$$, so that $$h \in f\big (j\big )$$, as desired. The Claim has therefore been proven. We now consider the set $$\Delta \cup \bigcup \left \{ \Delta _{j}\mid j \in Ag \right \}$$ and show its consistency. Indeed, if this set is inconsistent, then, wlog, we would have a provable formula of the following form: $$\vdash_{\mathcal{CS}} \left(\Box A \wedge \bigwedge_{j \in Ag}\big[j\big]A_{j}\right) \rightarrow \bot,$$ (26) where $$\Box A \in \Delta$$, and for every j ∈ Ag we will have $$\big [j\big ]A_{j} \in \Delta _{j}$$. But then, choose for every j ∈ Ag an element ξj ∈ m such that $$\left(\forall A \in Form^{Ag}\right)\left(\big[j\big]A \in end\left(\xi_{j}\right) \Leftrightarrow \big[j\big]A \in \Delta_{j}\right).$$ This is possible, since we may simply choose an arbitrary $$h_{j} \in f\big (j\big )$$ and set ξj := m ⊓ hj. Then we will have $$\big [j\big ]A_{j} \in end\big (\xi _{j}\big )$$ for every j ∈ Ag. Next, consider Γn+1. Since $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{n + 1}\big )\big ]_{\equiv }$$ and $$\Box$$ is an S5-modality, we must have $$\left\{ \Diamond\big[j\big]A_{j}\mid j \in Ag \right\} \subseteq \Gamma_{n + 1},$$ whence, by Lemma 4.5: $$\bigwedge_{j \in Ag}\Diamond\big[j\big]A_{j} \in \Gamma_{n + 1},$$ and further, by (A3) and Lemma 4.4: $$\Diamond\bigwedge_{j \in Ag}\big[j\big]A_{j} \in \Gamma_{n + 1}.$$ Also, by definition of Δ and the fact that $$\big (\Gamma _{1},\ldots , \Gamma _{n + 1}\big ) \in m$$, we get successively $$\Box A \in \Gamma_{n + 1},$$ then, by Lemma 4.5 $$\Box A \wedge \Diamond\bigwedge_{j \in Ag}\big[j\big]A_{j} \in \Gamma_{n + 1},$$ and finally, by the fact that $$\Box$$ is an S5-modality: \begin{align} \Diamond\left(\Box A \wedge \bigwedge_{j \in Ag}\big[j\big]A_{j}\right) \in \Gamma_{n + 1}. \end{align} (27) From (26), together with (27), it follows by S5 reasoning for $$\Box$$ that ◊⊥ ∈ Γn+1, so that, again by S5 properties of $$\Box$$ and Lemma 4.4, it follows that ⊥ ∈ Γn+1, which is in contradiction with $$\mathcal{CS}$$-maxiconsistency of Γn+1. Hence $$\Delta \cup \bigcup \left \{ \Delta _{j}\mid j \in Ag \right \}$$ is $$\mathcal{CS}$$-consistent, and we can extend it to a $$\mathcal{CS}$$-maxiconsistent Ξ. By Lemma 8 and Δ ⊆ Ξ, we know that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Xi \big )$$ is an element and that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Xi \big ) \in m$$ so that $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{n}, \Xi \big )\big ]_{\equiv }$$. Using Lemma 12, we can choose a g ∈ Hm such that $$m \sqcap g = \big (\Gamma _{1},\ldots , \Gamma _{n}, \Xi \big )$$. We also know that for every j ∈ Ag, there is a history $$h_{j} \in f\big (j\big )$$ such that m ⊓ hj = ξj by the choice of ξj. Therefore, for every j ∈ Ag, $$Choic{e^{m}_{j}}\big (h_{j}\big ) = f\big (j\big )$$. Also, if $$\big [j\big ]A \in end\big (\xi _{j}\big ) = end\big (m \sqcap h_{j}\big )$$, then $$\big [j\big ]A \in \Delta _{j}$$, hence $$[j]A \in \Xi = end\big (m \sqcap g\big )$$, therefore, by (A1), $$A \in end\big (g \sqcap m\big )$$. Thus we get that $$g \in \bigcap _{j \in Ag}Choic{e^{m}_{j}}\big (h_{j}\big ) = \bigcap _{j \in Ag}f\big (j\big )$$ so that the independence of agents is verified. 4.3 R and $$\mathcal{E}$$ We now define the justifications-related components of our canonical model. We first define R as follows: $$R\big (\big [\big (\Gamma _{1},\ldots ,\Gamma _{n}\big )\big ]_{\equiv }, m^{\prime}\big )\Leftrightarrow \big (m^{\prime} \in Tree \setminus \{ \dagger ,\ddagger \}\big )\&\left (\forall \tau \in m^{\prime}\right )\left (\forall A \in Form^{Ag}\right )\big (KA \in \Gamma _{n} \Rightarrow KA \in end(\tau )\big )$$; R(†, m), for all m ∈ Tree; R(‡, m) ⇔ m = ‡. Since our canonical model is unirelational, we will assume that Re is the same as R and will use the simplified satisfaction clause for t:A. Now, for the definition of $$\mathcal{E}$$: For all t ∈ Pol: $$\mathcal{E}(\dagger , t) = \mathcal{E}(\ddagger , t) = \left \{ A \in Form^{Ag} \mid \vdash _{\mathcal{CS}} t{:} A \right \}$$; For all t ∈ Pol and m ∈ Tree ∖ {†, ‡}: $$\left(\forall A \in Form^{Ag}\right)\big(A \in \mathcal{E}(m, t) \Leftrightarrow \big(\forall \xi \in m\big)\big(t{:} A \in end\big(\xi\big)\big)\big).