Tableaux for a combination of propositional dynamic logic and epistemic logic with interactions

Tableaux for a combination of propositional dynamic logic and epistemic logic with interactions Abstract Epistemic propositional dynamic logic (EPDL) is a combination of epistemic logic and propositional dynamic logic. Two properties, perfect recall and no miracles, capture the interactions between actions and knowledge. In this article, we present a tableau-based decision procedure for deciding the satisfiability of single-agent EPDL formula in models with perfect recall and no miracles. 1 Introduction Propositional dynamic logic (PDL) is an important logic for reasoning about programs or actions [6]. Epistemic logic (EL) is a modal logic concerned with reasoning about informational aspects of agent, in particular, agent’s belief and knowledge [11]. Thus the combination of EL and PDL is a powerful tool for reasoning about interactions between knowledge and actions. In this article, we call the combination of EL and PDL ‘epistemic propositional dynamic logic’ (EPDL). There are two ways to combine PDL and EL: product and fusion [7]. In this article, we combine PDL and EL by way of fusion. It is shown that the fusion of two modal logics inherit properties such as completeness, the finite model property and decidability from the individual logics [12, 13, 26]. However, it is much more complex if the fusion is extended with interactions of these two logics. The most frequently discussed interactions in the combinations of PDL and EL are perfect recall, no learning and Church-Rosser axiom. The conjunction of perfect recall and no learning is also called commutativity in [7]. Perfect recall ($$\texttt{PR}$$) is commonly formulated by the axiom schema $$\left\langle {{a}} \right\rangle \hat{K}\phi\to \hat{K}\left\langle {{a}} \right\rangle \phi$$ where $$\left\langle {{a}} \right\rangle$$ is a PDL modality and $$\hat{K}$$ is the epistemic modality. It expresses the persistence of the agent’s knowledge after the execution of an action. It can be described by the first-order sentence $$\forall x\forall y\forall z(x{\stackrel{a}{\rightarrow}}y\land y\sim z\to \exists u(x\sim u\land u{\stackrel{{a}}{\rightarrow}}z))$$ where $${\stackrel{{a}}{\rightarrow}}$$ is the binary relation that interprets the atomic action $$a$$ and $$\sim$$ is the equivalence relation that interprets the EL’s modality $$K$$. Perfect recall can be depicted as follows: No learning ($$\texttt{NL}$$) is given by the axiom schema $$\hat{K}\left\langle {{a}} \right\rangle \phi\to\left\langle {{a}} \right\rangle \hat{K}\phi$$. The formula says the agent knows the result of his action in advance. In other words, there is no learning. $$\texttt{NL}$$ can be described by the first-order sentence $$\forall x\forall y\forall z(x\sim y\land y{\stackrel{{a}}{\rightarrow}} z\to \exists u(x{\stackrel{{a}}{\rightarrow}} u\land u\sim z))$$ and depicted as follows: Church-Rosser axiom ($$\texttt{CR}$$) is the axiom schema $$\left\langle {{a}} \right\rangle K\phi\to K\left\langle {{a}} \right\rangle \phi$$. It says that if an agent is possible to know $$\phi$$ by performing an action $$a$$ then he knows that it is possible to achieve $$\phi$$ by doing $$a$$. $$\texttt{CR}$$ can be described by the first-order sentence $$\forall x\forall y\forall z(x\sim y\land x{\stackrel{{a}}{\rightarrow}} z\to \exists u(y{\stackrel{{a}}{\rightarrow}} u\land u\sim z))$$ and depicted as follows: The fusions of PDL and different versions of EL (such as K45, KD45 and S5) extended with various choices of $$\texttt{PR}$$, $$\texttt{NL}$$ and $$\texttt{CR}$$ are well studied in [7, 17–19]. For example, it has been shown that the fusion of PDL and S5 extended with $$\texttt{PR}$$ and $$\texttt{NL}$$ coincides with the product of PDL and S5, and that the fusion of PDL and S5 extended with $$\texttt{PR}$$ and $$\texttt{NL}$$ has the 2-exponential finite model property. It has also been proved that the fusion of PDL and S5 extended with $$\texttt{NL}$$ is the same as the fusion of PDL and S5 extended with $$\texttt{CR}$$, which is consistent with the fact that $$\texttt{NL}$$ and $$\texttt{CR}$$ are equal to each other if the binary relation $$\sim$$ is an equivalence relation. Besides $$\texttt{PR}$$, $$\texttt{NL}$$ and $$\texttt{CR}$$, no miracles is also an important property which captures the interaction between agent’s actions and knowledge [21–23]. No miracles ($$\texttt{NM}$$) is syntactically given by the axiom schema $$\hat{K}\left\langle {{a}} \right\rangle \phi\to [a]\hat{K}\phi$$. Please note that the structure of the $$\texttt{NM}$$ axiom differs from the above $$\texttt{NL}$$ axiom in the form of the first modality: in $$\texttt{NM}$$ it is a box modality $$[a]$$ while in $$\texttt{NL}$$ it is a diamond modality $$\left\langle {{a}} \right\rangle$$. This is because not all actions are executable at the current world. This subtle difference makes $$\texttt{NM}$$ more suitable to capture the interaction between knowledge and actions and $$\texttt{NL}$$ more suitable to describe the interaction between knowledge and time. A more detailed discussion of the difference between $$\texttt{NM}$$ and $$\texttt{NL}$$ can be found in [23]. Intuitively, no miracles expresses that there is no ‘miracles’ situation such that the agent cannot distinguish two states initially but nevertheless he can distinguish the states resulting from executing the same action on these two states. $$\texttt{NM}$$ can be described by the first-order sentence $$\forall x\forall y\forall z\forall u((x\sim y\land x{\stackrel{{a}}{\rightarrow}} z\land y{\stackrel{{a}}{\rightarrow}}u)\to u\sim z)$$ and depicted as follows: While the combinations of PDL and EL extended with $$\texttt{PR}$$, $$\texttt{NL}$$ or $$\texttt{CR}$$ are well studied, the combination extended with $$\texttt{NM}$$ and these properties has not been investigated sufficiently. In this article, we will focus on the fusion of PDL and EL, which is called EPDL in this article, extended with $$\texttt{PR}$$ and $$\texttt{NM}$$. We observe that in real life there are certain knowledge changes over actions for which it is very natural to think in terms of EPDL models with the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. The following are two examples. First, let us consider a simplified version of the Monty Hall problem. There are three doors: behind one door is a car; behind the others are goats. Initially the agent does not know what is behind the doors. Now an action happens; specifically, one door with a goat is opened. Subsequently, the agent knows that the car is not behind the opened door, but he still does not know behind which door is the car. We use $$A$$ to denote that the car is behind the first door, and $$Ab$$ to denote that the car is behind the first door and the second door is opened. The action $$a$$ means to open the first door. The others are similar. The model depicted in Figure 1 represents the knowledge development in this example, and it has the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. Figure 1. View largeDownload slide Initially, the agent cannot distinguish states A, B and C. Figure 1. View largeDownload slide Initially, the agent cannot distinguish states A, B and C. In the second example, EPDL with $$\texttt{PR}$$ and $$\texttt{NM}$$ is a natural way to model the conformant planning problems (cf. [25]). Conformant planning is the problem of finding a linear plan (a sequence of actions) that is guaranteed to achieve a goal in presence of uncertainty about the initial state (cf. [20]). Considering the example depicted in Figure 2, the initial uncertainty set is $$\{s_1,s_2\}$$ and the goal set is $$\{s_3,s_4\}$$. A knowledge state is a subset of the state space, which records the uncertainty during the execution of a plan, e.g., $$\{s_1, s_2\}$$ is an initial knowledge state. To make sure a goal is achieved eventually, it is crucial to track the transitions of knowledge states during the execution of the plan. Figure 3, which sketches an EPDL model with $$\texttt{PR}$$ and $$\texttt{NM}$$, displays the knowledge development over actions. From Figure 3, we can see that the action sequence $$aa$$ is a solution.1 In [25], it is shown that the existence of a solution for a conformant planning problem can be expressed in the language of EPDL. Therefore, the satisfiability of EPDL with $$\texttt{PR}$$ and $$\texttt{NM}$$ is interesting. Figure 2. View largeDownload slide A conformant planning problem. Figure 2. View largeDownload slide A conformant planning problem. Figure 3. View largeDownload slide Knowledge development in the conformant planning problem. Figure 3. View largeDownload slide Knowledge development in the conformant planning problem. This article presents a tableau-based decision procedure for deciding satisfiability of single-agent EPDL in models with the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. EPDL is a 2-dimensional logic in the sense that it has two kinds of modalities: PDL modalities and epistemic modality. For the construction of the tableau for this 2-dimensional logic, our method is a combination of the tableau-construction method of temporal epistemic logic (TEL) used in [1, 2, 4, 8, 24] and the tableau-construction method of PDL used in [9, 16]. TEL, which is a logic for reasoning about agents’ knowledge evolution over time [5, 15], is also a 2-dimensional logic: it has both temporal modalities and epistemic modality. The construction of the tableau for TEL is based on bubbles, and the tableau of TEL is a bubble graph. On the other hand, the structure of PDL modalities is much more complex compared to the structure of temporal modalities. We deal with PDL modalities with the method used for building the tableau for PDL. The article is organized as follows. Section 2 introduces the language, model and semantics of EPDL. Section 3 presents the tableau procedure for EPDL with $$\texttt{PR}$$ and $$\texttt{NM}$$. Section 4 proves the soundness and completeness of the tableau procedure, and we conclude in Section 5 and point to future works. 2 Epistemic propositional dynamic logic This section will present the language, model and semantics of EPDL, and define the properties of perfect recall and no miracles. The language of EPDL is constructed by combining knowledge and program operators. Definition 1 (Language) Let $$\mathbf{\Phi_0}$$ and $$\mathbf{\Pi_0}$$ be two countably infinite sets of propositions and actions, respectively. The language of EPDL is defined in BNF as follows:   \begin{align*} \phi &::= \top\mid p\mid \neg\phi\mid(\phi\wedge\phi)\mid [\pi]\phi \mid K\phi\\ \pi &::= a\mid ?\phi\mid (\pi;\pi)\mid (\pi+\pi)\mid \pi^*, \end{align*} where $$p\in\mathbf{\Phi_0}$$ and $$a\in\mathbf{\Pi_0}$$. We will often omit parentheses when doing so ought not cause confusion. The language has expressions of two sorts: formulas $$\phi$$ and programs $$\pi$$. The set of all programs is denoted $$\mathbf{\Pi}$$, and the set of all formulas is denoted $$\mathbf{\Phi}$$. As usual, we use the following abbreviations: $$\bot:=\neg\top$$, $$\phi\vee\psi:=\neg(\neg \phi\wedge\neg \psi),\phi\rightarrow\psi:=\neg\phi\vee\psi, \left\langle {{a}} \right\rangle \phi:=\neg [a]\neg\phi, \hat{K}\phi:=\neg K\neg\phi$$. Definition 2 (Model) A model $$\mathcal{M}$$ is a tuple $$\left\langle {{S^\mathcal{M},\{R^\mathcal{M}_a\mid a\in\mathbf{\Pi_0}\},R^\mathcal{M},V^\mathcal{M}}} \right\rangle$$, where $$S^\mathcal{M}$$ is a nonempty set of states, $$R^\mathcal{M}_a$$ is a binary relation on $$S^\mathcal{M}$$, $$R^\mathcal{M}$$ is an equivalence relation on $${S^\mathcal{M}}$$ and $$V^\mathcal{M}:\mathbf{\Phi_0}\to\mathcal{P}({S^\mathcal{M}})$$ is a function. A pointed model is a pair $$(\mathcal{M},s)$$ consisting of a model $$\mathcal{M}$$ and a state $$s\in S^\mathcal{M}$$. Given a model $$\mathcal{M}$$, we also write $$(s,t)\in R^\mathcal{M}_a$$ as $$s{\stackrel{a}{\rightarrow}}_{\mathcal{M}}t$$ or $$t\in R^\mathcal{M}_a(s)$$, and write $$(s,t)\in R^\mathcal{M}_a$$ as $$s\sim_{\mathcal{M}}t$$ or $$t\in R^\mathcal{M}(s)$$. If the model $$\mathcal{M}$$ is obvious from the context, we omit it as an index. Definition 3 (Semantics) Given a pointed model $$(\mathcal{M},s)$$ and a formula $$\phi$$, we write $$\mathcal{M},s\vDash\phi$$ to mean $$(\mathcal{M},s)$$ satisfies $$\phi$$. The satisfaction relation $$\vDash$$, shown in Table 1, is defined as usual by combining the semantics of EL and that of PDL. A formula $$\phi$$ is satisfiable if $$\mathcal{M},s\vDash\phi$$ for some model $$\mathcal{M}$$ and a state $$s\in S^\mathcal{M}$$. Table 1 Semantics $$\mathcal{M},s\vDash\top$$      $$\mathcal{M},s\vDash p$$  $$\iff$$  $$s\in V(p)$$  $$\mathcal{M},s\vDash \neg\phi$$  $$\iff$$  $$\mathcal{M},s\nvDash \phi$$  $$\mathcal{M},s\vDash \phi\wedge\psi$$  $$\iff$$  $$\mathcal{M},s\vDash\phi \text{ and } \mathcal{M},s\vDash \phi$$  $$\mathcal{M},s\vDash K\phi$$  $$\iff$$  $$s\sim t \text{ implies } \mathcal{M},t\vDash \phi$$  $$\mathcal{M},s\vDash [\pi]\phi$$  $$\iff$$  $$s{\stackrel{{\pi}}{\rightarrow}}t \text{ implies } \mathcal{M},t\vDash\phi$$  $${\stackrel{{a}}{\rightarrow}}$$  =  $$R^\mathcal{M}_a$$  $${\stackrel{{?\phi}}{\rightarrow}}$$  =  $$\{(s,s)\mathcal{M}id \mathcal{M},s\vDash\phi\}$$  $${\stackrel{{\pi_1+\pi_2}}{\rightarrow}}$$  =  $${\stackrel{\pi_1}{\rightarrow}}\cup{\stackrel{\pi_2}{\rightarrow}}$$  $${\stackrel{{\pi_1;\pi_2}}{\rightarrow}}$$  =  $${\stackrel{{\pi_1}}{\rightarrow}}\circ{\stackrel{{\pi_2}}{\rightarrow}}$$  $${\stackrel{{\pi^*}}{\rightarrow}}$$  =  $$\bigcup_{n\geq 0}({\stackrel{{\pi}}{\rightarrow}})^n$$  $$\mathcal{M},s\vDash\top$$      $$\mathcal{M},s\vDash p$$  $$\iff$$  $$s\in V(p)$$  $$\mathcal{M},s\vDash \neg\phi$$  $$\iff$$  $$\mathcal{M},s\nvDash \phi$$  $$\mathcal{M},s\vDash \phi\wedge\psi$$  $$\iff$$  $$\mathcal{M},s\vDash\phi \text{ and } \mathcal{M},s\vDash \phi$$  $$\mathcal{M},s\vDash K\phi$$  $$\iff$$  $$s\sim t \text{ implies } \mathcal{M},t\vDash \phi$$  $$\mathcal{M},s\vDash [\pi]\phi$$  $$\iff$$  $$s{\stackrel{{\pi}}{\rightarrow}}t \text{ implies } \mathcal{M},t\vDash\phi$$  $${\stackrel{{a}}{\rightarrow}}$$  =  $$R^\mathcal{M}_a$$  $${\stackrel{{?\phi}}{\rightarrow}}$$  =  $$\{(s,s)\mathcal{M}id \mathcal{M},s\vDash\phi\}$$  $${\stackrel{{\pi_1+\pi_2}}{\rightarrow}}$$  =  $${\stackrel{\pi_1}{\rightarrow}}\cup{\stackrel{\pi_2}{\rightarrow}}$$  $${\stackrel{{\pi_1;\pi_2}}{\rightarrow}}$$  =  $${\stackrel{{\pi_1}}{\rightarrow}}\circ{\stackrel{{\pi_2}}{\rightarrow}}$$  $${\stackrel{{\pi^*}}{\rightarrow}}$$  =  $$\bigcup_{n\geq 0}({\stackrel{{\pi}}{\rightarrow}})^n$$  Definition 4 (Perfect recall and no miracles) A model $$\mathcal{M}$$ has the property of Perfect Recall ($$\texttt{PR}$$) if for all $$a\in\mathbf{\Pi_0}$$ and all $$s,t,t'\in S^\mathcal{M}$$, $$s{\stackrel{{a}}{\rightarrow}}t$$ and $$t\sim t'$$ imply that there exists $$s'\in S^\mathcal{M}$$ such that $$s\sim s'$$ and $$s'{\stackrel{{a}}{\rightarrow}} t'$$. No Miracles ($$\texttt{NM}$$) if for all $$a\in\mathbf{\Pi_0}$$ and all $$s,s',t,t'\in S^\mathcal{M}$$, $$s{\stackrel{{a}}{\rightarrow}}t$$, $$s\sim s'$$ and $$s'{\stackrel{{a}}{\rightarrow}}t'$$ imply $$t\sim t'$$. In the rest of this article, we will always assume that models have the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. Moreover, since it follows by bisimulation invariance that EPDL has the tree model property (cf. [3]), we also assume all the models in this paper have the tree property; i.e., if $$s{\stackrel{{a}}{\rightarrow}}t$$ and $$s'{\stackrel{{a}}{\rightarrow}}t$$, then $$s=s'$$ for each $$a\in\mathbf{\Pi_0}$$. Next, we will focus on the problem whether a given formula is satisfiable in models with $$\texttt{PR}$$ and $$\texttt{NM}$$. This problem is tackled by building a tableau from an input formula and deciding if the tableau is open or not. The key is sorting out how to build a proper tableau from an input formula. 