$$ We start by mentioning a straightforward corollary to the above definition: Lemma 14 For all m ∈ Tree and t ∈ Pol it is true that $$\left \{ A \in Form^{Ag} \mid \vdash _{\mathcal{CS}} t{:} A \right \} \subseteq \mathcal{E}(m,t)$$. Proof. This holds simply by the definition of $$\mathcal{E}$$ when m ∈ {†, ‡}. If m ∈ Tree ∖ {†, ‡}, then, for every ξ ∈ m, $$end\big (\xi \big )$$ is a $$\mathcal{CS}$$-maxiconsistent subset of FormAg and must contain every provable formula. Note that it follows from Lemma 14 that the abovedefined function $$\mathcal{E}$$ satisfies the $$\mathcal{CS}$$-normality condition on jstit models. Lemma 15 The relation R, as defined above, is a preorder on Tree and, together with $$\unlhd$$, verifies the future always matters constraint. Proof. Transitivity of R is immediate from its definition. We show that R is reflexive. If $$m = \big [\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Gamma \big )\big ]_{\equiv }$$ and KA ∈ Γ, then, by (T3) and maxiconsistency of Γ, we will also have that $$\Box KA \in \Gamma$$, whence, by definition of ≡, we know that KA ∈ end(τ) for every $$\tau \in \big [\big (\Gamma _{1},\ldots ,\Gamma _{n}, \Gamma \big )\big ]_{\equiv }$$. Next, we look into future always matters constraint. Assume m ∈ Tree. If m = †, then it is connected to all the elements in Tree by both $$\unlhd$$ and R, and if m = ‡, then it is connected only to itself by both $$\unlhd$$ and R, so these moments cannot falsify the constraint. So let us assume that m is a class of equivalence generated by some element, say $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\big ]_{\equiv }$$. If m = m$$^{\prime}$$, then we are done by reflexivity of R. On the other hand, if $$m \lhd m^{\prime}$$ then we may assume, wlog, that $$m^{\prime} = \big [\big (\Gamma _{1},\ldots , \Gamma _{k}\big )\big ]_{\equiv }$$ for some k > n. But then take an arbitrary A ∈ FormAg. If KA ∈ Γn, then, since $$\big (\Gamma _{1},\ldots , \Gamma _{k}\big )$$ is an element, KA ∈ Γk. By (T3) and $$\mathcal{CS}$$-maxiconsistency of Γk, we will also have $$\Box KA \in \Gamma _{k}$$. Now, by definition of ≡, we get KA ∈ end(τ) for any given τ ∈ m′. It follows then that $$R\left (m, m^{\prime}\right )$$ as desired. We now check the semantical constraints for $$\mathcal{E}$$: Lemma 16 The function $$\mathcal{E}$$, as defined above, satisfies both monotonicity of evidence and evidence closure properties. Proof. We start with monotonicity of evidence. Assume $$R\left (m,m^{\prime}\right )$$ and t ∈ Pol. If $$m \in \left \{ \dagger , \ddagger \right \}$$ then, by Lemma 14, $$\mathcal{E}(m,t) = \left \{ A \in Form^{Ag} \mid \vdash _{\mathcal{CS}} t{:} A \right \} \subseteq \mathcal{E}\left (m^{\prime},t\right )$$ for any m′∈ Tree. Assume, further, that m is an equivalence class of elements. Let t ∈ Pol and A ∈ FormAg be such that $$A \in \mathcal{E}(m,t)$$. Then, for every ξ ∈ m, t:A ∈ end(ξ) and, by (T1), also Kt:A ∈ end(ξ). Therefore, by $$R\left (m,m^{\prime}\right )$$, we get that, for every τ ∈ m$$^{\prime}$$, Kt:A ∈ end(τ), so that by (A7) and maxiconsistency of every end(τ), also t:A ∈ end(τ). Therefore, $$A \in \mathcal{E}\left (m^{\prime},t\right )$$, as desired. We turn now to the closure conditions. We verify the first two conditions, and the third one can be verified in a similar way, restricting attention to t rather than considering both s and t. Let s, t ∈ Pol. We need to consider two cases: Case 1. m ∈ {†, ‡}. If $$A \in \mathcal{E}(m,s)$$, then $$\vdash _{\mathcal{CS}} s{:} A$$. Therefore, by (A6), we must also have $$\vdash _{\mathcal{CS}} (s + t){:}\ A$$ so that $$A \in \mathcal{E}(m,s + t)$$. Similarly, if $$A \in \mathcal{E}(m,t)$$, then also $$A \in \mathcal{E}(m,s + t)$$ and the closure constraint (b) is verified. If, on the other hand, it is true that for some A, B ∈ FormAg we have both $$A \rightarrow B \in \mathcal{E}(m,s)$$ and $$A \in \mathcal{E}(m,t)$$, then, again, this means that both $$\vdash _{\mathcal{CS}} s{:} A \rightarrow B$$ and $$\vdash _{\mathcal{CS}} t{:} A$$. By (A4), it follows that $$\vdash _{\mathcal{CS}} s\cdot t{:} B$$ and, therefore, also $$B \in \mathcal{E}(m,s\cdot t)$$, so that the closure condition (a) is also verified. Case 2. m ∈ Tree ∖{†, ‡}. If A ∈ FormAg and $$A \in \mathcal{E}(m,s)$$, then, for every ξ ∈ m, $$s{:} A \in end\big (\xi \big )$$, and, by (A6) and $$\mathcal{CS}$$-maxiconsistency of every $$end\big (\xi \big )$$, we get that s + t:A ∈ end(ξ). Therefore, $$A \in \mathcal{E}(m,s + t)$$. Similarly, if $$A \in \mathcal{E}(m,t)$$, then $$A \in \mathcal{E}(m,s + t)$$ as well, and closure condition (b) is verified. On the other hand, if A, B ∈ FormAg and we have both $$A \rightarrow B \in \mathcal{E}(m,s)$$ and $$A \in \mathcal{E}(m,t)$$, then, for every ξ ∈ m, we have $$t{:} A, s{:}\big (A \rightarrow B\big ) \in end\big (\xi \big )$$. By (A4) and $$\mathcal{CS}$$-maxiconsistency of every $$end\big (\xi \big )$$, we get that $$s \cdot t{:} B \in end\big (\xi \big )$$, thus $$B \in \mathcal{E}(m,s \cdot t)$$, and closure condition (a) is verified. 4.4 Act and V It only remains to define Act and V for our canonical model, and we define them as follows: $$\big (m,h\big ) \in V\big (p\big ) \Leftrightarrow p \in end\big (m \sqcap h\big )$$, for all p ∈ V ar; $$Act\left (\dagger , \big (\dagger ,\ddagger \big )\right ) = Act\left (\ddagger , \big (\dagger ,\ddagger \big )\right ) = \emptyset$$; $$Act\left (\dagger , \big (\dagger ,[\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots , \big )\right ) = \left \{ t \in Pol\mid \Box Et \in end\big (\xi _{1}\big ) \right \}$$; $$Act\big (m,h\big ) = \left \{ t \in Pol \mid Et \in end\big (m \sqcap h\big ) \right \}$$, if m ∈ Tree ∖ {†, ‡}. We first establish a technical claim: Lemma 17 If m ∈ Tree ∖ {†, ‡} and t ∈ Pol, then $$t \in Act_{m} \Leftrightarrow \Box Et \in \bigcap_{\xi \in m}end\big(\xi\big).$$ Proof. (⇒) Assume that t ∈ Actm. Since m ∈ Tree ∖ {†, ‡}, then, for appropriate Γ1, …, Γn, Γ ⊆ FormAg and n ≥ 0 we will have $$m = \left [\big (\Gamma _{1},\ldots ,\Gamma _{n},\Gamma \big )\right ]_{\equiv }$$. Note that, by Lemma 12, we know that for every ξ ∈ m we can choose a history h ∈ Hm such that ξ = m ⊓ h. It follows, then, by our assumption, that $$Et \in end\big (\xi \big )$$ for all ξ ∈ m. But this means that the set $$\Xi = \big\{ \Box B \mid \Box B \in \Gamma \big\} \cup \big\{ \neg Et \big\}$$ must be $$\mathcal{CS}$$-inconsistent. Suppose otherwise. Then we can extend Ξ to a $$\mathcal{CS}$$-maxiconsistent Δ. By Lemma 8 and Ξ ⊆ Δ we get that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Delta \big )$$ is an element and that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Delta \big ) \equiv \big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )$$, so that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Delta \big ) \in m$$, but also that $$\neg Et \in \Delta = end\left (\big (\Gamma _{1},\ldots , \Gamma _{n}, \Delta \big )\right )$$ which contradicts the earlier established fact that $$Et \in end\big (\xi \big )$$ for all ξ ∈ m. The obtained contradiction shows that Ξ must be $$\mathcal{CS}$$-inconsistent, and this further means that there are some $$\Box B_{1},\ldots ,\Box B_{n} \in \Gamma _{1}$$ for which we have $$\vdash_{\mathcal{CS}} \big(\Box B_{1}\wedge\ldots \wedge \Box B_{n}\big) \rightarrow Et,$$ which, by S5 reasoning for $$\Box$$, would further mean that $$\vdash_{\mathcal{CS}} \big(\Box B_{1}\wedge\ldots \wedge \Box B_{n}\big) \rightarrow \Box Et,$$ so that, by $$\mathcal{CS}$$-maxiconsistency of Γ we know that $$\Box Et \in \Gamma$$, whence, again by $$\mathcal{CS}$$-maxiconsistency of Γ and (A1), we get that $$\Box \Box Et \in \Gamma = end\big (\big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )\big )$$. Now, if ξ ∈ m is arbitrary, then we must have $$\xi \equiv \big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )$$, whence by definition of ≡ we obtain that $$\Box Et \in end\big (\xi \big )$$ and, further, $$\Box Et \in \bigcap _{\xi \in m}end\big (\xi \big )$$. (⇐) If $$\Box Et \in \bigcap _{\xi \in m}end\big (\xi \big )$$, then, for every h ∈ Hm, m ⊓ h ∈ m, so that, by (A1), we must have $$Et \in end\big (m \sqcap h\big )$$, and, by definition of Act, $$t \in Act\big (m,h\big )$$. We now check that the remaining semantic constraints on jstit models are satisfied: Lemma 18 The canonical model, as defined above, satisfies the constraints as to the expansion of presented proofs, no new proofs guaranteed, presenting new proofs makes histories divide and epistemic transparency of presented proofs. Proof. We consider the expansion of presented proofs first. Let $$m^{\prime} \lhd m$$ and let h ∈ Hm. Then $$m^{\prime} \neq \ddagger$$, since ‡ has no $$\lhd$$-successors. If m$$^{\prime}$$ = † and m = ‡, then h must be (†, ‡) and we have $$Act\!\left (\dagger , \big (\dagger ,\ddagger \big )\right ) = Act\!\left (\ddagger , \big (\dagger ,\ddagger \big )\right ) = \emptyset$$, so that the expansion of presented proofs holds. If m$$^{\prime}$$ = † and m is an equivalence class of elements, then h is of the form $$\big (\dagger ,[\xi _{1}]_{\equiv },\ldots , [\xi _{n}]_{\equiv },\ldots , \big )$$, and, for some natural n, m ⊓ h = ξn. Also, we must have $$\xi _{n} = \big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ for appropriate Γ1, …, Γn, whereas ξ1 must be the initial segment of ξn of length 1, so that we have $$\xi _{1} = \big (\Gamma _{1}\big )$$. Let t ∈ Pol be arbitrary. If $$t \in Act\big (\dagger , h\big )$$, then $$\Box Et \in \Gamma _{1}$$, therefore, by (A1) and $$\mathcal{CS}$$-maxiconsistency of Γ1, Et ∈ Γ1. Hence we must have $$\Box Et \in \Gamma _{n}$$, since $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ is an element. Again, using (A1) plus $$\mathcal{CS}$$-maxiconsistency of Γn, we infer that Et ∈ Γn, or, equivalently $$Et \in end\big (\xi _{n}\big ) = end\big (m \sqcap h\big )$$, which means that $$t \in Act\big (m,h\big )$$. Finally, if m$$^{\prime}$$ is an equivalence class of elements, then m is also an equivalence class of elements. In this case, m ⊓ h must be of the form $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ for the respective Γ1, …, Γn ⊆ FormAg. But then, for some k < n, m$$^{\prime}$$⊓ h must be of the form $$\big (\Gamma _{1},\ldots , \Gamma _{k}\big )$$. Let t ∈ Pol be arbitrary. If $$t \in Act\big (m^{\prime},h\big )$$, then $$Et \in end\big (m^{\prime} \sqcap h\big ) = \Gamma _{k}$$. Since $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$ is an element, we must have then $$\Box Et \in \Gamma _{n}$$. Further, using (A1) plus maxiconsistency of Γn, we infer that Et ∈ Γn or, equivalently $$Et \in end\big (\xi _{n}\big ) = end\big (m \sqcap h\big )$$, which means that $$t \in Act\big (m,h\big )$$. We consider next the no new proofs guaranteed constraint. Let m ∈ Tree. If m ∈ {†, ‡}, then Actm = ∅ and the constraint is trivially satisfied. Assume that m ∈ Tree ∖ {†, ‡}. We need to distinguish then between two cases: Case 1. The length of m equals 1. Then m is of the form $$\left [\big (\Gamma \big )\right ]_{\equiv }$$ for the respective Γ ⊆ FormAg. Let t ∈ Pol be such that t ∈ Actm and choose, by Lemma 12, an h ∈ Hm such that $$\big (\Gamma \big ) = m \sqcap h$$. By Lemma 17, we know that $$\Box Et \in \Gamma = end\big (m \sqcap h\big )$$, therefore, by definition of Act, we have $$t \in Act(\dagger , h) \subseteq \bigcup _{m^{\prime} \lhd m, h \in H_{m}}\left (Act\big (m^{\prime},h\big )\right )$$, as desired. Case 2. The length of m is greater than 1. Then m must be of the form $$\left [\big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )\right ]_{\equiv }$$ for the respective Γ1, …, Γn, Γ ⊆ FormAg and n > 0. Furthermore, $$m^{\prime} = \left [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\right ]_{\equiv }$$ is the immediate $$\lhd$$-predecessor of m. Let t ∈ Pol be such that t ∈ Actm and choose, by Lemma 12, an h ∈ Hm such that $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big ) = m \sqcap h$$. By Lemma 17, we know that $$\Box Et \in \Gamma = end\big (m \sqcap h\big )$$. Since $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )$$ is an element, Γn must be $$\mathcal{CS}$$-maxiconsistent; therefore, by Lemma 4.2, we either have Et ∈ Γn, or ¬Et ∈ Γn. In the latter case we would have $$\neg \Box Et \in \Gamma$$, which, given that also $$\Box Et \in \Gamma$$, is in contradiction with $$\mathcal{CS}$$-maxiconsistency of Γ. Therefore, we must have Et ∈Γn. Moreover, since $$\big (\Gamma _{1},\ldots , \Gamma _{n}, \Gamma \big )$$ is in the representation of h, it is clear that its greatest proper initial segment, $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big )$$, must also be in the representation of h. Therefore, we get that $$\big (\Gamma _{1},\ldots , \Gamma _{n}\big ) = m^{\prime} \sqcap h$$ so that $$\Gamma _{n} = end\big (m^{\prime} \sqcap h\big )$$. It follows, by definition of Act, that $$t \in Act\big (m^{\prime},h\big )$$, and, given $$m^{\prime} \lhd m$$, also that $$t \in \bigcup _{m^{\prime} \lhd m, h \in H_{m}}\big (Act\big (m^{\prime},h\big )\big )$$, as desired. We turn next to the presenting a new proof makes histories divide constraint. Consider an m ∈ Tree and arbitrary h, h$$^{\prime}$$ ∈ Hm′ such that h ≈mh$$^{\prime}$$. We may assume that $$h \neq h^{\prime}$$ since otherwise the constraint is trivially satisfied. Therefore, there must be an $$m^{\prime} \rhd m$$ such that m′∈ h ∩ h$$^{\prime}$$. Then $$m \neq \ddagger$$, since ‡ has no $$\lhd$$-successors. If m = † and m$$^{\prime}$$ = ‡, then we must have h = h$$^{\prime}$$ = (†, ‡) and the constraint is verified trivially. If m = † and $$m^{\prime} \neq \ddagger$$, then m$$^{\prime}$$ must be an equivalence class of elements, say, $$m^{\prime} = \left [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\right ]_{\equiv }$$. Since m$$^{\prime}$$ ∈ h ∩ h$$^{\prime}$$, then for some ξ, τ ∈ m$$^{\prime}$$ we must have ξ = m$$^{\prime}$$⊓ h and τ = m$$^{\prime}$$⊓ h$$^{\prime}$$, respectively. By definition of ≡ we know that for some maxiconsistent Δ, Ξ ⊆ FormAg we will have both $$\xi = \big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Delta \big )$$ and $$\tau = \big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Xi \big )$$, and we will also have ξ ≡ τ. We now have two cases to consider: Case 1. n > 1. Then $$\big (\Gamma _{1}\big )$$ is a proper initial segment of both ξ and τ which means that, by $$\left [\big (\Gamma _{1}\big )\right ]_{\equiv } \lhd m^{\prime}$$ and no backward branching, we must have $$\left [\big (\Gamma _{1}\big )\right ]_{\equiv } \in h \cap h^{\prime}$$. Since $$\big (\Gamma _{1},\ldots , \Gamma _{n - 1}\big )$$ is in the representations of both h and h$$^{\prime}$$, we know that also $$\big (\Gamma _{1}\big )$$ must be in both representations so that $$\big(\Gamma_{1}) = \big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h = \big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h^{\prime}.$$ It follows immediately that $$\Gamma_{1} = end\left(\big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h\right) = end\left(\big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h^{\prime}\right),$$ and, further, that $$Act\big(\dagger, h\big)\!