3 Tableau for EPDL with PR and NM This section will present how to construct the tableau from an input formula. To deal with the complication arising with interacting actions and knowledge, the tableau procedure will act on bubbles. A bubble, which represents an equivalence class, is a set of states, and it is epistemically sufficient and knowledge-consistent. It will be seen that each relevant formula of the form $$\neg K\phi$$ is realized within bubbles. The tableau procedure, then, will only need to focus on the realization of formulas in the shape of $$\neg[\pi]\phi$$ and the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. Before the tableau procedure, we introduce some terminology. Definition 5 (Fisher–Ladner closure) The Fisher-Ladner closure of a formula $$\phi$$, denoted $$\mathcal{FL}(\phi)$$, is the minimal set satisfying: $$\phi\in \mathcal{FL}(\phi)$$; if $$\psi\in \mathcal{FL}(\phi)$$ then $$\neg\psi\in\mathcal{FL}(\phi)$$, provided $$\psi$$ does not start with $$\neg$$; if $$\neg\psi$$, $$\psi\wedge\chi$$, $$K\psi$$ or $$[\pi]\psi$$ are in $$\mathcal{FL}(\phi)$$ then $$\psi,\chi\in\mathcal{FL}(\phi)$$; if $$[\pi_1;\pi_2]\psi\in\mathcal{FL}(\phi)$$ then $$[\pi_1][\pi_2]\psi\in\mathcal{FL}(\phi)$$; if $$[\pi_1+\pi_2]\psi\in\mathcal{FL}(\phi)$$ then both $$[\pi_1]\psi$$ and $$[\pi_2]\psi$$ are in $$\mathcal{FL}(\phi)$$; if $$[?\chi]\psi\in\mathcal{FL}(\phi)$$ then $$\chi\in\mathcal{FL}(\phi)$$; if $$[\pi^*]\psi\in\mathcal{FL}(\phi)$$ then $$[\pi][\pi^*]\psi\in\mathcal{FL}(\phi)$$. Given a formula set $$\Gamma$$, we also use $$\mathcal{FL}({\Gamma})$$ to mean $$\bigcup_{\phi\in\Gamma}\mathcal{FL}(\phi)$$. Please note that $$\mathcal{FL}(\phi)$$ is finite for every $$\phi\in\mathbf{\Phi}$$. What is more, we categorize formulas as $$\boldsymbol{\alpha}/\boldsymbol{\beta}$$-formulas, each with two components, as shown in Table 2. Table 2 $$\boldsymbol{\alpha}$$- and $$\boldsymbol{\beta}$$-formulas $$\boldsymbol{\alpha}$$  $$\neg\neg\phi$$  $$\phi\wedge\psi$$  $$[\pi_1;\pi_2]\phi$$  $$\neg [\pi_1;\pi_2]\phi$$  $$\neg[?\psi]\phi$$  $$[\pi_1+\pi_2]\phi$$  $$[\pi^*]\phi$$  $$K\phi$$  $$\boldsymbol{\alpha}_1$$  $$\phi$$  $$\phi$$  $$[\pi_1][\pi_2]\phi$$  $$\neg [\pi_1][\pi_2]\phi$$  $$\psi$$  $$[\pi_1]\phi$$  $$\phi$$  $$\phi$$  $$\boldsymbol{\alpha}_2$$    $$\psi$$      $$\neg\phi$$  $$[\pi_2]\phi$$  $$[\pi][\pi^*]\phi$$    $$\boldsymbol{\beta}$$  $${\neg(\phi\wedge\psi)}$$  $${\neg[\pi_1+\pi_2]\phi}$$  $${{[?\psi]\phi}}$$  $${\neg[\pi^*]\phi}$$  $$\boldsymbol{\beta_1}$$  $${\neg\phi}$$  $${\neg[\pi_1]\phi}$$  $${\neg\psi}$$  $${\neg\phi}$$  $$\boldsymbol{\beta_2}$$  $${\neg\psi}$$  $${\neg[\pi_2]\phi}$$  $${\phi}$$  $${\phi,\neg[\pi][\pi^*]\phi}$$  $$\boldsymbol{\alpha}$$  $$\neg\neg\phi$$  $$\phi\wedge\psi$$  $$[\pi_1;\pi_2]\phi$$  $$\neg [\pi_1;\pi_2]\phi$$  $$\neg[?\psi]\phi$$  $$[\pi_1+\pi_2]\phi$$  $$[\pi^*]\phi$$  $$K\phi$$  $$\boldsymbol{\alpha}_1$$  $$\phi$$  $$\phi$$  $$[\pi_1][\pi_2]\phi$$  $$\neg [\pi_1][\pi_2]\phi$$  $$\psi$$  $$[\pi_1]\phi$$  $$\phi$$  $$\phi$$  $$\boldsymbol{\alpha}_2$$    $$\psi$$      $$\neg\phi$$  $$[\pi_2]\phi$$  $$[\pi][\pi^*]\phi$$    $$\boldsymbol{\beta}$$  $${\neg(\phi\wedge\psi)}$$  $${\neg[\pi_1+\pi_2]\phi}$$  $${{[?\psi]\phi}}$$  $${\neg[\pi^*]\phi}$$  $$\boldsymbol{\beta_1}$$  $${\neg\phi}$$  $${\neg[\pi_1]\phi}$$  $${\neg\psi}$$  $${\neg\phi}$$  $$\boldsymbol{\beta_2}$$  $${\neg\psi}$$  $${\neg[\pi_2]\phi}$$  $${\phi}$$  $${\phi,\neg[\pi][\pi^*]\phi}$$  Definition 6 (State) A state $$\Delta$$ is a set of formulas satisfying the following conditions. $$\Delta$$ is not patently inconsistent, i.e. it does not contain both $$\phi$$ and $$\neg\phi$$; $$\boldsymbol{\alpha}\in\Delta$$ implies $$\boldsymbol{\alpha}_1\in\Delta$$ and $$\boldsymbol{\alpha}_2\in\Delta$$; $$\boldsymbol{\beta}\in\Delta$$ implies $$\boldsymbol{\beta}_1\in\Delta$$ or $$\boldsymbol{\beta}_2\in\Delta$$; For each $$K\phi\in\mathcal{FL}({\psi})$$ where $$\psi\in\Delta$$, either $$K\phi\in\Delta$$ or $$\neg K\phi\in \Delta$$. Given a finite set of formulas $$\Gamma$$, let $$\mathtt{S}({\Gamma})$$ be the set of states generated by $$\Gamma$$; namely $$\mathtt{S}({\Gamma})=\{\Delta\mid \Delta$$ is the minimal state such that $$\Gamma\subseteq \Delta \}$$. When $$\Gamma=\{\phi \}$$, we also write it as $$\mathtt{S}({\phi})$$ for abbreviation. For each $$\Delta\in\mathtt{S}({\Gamma})$$, we have it that $$\Delta\subseteq \mathcal{FL}({\Gamma})$$. Moreover, we use the following abbreviations: given a formula set $$\Delta$$, let $$K(\Delta)=\{K\phi\mid K\phi\in\Delta \}$$, $$Epi(\Delta)=K(\Delta)\cup\{\neg K\phi\mid\neg K\phi\in\Delta \}$$, and $$[a]^-(\Delta)=\{\phi\mid [a]\phi\in\Delta \}$$. Definition 7 (Bubble) A bubble$$B$$ is a finite and non-empty set of states such that: $$B$$ is epistemically sufficient, i.e. for each $$\Delta\in B$$ and each $$\neg K\phi\in \Delta$$, there exists a set $$\Delta'\in B$$ such that $$\neg\phi\in \Delta'$$; $$B$$ is knowledge-consistent, i.e. $$K(\Delta)=K(\Delta')$$ for all $$\Delta,\Delta'\in B$$; Any finite and non-empty set of states is called a pre-bubble. Definition 8 (Satisfiability of (pre-)bubble) Given a (pre-)bubble $$B$$ and a pointed model $$(\mathcal{M},s)$$, let $$[s]$$ be the equivalence class containing $$s$$, i.e. $$\{s'\in S^\mathcal{M}\mid s\sim s' \}$$. If there is a surjective function $$f:[s]\to B$$ such that $$\mathcal{M},s\vDash f(s)$$, we say $$B$$ is satisfied by $$(\mathcal{M},s)$$ and $$f$$, written as $$\mathcal{M},s\vDash^f B$$. If there are such a pointed model and such a function, we say $$B$$ is satisfiable. Intuitively, a state represents a possible world in a model. A bubble is a set of states in the same equivalence class. For each state $$\Delta$$ in a bubble $$B$$, we also denote it as $$\Delta_B$$. Given model $$\mathcal{M}$$ with the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$, it can be seen as a deterministic bubble tree, as shown in Figure 4. Each state $$\Delta_i$$ in Figure 4b represents the set of formulas that the state $$s_i$$ in Figure 4a satisfies. Except for the top bubble, every state in the other bubbles has an annotation which indicates its predecessor. Figure 4. View largeDownload slide (a) Model $$\mathcal{M}$$; (b) deterministic bubble tree. Figure 4. View largeDownload slide (a) Model $$\mathcal{M}$$; (b) deterministic bubble tree. A tableau $$\mathcal{T}$$ is a graph with bubbles and labelled edges $${\stackrel{{a}}{\rightarrow}}$$ where $$a\in\mathbf{\Pi_0}$$. As mentioned above, a bubble is a set of states. If $$B{\stackrel{{a}}{\rightarrow}}B'$$, we annotate each state $$\Delta'\in B'$$ with a state $$\Delta\in B$$, indicating that $$\Delta{\stackrel{{a}}{\rightarrow}}\Delta'$$; namely $$\Delta$$ is an $$a$$-predecessor of $$\Delta'$$. The states in the first layer will be annotated with $$\emptyset$$ since they have no predecessors. Let $$B$$ and $$B'$$ be two bubbles from a tableau such that $$B{\stackrel{{a}}{\rightarrow}}B'$$. The property $$\texttt{PR}$$ requires that each state in $$B'$$ has an $$a$$-predecessor in $$B$$. In other words, there must be ‘enough’ states in $$B$$. From the construction of the tableau below, we will see that each state $$\Delta'\in B'$$ is annotated with a state $$\Delta\in B$$ and $$[a]^-(\Delta)\subseteq \Delta'$$. The property $$\texttt{NM}$$ requires that all formulas of the form $$\neg[a]\phi$$ from a bubble $$B$$ will be realized in the same bubble $$B'$$, which is also guaranteed from the construction of the tableau. The tableau procedure for EPDL with $$\texttt{PR}$$ and $$\texttt{NM}$$ consists of two phases: a construction phase and an elimination phase. In the construction phase, a pre-tableau with pre-bubbles and bubbles is established. In the elimination phase, the pre-tableau is pruned to an initial tableau and then a final tableau. 3.1 Construction of the pre-tableau Each state $$\Delta\in\mathtt{S}({\phi})$$ is a potential expansion of $$\phi$$. The construction is based on (pre-)bubbles. Therefore, we start with pre-bubbles containing $$\Delta\in\mathtt{S}({\phi})$$. To keep the property $$\texttt{PR}$$, we need to make sure that at least one of the initial pre-bubbles has enough states. For each state, if it is in the same equivalence class with $$\Delta$$, then $$Epi(\Delta)$$ is the minimal set of formulas that it contains. Therefore, there exists $$P\subseteq \mathtt{S}({Epi(\Delta)})$$ such that $$\{\Delta \}\cup P$$ potentially has enough states. The idea of the construction from an input $$\phi$$ is that initially we make $$\{\Delta \}\cup P$$ a pre-bubble for each $$\Delta\in\mathtt{S}({\phi})$$ and each $$P\subseteq \mathtt{S}({Epi(\Delta)})$$. Since $$Epi(\Delta)$$ is the minimal set of formulas that a state in the same equivalence class with $$\Delta$$ needs to satisfy, at least one of these initial pre-bubbles potentially has enough states. To continue the construction, we need to realize formulas of the forms $$\neg K\psi$$ and $$\neg[a]\psi$$ and keep the properties $$\texttt{PR}$$ and $$\texttt{NM}$$. The procedure ExtendToBubbles extends a pre-bubble to all potential bubbles and also realizes all formulas of the form $$\neg K\psi$$. The procedure SuccessorPreBubbles realizes all formulas of the form $$\neg[a]\psi$$. What is more, $$\texttt{PR}$$ and $$\texttt{NM}$$ are guaranteed in these procedures. The construction of the pre-tableau for $$\phi$$ works as follows. (1) For each $$\Delta\in\mathtt{S}({\phi})$$ and each $$P\subseteq \mathtt{S}({Epi(\Delta)})$$, make the pre-bubble $$\{\Delta \}\cup P$$ as a node and annotate each state in it with $$\emptyset$$. (2) Expand each pre-bubble $$A$$ into bubbles by calling the procedure ExtendToBubbles(A), which is presented in Algorithm 3. For each bubble $$B'\in$$ExtendToBubbles$${(A)}$$, add $$B'$$ as a node if $$B'$$ is not already there. We then produce an arrow $$A{\dashrightarrow} B'$$. (3) For each bubble $$B$$, if there exists a formula $$\neg[a]\psi\in\Delta$$ for some $$\Delta\in B$$ and some $$a\in\mathbf{\Pi_0}$$, produce successor pre-bubbles by calling the procedure SuccessorPreBubbles$${(B,a)}$$, which is presented in Algorithm 4. For each pre-bubble $$A\in$$SuccessorPreBubbles$${(B,a)}$$, add $$A$$ as a node if $$A$$ is not already there. We then produce an arrow $$B{\stackrel{{a}}{\rightarrow}}A$$. (4) Repeat steps 2 and 3 until no new bubbles or pre-bubbles are created. View largeDownload slide View largeDownload slide View largeDownload slide View largeDownload slide View largeDownload slide View largeDownload slide View largeDownload slide View largeDownload slide Since each bubble or pre-bubble in the construction is a subset of $$\mathcal{P}({\mathcal{FL}(\phi)})$$, the pre-tableau for $$\phi$$ is a finite graph, and the construction of the pre-tableau will terminate. The following are all the procedures, which are adaptations from [1], and we will also show that each procedure functions well. KnowledgeConsistent The procedure KnowledgeConsistent, presented in Algorithm 1, takes a pre-bubble $$A$$ as input and returns a set of knowledge-consistent pre-bubbles. To make the pre-bubble $$A$$ knowledge-consistent, it seems that each $$\Delta\in A$$ only need to be extended to $$\Delta\cup\Sigma$$ for some $$\Sigma\in\mathtt{S}({Epi(A)\setminus Epi(\Delta)})$$. However, this is not sufficient. To preserve the satisfiability of the pre-bubble $$A$$, we need to extend each $$\Delta\in A$$ to a subset of $$\{\Delta\cup \Sigma\mid \Sigma\in\mathtt{S}({Epi(A)\setminus Epi(\Delta)})\}$$. For example, given a pre-bubble $$A=\{\Delta_1,\Delta_2 \}$$ where $$\Delta_1=\{K(p\lor q\lor r),p\lor q\lor r,p \}$$ and $$\Delta_2=\{\neg p \}$$, a model $$\mathcal{M}$$ is depicted in Figure 5. Let function $$f$$ be $$f=\{s\mapsto \Delta_1,t\mapsto \Delta_2,v\mapsto \Delta_2 \}$$. Then we have $$\mathcal{M},s\vDash^f A$$. Let $$\Gamma_1=\{K(p\lor q\lor r),p\lor q\lor r,q,\neg p \}$$ and $$\Gamma_2=\{K(p\lor q\lor r),p\lor q\lor r,r,\neg p \}$$. To preserve the satisfiability of $$A$$ on model $$\mathcal{M}$$, the pre-bubble $$A_3$$ must be returned by the procedure KnowledgeConsistent$${(A)}$$, as is shown in Figure 6. Figure 5. View largeDownload slide Model Figure 5. View largeDownload slide Model Figure 6. View largeDownload slide KnowledgeConsistent$$(A)$$. Figure 6. View largeDownload slide KnowledgeConsistent$$(A)$$. The procedure KnowledgeConsistent has the following properties. Proposition 1 If $$A$$ is a pre-bubble, each $$A'\in$$KnowledgeConsistent$${(A)}$$ is knowledge-consistent. Proof. Assume that there is a pre-bubble $$A'\in$$KnowledgeConsistent$${(A)}$$ such that $$A'$$ is not knowledge-consistent. It follows that there are $$\Omega,\Omega'\in A'$$ and $$K\phi$$ such that $$K\phi\in\Omega$$ and $$K\phi\not\in\Omega'$$. By Algorithm 1, we have $$Epi(A)\subseteq \Omega$$ and $$Epi(A)\subseteq \Omega'$$. Let $$\Omega=\Delta_i\cup \Sigma$$ for some $$\Delta_i\in A$$ and some $$\Sigma\in\mathtt{S}({Epi(A)\setminus Epi(\Delta_i)})$$, and $$\Omega'=\Delta_j\cup \Sigma'$$ for some $$\Delta_j\in A$$ and some $$\Sigma'\in\mathtt{S}({Epi(A)\setminus Epi(\Delta_j)})$$. Since $$K\phi\in\Delta_i$$ implies $$K\phi\in \Omega'$$, it follows that $$K\phi\in \Sigma$$. It follows that there exists $$\psi\in Epi(A)$$ such that $$K\phi\in sub(\psi)$$. Since $$Epi(A)\subseteq \Omega'$$ and $$\Omega'$$ is a state, we have $$\neg K\phi\in \Omega'$$. Similarly, $$\neg K\phi\in\Delta_j$$ implies $$\neg K\phi\in \Omega$$. Thus we have $$\neg K\phi\in\Sigma'$$. Then there exists $$\Delta\in A$$ and $$\psi'\in Epi(\Delta)$$ such that $$K\phi\in sub(\psi')$$. Since $$\Delta$$ is a state, either $$K\phi\in \Delta$$ or $$\neg K\phi\in\Delta$$. This implies that $$K\phi\in Epi(A)$$ or $$\neg K\phi\in Epi(A)$$. This is contradictory with the facts that $$Epi(A)\subseteq \Omega$$, $$Epi(A)\subseteq \Omega'$$, $$K\phi\in \Omega$$ and $$\neg K\phi\in\Omega'$$. Thus each $$A'\in$$KnowledgeConsistent$${(A)}$$ is knowledge-consistent. ■ Proposition 2 If $$A$$ is a pre-bubble and $$\mathcal{M},s\vDash^f A$$, there exists a pre-bubble $$A'\in$$KnowledgeConsistent$${(A)}$$ and a function $$f':[s]\to A'$$ such that $$\mathcal{M},s\vDash^{f'}A'$$, $$f(s)\subseteq f'(s)$$, and $$f'(s)$$ has the same annotation as $$f(s)$$. Proof. Since $$\mathcal{M},s\vDash^f A$$, we have it that, for each $$u\in[s]$$, $$\mathcal{M},u\vDash f(u)$$ and $$\mathcal{M},u\vDash