=\! \left\{ t \in Pol \mid \Box Et\! \in\! end\left(\big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h\right) \right\}\! =\! \left\{ t\! \in\! Pol \mid \Box Et\! \in\! end\big(\big[\big(\Gamma_{1}\big)\big]_{\equiv} \sqcap h^{\prime}\big) \right\}\! =\! Act\big(\dagger, h^{\prime}\big),$$ and we are done. Case 2. n = 1. Then we must have (Δ) = ξ ≡ τ = (Ξ) so that, by definition of ≡ and S5 properties of $$\Box$$ we get, among other things, that $$\Box Et \in \Delta = end\big(m^{\prime} \sqcap h\big) \Leftrightarrow \Box Et \in \Xi =end\big(m^{\prime} \sqcap h^{\prime}\big)$$ for every t ∈ Pol. Whence it clearly follows that $$Act\big(\dagger, h\big) = \left\{ t \in Pol \mid \Box Et \in end\big(m^{\prime} \sqcap h\big) \right\} = \left\{ t \in Pol \mid \Box Et \in end\big(m^{\prime} \sqcap h^{\prime}\big) \right\} = Act\big(\dagger, h^{\prime}\big),$$ and we are done. The only case that is left now is when m is an equivalence class of elements. In this case m$$^{\prime}$$ is an equivalence class of elements either, so that, say, $$m^{\prime} = \big [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\big ]_{\equiv }$$. Since m$$^{\prime}$$ ∈ h ∩ h$$^{\prime}$$, then for some ξ, τ ∈ m$$^{\prime}$$ we must have ξ = m$$^{\prime}$$⊓ h and τ = m$$^{\prime}$$⊓ h$$^{\prime}$$, respectively. By definition of ≡ we know that for some $$\mathcal{CS}$$-maxiconsistent Δ, Ξ ⊆ FormAg we will have both $$\xi = \big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Delta \big )$$ and $$\tau = \big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Xi \big )$$, and, moreover, we will have ξ ≡ τ. Since $$m \lhd m^{\prime}$$ and m ∈ h ∩ h$$^{\prime}$$, then m ⊓ h must be a proper initial segment of $$\big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Delta \big )$$, whereas m ⊓ h$$^{\prime}$$ must be a proper initial segment of $$\big (\Gamma _{1},\ldots , \Gamma _{n - 1}, \Xi \big )$$ and both segments must be the elements in m so that they must be of the same length. It follows then, that for some k < n we will have $$m \sqcap h = \big(\Gamma_{1},\ldots, \Gamma_{k}\big) = m \sqcap h^{\prime},$$ whence $$end\big(m \sqcap h\big) = \Gamma_{k} = end\big(m \sqcap h^{\prime}\big).$$ The latter equation entails that $$Act\big(m, h\big) = \left\{ t \in Pol \mid Et \in end\big(m \sqcap h\big) \right\} = \left\{ t \in Pol \mid Et \in end\big(m \sqcap h^{\prime}\big) \right\} = Act\big(m,h^{\prime}\big),$$ as desired. It remains to check the epistemic transparency of presented proofs constraint. Assume that m, m′∈ Tree are such that $$R\big (m,m^{\prime}\big )$$. If we have m ∈ {†, ‡}, then, by definition, we must have Actm = ∅, and the constraint is verified in a trivial way. If, on the other hand, m ∈ Tree ∖ {†, ‡}, then $$m = \big [\big (\Gamma _{1},\ldots , \Gamma _{n}\big )\big ]_{\equiv }$$ for appropriate Γ1, …, Γn ⊆ FormAg and n > 0. By R(m, m$$^{\prime}$$), we must also have m$$^{\prime}$$ ∈ Tree ∖ {†, ‡}. Now, if for some t ∈ Pol we have t ∈ Actm, then, by Lemma 17, we will also have $$\Box Et \in \Gamma _{n}$$, whence, by (A9) and $$\mathcal{CS}$$-maxiconsistency of Γn, that $$K\Box Et \in \Gamma _{n}$$. The latter means, by definition of R, that $$K\Box Et \in \bigcap _{\tau \in m^{\prime}}end(\tau )$$. $$\mathcal{CS}$$-maxiconsistency of end(τ) for every τ ∈ m$$^{\prime}$$, together with (A7), yields then $$\Box Et \in \bigcap _{\tau \in m^{\prime}}end(\tau )$$ so that, by Lemma 17 again, it follows that t ∈ Actm′ and thus we are done. 4.5 The truth lemma It follows from Lemmas 9, 13, 15, 16 and 18 that our abovedefined canonical model $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ is a unirelational jstit model for Ag. By Corollary 2 we know that $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ is based on discrete time and by Lemma 14 we know that $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ is $$\mathcal{CS}$$-normal. Summing everything up, we get that $$\mathcal{M}^{Ag}_{\mathcal{CS}} \in Mod^{\downarrow }_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{discr}\right )$$. Now we need to supply a truth lemma: Lemma 19 Let A ∈ FormAg, let m ∈ Tree ∖ {†, ‡}, and let h ∈ Hm. Then $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models A \Leftrightarrow A \in end\big(m \sqcap h\big).$$ Proof. As is usual, we prove the lemma by induction on the construction of A. The basis of induction with A = p ∈ V ar we have by definition of V, whereas Boolean cases for the induction step are trivial. We treat the modal cases: Case 1. $$A = \Box B$$. If $$\Box B \in end\big (m \sqcap h\big )$$, then note that for every h′∈ Hm we must have m ⊓ h′ ∈ m so that m ⊓ h′ ≡ m ⊓ h. By definition of ≡ and the fact that m ∈ Tree ∖ {†, ‡}, we must have then $$B \in end\big (m \sqcap h^{\prime}\big )$$ for all h′∈ Hm and thus, by induction hypothesis, we obtain that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models \Box B$$. If, on the other hand, $$\Box B \notin end(m \sqcap h)$$, we must have then $$m \sqcap h = \big (\Gamma _{1},\ldots ,\Gamma _{n},\Gamma \big )$$ for some appropriate n ≥ 0 and Γ1, …, Γn, Γ so that Γ = end(m ⊓ h). Then the set $$\Xi = \left\{ \Box C \mid \Box C \in \Gamma \right\} \cup \left\{ \neg B \right\}$$ must be $$\mathcal{CS}$$-consistent, since otherwise we would have $$\vdash_{\mathcal{CS}} \big(\Box C_{1}\wedge\ldots\wedge\Box C_{n}\big) \rightarrow B$$ for some $$\Box C_{1},\ldots ,\Box C_{n} \in \Gamma$$, whence, since $$\Box$$ is an S5-modality, we would get $$\vdash_{\mathcal{CS}} \big(\Box C_{1}\wedge\ldots\wedge\Box C_{n}\big) \rightarrow \Box B,$$ which would mean that $$\Box B \in \Gamma$$, contrary to our assumption. Therefore, Ξ is $$\mathcal{CS}$$-consistent and we can extend Ξ to a $$\mathcal{CS}$$-maxiconsistent Δ ∈ FormAg. Of course, in this case B ∉ Δ. By Lemma 8 and Ξ ⊆ Δ, we know that $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big )$$ is actually an element and that $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Gamma \big ) \equiv \big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big )$$. By Lemma 12, for some h′ ∈ Hm we will have $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big ) = m \sqcap h^{\prime}$$ and, therefore, $$\Delta = end\big (m \sqcap h^{\prime}\big )$$. Since B ∉ Δ, it follows, by induction hypothesis, that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h^{\prime} \not \models B$$, hence $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \not \models \Box B$$ as desired. Case 2. $$A = \big [j\big ]B$$ for some j ∈ Ag. Then, if $$\big [j\big ]B \in end(m \sqcap h)$$, by definition of Choice and the fact that m ∈ Tree ∖ {†, ‡} we must have $$Choic{e^{m}_{j}}\big(h\big) = \left\{ h^{\prime} \in H_{m} \mid \left(\forall C \in Form^{Ag}\right)\left(\big[j\big]C \in end\big(m \sqcap h\big) \Rightarrow C \in end\big(m \sqcap h^{\prime}\big)\right)\right\}.$$ Therefore, if $$h^{\prime} \in Choic{e^{m}_{j}}\big (h\big )$$, then we must have $$B \in end\big (m \sqcap h^{\prime}\big )$$, and, by induction hypothesis, $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h^{\prime} \models B$$, so that we get $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models \big [j\big ]B$$. On the other hand, if $$\big [j\big ]B \notin end\big (m \sqcap h\big )$$, then we must have $$m \sqcap h = \big (\Gamma _{1},\ldots ,\Gamma _{n},\Gamma \big )$$ for some appropriate n ≥ 0 and Γ1, …, Γn, Γ so that $$\Gamma = end\big (m \sqcap h\big )$$. Then the set $$\Xi = \left\{ \big[j\big]C \mid \big[j\big]C \in \Gamma \right\} \cup \left\{ \neg B \right\}$$ must be $$\mathcal{CS}$$-consistent, since otherwise we would have $$\vdash_{\mathcal{CS}} \big(\big[j\big]C_{1}\wedge\ldots\wedge\big[j\big]C_{n}\big) \rightarrow B$$ for some $$\big [j\big ]C_{1},\ldots ,\big [j\big ]C_{n} \in \Gamma$$, whence, since $$\big [j\big ]$$ is an S5-modality, we would get $$\vdash_{\mathcal{CS}} \left(\big[j\big]C_{1}\wedge\ldots\wedge\big[j\big]C_{n}\right) \rightarrow \big[j\big]B,$$ which would mean that $$\big [j\big ]B \in \Gamma$$, contrary to our assumption. Therefore, Ξ is $$\mathcal{CS}$$-consistent and we can extend Ξ to a $$\mathcal{CS}$$-maxiconsistent Δ ⊆ FormAg. Of course, in this case B ∉ Δ. Furthermore, note that if D ∈ FormAg is such that $$\Box D \in \Gamma$$, then, by (A1), (A2) and $$\mathcal{CS}$$-maxiconsistency of Γ, we know that $$\big [j\big ]\Box D \in \Gamma$$, so that also $$\big [j\big ]\Box D \in \Delta$$, and hence, by (A1) and $$\mathcal{CS}$$-maxiconsistency of Δ, $$\Box D \in \Delta$$. We have thus shown that $$\left \{ \Box D \mid \Box D \in \Gamma \right \} \subseteq \Delta$$. Again, by Lemma 8 it follows that $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big )$$ is an element and that $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Gamma \big ) \equiv \big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big )$$. By Lemma 12, for some h′∈ Hm we will have $$\big (\Gamma _{1},\ldots ,\Gamma _{n},\Delta \big ) = m \sqcap h^{\prime}$$ and, therefore, $$\Delta = end\big (m \sqcap h^{\prime}\big )$$. Also, since Δ contains all the $$\big [j\big ]$$-modalized formulas from Γ, we know that for any such h′ we will have $$h^{\prime} \in Choic{e^{m}_{j}}\big (h\big )$$. Since B ∉ Δ, it follows, by induction hypothesis, that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,\,h^{\prime} \not \models B$$, hence $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \not \models \big [j\big ]B$$ as desired. Case 3. A = KB. Assume that $$KB \in end\big (m \sqcap h\big )$$. We clearly have then $$m = \big [\big (m \sqcap h\big )\big ]_{\equiv }$$. Hence, by definition of R and the fact that m ∈ Tree ∖ {†, ‡} we must have for every m′∈ Tree: $$R\big(m,m^{\prime}\big) \Rightarrow \big(\forall \tau \in m^{\prime}\big)\big(\forall C \in Form^{Ag}\big)\big(KC \in end\big(m \sqcap h\big) \Rightarrow KC \in end(\tau)\big).$$ Therefore, if $$R\big (m,m^{\prime}\big )$$ and h′∈ Hm′ is arbitrary, then, of course, (m′⊓ h′) ∈ m′ so that $$KB \in end\big (m^{\prime} \sqcap h^{\prime}\big )$$, and, further, $$B \in end\big (m^{\prime} \sqcap h^{\prime}\big )$$ by S4 reasoning for K. Therefore, by induction hypothesis, we get that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m^{\prime},h^{\prime} \models B$$, whence $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models KB$$. On the other hand, if $$KB \notin end\big (m \sqcap h\big )$$, then consider the set $$\Xi = \big\{ KC \mid KC \in end\big(m \sqcap h\big) \big\} \cup \big\{ \neg\Box B \big\}.$$ This set must be $$\mathcal{CS}$$-consistent, since otherwise we would have $$\vdash_{\mathcal{CS}} \big(KC_{1}\wedge\ldots\wedge KC_{n}\big) \rightarrow \Box B$$ for some KC1, …, KCn ∈Γ, whence, since K is an S4-modality, we would get $$\vdash_{\mathcal{CS}} \big(KC_{1}\wedge\ldots\wedge KC_{n}\big) \rightarrow K\Box B,$$ which would mean that $$K\Box B \in end\big (m \sqcap h\big )$$, hence, by (A1), (A7) and $$\mathcal{CS}$$-maxiconsistency of $$end\big (m \sqcap h\big )$$, that $$KB \in end\big (m \sqcap h\big )$$, contrary to our assumption. Therefore, Ξ is $$\mathcal{CS}$$-consistent and we can extend Ξ to a $$\mathcal{CS}$$-maxiconsistent Δ ⊆ FormAg. Of course, in this case $$\Box B \notin \Delta$$. We will have then that $$\big (\Delta \big )$$ is an element. So we set $$m^{\prime} = \big [\big (\Delta \big )\big ]_{\equiv }$$. Now choose an arbitrary (Δ′) ∈ m′. We will have, by the choice of m′, that $$\big (\Delta ^{\prime}\big ) \equiv \big (\Delta \big )$$. Then every boxed formula from Δ will be in Δ′. In particular, whenever $$KC \in end\big (m \sqcap h\big )$$ for a given C ∈ FormAg, it follows that KC ∈ Δ, whence $$\Box KC \in \Delta$$ and thus $$KC \in \Delta ^{\prime} = end\big (\big (\Delta ^{\prime}\big )\big )$$, by (T3) and maxiconsistency of Δ. Since $$\big (\Delta ^{\prime}\big )$$ was chosen as an arbitrary element of m′, this means that whenever KC ∈ end(m ⊓ h), it is also the case that $$KC \in \bigcap _{\tau \in m^{\prime}}end(\tau )$$ so that we must have R(m, m′). On the other hand, since $$\Box B \notin \Delta$$, then, by Case 1, there must be a τ ∈ m′ such that B ∉ end(τ). But then, by Lemma 12, we can choose an h′ ∈ Hm′ in such a way that τ = m′ ⊓ h′, and we get that $$B \notin end\big (m^{\prime} \sqcap h^{\prime}\big )$$. Therefore, by induction hypothesis, we get $$\mathcal{M}^{Ag}_{\mathcal{CS}},m^{\prime},h^{\prime} \not \models B$$. In view of the fact that also $$R\big (m,m^{\prime}\big )$$, this means that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \not \models KB$$ as desired. Case 4. A = t:B for some t ∈ Pol. If $$t{:} B \in end\big (m\ \sqcap\ h\big )$$, then, by $$\mathcal{CS}$$-maxiconsistency of $$end\big (m\ \sqcap\ h\big )$$ and (T2), we must have $$\Box t{:} B \in end\big (m \sqcap h\big )$$. Now if ξ ∈ m, then we must have, of course, ξ ≡ m ⊓ h, whence $$t{:} B \in end\big (\xi \big )$$. Therefore, we must have $$B \in \mathcal{E}(m,t)$$. Also, by maxiconsistency of $$end\big (m\ \sqcap \ h\big )$$ and (A5), we will have $$KB \in end\big (m\ \sqcap \ h\big )$$. Therefore, by Case 3, we will have that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models KB$$ and further, by $$B \in \mathcal{E}(m,t)$$, that $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models t{:} B$$. On the other hand, if $$t{:} B \notin end\big (m \sqcap h\big )$$, then since clearly m ⊓ h ∈ m, we must have $$B \notin \mathcal{E}(m,t)$$, whence $$\mathcal{M}^{Ag}_{\mathcal{CS}},m,h \not \models t{:} B$$. Case 5. A = Et for some t ∈ Pol. Then, given that m ∈ Tree ∖ {†, ‡}, we have, simply by definition of Act, that $$Et \in end\big(m \sqcap h\big) \Leftrightarrow t \in Act\big(m,h\big) \Leftrightarrow \mathcal{M}^{Ag}_{\mathcal{CS}},m,h \models Et.$$ This finishes the list of the modal induction cases at hand, and thus the proof of our truth lemma is complete. 5 The main result We are now in a position to prove Theorem 2. One direction of the theorem was proved as Corollary 1. In the other direction, assume that Γ ⊆ FormAg is $$\mathcal{CS}$$-consistent. Then, by Lemma 4.1, Γ can be extended to a $$\mathcal{CS}$$-maxiconsistent Δ. Now, consider $$\mathcal{M}^{Ag}_{\mathcal{CS}} = \big \langle Tree, \unlhd , Choice, Act, R, \mathcal{E}, V\big \rangle$$, the canonical model defined in Section 4. Since $$\mathcal{M}^{Ag}_{\mathcal{CS}} \in Mod^{\downarrow }_{\mathcal{CS}}\left (\mathcal{C}^{Ag}_{discr}\right )$$, we know that $$\mathcal{M}^{Ag}_{\mathcal{CS}} \in Mod^{\downarrow }_{\mathcal{CS}}\left (\mathcal{C}\right ) \subseteq Mod_{\mathcal{CS}}\left (\mathcal{C}\right )$$. The structure $$\big (\Delta \big )$$ is an element, therefore $$\big [\big (\Delta \big )\big ]_{\equiv } \in Tree$$. By Lemma 12, there is a history $$h \in H_{[(\Delta )]_{\equiv }}$$ such that $$\big (\Delta \big ) = \big [\big (\Delta \big )\big ]_{\equiv } \sqcap h$$. For this h, we will also have $$\Delta = end\big (\big [\big (\Delta \big )\big ]_{\equiv } \sqcap h\big )$$. By Lemma 19, we therefore get that $$\mathcal{M}^{Ag}_{\mathcal{CS}}, \big[\big(\Delta\big)\big]_{\equiv}, h \models \Delta \supseteq \Gamma,$$ and thus Γ is shown to be satisfiable in a jstit model based on frame from $$\mathcal{C}$$. We observe that it follows from the above proof that the canonical model $$\mathcal{M}^{Ag}_{\mathcal{CS}}$$ defined in Section 4 is $$\mathcal{CS}$$-universal in the sense that it satisfies every $$\mathcal{CS}$$-consistent subset of FormAg. As standard corollaries to Theorem 2, we get the respective weak completeness and compactness results. 