epi_u$$, where $$f(u)\in A$$ and $$epi_u=Epi(A)\setminus Epi(f(u))$$. Thus, in Algorithm 1, there exists $$\Sigma_u\in \mathtt{S}({epi'})$$ such that $$\mathcal{M},u\vDash \Sigma_u$$. Therefore we have $$\mathcal{M},u\vDash f(u)\cup\Sigma_u$$, and it follows by Algorithm 1 that $$f(u)\cup\Sigma_u\in Alt_i$$ for some $$i$$, and $$f(u)\cup\Sigma_u$$ has the same annotation as $$f(u)$$. Let $$D_\Delta=\{\Delta\cup\Sigma_u\mid u\in [s], f(u)=\Delta \}$$ for each $$\Delta\in A$$. Thus, $$D_\Delta\in P(Alt_i)$$ and $$A'=\cup \{D_\Delta\mid \Delta\in A \}\in$$KnowledgeConsistent$${(A)}$$. We define the function $$f':[s]\to A'$$ as $$f'(u)=f(u)\cup\Sigma_u$$ for each $$u\in[s]$$. It follows that $$\mathcal{M},u\vDash f'(u)$$ and $$f'$$ is surjective. Therefore, $$\mathcal{M},s\vDash^{f'} A'$$, $$f(s)\subseteq f'(s)$$, and $$f'(s)$$ has the same annotation as $$f(s)$$. ■ EpistemicallySufficient The procedure EpistemicallySufficient, which is presented in Algorithm 2, takes a pre-bubble $$A$$ as input and returns a set of epistemically sufficient pre-bubbles. To make $$A$$ epistemically sufficient, it seems that for each unrealized $$\neg K\phi$$ from $$A$$, we need to extend $$A$$ by adding some $$\Sigma\in\mathtt{S}({\neg\phi})$$. However, there are several problems. First, if $$A$$ is knowledge-consistent, to preserve the knowledge consistency, it seems that we should add some $$\Sigma\in\mathtt{S}({Epi(A)\cup\{\neg\phi\}})$$ to $$A$$. Secondly, if $$B{\stackrel{{a}}{\rightarrow}}A$$ it follows by $$\texttt{PR}$$ that each $$\Sigma$$ which is intended to be added in $$A$$ needs to have an $$a$$-predecessor $$\Delta\in B$$. This means we should add $$\Delta'\cup \Sigma$$ in $$A$$ where $$\Delta'\in \mathtt{S}({[a]^-(\Delta)})$$ rather than add only $$\Sigma$$ in $$A$$. We assume that $$A$$ has enough states, which means that all such $$\Delta'$$ are already in $$A$$. Finally, there might be more than one unrealized formulas $$\neg K\phi_1\cdots\neg K\phi_n$$ from $$A$$ such that they are realized by the same state. In this situation, to preserve the satisfiability of $$A$$, we might need to extend $$A$$ by adding $$\Delta'\cup \Sigma_i\cdots \Sigma_n$$ in $$A$$ where $$\Delta'\in A$$ and $$\Sigma_i\in\mathtt{S}({Epi(A)\cup\neg\phi_i})$$. For example, given pre-bubble $$A=\{\Delta,\Gamma \}$$ where $$\Delta=\{\hat{K}p,\hat{K}q \}$$ and $$\Gamma=\{\hat{K}p,\hat{K}q,r\}$$, the model $$\mathcal{M}$$ is depicted as Figure 7, and let function $$f$$ be $$\{s\mapsto \Delta,t\mapsto\Gamma \}$$. It is obvious that $$\mathcal{M},s\vDash^f A$$. To preserve the satisfiability of $$A$$ on $$\mathcal{M}$$, the procedure EpistemicallySufficient(A), as shown in Figure 8, needs to return a bubble such that it only has two states, and one of them realizes both $$\hat{K}p$$ and $$\hat{K}q$$. Figure 7. View largeDownload slide Model Figure 7. View largeDownload slide Model Figure 8. View largeDownload slide EpistemicallySufficient$$(A)$$  $$\matrix{ {{\Delta _1} = \{ \hat Kp,\hat Kq,p\} \qquad {\Gamma _1} = \{ \hat Kp,\hat Kq,r,p\} } \hfill \cr {{\Delta _2} = \{ \hat Kp,\hat Kq,q\} \qquad {\Gamma _2} = \{ \hat Kp,\hat Kq,r,q\} } \hfill \cr {{\Delta _3} = \{ \hat Kp,\hat Kq,p,q\} \qquad {\Gamma _3} = \{ \hat Kp,\hat Kq,r,p,q\} .} \hfill \cr }$$ Figure 8. View largeDownload slide EpistemicallySufficient$$(A)$$  $$\matrix{ {{\Delta _1} = \{ \hat Kp,\hat Kq,p\} \qquad {\Gamma _1} = \{ \hat Kp,\hat Kq,r,p\} } \hfill \cr {{\Delta _2} = \{ \hat Kp,\hat Kq,q\} \qquad {\Gamma _2} = \{ \hat Kp,\hat Kq,r,q\} } \hfill \cr {{\Delta _3} = \{ \hat Kp,\hat Kq,p,q\} \qquad {\Gamma _3} = \{ \hat Kp,\hat Kq,r,p,q\} .} \hfill \cr }$$ The procedure EpistemicallySufficient has the following properties. Proposition 3 If $$A$$ is a pre-bubble, each $$A'\in$$EpistemicallySufficient$${(A)}$$ is epistemically sufficient. Proof. By Algorithm 2, it follows that for each $$\neg K\phi\in\Delta\in A$$, there is $$\Omega\in A'$$ such that $$\neg\phi\in \Omega$$. Next we will show that for each $$\neg K\phi\in\Omega\in A'$$, there is $$\Delta\in A$$ such that $$\neg K\phi\in\Delta$$. This implies that each $$A'\in$$EpistemicallySufficient$${(A)}$$ is epistemically sufficient. If $$\neg K\phi\in\Omega\in A'$$ and $$\Omega\not\in A$$, we know that $$\Omega=\Delta\cup \Sigma_1\cdots\Sigma_k$$ where $$\Delta\in A$$ and, for each $$1\leq i\leq k$$, $$\Sigma_i\in\mathtt{S}({Epi(A)\cup\{\neg\psi_i \}})$$ for some $$\neg K\psi_i\in\Delta''$$ and $$\Delta''\in A$$. If $$\neg K\phi\in\Sigma_i$$ for some $$1\leq i\leq k$$, there exists $$\chi\in\Delta'$$ for some $$\Delta'\in A$$ such that $$K\phi\in sub(\chi)$$. Since $$\Delta'$$ is a state, we have $$K\phi\in\Delta'$$ or $$\neg K\phi\in \Delta'$$. If $$K\phi\in \Delta'$$, it follows $$K\phi\in Epi(A)$$ and then $$K\phi\in\Omega$$. This is contradictory with $$\neg K\phi\in\Omega$$. Thus, $$\neg K\phi\in \Delta'$$. ■ Proposition 4 If a pre-bubble $$A$$ is knowledge-consistent, each pre-bubble $$A' \in$$EpistemicallySufficient$${(A)}$$ is also knowledge-consistent. Proof. We only need to show that $$K\phi\in\Omega$$ implies $$K\phi\in\Omega'$$ for all $$\Omega,\Omega'\in A'$$. By Algorithm 2, it follows that, for each $$\Omega\in A'$$, there is $$\Delta\in A$$ such that $$\Delta\subseteq \Omega$$. Since $$A$$ is knowledge-consistent, we only need to show that $$K\phi\in\Omega$$ implies $$K\phi\in K(A)$$. If $$\Omega\not\in A$$, we know $$\Omega=\Delta\cup \Sigma_1\cdots\Sigma_k$$ where $$\Delta\in A$$ and, for each $$1\leq i\leq k$$, $$\Sigma_i\in\mathtt{S}({Epi(A)\cup\{\neg\psi_i \}})$$ for some $$\neg K\psi_i\in\Delta''$$ and $$\Delta''\in A$$. It follows $$Epi(A)\subseteq \Omega$$. If $$K\phi\in\Sigma_i$$ for some $$1\leq i\leq k$$, there exists $$\Delta'\in A$$ and $$\psi'\in \Delta'$$ such that $$K\phi\in sub(\psi')$$. Since $$\Delta'$$ is a state, we have $$K\phi\in\Delta'$$ or $$\neg K\phi\in\Delta'$$. If $$\neg K\phi\in\Delta'$$, it follows by $$Epi(A)\subseteq \Omega$$ that $$\neg K\phi\in \Omega$$. This is contradictory with $$K\phi\in\Omega$$. Thus, $$K\phi\in \Delta'$$, and consequently $$K\phi\in K(A)$$. ■ Proposition 5 If $$A$$ is a pre-bubble and $$\mathcal{M},s\vDash^f A$$, there exists a pre-bubble $$A'\in$$EpistemicallySufficient$${(A)}$$ and a function $$f':[s]\to A'$$ such that $$\mathcal{M},s\vDash^{f'}A'$$, $$f(s)\subseteq f'(s)$$, and $$f'(s)$$ has the same annotation as $$f(s)$$. Proof. Since $$\mathcal{M},s\vDash^f A$$, we have that for each $$u\in[s]$$, $$\mathcal{M},u\vDash Epi(A)$$ and $$\mathcal{M},u\vDash f(u)$$. Moreover, since $$f$$ is surjective, this means that for each $$\neg K\phi\in Epi(A)$$ there exists $$u\in[s]$$ such that $$\mathcal{M},u\vDash\neg K\phi$$. It follows that there is some $$s'\in[s]$$ such that $$\mathcal{M},s'\vDash\neg\phi$$. Let $$epi$$ be the set $$\{\neg K\phi\in Epi(A)\mid \neg\phi\not\in\Delta$$ for any $$\Delta\in A \}$$. For each $$u\in[s]$$, let $$\sigma(u)=\{\neg\phi\mid \neg K\phi\in epi$$ and $$\mathcal{M},u\vDash\neg\phi \}$$. It is obvious that $$\mathcal{M},u\vDash f(u)\land\sigma(u)$$ for each $$u\in[s]$$. If $$\sigma(u)\neq\emptyset$$, it follows that $$\mathcal{M},u\vDash f(u)\land\Sigma_1\cdots\Sigma_{|\sigma(u)|}$$ where $$\Sigma_i\in \mathtt{S}({Epi(A)\cup\{\neg\phi_i \}})$$ for each $$1\leq i\leq |\sigma(u)|$$. We define $$\Omega_u$$ as $$\Omega_{u}=f(u)\cup\Sigma_1\cdots\Sigma_{|\sigma(u)|}$$. It is obvious that $$\mathcal{M},u\vDash\Omega_{u}$$. By Algorithm 2, it follows that $$\Omega_{u}$$ is in $$Alt$$ and $$\Omega_{u}$$ has the same annotation as $$\Delta$$. If $$\sigma(u)=\emptyset$$, let $$\Omega_{u}=f(u)$$. Therefore, $$\mathcal{M},u\vDash\Omega_{u}$$ for each $$u\in[s]$$. Let a pre-bubble $$A'$$ be the set $$\{\Omega_{u}\mid u\in [s] \}$$. By Algorithm 2, we have $$A'\in$$EpistemicallySufficient$${(A)}$$. We define $$f':[s]\to A'$$ as $$f'(u)=\Omega_{u}$$ for each $$u\in[s]$$. Therefore, $$\mathcal{M},s\vDash^{f'} A'$$, $$f(s)\subseteq f'(s)$$, and $$f'(s)$$ has the same annotation as $$f(s)$$. ■ ExtendToBubbles The procedure ExtendToBubbles, which is presented in Algorithm 3, extends a pre-bubble $$A$$ to bubbles. To do so, we firstly make the pre-bubble $$A$$ knowledge-consistent with the procedure KnowledgeConsistent$${(A)}$$ and then make each knowledge-consistent pre-bubble $$A'\in$$KnowledgeConsistent$${(A)}$$ epistemically sufficient with the procedure EpistemicallySufficient$${(A')}$$. The properties of the procedures KnowledgeConsistent and EpistemicallySufficient guarantee that each pre-bubble $$B\in$$ExtendToBubbles$${(A)}$$ is a bubble extension of $$A$$. Proposition 6 If $$A$$ is a pre-bubble, each $$B\in$$ExtendToBubbles$${(A)}$$ is a bubble. Proof. It follows from Propositions 1, 3 and 4. ■ Proposition 7 Given a pre-bubble $$A$$ and a bubble $$B\in$$ExtendToBubbles$${(A)}$$, we have the following results: For each $$\Delta\in A$$, there is a state $$\Delta'\in B$$ such that $$\Delta\subseteq \Delta'$$ and they have the same annotation; For each $$\Delta'\in B$$, there is a state $$\Delta\in A$$ such that $$\Delta\subseteq \Delta'$$ and they have the same annotation. Proof. It is obvious from Algorithms 1, 2 and 3. ■ Proposition 8 If $$A$$ is a pre-bubble and $$\mathcal{M},s\vDash^f A$$, there exists a bubble $$B\in$$ExtendToBubbles$${(A)}$$ and a function $$f':[s]\to B$$ such that $$\mathcal{M},s\vDash^{f'}B$$, $$f(s)\subseteq f'(s)$$, and $$f'(s)$$ has the same annotation as $$f(s)$$. Proof. It follows from Propositions 2 and 5. ■ SuccessorPreBubbles The procedure SuccessorPreBubbles, presented in Algorithm 4, takes a bubble $$B$$ and an action $$a\in\mathbf{\Pi_0}$$ as input and returns a set of $$a$$-successor pre-bubbles for $$B$$. If a pre-bubble $$A$$ is returned, it follows by $$\texttt{NM}$$ that all $$\neg[a]\phi$$ from $$B$$ is realized in $$A$$. Similar to the case in EpistemicallySufficient, formulas $$\neg[a]\phi_1\cdots\neg[a]\phi_n\in\Delta\in B$$ might be realize by the same state, so there might be state $$\Sigma\in\mathtt{S}({[a]^-(\Delta)\cup\{\neg\phi_1\cdots\neg\phi_n \}})$$ in some returned successor pre-bubble. Furthermore, the property $$\texttt{PR}$$ requires that each bubble has enough states. Step 1 of the construction of the pre-tableau guarantees that at least one bubble in the first layer has enough states. To preserve $$\texttt{PR}$$, our strategy is that, if the bubble $$B$$ has enough states and $$\{\Omega_1,\cdots,\Omega_k\}$$ is the state set realizing all $$\neg[a]\phi$$ from $$B$$, let the pre-bubble $$\{\Omega_1,\cdots,\Omega_k\}\cup P$$ be in the returned set for each $$P\subseteq H$$ where $$H=\bigcup_{\Delta\in B}\mathtt{S}({[a]^-(\Delta)})$$. This guarantees that one returned pre-bubble has enough states. For example, given a bubble $$B=\{\Delta,\Gamma \}$$ in which $$\Delta=\{\left\langle {{a}} \right\rangle p,\hat{K}q,\left\langle {{a}} \right\rangle \hat{K}q\}$$ and $$\Gamma=\{[a]r,\hat{K}q,q\}$$, the pre-bubbles returned by the procedure SuccessorPreBubbles$$(B,a)$$ are shown in Figure 9. Figure 9. View largeDownload slide SuccessorPreBubbles (B,a)   $$\matrix{ {{\Delta _1} = \{ p\} \qquad {\Delta _ \bullet } = \{ \top \} } \hfill \cr {{\Delta _2} = \{ \hat Kq\} \qquad {\Gamma _ \bullet } = \{ r\} } \hfill \cr {{\Delta _3} = \{ p,\hat Kq\} .} \hfill \cr }$$ Figure 9. View largeDownload slide SuccessorPreBubbles (B,a)   $$\matrix{ {{\Delta _1} = \{ p\} \qquad {\Delta _ \bullet } = \{ \top \} } \hfill \cr {{\Delta _2} = \{ \hat Kq\} \qquad {\Gamma _ \bullet } = \{ r\} } \hfill \cr {{\Delta _3} = \{ p,\hat Kq\} .} \hfill \cr }$$ Proposition 9 Given a bubble $$B$$ and $$A\in$$SuccessorPreBubbles$${(B,a)}$$, we have the following results: For each $$\Delta\in B$$ and each $$\neg[a]\phi\in\Delta$$, there is a state $$\Delta'\in A$$ such that $$\neg\phi\in\Delta'$$ and the annotation of $$\Delta'$$ is $$\Delta$$. If $$\Delta'\in A$$ and the annotation of $$\Delta'$$ is a state $$\Delta$$, then we have $$\Delta\in B$$ and $$[a]^-(\Delta)\subseteq \Delta'$$. Proof. It is obvious from Algorithm 4. ■ Proposition 10 Let $$B$$ be a bubble. If $$\mathcal{M},s\vDash^f B$$, $$\neg[a]\phi\in f(s)$$, $$t\in R^\mathcal{M}_a(s)$$ and $$\mathcal{M},t\vDash\phi$$, then there are a pre-bubble $$A\in$$SuccessorPreBubbles$${(B,a)}$$ and a function $$f':[t]\to A$$ such that $$f'(t)$$ is annotated with $$f(s)$$, $$\neg\phi\in f'(t)$$, and $$\mathcal{M},t\vDash^{f'} A$$. Proof. Because of our assumption at the end of Section 2, each $$v\in [t]$$ has a unique $$a$$-predecessor, written as $$v^-$$. It is obvious that $$t^-=s$$. Due to $$\mathcal{M},s\vDash^f B$$ and the property of $$\texttt{NM}$$, it follows that for each $$\neg[a]\psi\in\Delta\in B$$ there exists $$v\in[t]$$ such that $$\mathcal{M},v\vDash[a]^-(\Delta)\land \neg\psi$$. Let $$\sigma(v)=[a]^-(f(v^-))\cup\{\neg\psi\mid \neg[a]\psi\in f(v^-)$$ and $$\mathcal{M},v\vDash \neg\psi \}$$. It is obvious that $$\neg\phi\in \sigma(t_0)$$. If $$\sigma(v)\neq\emptyset$$, let $$\sigma(v)=[a]^-(f(v^-))\cup\{\neg\psi_1,\cdots,\neg\psi_k\}$$. This implies that there are $$\Delta'\in\mathtt{S}({[a]^-(f(v^-))})$$ and $$\Sigma_j\in\mathtt{S}({\neg\phi_j})$$, where $$1\leq j\leq k$$, such that $$\mathcal{M},v\vDash\Delta'\cup\Sigma_1\dots\Sigma_k$$. We define $$\Omega_{v}=\Delta'\cup\Sigma_1\dots\Sigma_k$$. If $$\sigma(v)=\emptyset$$, we define $$\Omega_{v}=\{\top \}$$. Moreover, let $$\Omega_{v}$$ be annotated with $$f(v^-)$$. Please note that $$\neg\phi\in\Omega_{t}$$ and $$\Omega_{t}$$ is annotated with $$f(s)$$. It follows by Algorithm 4 that $$\Omega_{v}\in Alt$$. Let $$A=\{\Omega_{v}\mid v\in [t] \}$$. Then we have $$A\in$$SuccessorPreBubbles$${(B,a)}$$. We define $$f':[t]\to A$$ as $$f'(v)=\Omega_{v}$$. It follows that $$\mathcal{M},v\vDash f'(v)$$ for each $$v\in [t]$$ and $$f'$$ is surjective. Thus, $$\mathcal{M},t\vDash^{f'}A$$. Moreover, we also have it that $$f'(t)$$ is annotated with $$f(s)$$ and that $$\neg\phi\in f'(t)$$. ■ 3.2 Construction of the initial and final tableau From the construction of the pre-tableau, we know that if the pre-bubble $$A$$ is an $$a$$-successor of a bubble $$B$$ for some $$a\in{\mathbf{\Pi_0}}$$ and the bubble $$B'$$ is a bubble-extension of $$A$$, namely $$B{\stackrel{a}{\rightarrow}}A{\dashrightarrow} B'$$, then it follows by Propositions 7 and 9 that the annotation of each state $$\Delta'\in B'$$ is some state $$\Delta\in B$$. The initial tableau is produced by removing the pre-bubbles and redirecting the arrows in the pre-tableau. For example, if $$B{\stackrel{a}{\rightarrow}}A{\dashrightarrow} B'$$, we then delete the pre-bubble $$A$$ and add an $$a$$-arrow from $$B$$ to $$B'$$, namely $$B{\stackrel{a}{\rightarrow}}B'$$. Since each bubble is a finite set of states and each state is a subset of $$\mathcal{FL}({\phi})$$, the initial tableau is a finite graph. Next, we construct the final tableau based on the initial tableau. Before this, we firstly introduce the notion of a fulfilling chain. Definition 9 (Fulfilling chain) Let $$B$$ be a bubble of a tableau $${\mathcal{T}}$$ and $$\neg[\pi]\phi\in\Delta\in B$$. A fulfilling chain for $$(\neg[\pi]\phi,\Delta,B)$$ is defined as follows. $$(\Delta_0,B_0){\stackrel{a_1}{\rightarrow}}(\Delta_1,B_1)$$ is a fulfilling chain for $$(\neg[a_1]\phi,\Delta_0,B_0)$$ if $$B_0{\stackrel{a_1}{\rightarrow}}B_1$$, $${\neg\phi\in\Delta_1}$$ and the annotation of $$\Delta_1\in B_1$$ is $$\Delta_0\in B_0$$; $$(\Delta_0,B_0)$$ is a fulfilling chain for $$(\neg[?\psi]\phi,\Delta_0,B_0)$$ if $$\neg[?\psi]\phi,{\neg\phi}\in\Delta_0$$; $$(\Delta_0,B_0){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_n,B_n)$$ is a fulfilling chain for $$(\neg[\pi_1;\pi_2]\phi,\Delta_0,B_0)$$ if there exists $$0\leq i\leq n$$ such that $$(\Delta_0,B_0){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_i,B_i)$$ is a fulfilling chain for $$(\neg[\pi_1][\pi_2]\phi,\Delta_0,B_0)$$ and $$(\Delta_i,B_i){\stackrel{a_{i+1}}{\rightarrow}}\cdots(\Delta_n,B_n)$$ is a fulfilling chain for $$(\neg[\pi_2]\phi,\Delta_0,B_0)$$; $$(\Delta_0,B_0){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_n,B_n)$$ is a fulfilling chain for $$(\neg[\pi_1+\pi_2]\phi,\Delta_0,B_0)$$ if it is a fulfilling chain for $$(\neg[\pi_1]\phi,\Delta_0,B_0)$$ or $$(\neg[\pi_2]\phi,\Delta_0,B_0)$$; $$(\Delta_0,B_0){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_n,B_n)$$ is a fulfilling chain for $$(\neg[\pi^*]\phi,\Delta_0,B_0)$$ if $$\neg[\pi^*]\phi\in\Delta_0,{\neg\phi}\in \Delta_n$$, and if $$n>0$$ then there are $$0= i_0\leq \cdots\leq i_{k}= n$$ such that $$(\Delta_{i_j},B_{i_j}){\stackrel{a_{i_j}}{\rightarrow}}\cdots(\Delta_{i_{j+1}},B_{i_{j+1}})$$ is a fulfilling chain for $$(\neg[{\pi}][{\pi^*}]\phi,\Delta_{i_j},B_{i_j})$$ for all $$0\leq j\leq k-1$$. The final tableau $${\mathcal{T}}$$ is obtained by the following procedure. (1) Let $${\mathcal{T}}_0$$ be the initial tableau. (2) Let $$B$$ be a bubble of $${\mathcal{T}}_n$$. If there is $$\neg[\pi]\phi\in\Delta\in B$$ such that there is no fulfilling chain for $$(\neg[\pi]\phi,\Delta,B)$$, then let $${\mathcal{T}}_{n+1}={\mathcal{T}}_n\setminus\{B\}$$. (3) Repeat the step 2 until no bubble is to be deleted, i.e. $${\mathcal{T}}_{n+1}={\mathcal{T}}_n$$. Definition 10 The final tableau $${\mathcal{T}}$$ for $$\phi$$ is open if there is a bubble $$B$$ in $${\mathcal{T}}$$ and $$\Delta\in B$$ such that $$\phi\in\Delta$$. 4 Soundness and completeness In this section, we will show that the tableau constructed in the previous section is proper, i.e. it is sound and complete. 4.1 Soundness To show the soundness of the procedure, we need to show that the final tableau for $$\phi$$ is open if $$\phi$$ is satisfiable. We will show that, in the initial tableau, there exists a bubble that contains a state including $$\phi$$ and this bubble is to survive in the final tableau. Proposition 11 If $$\phi$$ is satisfiable, there exists a bubble $$B$$ in the initial tableau and a state $$\Delta\in B$$ such that $$\phi\in\Delta$$ and $$B$$ is satisfiable. Proof. Assume that $$\mathcal{M},s\vDash\phi$$. It follows that $$\mathcal{M},s\vDash\Delta_s$$ for some $$\Delta_s\in{\mathtt{S}}({\phi})$$. For each $$u\in[s]$$ and $$u\neq s$$, we have $$\mathcal{M},u\vDash Epi(\Delta_s)$$. This implies that $$\mathcal{M},u\vDash \Delta_u$$ for some $$\Delta_u\in{\mathtt{S}}({Epi(\Delta_s)})$$. Let $$A=\{\Delta_u\mid u\in[s],u\neq s \}\cup\{\Delta_s \}$$. Then it follows by the step 1 of the construction of the pre-tableau that $$A$$ is a pre-bubble in the pre-tableau. We define $$f:[s]\to A$$ as $$f(s)=\Delta_s$$ and $$f(u)=\Delta_u$$ for each $$u\in[s]$$ and $$u\neq s$$. Therefore we have $$\mathcal{M},s\vDash^f A$$ and $$\phi\in f(s)$$. It follows by Proposition 8 that $$\mathcal{M},s\vDash^{f'}B$$ and $$f(s)\subseteq f'(s)$$ for some function $$f'$$ and some bubble $$B\in$$ExtendToBubbles$${(A)}$$. ■ Definition 11 (Witness chain) Let $$\mathcal{M},s$$ be a pointed model. A witness chain for $$(\mathcal{M},s,\neg[{\pi}]\phi)$$ is defined as follows. $$(s_0,\{\neg[a_1]\phi \}){\stackrel{a_1}{\rightarrow}}(s_1,\{\neg\phi\})$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{a_1}]\phi)$$ if $$s_0{\stackrel{a_1}{\rightarrow}}s_1$$, $$\mathcal{M},s_0\vDash\neg[a_1]\phi$$ and $$\mathcal{M},s_1\vDash\neg\phi$$; $$(s,\{\neg[?\psi]\phi,\neg\phi\})$$ is a witness chain for $$(\mathcal{M},s,\neg[{?\psi}]\phi)$$ if $$\mathcal{M},s\vDash\neg[?\psi]\phi$$; $$(s_0,\Gamma\cup\{\neg[\pi_1;\pi_2]\phi \}){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{\pi_1;\pi_2}]\phi)$$ if there exists $$0\leq i\leq n$$ such that $$(s_0,\Gamma){\stackrel{a_1}{\rightarrow}}\cdots(s_i,\Gamma_i)$$ is a witness chain for $$(\mathcal{M},s_0,\neg[\pi_1][\pi_2]\phi)$$ and $$(s_i,\Gamma_i){\stackrel{a_i}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_i,\neg[\pi_2]\phi)$$. $$(s_0,\Gamma\cup\{\neg[\pi_1+\pi_2]\phi \}){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{\pi_1+\pi_2}]\phi)$$ if $$(s_0,\Gamma){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{\pi_1}]\phi)$$ or $$(\mathcal{M},s_0,\neg[{\pi_2}]\phi)$$. $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n\cup\{\neg\phi\})$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{\pi^*}]\phi)$$ if $$\neg[\pi^*]\phi\in\Gamma_0$$, $$\mathcal{M},s_n\vDash\neg\phi$$, and if $$n>0$$ then there are $$0= i_0\leq \cdots\leq i_{k}= n$$ such that $$(s_{i_j},\Gamma_{i_j}\setminus\{\neg[\pi^*]\phi\}),\cdots,(s_{i_{j+1}},\Gamma_{i_{j+1}})$$ is a witness chain for $$(\mathcal{M},s_{i_j},\neg[{\pi}][{\pi^*}]\phi)$$ for all $$0\leq j\leq k-1$$. By the definition above, it is easy to show the following propositions. Proposition 12 If $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s,\neg[{\pi}]\phi)$$ then $$s_0=s$$, $$\neg[{\pi}]\phi\in\Gamma_0$$, $$\neg\phi\in\Gamma_n$$ and $$\mathcal{M},s_i\vDash\Gamma_i$$ for all $$0\leq i\leq n$$. Proposition 13 If $$\mathcal{M},s\vDash^f B$$, $$\neg[\pi]\phi\in f(s)$$ then there is a witness chain $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ for $$(\mathcal{M},s,\neg[\pi]\phi)$$ such that $$\Gamma_0\subseteq f(s)$$. Proposition 14 Let $$B$$ be a bubble in the initial tableau $${\mathcal{T}}_0$$. If $$\mathcal{M},s\vDash^f B$$, $$\neg[a]\phi\in f(s)$$, and $$\mathcal{M},t\vDash\neg\phi$$ for some $$t\in R^\mathcal{M}_a(s)$$, then there are a bubble $${B'}$$ in $${\mathcal{T}}_0$$ and a function $$f':[t]\to B'$$ such that $$\mathcal{M},t\vDash^{f'} B'$$ and $$(f(s),B){\stackrel{a}{\rightarrow}}(f'(t),B')$$ is a fulfilling chain for $$(\neg[a]\phi,f(s),B)$$. Proof. First, it follows by Proposition 10 that there exists a pre-bubble $$A\in$$SuccessorPreBubbles$${(B,a)}$$ and a function $$f'':[t]\to A$$ such that $$\mathcal{M},t\vDash^{f''} A$$, $$\neg\phi\in f''(t)$$ and $$f''(t)$$ is annotated with $$f(s)$$. It follows by Proposition 8 that there are a bubble $$B'\in$$ExtendToBubbles$${(A)}$$ and a function $$f':[t]\to B'$$ such that $$\mathcal{M},t\vDash^{f'} B'$$, $$f''(t)\subseteq f'(t)$$, and the annotation of $$f'(t)$$ (which is the same as $$f''(t)$$) is $$f(s)$$. Since $$B{\stackrel{a}{\rightarrow}}A{\dashrightarrow} B'$$, it follows that $$B{\stackrel{a}{\rightarrow}}B'$$ in the tableau $${\mathcal{T}}_0$$. Since the annotation of $$f'(t)\in B'$$ is $$f(s)\in B$$ and $$\neg\phi\in f''(t)\subseteq f'(t)$$, we have $$\neg\phi\in f'(t)$$. Therefore, $$(f(s),B){\stackrel{a}{\rightarrow}}(f'(t),B')$$ is a fulfilling chain for $$(\neg[a]\phi,f(s),B)$$. ■ Proposition 15 Let $$B$$ be a bubble of the initial tableau $${\mathcal{T}}_0$$. If $$\mathcal{M},s\vDash^f B$$, $$\neg[\pi]\phi\in f(s)$$, and $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s,\neg[\pi]\phi)$$ such that $$\Gamma_0\subseteq f(s)$$, then there are a sequence of bubbles $${B_1}\cdots B_n$$ of $${\mathcal{T}}_0$$ and a sequence of functions $$f_1\cdots f_n$$ such that $$f_i:[s_i]\to B_i$$, $$\mathcal{M},s_i\vDash^{f_i} B_i$$, $$\Gamma_i\subseteq f_i(s_i)$$ for all $$1\leq i\leq n$$, and $$(f(s),{B}){\stackrel{a_1}{\rightarrow}}\cdots (f_n(s_n),B_n)$$ is a fulfilling chain for $$(\neg[\pi]\phi,f(s),B)$$. Proof. We prove it by induction on $$\pi$$. In the case of $$\neg[a]\phi$$, it is obvious by Proposition 14. Case of $$\neg[?\psi]\phi$$: It follows by $$\neg[?\psi]\phi\in f(s)$$ that both $$\psi$$ and $$\neg \phi$$ are in $$f(s)$$. Therefore, $$(f(s),B)$$ is a fulfilling chain for $$(\neg[?\psi]\phi,f(s),B)$$. Case of $$\neg [\pi_1;\pi_2]\phi$$: Since we have that $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s,\neg[\pi_1;\pi_2]\phi)$$, it follows that there exists $$i$$ such that $$(s_0,\Gamma_0\setminus\{\neg[\pi_1;\pi_2]\phi\}){\stackrel{a_1}{\rightarrow}}\cdots(s_i,\Gamma_i)$$ is a witness chain for $$(\mathcal{M},s,\neg[\pi_1][\pi_2]\phi)$$ and $$(s_i,\Gamma_i){\stackrel{a_i}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_i,\neg[\pi_2]\phi)$$. It follows by $$\neg [\pi_1;\pi_2]\phi\in f(s)$$ that $$\neg{[\pi_1][\pi_2]}\phi\in f(s)$$. By induction on $$\pi_1$$, there are a bubble sequence $${B_1}\cdots B_i$$ and a function sequence $$f_1\cdots f_i$$ such that $$f_j:[s_j]\to B_j$$, $$\mathcal{M},s_j\vDash^{f_j} B_j$$, $$\Gamma_j\subseteq f_j(s_j)$$ for all $$1\leq j\leq i$$, and $$(f(s),{B}){\stackrel{a_1}{\rightarrow}}\cdots(f_i(s_i),B_i)$$ is a fulfilling chain for $$(\neg[\pi_1][\pi_2]\phi,f(s),B)$$. It follows by Proposition 12 that $$\neg[\pi_2]\phi\in f_i(s_i)$$. Then, by induction on $$\pi_2$$, there are a bubble sequence $${B_{i+1}}\cdots B_n$$ and a function sequence $$f_{i+1}\cdots f_n$$ such that $$f_j:[s_j]\to B_j$$, $$\mathcal{M},s_j\vDash^{f_j} B_j$$, $$\Gamma_j\subseteq f_j(s_j)$$ for all $$i+1\leq j\leq i$$, and $$(f(s_i),{B_i}){\stackrel{a_i}{\rightarrow}}\cdots(f_n(s_n),B_n)$$ is a fulfilling chain for $$(\neg[\pi_2]\phi,f(s),B)$$. Therefore, $$(f(s),{B}){\stackrel{a_i}{\rightarrow}}\cdots(f_n(s_n),B_n)$$ is a fulfilling chain for $$(\neg[\pi_1;\pi_2]\phi,f(s),B)$$. Case of $$\neg [\pi_1+\pi_2]\phi$$: Since $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s,\neg[\pi_1+\pi_2]\phi)$$, it follows that $$(s_0,\Gamma_0\setminus\{\neg[\pi_1+\pi_2]\phi\}){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s,\neg[\pi_1]\phi)$$ or $$(\mathcal{M},s,\neg[\pi_1]\phi)$$. Assuming it is a witness chain for $$(\mathcal{M},s,\neg[\pi_1]\phi)$$, it follows by Proposition 12 that $$\neg[\pi_1]\phi\in \Gamma_0$$. Because of $$\Gamma_0\subseteq f(s)$$, it follows that $$\neg[\pi_1]\phi\in f(s)$$. By induction on $$\pi_1$$, there are a bubble sequence $${B_{1}}\cdots B_n$$ and a function sequence $$f_{1}\cdots f_n$$ such that $$f_j:[s_j]\to B_j$$, $$\mathcal{M},s_j\vDash^{f_j} B_j$$ and $$\Gamma_j\subseteq f_j(s_j)$$ for all $$1\leq j\leq n$$, and $$(f(s),{B}){\stackrel{a_1}{\rightarrow}}\cdots(f_n(s_n),B_n)$$ is a fulfilling chain for $$(\neg[\pi_1]\phi,f(s),B)$$. Therefore, it is a fulfilling chain for $$(\neg[\pi_1+\pi_2]\phi,f(s),B)$$. Case of $$\neg [\pi^*]\phi$$: If $$n=0$$, then $$(f(s),B)$$ is a fulfilling chain for $$(\neg[\pi^*]\phi,f(s),B)$$. If $$n>0$$, it follows that there are $$0= i_0\leq \cdots\leq i_{k}= n$$ such that $$(s_{i_j},\Gamma_{i_j}\setminus\{\neg[\pi^*]\phi\}){\stackrel{a_{i_j}}{\rightarrow}}\cdots(s_{i_{j+1}},\Gamma_{i_{j+1}})$$ is a witness chain for $$(\mathcal{M},s_{i_j},\neg[{\pi}][{\pi^*}]\phi)$$ for all $$0\leq j\leq k-1$$. Therefore, $$(s_{i_0},\Gamma_{i_0}\setminus\{\neg[\pi^*]\phi\}),\cdots,(s_{i_{1}},\Gamma_{i_{1}})$$ is a witness chain for $$(\mathcal{M},s_{i_0},\neg[{\pi}][{\pi^*}]\phi)$$ (please note that $$i_0=0$$). It follows by Proposition 12 that $$\neg[\pi][\pi^*]\phi\in \Gamma_{i_j}$$ for all $$0\leq j\leq k-1$$. Because of $$\Gamma_{i_0}\subseteq f(s)$$, it follows that $$\neg[\pi][\pi^*]\phi\in f(s)$$. By induction on $$\pi$$, there are a bubble sequence $$B_1\cdots{B_{i_1}}$$ and a function sequence $$f_1\cdots f_{i_1}$$ such that $$f_{j}:[s_{j}]\to B_{j}$$, $$\mathcal{M},s_{j}\vDash^{f_{j}} B_{j}$$, $$\Gamma_{j}\subseteq f_{j}(s_{j})$$ for all $$1\leq j\leq i_1$$, and $$(f(s),{B}){\stackrel{a_{1}}{\rightarrow}}\cdots(f_{i_1}(s_{i_1})_{B_{i_1}})$$. Similarly, we can do this for all $$0\leq j\leq k-1$$. Therefore, $$\neg[\pi^*]\phi\in\Gamma_{i_{j+1}}\subseteq f_{i_{j+1}}(s_{i_{j+1}})$$, and $$(f_{i_j}(s_{i_j}),{B_{i_j}}){\stackrel{a_{i_j+1}}{\rightarrow}}\cdots (f_{i_{j+1}}(s_{i_{j+1}}),{B_{i_{j+1}}})$$ is a fulfilling chain for $$(\neg[\pi][\pi^*]\phi,f_{i_j}(s_{i_j}),{B_{i_j}})$$ for all $$0\leq j\leq k-1$$. Moreover, since $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{\pi^*}]\phi)$$, it follows that $$\mathcal{M},s_n\vDash\neg\phi$$. Because of $$\mathcal{M},s_n\vDash^{f_n}B_n$$ and $$\neg[\pi^*]\phi\in f_n(s_n)$$, it follows that $$\neg\phi\in f_n(s_n)$$. Otherwise, by Table 2, we will have $$\phi\in f_n(s_n)$$. This is contradictory with $$\mathcal{M},s_n\vDash\neg\phi$$ and $$\mathcal{M},s\vDash f_n(s_n)$$. Therefore, $$(f(s),B){\stackrel{a_1}{\rightarrow}}\cdots(f_n(s_n),B_n)$$ is a fulfilling chain for $$(\neg[\pi^*]\phi,f(s),B)$$. ■ Lemma 1 For each bubble $$B$$ in the initial tableau, if $$B$$ is satisfiable, then $$B$$ is to survive in the final tableau. Proof. We only need to show that if $$B$$ is satisfiable then $$B$$ is to survive in $${\mathcal{T}}_{i}$$ for each $$i\in\mathbb{N}$$. We prove it by induction on $$i$$. It is obvious if $$i=0$$. Next we only need to show that if $$B$$ survives in $${\mathcal{T}}_i$$ and is satisfiable, then $$B$$ will also survive in $${\mathcal{T}}_{i+1}$$. Therefore, we need to show that for each $$\neg[\pi]\phi\in\Delta\in B$$ there is a fulfilling chain of $${\mathcal{T}}_i$$ for $$(\neg[\pi]\phi,\Delta,B)$$. If $$\neg[\pi]\phi\in\Delta\in B$$ and $$B$$ is satisfiable, it follows by Propositions 13 and 15 that there is a fulfilling chain $$(\Delta,B){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_n,B_n)$$ of $${\mathcal{T}}_0$$ for $$(\neg[\pi]\phi,\Delta,B)$$ such that each $$B_i$$ is satisfiable for all $$1\leq j\leq n$$. By induction on $$i$$, each $$B_j (1\leq j\leq n)$$ is to survive in $${\mathcal{T}}_i$$. Therefore, $$(\Delta,B){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_n,B_n)$$ is also a chain of $${\mathcal{T}}_i$$, and then $$B$$ survives in $${\mathcal{T}}_{i+1}$$. Thus, we have shown that if $$B$$ is satisfiable, then $$B$$ is to survive in $${\mathcal{T}}_{i}$$ for each $$i\in\mathbb{N}$$, namely $$B$$ is to survive in the final tableau. ■ Theorem 1 (Soundness) If $$\phi$$ is satisfiable, the final tableau $${\mathcal{T}}$$ for $$\phi$$ is open. Proof. It follows from Proposition 11 and Lemma 1. ■ 4.2 Completeness To show the completeness of the procedure, we need to show that there is a model for the input formula $$\phi$$ if the final tableau for $$\phi$$ is open. First, we construct a deterministic tree of bubbles based on the open final tableau such that each $$\neg[\pi]\psi$$ from each bubble is fulfilled on this tree. Next, this deterministic bubble tree can be turned into a model with $$\texttt{PR}$$ and $$\texttt{NM}$$ on which $$\phi$$ is satisfied. Please note that in this section we always assume the final tableau is open. Definition 12 (Bubble tree) Let $${\langle {S,\{R_a\mid a\in {\mathbf{\Pi_0}}\}} \rangle}$$ be a tree in which $$S\subseteq{\mathbb{N}}$$. $${L}$$ is a function on $$S$$ which labels each $$n\in S$$ a bubble from the final tableau $${\mathcal{T}}$$. A labelled bubble tree (or just bubble tree) $${\mathfrak{T}}$$ based on $${\mathcal{T}}$$ is a tuple $${\mathfrak{T}}={\langle {S,\{R_a\mid a\in {\mathbf{\Pi_0}}\},{L}} \rangle}$$ such that $$n{\stackrel{a}{\rightarrow}}k$$ implies $$l(n){\stackrel{a}{\rightarrow}}l(k)$$ in $${\mathcal{T}}$$ for each $$a\in{\mathbf{\Pi_0}}$$. We refer to the components of $${\mathfrak{T}}$$ as $$S^{\mathfrak{T}}$$, $$R^{\mathfrak{T}}_a$$ and $${L}^{\mathfrak{T}}$$. A bubble tree $${\mathfrak{T}}$$ is deterministic if $$n{\stackrel{a}{\rightarrow}}k$$ and $$n{\stackrel{a}{\rightarrow}}l$$ imply $$k=l$$. Definition 13 (Expanding a bubble tree) Given a bubble tree $${\mathfrak{T}}$$ based on $${\mathcal{T}}$$, let $$n$$ be a node on the tree and $$\sigma=B_0{\stackrel{a_1}{\rightarrow}}\cdots B_k$$ be a chain from $${\mathcal{T}}$$ such that $${L}(n)=B_0$$. We expand $${\mathfrak{T}}$$ at $$n$$ by $$\sigma$$ as below. If there is $$i\in S^{\mathfrak{T}}$$ such that $$n{\stackrel{a_1}{\rightarrow}}i$$ and $${L}^{\mathfrak{T}}(i)=B_1$$, we continue expanding $${\mathfrak{T}}$$ at $$i$$ by $$B_1{\stackrel{a_2}{\rightarrow}}\cdots B_k$$. If not, we first add a new number $$j$$, add an arrow $$n{\stackrel{a_1}{\rightarrow}}j$$, and label $$j$$ as $${B_1}$$. Next, we continue expanding it at the node $$j$$ by $$B_1{\stackrel{a_2}{\rightarrow}}\cdots B_k$$. Proposition 16 let $$k$$ be a node on a finite deterministic bubble tree $${\mathfrak{T}}$$ and $${L}(k)=B$$. For each $$\neg[\pi]\phi\in\Delta\in B$$, there is a fulfilling chain $$(\Delta,B){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_n,B_n)$$ for $$(\neg[\pi]\phi,\Delta,B)$$ such that expanding $${\mathfrak{T}}$$ at $$k$$ by $$B_0{\stackrel{a_1}{\rightarrow}}\cdots B_n$$ is still a finite deterministic bubble tree. Proof. We prove it by induction on $$\pi$$. We only focus on the cases of $$\neg[a]\phi$$ and $$\neg[\pi^*]\phi$$; the others are obvious. In the case of $$\neg[a]\phi$$, if there is a node $$j$$ on the tree such that $$k{\stackrel{a}{\rightarrow}}j$$, then it follows that $${L}(k){\stackrel{a}{\rightarrow}}{L}(j)$$ in the final tableau. It follows by Propositions 9 and 7 that there is some $$\Delta'\in{L}(j)$$ such that $$\Delta'$$’s notation is $$\Delta\in B={L}(k)$$. Therefore, $$(\Delta,{L}(k)){\stackrel{a}{\rightarrow}}(\Delta',{L}(j))$$ is a fulfilling chain for $$(\neg[a]\phi,\Delta,{L}(k))$$. If there is no such $$j$$ on the tree, it follows that there is a fulfilling chain $$(\Delta,{L}(k)){\stackrel{a}{\rightarrow}}(\Delta',B')$$ in the final tableu for $$(\neg[a]\phi,\Delta,{L}(k))$$. Then we expand $${\mathfrak{T}}$$ at $$k$$ by $${L}(k){\stackrel{a}{\rightarrow}}B'$$. It is obvious that it is still a finite deterministic tree after the expansion. In the case of $$\neg[\pi^*]\phi\in\Delta\in B$$, there are two situations: $$\neg\phi\in\Delta$$ or $$\neg[\pi][\pi^*]\phi\in\Delta$$. If $$\neg\phi\in\Delta$$, it is obvious. If $$\neg[\pi][\pi^*]\phi\in\Delta$$, it follows by induction on $$\pi$$ that there is a fulfilling chain $$(\Delta,B){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_n,B_n)$$ for $$(\neg[\pi][\pi^*]\phi,\Delta,B)$$ such that expanding $${\mathfrak{T}}$$ at $$k$$ by $$B_0{\stackrel{a_1}{\rightarrow}}\cdots B_n$$ is still a finite deterministic bubble tree. If $$\neg\phi$$ is not in $$\Delta_n\in B_n$$, it follows by $$\neg[\pi^*]\phi\in\Delta_n$$ that $$\neg[\pi][\pi^*]\phi\in\Delta_n$$. Therefore, we can repeat this procedure until $$\neg\phi\in\Delta_n\in B_n$$ or $$B_n$$ is labelled on a leaf node. If $$\neg[\pi^*]\phi$$ is still not fulfilled and $$B_n$$ is labelled on a leaf node $$l$$, since the final tableau is open, it follows that there is a fulfilling chain $$(\Delta_n,{L}(l)){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_h,B_h)$$ for $$(\neg[\pi^*]\phi,\Delta_n,{L}(l))$$, which is a chain of the final tableau. We can than continue expanding the expanded tree with $${L}(l){\stackrel{a_1}{\rightarrow}}\cdots B_h$$. It is still a finite deterministic tree after all these expansion, and $$(\Delta,{L}(k)){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_h,B_h)$$ is a fulfilling chain for $$(\neg[\pi^*],\Delta,{L}(k))$$. ■ Let the final tableau $${\mathcal{T}}$$ for $$\phi$$ be open, and $$\phi\in\Delta$$ for some state $$\Delta\in B$$ and some bubble $$B$$ of $${\mathcal{T}}$$. Next we will construct a deterministic bubble tree $${\mathfrak{T}}^\phi$$ based on the the final tableau $${\mathcal{T}}$$ for $$\phi$$ and then construct a model $$\mathcal{M}^\phi$$ based on $${\mathfrak{T}}^\phi$$. $${\mathfrak{T}}^\phi$$ is constructed as follows. (1) Let $$0$$ be the root and $${L}(0)=B$$, and mark the root as unfulfilled. (2) Let $$n$$ be the first unfulfilled node with the minimal level of the tree. We expand the tree as follows: – For each $$\neg[\pi]\phi\in \Delta\in {L}(n)$$, let $$\sigma$$ be a fulfilling chain for $$(\neg[\pi]\phi,\Delta,{L}(n))$$ such that expanding the tree at $$n$$ by $$\sigma$$ will keep the tree deterministic and finite. Proposition 16 makes sure that there exists such a fulfilling chain. We expand the tree at $$n$$ by $$\sigma$$ and mark the new nodes as unfulfilled. – After fulfilling all the $$\neg[\pi]\phi\in\Delta$$ for all $$\Delta\in {L}(n)$$, mark $$n$$ as fulfilled. (3) Repeat the step 2 until there are no unfulfilled nodes. Since each step of the construction keeps the tree deterministic, we have the following result. Proposition 17 $${\mathfrak{T}}^\phi$$ is a deterministic bubble tree. Proposition 18 For each node $$n$$ of $${\mathfrak{T}}^\phi$$ and each $$\neg[\pi]\psi\in\Delta\in {L}(n)$$, there exists a fulfilling chain $$(\Delta_0,{L}(n_0)){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_n,{L}(n_k))$$ for $$(\neg[\pi]\psi,\Delta,{L}(n))$$ such that $$n_0{\stackrel{a_1}{\rightarrow}}\cdots n_k$$. Proof. Let the level of node $$n$$ be $$m$$. Since there are only finite states in each bubble and finite formulas in each state, each node of $${\mathfrak{T}}^\phi$$ has a finite number of children. This means that $${\mathfrak{T}}^\phi$$ is finitely branching. Therefore, we will deal with $$\neg[\pi]\phi\in\Delta\in {L}(n)$$ in some step of the construction. It follows by the construction and Definition 13 that there exists a fulfilling chain $$(\Delta_0,{L}(n_0)){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_n,{L}(n_k))$$ for $$(\neg[\pi]\psi,\Delta,{L}(n))$$ such that $$n_0{\stackrel{a_1}{\rightarrow}}\cdots n_k$$. ■ Next, we define the model $$\mathcal{M}^\phi$$ based on $${\mathfrak{T}}^\phi$$. Definition 14 The model $$\mathcal{M}^\phi$$ is defined as follows. $$S=\{(\Delta,n)\mid n$$ is a node of $${\mathfrak{T}}^\phi$$ and $$\Delta\in {L}(n) \}$$ $${\stackrel{a}{\rightarrow}}=\{((\Delta,n),(\Delta',n'))\mid n{\stackrel{a}{\rightarrow}}n'$$ and $$\Delta'$$’s annotation is $$\Delta. \}$$ $$\sim=\{((\Delta,n),(\Delta',n'))\mid n=n' \}$$ $$V(p)=\{(\Delta,n)\mid p\in\Delta \}$$ for each $$p\in\mathcal{FL}({\phi})$$ Proposition 19 The model $$\mathcal{M}^\phi$$ has the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. Proof. First we will show $$\mathcal{M}^\phi$$ has the property of $$\texttt{PR}$$. If $$(\Delta,n){\stackrel{a}{\rightarrow}}(\Delta',n')$$ and $$(\Delta',n')\sim(\Theta',k')$$, it follows that $$k'=n'$$, $$n{\stackrel{a}{\rightarrow}}n'$$ and $$\Theta'\in {L}(n')$$. It follows by Propositions 9 and 7 that $$B{\stackrel{a}{\rightarrow}}B'$$ implies the annotation of each state in $$B'$$ is a state in $$B$$. Therefore, the annotation of $$\Theta'$$ is a state $$\Theta\in {L}(n)$$. It follows that $$(\Delta,n)\sim(\Theta,n)$$ and $$(\Theta,n){\stackrel{a}{\rightarrow}}(\Delta',k')$$. Secondly if $$(\Delta,n){\stackrel{a}{\rightarrow}}(\Delta',n')$$, $$(\Delta,n)\sim(\Theta,n)$$, and $$(\Theta,n){\stackrel{a}{\rightarrow}}(\Theta',k)$$, we need to show that $$(\Delta',n')\sim(\Theta',k)$$. We then have $$n{\stackrel{a}{\rightarrow}}n'$$ and $$n{\stackrel{a}{\rightarrow}}k$$. It follows by Proposition 17 that $$n'=k$$. Thus we have $$(\Delta',n')\sim(\Theta',k)$$. This means the model $$\mathcal{M}^\phi$$ has the property of $$\texttt{NM}$$. ■ Lemma 2 Given a formula $$\phi$$ and the model $$\mathcal{M}^\phi$$, we have the following results. (1) For each $$[\pi]\psi\in\mathcal{FL}({\phi})$$, if $$[\pi]\phi\in\Delta$$ and $$(\Delta,n){\stackrel{\pi}{\rightarrow}}(\Delta',n')$$, then $$\psi\in\Delta'$$; (2) For each $$\neg[\pi]\psi\in\mathcal{FL}({\phi})$$, if $$(\Delta_0,{L}(n_0)){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_k,{L}(n_k))$$ is a fulfilling chain for $$(\neg[\pi]\psi,\Delta_0,{L}(n_0))$$ such that $$n_0{\stackrel{a_1}{\rightarrow}}\cdots n_k$$, then $$(\Delta_0,n_0){\stackrel{\pi}{\rightarrow}}(\Delta_k,n_k)$$; (3) For each $$\psi\in\mathcal{FL}({\phi})$$, if $$\psi\in \Delta\in{L}(n)$$, then $$\mathcal{M}^\phi, (\Delta,n)\vDash\psi$$. Proof. We prove it by simultaneous induction on 1, 2 and 3. For 1, we only focus on the cases of $$[a]\psi$$, $$[?\chi]\psi$$ and $$[\pi^*]\psi$$; the other cases are obvious. In the case of $$[a]\psi\in\Delta$$, it follows by $$(\Delta,n){\stackrel{a}{\rightarrow}}(\Delta',n')$$ that $${L}(n){\stackrel{a}{\rightarrow}}{L}(n')$$ and that $$\Delta'$$’s annotation is $$\Delta$$. Since each state only has one annotation in the tableau, it follows by Propositions 9 and 7 that $$[a]^-(\Delta)\subseteq \Delta'$$. Thus, we have $$\psi\in\Delta'$$. In the case of $$[?\chi]\psi\in\Delta$$, it follows that either $$\neg\chi\in\Delta$$ or $$\psi\in\Delta$$. It follows by $$(\Delta,n){\stackrel{\chi}{\rightarrow}}(\Delta',n')$$ that $$\mathcal{M}^\phi,(\Delta,n)\vDash\chi$$. Assume $$\neg\chi\in\Delta$$. Then it follows by IH of 3 that $$\mathcal{M}^\phi,(\Delta,n)\vDash\neg\chi$$. This results in a contradiction. Therefore, $$\neg\chi\not\in\Delta$$, and consequently $$\psi\in\Delta$$. In the case of $$[\pi^*]\psi\in\Delta$$, it follows by $$(\Delta,n){\stackrel{\pi^*}{\rightarrow}}(\Delta',n')$$ that $$(\Delta,n){\stackrel{\pi^k}{\rightarrow}}(\Delta',n')$$ for some $$k\in\mathbb{N}$$. Since $$[\pi^*]\psi\in\Delta_i$$ implies that $$\psi,[\pi][\pi^*]\psi\in \Delta_i$$ for each state $$\Delta_i$$, where $$0\leq i\leq k$$. By induction on $$\pi$$, it follows that $$\psi\in\Delta'$$. For 2, we only focus on the cases of $$\neg[?\chi]\psi$$ and $$\neg[\pi^*]\psi$$; the other cases are obvious. In the case of $$\neg[?\chi]\psi$$, the fulfilling chain for $$(\neg[?\chi]\psi,\Delta_0,{L}(n_0))$$ is $$(\Delta_0,{L}(n_0))$$. It follows that $$\chi\in\Delta_0\in{L}(n_0)$$. By IH of 3, we have $$\mathcal{M}^\phi,(\Delta_0,n_0)\vDash\chi$$. Thus, we have $$(\Delta_0,n_0){\stackrel{\chi}{\rightarrow}}(\Delta_0,n_0)$$. In the case of $$\neg[\pi^*]\psi$$, it is obvious if $$k=0$$. If $$k>0$$, there are $$0= i_0\leq \cdots\leq i_{h}= k$$ such that $$(\Delta_{i_j},{L}(n_{i_j})){\stackrel{a_{i_j}}{\rightarrow}}\cdots(\Delta_{i_{j+1}},{L}(n_{i_{j+1}}))$$ is a fulfilling chain for $$(\neg[{\pi}][{\pi^*}]\psi,\Delta_{i_j},{L}(n_{i_j}))$$ for all $$0\leq j\leq h-1$$. Moreover, we have $$n_{i_j}{\stackrel{a_{i_j}}{\rightarrow}}\cdots n_{i_{j+1}}$$ for all $$0\leq j\leq h-1$$. By induction on $$\pi$$, it follows that $$(\Delta_{i_j},n_{i_j}){\stackrel{\pi}{\rightarrow}}(\Delta_{i_{j+1}},n_{i_{j+1}})$$ for all $$0\leq j\leq h-1$$. Since $$i_0=0$$ and $$i_h=k$$, we have $$(\Delta_0,n_0){\stackrel{\pi^*}{\rightarrow}}(\Delta_k,n_k)$$. For 3, we only focus on the cases of $$[\pi]\psi$$, $$\neg[\pi]\psi$$, $$K\psi$$ and $$\neg K\psi$$; it is obvious in the cases of $$p$$, $$\neg p$$, $$\psi\land\psi'$$ and $$\neg(\psi\land\psi')$$. In the case of $$[\pi]\psi\in \Delta\in{L}(n)$$, it follows by IH of 1 that $$\psi\in\Delta'\in{L}(n')$$ for each $$(\Delta',n')$$ such that $$(\Delta,n){\stackrel{\pi}{\rightarrow}}(\Delta',n')$$. By induction on $$\psi$$, we have $$\mathcal{M}^\phi,(\Delta',n')\vDash\psi$$. It follows that $$\mathcal{M}^\phi,(\Delta,n)\vDash[\pi]\psi$$. In the case of $$\neg[\pi]\psi\in\Delta\in{L}(n)$$, it follows by Proposition 18 that there exists a fulfilling chain $$(\Delta_0,{L}(n_0)){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_n,{L}(n_k))$$ for $$(\neg[\pi]\psi,\Delta,{L}(n))$$ such that $$n_0{\stackrel{a_1}{\rightarrow}}\cdots n_k$$. By IH of 2, it follows that $$(\Delta,n){\stackrel{\pi}{\rightarrow}}(\Delta_k,n_k)$$. We also have that $$\neg\psi\in \Delta_k\in{L}(n_k)$$ because it is a fulfilling chain. By induction on $$\neg\psi$$, it follows that $$\mathcal{M}^\phi,(\Delta_k,n_k)\vDash\neg\psi$$. Thus, we have $$\mathcal{M}^\phi,(\Delta,n)\vDash\neg[\pi]\psi$$. In the case of $$K\psi\in\Delta\in{L}(n)$$, since $${L}(n)$$ is a bubble, it follows that $$K\psi,\psi\in \Delta'$$ for each $$\Delta'\in {L}(n)$$. It follows that $$\psi\in\Delta'$$ for each $$(\Delta',n)$$ such that $$(\Delta,n)\sim(\Delta',n)$$. By induction on $$\psi$$, it follows that $$\mathcal{M}^\phi,(\Delta',n)\vDash\psi$$, and then we have $$\mathcal{M}^\phi,(\Delta,n)\vDash K\psi$$. In the case of $$\neg K\psi\in\Delta\in{L}(n)$$, since $${L}(n)$$ is a bubble, it follows that $$\neg\psi\in\Delta'$$ for some $$\Delta'\in {L}(n)$$. By induction on $$\psi$$, it follows that $$\mathcal{M}^\phi,(\Delta',n)\vDash\neg\psi$$, and then we have $$\mathcal{M}^\phi,(\Delta,n)\vDash \neg K\psi$$. ■ Theorem 2 (Completeness) If the final tableau $${\mathcal{T}}$$ for $$\phi$$ is open, $$\phi$$ is satisfiable. Proof. If the final tableau $${\mathcal{T}}$$ for $$\phi$$ is open, let $$\phi\in\Delta\in B$$ for some state $$\Delta\in B$$ and some bubble $$B$$. We can construct the deterministic tree $${\mathfrak{T}}^\phi$$ with the root $$B$$ and then construct the model $$\mathcal{M}^\phi$$. It follows by Proposition 19 that $$\mathcal{M}^\phi$$ has the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. Moreover, it follows by Lemma 2 that $$\mathcal{M}^\phi,(\Delta,0)\vDash\phi$$. ■ 5 Conclusion This article presented a tableau-based procedure for EPDL with the properties of perfect recall and no miracles. Perfect recall and no miracles are important properties for capturing the interactions between knowledge and actions. In [1, 2], a tableau-based decision procedure for TEL with no forgetting and no learning was developed. The property of perfect recall in EPDL makes no great difference to the property of no forgetting in TEL. However, the property of no miracles in EPDL is more complex than the property of no learning in TEL. Compared to TEL, EPDL has branching actions, and accessibility relations are usually not serial. This makes the tableau procedure for EPDL with no miracles more complex than the tableau procedure for TEL with no learning. There are a number of issues to explore in future research. The decision procedure presented in this article is not guaranteed to have an optimal complexity. Relevantly, it is shown in [10] that single-agent TEL with perfect recall and no learning is EXPSPACE. What is more, it is shown in [7] that EPDL with perfect recall and no learning is decidable in coN2EXPTIME. We can improve our procedure and compare the complexity to see whether no miracles makes a real difference with no learning on complexity. As we mentioned at the beginning, the axiomatization and decidability of EPDL extended with various combinations of perfect recall, no learning and Church-Rosser properties are well studied in [18]. Therefore, another interesting direction is to investigate these issues of EPDL extended with no miracles and the other properties. Moreover, we can also extend and adapt this tableau construction for EPDL extended with the other combination of perfect recall, no miracles and no learning and Church-Rosser properties. Acknowledgements The author thanks Barteld Kooi and the two anonymous reviewers of this journal for their helpful comments on the earlier versions of the article. The author is grateful to Max Bialek who proofread the article. Footnotes *A preliminary version of this article appeared in the proceedings of LORI-V [14]. 1The sequences $$b$$, $$ba$$ and $$aaa$$ are not solutions since they are not executable at $$s_2$$. References [1] Ajspur M.. Tableau-based Decision Procedures for Epistemic and Temporal Epistemic Logics . PhD Thesis, Doctoral School of Communication, Business and Information Technologies, Roskilde University, Denmark, 10 2013. [2] Ajspur M. and Goranko V.. 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EPTCS, 2016. Google Scholar CrossRef Search ADS   [26] Zakharyaschev M. Wolter F. and Chagrov A.. Advanced modal logic. In Handbook of Philosophical Logic , Vol. 3, Gabbay D. M. and Guenthner F. eds, pp. 83– 266. Springer, 2001. Google Scholar CrossRef Search ADS   © The Author, 2017. 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/about_us/legal/notices) For permissions, please e-mail: journals. permissions@oup.com http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Journal of Logic and Computation Oxford University Press