6 Conclusions We have shown that the Hilbert-style axiomatic system ΣD is strongly complete w.r.t. the set of JA-STIT formulas valid over jstit models based on discrete times and that this set does not change as long as one adds further models to this class, provided that these models are based on mixed successor stit frames. We have also shown that this result is stable modulo constant specifications that might be added to ΣD. The fact that claiming the full set of ΣD-theorems for the class of models based on a given stit frame C does not ensure that the C will be based on discrete time lights up some essential limitations of expressive powers of JA-STIT and calls for further investigation. Therefore, in the second part of the paper we will look in the reverse direction, and instead of finding an axiomatic system for a given class of frames we will identify classes of frames defined by ΣD. We will present two main results, one for the stit frame as defined above and another one for jstit frames, which extend stit frames with the epistemic accessibility relations R and Re. The presence of these relations happens to be a game changer in that the form of frame definability result for jstit frames is very different from the respective theorem for stit frames. Funding The author would like to acknowledge financial support from the Deutsche Forschungsgemeinschaft, DFG, project WA 936/11-1. Acknowledgements The author would like to thank the anonymous reviewer of this paper, who made a lot of useful comments which led to some corrections and other improvements in the presentation. Footnotes 1  Backus-Naur Form. 2  A more common notation ≤ is not convenient for us since we also widely use ≤ in this paper to denote the natural order relation between elements of ω. 3  In other words, our stit frames are just a propositional version of what [4] calls BT+AC structures. A more inclusive notion of jstit frame will be considered in some detail in Part II of this paper. 4  One exception is the class of colinear frames only mentioned in this paper for technical reasons, although our axiomatization is sound also w.r.t. this class. 5  See, e.g. [4, Ch. 17], although ΣD uses a simpler format closer to that given in [3, Section 2.3]. 6  The format for the variable assignment V is slightly different, but this is of no consequence for the present setting. References [1] S. Artemov and M. Fitting . Justification Logic , E. N. Zalta ed. , winter 2016 edn. The Stanford Encyclopedia of Philosophy . Metaphysics Research Lab, Stanford University, 2016 . [2] S. N. Artemov and E. Nogina . Introducing justification into epistemic logic . Journal of Logic and Computation , 15 , 1059 -- 1073 , 2005 . Google Scholar CrossRef Search ADS [3] P. Balbiani , A. Herzig and N. Troquard . Alternative axiomatics and complexity of deliberative stit theories . Journal of Philosophical Logic , 37 , 387 -- 406 , 2008 . Google Scholar CrossRef Search ADS [4] N. Belnap , M. Perloff , and M. Xu . Facing the Future: Agents and Choices in Our Indeterminist World . Oxford University Press , 2001 . [5] J. Horty . Agency and Deontic Logic . Oxford University Press , USA , 2001 . Google Scholar CrossRef Search ADS [6] G. Olkhovikov . Stit logic of justification announcements: a completeness result . Journal of Logic and Computation , (to appear) , 2018 . Google Scholar CrossRef Search ADS [7] G. Olkhovikov and H. Wansing . Inference as doxastic agency. Part I: the basics of justification stit logic. Studia Logica, 2018 . Online first : https://doi.org/10.1007/s11225-017-9779-z. [8] G. Olkhovikov and H. Wansing . Inference as doxastic agency. Part II: ramifications and refinements . Australasian Journal of Philosophy , 14 , 408 -- 438 , 2017 . © The Author(s) 2018. Published by Oxford University Press. All rights reserved. For Permissions, please email: journals.permissions@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/open_access/funder_policies/chorus/standard_publication_model)

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Logic Journal of the IGPLOxford University Press

Published: Feb 27, 2018

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