Tableaux for a combination of propositional dynamic logic and epistemic logic with interactions

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Abstract

Abstract Epistemic propositional dynamic logic (EPDL) is a combination of epistemic logic and propositional dynamic logic. Two properties, perfect recall and no miracles, capture the interactions between actions and knowledge. In this article, we present a tableau-based decision procedure for deciding the satisfiability of single-agent EPDL formula in models with perfect recall and no miracles. 1 Introduction Propositional dynamic logic (PDL) is an important logic for reasoning about programs or actions [6]. Epistemic logic (EL) is a modal logic concerned with reasoning about informational aspects of agent, in particular, agent’s belief and knowledge [11]. Thus the combination of EL and PDL is a powerful tool for reasoning about interactions between knowledge and actions. In this article, we call the combination of EL and PDL ‘epistemic propositional dynamic logic’ (EPDL). There are two ways to combine PDL and EL: product and fusion [7]. In this article, we combine PDL and EL by way of fusion. It is shown that the fusion of two modal logics inherit properties such as completeness, the finite model property and decidability from the individual logics [12, 13, 26]. However, it is much more complex if the fusion is extended with interactions of these two logics. The most frequently discussed interactions in the combinations of PDL and EL are perfect recall, no learning and Church-Rosser axiom. The conjunction of perfect recall and no learning is also called commutativity in [7]. Perfect recall ($$\texttt{PR}$$) is commonly formulated by the axiom schema $$\left\langle {{a}} \right\rangle \hat{K}\phi\to \hat{K}\left\langle {{a}} \right\rangle \phi$$ where $$\left\langle {{a}} \right\rangle$$ is a PDL modality and $$\hat{K}$$ is the epistemic modality. It expresses the persistence of the agent’s knowledge after the execution of an action. It can be described by the first-order sentence $$\forall x\forall y\forall z(x{\stackrel{a}{\rightarrow}}y\land y\sim z\to \exists u(x\sim u\land u{\stackrel{{a}}{\rightarrow}}z))$$ where $${\stackrel{{a}}{\rightarrow}}$$ is the binary relation that interprets the atomic action $$a$$ and $$\sim$$ is the equivalence relation that interprets the EL’s modality $$K$$. Perfect recall can be depicted as follows: No learning ($$\texttt{NL}$$) is given by the axiom schema $$\hat{K}\left\langle {{a}} \right\rangle \phi\to\left\langle {{a}} \right\rangle \hat{K}\phi$$. The formula says the agent knows the result of his action in advance. In other words, there is no learning. $$\texttt{NL}$$ can be described by the first-order sentence $$\forall x\forall y\forall z(x\sim y\land y{\stackrel{{a}}{\rightarrow}} z\to \exists u(x{\stackrel{{a}}{\rightarrow}} u\land u\sim z))$$ and depicted as follows: Church-Rosser axiom ($$\texttt{CR}$$) is the axiom schema $$\left\langle {{a}} \right\rangle K\phi\to K\left\langle {{a}} \right\rangle \phi$$. It says that if an agent is possible to know $$\phi$$ by performing an action $$a$$ then he knows that it is possible to achieve $$\phi$$ by doing $$a$$. $$\texttt{CR}$$ can be described by the first-order sentence $$\forall x\forall y\forall z(x\sim y\land x{\stackrel{{a}}{\rightarrow}} z\to \exists u(y{\stackrel{{a}}{\rightarrow}} u\land u\sim z))$$ and depicted as follows: The fusions of PDL and different versions of EL (such as K45, KD45 and S5) extended with various choices of $$\texttt{PR}$$, $$\texttt{NL}$$ and $$\texttt{CR}$$ are well studied in [7, 17–19]. For example, it has been shown that the fusion of PDL and S5 extended with $$\texttt{PR}$$ and $$\texttt{NL}$$ coincides with the product of PDL and S5, and that the fusion of PDL and S5 extended with $$\texttt{PR}$$ and $$\texttt{NL}$$ has the 2-exponential finite model property. It has also been proved that the fusion of PDL and S5 extended with $$\texttt{NL}$$ is the same as the fusion of PDL and S5 extended with $$\texttt{CR}$$, which is consistent with the fact that $$\texttt{NL}$$ and $$\texttt{CR}$$ are equal to each other if the binary relation $$\sim$$ is an equivalence relation. Besides $$\texttt{PR}$$, $$\texttt{NL}$$ and $$\texttt{CR}$$, no miracles is also an important property which captures the interaction between agent’s actions and knowledge [21–23]. No miracles ($$\texttt{NM}$$) is syntactically given by the axiom schema $$\hat{K}\left\langle {{a}} \right\rangle \phi\to [a]\hat{K}\phi$$. Please note that the structure of the $$\texttt{NM}$$ axiom differs from the above $$\texttt{NL}$$ axiom in the form of the first modality: in $$\texttt{NM}$$ it is a box modality $$[a]$$ while in $$\texttt{NL}$$ it is a diamond modality $$\left\langle {{a}} \right\rangle$$. This is because not all actions are executable at the current world. This subtle difference makes $$\texttt{NM}$$ more suitable to capture the interaction between knowledge and actions and $$\texttt{NL}$$ more suitable to describe the interaction between knowledge and time. A more detailed discussion of the difference between $$\texttt{NM}$$ and $$\texttt{NL}$$ can be found in [23]. Intuitively, no miracles expresses that there is no ‘miracles’ situation such that the agent cannot distinguish two states initially but nevertheless he can distinguish the states resulting from executing the same action on these two states. $$\texttt{NM}$$ can be described by the first-order sentence $$\forall x\forall y\forall z\forall u((x\sim y\land x{\stackrel{{a}}{\rightarrow}} z\land y{\stackrel{{a}}{\rightarrow}}u)\to u\sim z)$$ and depicted as follows: While the combinations of PDL and EL extended with $$\texttt{PR}$$, $$\texttt{NL}$$ or $$\texttt{CR}$$ are well studied, the combination extended with $$\texttt{NM}$$ and these properties has not been investigated sufficiently. In this article, we will focus on the fusion of PDL and EL, which is called EPDL in this article, extended with $$\texttt{PR}$$ and $$\texttt{NM}$$. We observe that in real life there are certain knowledge changes over actions for which it is very natural to think in terms of EPDL models with the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. The following are two examples. First, let us consider a simplified version of the Monty Hall problem. There are three doors: behind one door is a car; behind the others are goats. Initially the agent does not know what is behind the doors. Now an action happens; specifically, one door with a goat is opened. Subsequently, the agent knows that the car is not behind the opened door, but he still does not know behind which door is the car. We use $$A$$ to denote that the car is behind the first door, and $$Ab$$ to denote that the car is behind the first door and the second door is opened. The action $$a$$ means to open the first door. The others are similar. The model depicted in Figure 1 represents the knowledge development in this example, and it has the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. Figure 1. View largeDownload slide Initially, the agent cannot distinguish states A, B and C. Figure 1. View largeDownload slide Initially, the agent cannot distinguish states A, B and C. In the second example, EPDL with $$\texttt{PR}$$ and $$\texttt{NM}$$ is a natural way to model the conformant planning problems (cf. [25]). Conformant planning is the problem of finding a linear plan (a sequence of actions) that is guaranteed to achieve a goal in presence of uncertainty about the initial state (cf. [20]). Considering the example depicted in Figure 2, the initial uncertainty set is $$\{s_1,s_2\}$$ and the goal set is $$\{s_3,s_4\}$$. A knowledge state is a subset of the state space, which records the uncertainty during the execution of a plan, e.g., $$\{s_1, s_2\}$$ is an initial knowledge state. To make sure a goal is achieved eventually, it is crucial to track the transitions of knowledge states during the execution of the plan. Figure 3, which sketches an EPDL model with $$\texttt{PR}$$ and $$\texttt{NM}$$, displays the knowledge development over actions. From Figure 3, we can see that the action sequence $$aa$$ is a solution.1 In [25], it is shown that the existence of a solution for a conformant planning problem can be expressed in the language of EPDL. Therefore, the satisfiability of EPDL with $$\texttt{PR}$$ and $$\texttt{NM}$$ is interesting. Figure 2. View largeDownload slide A conformant planning problem. Figure 2. View largeDownload slide A conformant planning problem. Figure 3. View largeDownload slide Knowledge development in the conformant planning problem. Figure 3. View largeDownload slide Knowledge development in the conformant planning problem. This article presents a tableau-based decision procedure for deciding satisfiability of single-agent EPDL in models with the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. EPDL is a 2-dimensional logic in the sense that it has two kinds of modalities: PDL modalities and epistemic modality. For the construction of the tableau for this 2-dimensional logic, our method is a combination of the tableau-construction method of temporal epistemic logic (TEL) used in [1, 2, 4, 8, 24] and the tableau-construction method of PDL used in [9, 16]. TEL, which is a logic for reasoning about agents’ knowledge evolution over time [5, 15], is also a 2-dimensional logic: it has both temporal modalities and epistemic modality. The construction of the tableau for TEL is based on bubbles, and the tableau of TEL is a bubble graph. On the other hand, the structure of PDL modalities is much more complex compared to the structure of temporal modalities. We deal with PDL modalities with the method used for building the tableau for PDL. The article is organized as follows. Section 2 introduces the language, model and semantics of EPDL. Section 3 presents the tableau procedure for EPDL with $$\texttt{PR}$$ and $$\texttt{NM}$$. Section 4 proves the soundness and completeness of the tableau procedure, and we conclude in Section 5 and point to future works. 2 Epistemic propositional dynamic logic This section will present the language, model and semantics of EPDL, and define the properties of perfect recall and no miracles. The language of EPDL is constructed by combining knowledge and program operators. Definition 1 (Language) Let $$\mathbf{\Phi_0}$$ and $$\mathbf{\Pi_0}$$ be two countably infinite sets of propositions and actions, respectively. The language of EPDL is defined in BNF as follows:   \begin{align*} \phi &::= \top\mid p\mid \neg\phi\mid(\phi\wedge\phi)\mid [\pi]\phi \mid K\phi\\ \pi &::= a\mid ?\phi\mid (\pi;\pi)\mid (\pi+\pi)\mid \pi^*, \end{align*} where $$p\in\mathbf{\Phi_0}$$ and $$a\in\mathbf{\Pi_0}$$. We will often omit parentheses when doing so ought not cause confusion. The language has expressions of two sorts: formulas $$\phi$$ and programs $$\pi$$. The set of all programs is denoted $$\mathbf{\Pi}$$, and the set of all formulas is denoted $$\mathbf{\Phi}$$. As usual, we use the following abbreviations: $$\bot:=\neg\top$$, $$\phi\vee\psi:=\neg(\neg \phi\wedge\neg \psi),\phi\rightarrow\psi:=\neg\phi\vee\psi, \left\langle {{a}} \right\rangle \phi:=\neg [a]\neg\phi, \hat{K}\phi:=\neg K\neg\phi$$. Definition 2 (Model) A model $$\mathcal{M}$$ is a tuple $$\left\langle {{S^\mathcal{M},\{R^\mathcal{M}_a\mid a\in\mathbf{\Pi_0}\},R^\mathcal{M},V^\mathcal{M}}} \right\rangle$$, where $$S^\mathcal{M}$$ is a nonempty set of states, $$R^\mathcal{M}_a$$ is a binary relation on $$S^\mathcal{M}$$, $$R^\mathcal{M}$$ is an equivalence relation on $${S^\mathcal{M}}$$ and $$V^\mathcal{M}:\mathbf{\Phi_0}\to\mathcal{P}({S^\mathcal{M}})$$ is a function. A pointed model is a pair $$(\mathcal{M},s)$$ consisting of a model $$\mathcal{M}$$ and a state $$s\in S^\mathcal{M}$$. Given a model $$\mathcal{M}$$, we also write $$(s,t)\in R^\mathcal{M}_a$$ as $$s{\stackrel{a}{\rightarrow}}_{\mathcal{M}}t$$ or $$t\in R^\mathcal{M}_a(s)$$, and write $$(s,t)\in R^\mathcal{M}_a$$ as $$s\sim_{\mathcal{M}}t$$ or $$t\in R^\mathcal{M}(s)$$. If the model $$\mathcal{M}$$ is obvious from the context, we omit it as an index. Definition 3 (Semantics) Given a pointed model $$(\mathcal{M},s)$$ and a formula $$\phi$$, we write $$\mathcal{M},s\vDash\phi$$ to mean $$(\mathcal{M},s)$$ satisfies $$\phi$$. The satisfaction relation $$\vDash$$, shown in Table 1, is defined as usual by combining the semantics of EL and that of PDL. A formula $$\phi$$ is satisfiable if $$\mathcal{M},s\vDash\phi$$ for some model $$\mathcal{M}$$ and a state $$s\in S^\mathcal{M}$$. Table 1 Semantics $$\mathcal{M},s\vDash\top$$      $$\mathcal{M},s\vDash p$$  $$\iff$$  $$s\in V(p)$$  $$\mathcal{M},s\vDash \neg\phi$$  $$\iff$$  $$\mathcal{M},s\nvDash \phi$$  $$\mathcal{M},s\vDash \phi\wedge\psi$$  $$\iff$$  $$\mathcal{M},s\vDash\phi \text{ and } \mathcal{M},s\vDash \phi$$  $$\mathcal{M},s\vDash K\phi$$  $$\iff$$  $$s\sim t \text{ implies } \mathcal{M},t\vDash \phi$$  $$\mathcal{M},s\vDash [\pi]\phi$$  $$\iff$$  $$s{\stackrel{{\pi}}{\rightarrow}}t \text{ implies } \mathcal{M},t\vDash\phi$$  $${\stackrel{{a}}{\rightarrow}}$$  =  $$R^\mathcal{M}_a$$  $${\stackrel{{?\phi}}{\rightarrow}}$$  =  $$\{(s,s)\mathcal{M}id \mathcal{M},s\vDash\phi\}$$  $${\stackrel{{\pi_1+\pi_2}}{\rightarrow}}$$  =  $${\stackrel{\pi_1}{\rightarrow}}\cup{\stackrel{\pi_2}{\rightarrow}}$$  $${\stackrel{{\pi_1;\pi_2}}{\rightarrow}}$$  =  $${\stackrel{{\pi_1}}{\rightarrow}}\circ{\stackrel{{\pi_2}}{\rightarrow}}$$  $${\stackrel{{\pi^*}}{\rightarrow}}$$  =  $$\bigcup_{n\geq 0}({\stackrel{{\pi}}{\rightarrow}})^n$$  $$\mathcal{M},s\vDash\top$$      $$\mathcal{M},s\vDash p$$  $$\iff$$  $$s\in V(p)$$  $$\mathcal{M},s\vDash \neg\phi$$  $$\iff$$  $$\mathcal{M},s\nvDash \phi$$  $$\mathcal{M},s\vDash \phi\wedge\psi$$  $$\iff$$  $$\mathcal{M},s\vDash\phi \text{ and } \mathcal{M},s\vDash \phi$$  $$\mathcal{M},s\vDash K\phi$$  $$\iff$$  $$s\sim t \text{ implies } \mathcal{M},t\vDash \phi$$  $$\mathcal{M},s\vDash [\pi]\phi$$  $$\iff$$  $$s{\stackrel{{\pi}}{\rightarrow}}t \text{ implies } \mathcal{M},t\vDash\phi$$  $${\stackrel{{a}}{\rightarrow}}$$  =  $$R^\mathcal{M}_a$$  $${\stackrel{{?\phi}}{\rightarrow}}$$  =  $$\{(s,s)\mathcal{M}id \mathcal{M},s\vDash\phi\}$$  $${\stackrel{{\pi_1+\pi_2}}{\rightarrow}}$$  =  $${\stackrel{\pi_1}{\rightarrow}}\cup{\stackrel{\pi_2}{\rightarrow}}$$  $${\stackrel{{\pi_1;\pi_2}}{\rightarrow}}$$  =  $${\stackrel{{\pi_1}}{\rightarrow}}\circ{\stackrel{{\pi_2}}{\rightarrow}}$$  $${\stackrel{{\pi^*}}{\rightarrow}}$$  =  $$\bigcup_{n\geq 0}({\stackrel{{\pi}}{\rightarrow}})^n$$  Definition 4 (Perfect recall and no miracles) A model $$\mathcal{M}$$ has the property of Perfect Recall ($$\texttt{PR}$$) if for all $$a\in\mathbf{\Pi_0}$$ and all $$s,t,t'\in S^\mathcal{M}$$, $$s{\stackrel{{a}}{\rightarrow}}t$$ and $$t\sim t'$$ imply that there exists $$s'\in S^\mathcal{M}$$ such that $$s\sim s'$$ and $$s'{\stackrel{{a}}{\rightarrow}} t'$$. No Miracles ($$\texttt{NM}$$) if for all $$a\in\mathbf{\Pi_0}$$ and all $$s,s',t,t'\in S^\mathcal{M}$$, $$s{\stackrel{{a}}{\rightarrow}}t$$, $$s\sim s'$$ and $$s'{\stackrel{{a}}{\rightarrow}}t'$$ imply $$t\sim t'$$. In the rest of this article, we will always assume that models have the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. Moreover, since it follows by bisimulation invariance that EPDL has the tree model property (cf. [3]), we also assume all the models in this paper have the tree property; i.e., if $$s{\stackrel{{a}}{\rightarrow}}t$$ and $$s'{\stackrel{{a}}{\rightarrow}}t$$, then $$s=s'$$ for each $$a\in\mathbf{\Pi_0}$$. Next, we will focus on the problem whether a given formula is satisfiable in models with $$\texttt{PR}$$ and $$\texttt{NM}$$. This problem is tackled by building a tableau from an input formula and deciding if the tableau is open or not. The key is sorting out how to build a proper tableau from an input formula. 3 Tableau for EPDL with PR and NM This section will present how to construct the tableau from an input formula. To deal with the complication arising with interacting actions and knowledge, the tableau procedure will act on bubbles. A bubble, which represents an equivalence class, is a set of states, and it is epistemically sufficient and knowledge-consistent. It will be seen that each relevant formula of the form $$\neg K\phi$$ is realized within bubbles. The tableau procedure, then, will only need to focus on the realization of formulas in the shape of $$\neg[\pi]\phi$$ and the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. Before the tableau procedure, we introduce some terminology. Definition 5 (Fisher–Ladner closure) The Fisher-Ladner closure of a formula $$\phi$$, denoted $$\mathcal{FL}(\phi)$$, is the minimal set satisfying: $$\phi\in \mathcal{FL}(\phi)$$; if $$\psi\in \mathcal{FL}(\phi)$$ then $$\neg\psi\in\mathcal{FL}(\phi)$$, provided $$\psi$$ does not start with $$\neg$$; if $$\neg\psi$$, $$\psi\wedge\chi$$, $$K\psi$$ or $$[\pi]\psi$$ are in $$\mathcal{FL}(\phi)$$ then $$\psi,\chi\in\mathcal{FL}(\phi)$$; if $$[\pi_1;\pi_2]\psi\in\mathcal{FL}(\phi)$$ then $$[\pi_1][\pi_2]\psi\in\mathcal{FL}(\phi)$$; if $$[\pi_1+\pi_2]\psi\in\mathcal{FL}(\phi)$$ then both $$[\pi_1]\psi$$ and $$[\pi_2]\psi$$ are in $$\mathcal{FL}(\phi)$$; if $$[?\chi]\psi\in\mathcal{FL}(\phi)$$ then $$\chi\in\mathcal{FL}(\phi)$$; if $$[\pi^*]\psi\in\mathcal{FL}(\phi)$$ then $$[\pi][\pi^*]\psi\in\mathcal{FL}(\phi)$$. Given a formula set $$\Gamma$$, we also use $$\mathcal{FL}({\Gamma})$$ to mean $$\bigcup_{\phi\in\Gamma}\mathcal{FL}(\phi)$$. Please note that $$\mathcal{FL}(\phi)$$ is finite for every $$\phi\in\mathbf{\Phi}$$. What is more, we categorize formulas as $$\boldsymbol{\alpha}/\boldsymbol{\beta}$$-formulas, each with two components, as shown in Table 2. Table 2 $$\boldsymbol{\alpha}$$- and $$\boldsymbol{\beta}$$-formulas $$\boldsymbol{\alpha}$$  $$\neg\neg\phi$$  $$\phi\wedge\psi$$  $$[\pi_1;\pi_2]\phi$$  $$\neg [\pi_1;\pi_2]\phi$$  $$\neg[?\psi]\phi$$  $$[\pi_1+\pi_2]\phi$$  $$[\pi^*]\phi$$  $$K\phi$$  $$\boldsymbol{\alpha}_1$$  $$\phi$$  $$\phi$$  $$[\pi_1][\pi_2]\phi$$  $$\neg [\pi_1][\pi_2]\phi$$  $$\psi$$  $$[\pi_1]\phi$$  $$\phi$$  $$\phi$$  $$\boldsymbol{\alpha}_2$$    $$\psi$$      $$\neg\phi$$  $$[\pi_2]\phi$$  $$[\pi][\pi^*]\phi$$    $$\boldsymbol{\beta}$$  $${\neg(\phi\wedge\psi)}$$  $${\neg[\pi_1+\pi_2]\phi}$$  $${{[?\psi]\phi}}$$  $${\neg[\pi^*]\phi}$$  $$\boldsymbol{\beta_1}$$  $${\neg\phi}$$  $${\neg[\pi_1]\phi}$$  $${\neg\psi}$$  $${\neg\phi}$$  $$\boldsymbol{\beta_2}$$  $${\neg\psi}$$  $${\neg[\pi_2]\phi}$$  $${\phi}$$  $${\phi,\neg[\pi][\pi^*]\phi}$$  $$\boldsymbol{\alpha}$$  $$\neg\neg\phi$$  $$\phi\wedge\psi$$  $$[\pi_1;\pi_2]\phi$$  $$\neg [\pi_1;\pi_2]\phi$$  $$\neg[?\psi]\phi$$  $$[\pi_1+\pi_2]\phi$$  $$[\pi^*]\phi$$  $$K\phi$$  $$\boldsymbol{\alpha}_1$$  $$\phi$$  $$\phi$$  $$[\pi_1][\pi_2]\phi$$  $$\neg [\pi_1][\pi_2]\phi$$  $$\psi$$  $$[\pi_1]\phi$$  $$\phi$$  $$\phi$$  $$\boldsymbol{\alpha}_2$$    $$\psi$$      $$\neg\phi$$  $$[\pi_2]\phi$$  $$[\pi][\pi^*]\phi$$    $$\boldsymbol{\beta}$$  $${\neg(\phi\wedge\psi)}$$  $${\neg[\pi_1+\pi_2]\phi}$$  $${{[?\psi]\phi}}$$  $${\neg[\pi^*]\phi}$$  $$\boldsymbol{\beta_1}$$  $${\neg\phi}$$  $${\neg[\pi_1]\phi}$$  $${\neg\psi}$$  $${\neg\phi}$$  $$\boldsymbol{\beta_2}$$  $${\neg\psi}$$  $${\neg[\pi_2]\phi}$$  $${\phi}$$  $${\phi,\neg[\pi][\pi^*]\phi}$$  Definition 6 (State) A state $$\Delta$$ is a set of formulas satisfying the following conditions. $$\Delta$$ is not patently inconsistent, i.e. it does not contain both $$\phi$$ and $$\neg\phi$$; $$\boldsymbol{\alpha}\in\Delta$$ implies $$\boldsymbol{\alpha}_1\in\Delta$$ and $$\boldsymbol{\alpha}_2\in\Delta$$; $$\boldsymbol{\beta}\in\Delta$$ implies $$\boldsymbol{\beta}_1\in\Delta$$ or $$\boldsymbol{\beta}_2\in\Delta$$; For each $$K\phi\in\mathcal{FL}({\psi})$$ where $$\psi\in\Delta$$, either $$K\phi\in\Delta$$ or $$\neg K\phi\in \Delta$$. Given a finite set of formulas $$\Gamma$$, let $$\mathtt{S}({\Gamma})$$ be the set of states generated by $$\Gamma$$; namely $$\mathtt{S}({\Gamma})=\{\Delta\mid \Delta$$ is the minimal state such that $$\Gamma\subseteq \Delta \}$$. When $$\Gamma=\{\phi \}$$, we also write it as $$\mathtt{S}({\phi})$$ for abbreviation. For each $$\Delta\in\mathtt{S}({\Gamma})$$, we have it that $$\Delta\subseteq \mathcal{FL}({\Gamma})$$. Moreover, we use the following abbreviations: given a formula set $$\Delta$$, let $$K(\Delta)=\{K\phi\mid K\phi\in\Delta \}$$, $$Epi(\Delta)=K(\Delta)\cup\{\neg K\phi\mid\neg K\phi\in\Delta \}$$, and $$[a]^-(\Delta)=\{\phi\mid [a]\phi\in\Delta \}$$. Definition 7 (Bubble) A bubble$$B$$ is a finite and non-empty set of states such that: $$B$$ is epistemically sufficient, i.e. for each $$\Delta\in B$$ and each $$\neg K\phi\in \Delta$$, there exists a set $$\Delta'\in B$$ such that $$\neg\phi\in \Delta'$$; $$B$$ is knowledge-consistent, i.e. $$K(\Delta)=K(\Delta')$$ for all $$\Delta,\Delta'\in B$$; Any finite and non-empty set of states is called a pre-bubble. Definition 8 (Satisfiability of (pre-)bubble) Given a (pre-)bubble $$B$$ and a pointed model $$(\mathcal{M},s)$$, let $$[s]$$ be the equivalence class containing $$s$$, i.e. $$\{s'\in S^\mathcal{M}\mid s\sim s' \}$$. If there is a surjective function $$f:[s]\to B$$ such that $$\mathcal{M},s\vDash f(s)$$, we say $$B$$ is satisfied by $$(\mathcal{M},s)$$ and $$f$$, written as $$\mathcal{M},s\vDash^f B$$. If there are such a pointed model and such a function, we say $$B$$ is satisfiable. Intuitively, a state represents a possible world in a model. A bubble is a set of states in the same equivalence class. For each state $$\Delta$$ in a bubble $$B$$, we also denote it as $$\Delta_B$$. Given model $$\mathcal{M}$$ with the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$, it can be seen as a deterministic bubble tree, as shown in Figure 4. Each state $$\Delta_i$$ in Figure 4b represents the set of formulas that the state $$s_i$$ in Figure 4a satisfies. Except for the top bubble, every state in the other bubbles has an annotation which indicates its predecessor. Figure 4. View largeDownload slide (a) Model $$\mathcal{M}$$; (b) deterministic bubble tree. Figure 4. View largeDownload slide (a) Model $$\mathcal{M}$$; (b) deterministic bubble tree. A tableau $$\mathcal{T}$$ is a graph with bubbles and labelled edges $${\stackrel{{a}}{\rightarrow}}$$ where $$a\in\mathbf{\Pi_0}$$. As mentioned above, a bubble is a set of states. If $$B{\stackrel{{a}}{\rightarrow}}B'$$, we annotate each state $$\Delta'\in B'$$ with a state $$\Delta\in B$$, indicating that $$\Delta{\stackrel{{a}}{\rightarrow}}\Delta'$$; namely $$\Delta$$ is an $$a$$-predecessor of $$\Delta'$$. The states in the first layer will be annotated with $$\emptyset$$ since they have no predecessors. Let $$B$$ and $$B'$$ be two bubbles from a tableau such that $$B{\stackrel{{a}}{\rightarrow}}B'$$. The property $$\texttt{PR}$$ requires that each state in $$B'$$ has an $$a$$-predecessor in $$B$$. In other words, there must be ‘enough’ states in $$B$$. From the construction of the tableau below, we will see that each state $$\Delta'\in B'$$ is annotated with a state $$\Delta\in B$$ and $$[a]^-(\Delta)\subseteq \Delta'$$. The property $$\texttt{NM}$$ requires that all formulas of the form $$\neg[a]\phi$$ from a bubble $$B$$ will be realized in the same bubble $$B'$$, which is also guaranteed from the construction of the tableau. The tableau procedure for EPDL with $$\texttt{PR}$$ and $$\texttt{NM}$$ consists of two phases: a construction phase and an elimination phase. In the construction phase, a pre-tableau with pre-bubbles and bubbles is established. In the elimination phase, the pre-tableau is pruned to an initial tableau and then a final tableau. 3.1 Construction of the pre-tableau Each state $$\Delta\in\mathtt{S}({\phi})$$ is a potential expansion of $$\phi$$. The construction is based on (pre-)bubbles. Therefore, we start with pre-bubbles containing $$\Delta\in\mathtt{S}({\phi})$$. To keep the property $$\texttt{PR}$$, we need to make sure that at least one of the initial pre-bubbles has enough states. For each state, if it is in the same equivalence class with $$\Delta$$, then $$Epi(\Delta)$$ is the minimal set of formulas that it contains. Therefore, there exists $$P\subseteq \mathtt{S}({Epi(\Delta)})$$ such that $$\{\Delta \}\cup P$$ potentially has enough states. The idea of the construction from an input $$\phi$$ is that initially we make $$\{\Delta \}\cup P$$ a pre-bubble for each $$\Delta\in\mathtt{S}({\phi})$$ and each $$P\subseteq \mathtt{S}({Epi(\Delta)})$$. Since $$Epi(\Delta)$$ is the minimal set of formulas that a state in the same equivalence class with $$\Delta$$ needs to satisfy, at least one of these initial pre-bubbles potentially has enough states. To continue the construction, we need to realize formulas of the forms $$\neg K\psi$$ and $$\neg[a]\psi$$ and keep the properties $$\texttt{PR}$$ and $$\texttt{NM}$$. The procedure ExtendToBubbles extends a pre-bubble to all potential bubbles and also realizes all formulas of the form $$\neg K\psi$$. The procedure SuccessorPreBubbles realizes all formulas of the form $$\neg[a]\psi$$. What is more, $$\texttt{PR}$$ and $$\texttt{NM}$$ are guaranteed in these procedures. The construction of the pre-tableau for $$\phi$$ works as follows. (1) For each $$\Delta\in\mathtt{S}({\phi})$$ and each $$P\subseteq \mathtt{S}({Epi(\Delta)})$$, make the pre-bubble $$\{\Delta \}\cup P$$ as a node and annotate each state in it with $$\emptyset$$. (2) Expand each pre-bubble $$A$$ into bubbles by calling the procedure ExtendToBubbles(A), which is presented in Algorithm 3. For each bubble $$B'\in$$ExtendToBubbles$${(A)}$$, add $$B'$$ as a node if $$B'$$ is not already there. We then produce an arrow $$A{\dashrightarrow} B'$$. (3) For each bubble $$B$$, if there exists a formula $$\neg[a]\psi\in\Delta$$ for some $$\Delta\in B$$ and some $$a\in\mathbf{\Pi_0}$$, produce successor pre-bubbles by calling the procedure SuccessorPreBubbles$${(B,a)}$$, which is presented in Algorithm 4. For each pre-bubble $$A\in$$SuccessorPreBubbles$${(B,a)}$$, add $$A$$ as a node if $$A$$ is not already there. We then produce an arrow $$B{\stackrel{{a}}{\rightarrow}}A$$. (4) Repeat steps 2 and 3 until no new bubbles or pre-bubbles are created. View largeDownload slide View largeDownload slide View largeDownload slide View largeDownload slide View largeDownload slide View largeDownload slide View largeDownload slide View largeDownload slide Since each bubble or pre-bubble in the construction is a subset of $$\mathcal{P}({\mathcal{FL}(\phi)})$$, the pre-tableau for $$\phi$$ is a finite graph, and the construction of the pre-tableau will terminate. The following are all the procedures, which are adaptations from [1], and we will also show that each procedure functions well. KnowledgeConsistent The procedure KnowledgeConsistent, presented in Algorithm 1, takes a pre-bubble $$A$$ as input and returns a set of knowledge-consistent pre-bubbles. To make the pre-bubble $$A$$ knowledge-consistent, it seems that each $$\Delta\in A$$ only need to be extended to $$\Delta\cup\Sigma$$ for some $$\Sigma\in\mathtt{S}({Epi(A)\setminus Epi(\Delta)})$$. However, this is not sufficient. To preserve the satisfiability of the pre-bubble $$A$$, we need to extend each $$\Delta\in A$$ to a subset of $$\{\Delta\cup \Sigma\mid \Sigma\in\mathtt{S}({Epi(A)\setminus Epi(\Delta)})\}$$. For example, given a pre-bubble $$A=\{\Delta_1,\Delta_2 \}$$ where $$\Delta_1=\{K(p\lor q\lor r),p\lor q\lor r,p \}$$ and $$\Delta_2=\{\neg p \}$$, a model $$\mathcal{M}$$ is depicted in Figure 5. Let function $$f$$ be $$f=\{s\mapsto \Delta_1,t\mapsto \Delta_2,v\mapsto \Delta_2 \}$$. Then we have $$\mathcal{M},s\vDash^f A$$. Let $$\Gamma_1=\{K(p\lor q\lor r),p\lor q\lor r,q,\neg p \}$$ and $$\Gamma_2=\{K(p\lor q\lor r),p\lor q\lor r,r,\neg p \}$$. To preserve the satisfiability of $$A$$ on model $$\mathcal{M}$$, the pre-bubble $$A_3$$ must be returned by the procedure KnowledgeConsistent$${(A)}$$, as is shown in Figure 6. Figure 5. View largeDownload slide Model Figure 5. View largeDownload slide Model Figure 6. View largeDownload slide KnowledgeConsistent$$(A)$$. Figure 6. View largeDownload slide KnowledgeConsistent$$(A)$$. The procedure KnowledgeConsistent has the following properties. Proposition 1 If $$A$$ is a pre-bubble, each $$A'\in$$KnowledgeConsistent$${(A)}$$ is knowledge-consistent. Proof. Assume that there is a pre-bubble $$A'\in$$KnowledgeConsistent$${(A)}$$ such that $$A'$$ is not knowledge-consistent. It follows that there are $$\Omega,\Omega'\in A'$$ and $$K\phi$$ such that $$K\phi\in\Omega$$ and $$K\phi\not\in\Omega'$$. By Algorithm 1, we have $$Epi(A)\subseteq \Omega$$ and $$Epi(A)\subseteq \Omega'$$. Let $$\Omega=\Delta_i\cup \Sigma$$ for some $$\Delta_i\in A$$ and some $$\Sigma\in\mathtt{S}({Epi(A)\setminus Epi(\Delta_i)})$$, and $$\Omega'=\Delta_j\cup \Sigma'$$ for some $$\Delta_j\in A$$ and some $$\Sigma'\in\mathtt{S}({Epi(A)\setminus Epi(\Delta_j)})$$. Since $$K\phi\in\Delta_i$$ implies $$K\phi\in \Omega'$$, it follows that $$K\phi\in \Sigma$$. It follows that there exists $$\psi\in Epi(A)$$ such that $$K\phi\in sub(\psi)$$. Since $$Epi(A)\subseteq \Omega'$$ and $$\Omega'$$ is a state, we have $$\neg K\phi\in \Omega'$$. Similarly, $$\neg K\phi\in\Delta_j$$ implies $$\neg K\phi\in \Omega$$. Thus we have $$\neg K\phi\in\Sigma'$$. Then there exists $$\Delta\in A$$ and $$\psi'\in Epi(\Delta)$$ such that $$K\phi\in sub(\psi')$$. Since $$\Delta$$ is a state, either $$K\phi\in \Delta$$ or $$\neg K\phi\in\Delta$$. This implies that $$K\phi\in Epi(A)$$ or $$\neg K\phi\in Epi(A)$$. This is contradictory with the facts that $$Epi(A)\subseteq \Omega$$, $$Epi(A)\subseteq \Omega'$$, $$K\phi\in \Omega$$ and $$\neg K\phi\in\Omega'$$. Thus each $$A'\in$$KnowledgeConsistent$${(A)}$$ is knowledge-consistent. ■ Proposition 2 If $$A$$ is a pre-bubble and $$\mathcal{M},s\vDash^f A$$, there exists a pre-bubble $$A'\in$$KnowledgeConsistent$${(A)}$$ and a function $$f':[s]\to A'$$ such that $$\mathcal{M},s\vDash^{f'}A'$$, $$f(s)\subseteq f'(s)$$, and $$f'(s)$$ has the same annotation as $$f(s)$$. Proof. Since $$\mathcal{M},s\vDash^f A$$, we have it that, for each $$u\in[s]$$, $$\mathcal{M},u\vDash f(u)$$ and $$\mathcal{M},u\vDash epi_u$$, where $$f(u)\in A$$ and $$epi_u=Epi(A)\setminus Epi(f(u))$$. Thus, in Algorithm 1, there exists $$\Sigma_u\in \mathtt{S}({epi'})$$ such that $$\mathcal{M},u\vDash \Sigma_u$$. Therefore we have $$\mathcal{M},u\vDash f(u)\cup\Sigma_u$$, and it follows by Algorithm 1 that $$f(u)\cup\Sigma_u\in Alt_i$$ for some $$i$$, and $$f(u)\cup\Sigma_u$$ has the same annotation as $$f(u)$$. Let $$D_\Delta=\{\Delta\cup\Sigma_u\mid u\in [s], f(u)=\Delta \}$$ for each $$\Delta\in A$$. Thus, $$D_\Delta\in P(Alt_i)$$ and $$A'=\cup \{D_\Delta\mid \Delta\in A \}\in$$KnowledgeConsistent$${(A)}$$. We define the function $$f':[s]\to A'$$ as $$f'(u)=f(u)\cup\Sigma_u$$ for each $$u\in[s]$$. It follows that $$\mathcal{M},u\vDash f'(u)$$ and $$f'$$ is surjective. Therefore, $$\mathcal{M},s\vDash^{f'} A'$$, $$f(s)\subseteq f'(s)$$, and $$f'(s)$$ has the same annotation as $$f(s)$$. ■ EpistemicallySufficient The procedure EpistemicallySufficient, which is presented in Algorithm 2, takes a pre-bubble $$A$$ as input and returns a set of epistemically sufficient pre-bubbles. To make $$A$$ epistemically sufficient, it seems that for each unrealized $$\neg K\phi$$ from $$A$$, we need to extend $$A$$ by adding some $$\Sigma\in\mathtt{S}({\neg\phi})$$. However, there are several problems. First, if $$A$$ is knowledge-consistent, to preserve the knowledge consistency, it seems that we should add some $$\Sigma\in\mathtt{S}({Epi(A)\cup\{\neg\phi\}})$$ to $$A$$. Secondly, if $$B{\stackrel{{a}}{\rightarrow}}A$$ it follows by $$\texttt{PR}$$ that each $$\Sigma$$ which is intended to be added in $$A$$ needs to have an $$a$$-predecessor $$\Delta\in B$$. This means we should add $$\Delta'\cup \Sigma$$ in $$A$$ where $$\Delta'\in \mathtt{S}({[a]^-(\Delta)})$$ rather than add only $$\Sigma$$ in $$A$$. We assume that $$A$$ has enough states, which means that all such $$\Delta'$$ are already in $$A$$. Finally, there might be more than one unrealized formulas $$\neg K\phi_1\cdots\neg K\phi_n$$ from $$A$$ such that they are realized by the same state. In this situation, to preserve the satisfiability of $$A$$, we might need to extend $$A$$ by adding $$\Delta'\cup \Sigma_i\cdots \Sigma_n$$ in $$A$$ where $$\Delta'\in A$$ and $$\Sigma_i\in\mathtt{S}({Epi(A)\cup\neg\phi_i})$$. For example, given pre-bubble $$A=\{\Delta,\Gamma \}$$ where $$\Delta=\{\hat{K}p,\hat{K}q \}$$ and $$\Gamma=\{\hat{K}p,\hat{K}q,r\}$$, the model $$\mathcal{M}$$ is depicted as Figure 7, and let function $$f$$ be $$\{s\mapsto \Delta,t\mapsto\Gamma \}$$. It is obvious that $$\mathcal{M},s\vDash^f A$$. To preserve the satisfiability of $$A$$ on $$\mathcal{M}$$, the procedure EpistemicallySufficient(A), as shown in Figure 8, needs to return a bubble such that it only has two states, and one of them realizes both $$\hat{K}p$$ and $$\hat{K}q$$. Figure 7. View largeDownload slide Model Figure 7. View largeDownload slide Model Figure 8. View largeDownload slide EpistemicallySufficient$$(A)$$  $$\matrix{ {{\Delta _1} = \{ \hat Kp,\hat Kq,p\} \qquad {\Gamma _1} = \{ \hat Kp,\hat Kq,r,p\} } \hfill \cr {{\Delta _2} = \{ \hat Kp,\hat Kq,q\} \qquad {\Gamma _2} = \{ \hat Kp,\hat Kq,r,q\} } \hfill \cr {{\Delta _3} = \{ \hat Kp,\hat Kq,p,q\} \qquad {\Gamma _3} = \{ \hat Kp,\hat Kq,r,p,q\} .} \hfill \cr }$$ Figure 8. View largeDownload slide EpistemicallySufficient$$(A)$$  $$\matrix{ {{\Delta _1} = \{ \hat Kp,\hat Kq,p\} \qquad {\Gamma _1} = \{ \hat Kp,\hat Kq,r,p\} } \hfill \cr {{\Delta _2} = \{ \hat Kp,\hat Kq,q\} \qquad {\Gamma _2} = \{ \hat Kp,\hat Kq,r,q\} } \hfill \cr {{\Delta _3} = \{ \hat Kp,\hat Kq,p,q\} \qquad {\Gamma _3} = \{ \hat Kp,\hat Kq,r,p,q\} .} \hfill \cr }$$ The procedure EpistemicallySufficient has the following properties. Proposition 3 If $$A$$ is a pre-bubble, each $$A'\in$$EpistemicallySufficient$${(A)}$$ is epistemically sufficient. Proof. By Algorithm 2, it follows that for each $$\neg K\phi\in\Delta\in A$$, there is $$\Omega\in A'$$ such that $$\neg\phi\in \Omega$$. Next we will show that for each $$\neg K\phi\in\Omega\in A'$$, there is $$\Delta\in A$$ such that $$\neg K\phi\in\Delta$$. This implies that each $$A'\in$$EpistemicallySufficient$${(A)}$$ is epistemically sufficient. If $$\neg K\phi\in\Omega\in A'$$ and $$\Omega\not\in A$$, we know that $$\Omega=\Delta\cup \Sigma_1\cdots\Sigma_k$$ where $$\Delta\in A$$ and, for each $$1\leq i\leq k$$, $$\Sigma_i\in\mathtt{S}({Epi(A)\cup\{\neg\psi_i \}})$$ for some $$\neg K\psi_i\in\Delta''$$ and $$\Delta''\in A$$. If $$\neg K\phi\in\Sigma_i$$ for some $$1\leq i\leq k$$, there exists $$\chi\in\Delta'$$ for some $$\Delta'\in A$$ such that $$K\phi\in sub(\chi)$$. Since $$\Delta'$$ is a state, we have $$K\phi\in\Delta'$$ or $$\neg K\phi\in \Delta'$$. If $$K\phi\in \Delta'$$, it follows $$K\phi\in Epi(A)$$ and then $$K\phi\in\Omega$$. This is contradictory with $$\neg K\phi\in\Omega$$. Thus, $$\neg K\phi\in \Delta'$$. ■ Proposition 4 If a pre-bubble $$A$$ is knowledge-consistent, each pre-bubble $$A' \in$$EpistemicallySufficient$${(A)}$$ is also knowledge-consistent. Proof. We only need to show that $$K\phi\in\Omega$$ implies $$K\phi\in\Omega'$$ for all $$\Omega,\Omega'\in A'$$. By Algorithm 2, it follows that, for each $$\Omega\in A'$$, there is $$\Delta\in A$$ such that $$\Delta\subseteq \Omega$$. Since $$A$$ is knowledge-consistent, we only need to show that $$K\phi\in\Omega$$ implies $$K\phi\in K(A)$$. If $$\Omega\not\in A$$, we know $$\Omega=\Delta\cup \Sigma_1\cdots\Sigma_k$$ where $$\Delta\in A$$ and, for each $$1\leq i\leq k$$, $$\Sigma_i\in\mathtt{S}({Epi(A)\cup\{\neg\psi_i \}})$$ for some $$\neg K\psi_i\in\Delta''$$ and $$\Delta''\in A$$. It follows $$Epi(A)\subseteq \Omega$$. If $$K\phi\in\Sigma_i$$ for some $$1\leq i\leq k$$, there exists $$\Delta'\in A$$ and $$\psi'\in \Delta'$$ such that $$K\phi\in sub(\psi')$$. Since $$\Delta'$$ is a state, we have $$K\phi\in\Delta'$$ or $$\neg K\phi\in\Delta'$$. If $$\neg K\phi\in\Delta'$$, it follows by $$Epi(A)\subseteq \Omega$$ that $$\neg K\phi\in \Omega$$. This is contradictory with $$K\phi\in\Omega$$. Thus, $$K\phi\in \Delta'$$, and consequently $$K\phi\in K(A)$$. ■ Proposition 5 If $$A$$ is a pre-bubble and $$\mathcal{M},s\vDash^f A$$, there exists a pre-bubble $$A'\in$$EpistemicallySufficient$${(A)}$$ and a function $$f':[s]\to A'$$ such that $$\mathcal{M},s\vDash^{f'}A'$$, $$f(s)\subseteq f'(s)$$, and $$f'(s)$$ has the same annotation as $$f(s)$$. Proof. Since $$\mathcal{M},s\vDash^f A$$, we have that for each $$u\in[s]$$, $$\mathcal{M},u\vDash Epi(A)$$ and $$\mathcal{M},u\vDash f(u)$$. Moreover, since $$f$$ is surjective, this means that for each $$\neg K\phi\in Epi(A)$$ there exists $$u\in[s]$$ such that $$\mathcal{M},u\vDash\neg K\phi$$. It follows that there is some $$s'\in[s]$$ such that $$\mathcal{M},s'\vDash\neg\phi$$. Let $$epi$$ be the set $$\{\neg K\phi\in Epi(A)\mid \neg\phi\not\in\Delta$$ for any $$\Delta\in A \}$$. For each $$u\in[s]$$, let $$\sigma(u)=\{\neg\phi\mid \neg K\phi\in epi$$ and $$\mathcal{M},u\vDash\neg\phi \}$$. It is obvious that $$\mathcal{M},u\vDash f(u)\land\sigma(u)$$ for each $$u\in[s]$$. If $$\sigma(u)\neq\emptyset$$, it follows that $$\mathcal{M},u\vDash f(u)\land\Sigma_1\cdots\Sigma_{|\sigma(u)|}$$ where $$\Sigma_i\in \mathtt{S}({Epi(A)\cup\{\neg\phi_i \}})$$ for each $$1\leq i\leq |\sigma(u)|$$. We define $$\Omega_u$$ as $$\Omega_{u}=f(u)\cup\Sigma_1\cdots\Sigma_{|\sigma(u)|}$$. It is obvious that $$\mathcal{M},u\vDash\Omega_{u}$$. By Algorithm 2, it follows that $$\Omega_{u}$$ is in $$Alt$$ and $$\Omega_{u}$$ has the same annotation as $$\Delta$$. If $$\sigma(u)=\emptyset$$, let $$\Omega_{u}=f(u)$$. Therefore, $$\mathcal{M},u\vDash\Omega_{u}$$ for each $$u\in[s]$$. Let a pre-bubble $$A'$$ be the set $$\{\Omega_{u}\mid u\in [s] \}$$. By Algorithm 2, we have $$A'\in$$EpistemicallySufficient$${(A)}$$. We define $$f':[s]\to A'$$ as $$f'(u)=\Omega_{u}$$ for each $$u\in[s]$$. Therefore, $$\mathcal{M},s\vDash^{f'} A'$$, $$f(s)\subseteq f'(s)$$, and $$f'(s)$$ has the same annotation as $$f(s)$$. ■ ExtendToBubbles The procedure ExtendToBubbles, which is presented in Algorithm 3, extends a pre-bubble $$A$$ to bubbles. To do so, we firstly make the pre-bubble $$A$$ knowledge-consistent with the procedure KnowledgeConsistent$${(A)}$$ and then make each knowledge-consistent pre-bubble $$A'\in$$KnowledgeConsistent$${(A)}$$ epistemically sufficient with the procedure EpistemicallySufficient$${(A')}$$. The properties of the procedures KnowledgeConsistent and EpistemicallySufficient guarantee that each pre-bubble $$B\in$$ExtendToBubbles$${(A)}$$ is a bubble extension of $$A$$. Proposition 6 If $$A$$ is a pre-bubble, each $$B\in$$ExtendToBubbles$${(A)}$$ is a bubble. Proof. It follows from Propositions 1, 3 and 4. ■ Proposition 7 Given a pre-bubble $$A$$ and a bubble $$B\in$$ExtendToBubbles$${(A)}$$, we have the following results: For each $$\Delta\in A$$, there is a state $$\Delta'\in B$$ such that $$\Delta\subseteq \Delta'$$ and they have the same annotation; For each $$\Delta'\in B$$, there is a state $$\Delta\in A$$ such that $$\Delta\subseteq \Delta'$$ and they have the same annotation. Proof. It is obvious from Algorithms 1, 2 and 3. ■ Proposition 8 If $$A$$ is a pre-bubble and $$\mathcal{M},s\vDash^f A$$, there exists a bubble $$B\in$$ExtendToBubbles$${(A)}$$ and a function $$f':[s]\to B$$ such that $$\mathcal{M},s\vDash^{f'}B$$, $$f(s)\subseteq f'(s)$$, and $$f'(s)$$ has the same annotation as $$f(s)$$. Proof. It follows from Propositions 2 and 5. ■ SuccessorPreBubbles The procedure SuccessorPreBubbles, presented in Algorithm 4, takes a bubble $$B$$ and an action $$a\in\mathbf{\Pi_0}$$ as input and returns a set of $$a$$-successor pre-bubbles for $$B$$. If a pre-bubble $$A$$ is returned, it follows by $$\texttt{NM}$$ that all $$\neg[a]\phi$$ from $$B$$ is realized in $$A$$. Similar to the case in EpistemicallySufficient, formulas $$\neg[a]\phi_1\cdots\neg[a]\phi_n\in\Delta\in B$$ might be realize by the same state, so there might be state $$\Sigma\in\mathtt{S}({[a]^-(\Delta)\cup\{\neg\phi_1\cdots\neg\phi_n \}})$$ in some returned successor pre-bubble. Furthermore, the property $$\texttt{PR}$$ requires that each bubble has enough states. Step 1 of the construction of the pre-tableau guarantees that at least one bubble in the first layer has enough states. To preserve $$\texttt{PR}$$, our strategy is that, if the bubble $$B$$ has enough states and $$\{\Omega_1,\cdots,\Omega_k\}$$ is the state set realizing all $$\neg[a]\phi$$ from $$B$$, let the pre-bubble $$\{\Omega_1,\cdots,\Omega_k\}\cup P$$ be in the returned set for each $$P\subseteq H$$ where $$H=\bigcup_{\Delta\in B}\mathtt{S}({[a]^-(\Delta)})$$. This guarantees that one returned pre-bubble has enough states. For example, given a bubble $$B=\{\Delta,\Gamma \}$$ in which $$\Delta=\{\left\langle {{a}} \right\rangle p,\hat{K}q,\left\langle {{a}} \right\rangle \hat{K}q\}$$ and $$\Gamma=\{[a]r,\hat{K}q,q\}$$, the pre-bubbles returned by the procedure SuccessorPreBubbles$$(B,a)$$ are shown in Figure 9. Figure 9. View largeDownload slide SuccessorPreBubbles (B,a)   $$\matrix{ {{\Delta _1} = \{ p\} \qquad {\Delta _ \bullet } = \{ \top \} } \hfill \cr {{\Delta _2} = \{ \hat Kq\} \qquad {\Gamma _ \bullet } = \{ r\} } \hfill \cr {{\Delta _3} = \{ p,\hat Kq\} .} \hfill \cr }$$ Figure 9. View largeDownload slide SuccessorPreBubbles (B,a)   $$\matrix{ {{\Delta _1} = \{ p\} \qquad {\Delta _ \bullet } = \{ \top \} } \hfill \cr {{\Delta _2} = \{ \hat Kq\} \qquad {\Gamma _ \bullet } = \{ r\} } \hfill \cr {{\Delta _3} = \{ p,\hat Kq\} .} \hfill \cr }$$ Proposition 9 Given a bubble $$B$$ and $$A\in$$SuccessorPreBubbles$${(B,a)}$$, we have the following results: For each $$\Delta\in B$$ and each $$\neg[a]\phi\in\Delta$$, there is a state $$\Delta'\in A$$ such that $$\neg\phi\in\Delta'$$ and the annotation of $$\Delta'$$ is $$\Delta$$. If $$\Delta'\in A$$ and the annotation of $$\Delta'$$ is a state $$\Delta$$, then we have $$\Delta\in B$$ and $$[a]^-(\Delta)\subseteq \Delta'$$. Proof. It is obvious from Algorithm 4. ■ Proposition 10 Let $$B$$ be a bubble. If $$\mathcal{M},s\vDash^f B$$, $$\neg[a]\phi\in f(s)$$, $$t\in R^\mathcal{M}_a(s)$$ and $$\mathcal{M},t\vDash\phi$$, then there are a pre-bubble $$A\in$$SuccessorPreBubbles$${(B,a)}$$ and a function $$f':[t]\to A$$ such that $$f'(t)$$ is annotated with $$f(s)$$, $$\neg\phi\in f'(t)$$, and $$\mathcal{M},t\vDash^{f'} A$$. Proof. Because of our assumption at the end of Section 2, each $$v\in [t]$$ has a unique $$a$$-predecessor, written as $$v^-$$. It is obvious that $$t^-=s$$. Due to $$\mathcal{M},s\vDash^f B$$ and the property of $$\texttt{NM}$$, it follows that for each $$\neg[a]\psi\in\Delta\in B$$ there exists $$v\in[t]$$ such that $$\mathcal{M},v\vDash[a]^-(\Delta)\land \neg\psi$$. Let $$\sigma(v)=[a]^-(f(v^-))\cup\{\neg\psi\mid \neg[a]\psi\in f(v^-)$$ and $$\mathcal{M},v\vDash \neg\psi \}$$. It is obvious that $$\neg\phi\in \sigma(t_0)$$. If $$\sigma(v)\neq\emptyset$$, let $$\sigma(v)=[a]^-(f(v^-))\cup\{\neg\psi_1,\cdots,\neg\psi_k\}$$. This implies that there are $$\Delta'\in\mathtt{S}({[a]^-(f(v^-))})$$ and $$\Sigma_j\in\mathtt{S}({\neg\phi_j})$$, where $$1\leq j\leq k$$, such that $$\mathcal{M},v\vDash\Delta'\cup\Sigma_1\dots\Sigma_k$$. We define $$\Omega_{v}=\Delta'\cup\Sigma_1\dots\Sigma_k$$. If $$\sigma(v)=\emptyset$$, we define $$\Omega_{v}=\{\top \}$$. Moreover, let $$\Omega_{v}$$ be annotated with $$f(v^-)$$. Please note that $$\neg\phi\in\Omega_{t}$$ and $$\Omega_{t}$$ is annotated with $$f(s)$$. It follows by Algorithm 4 that $$\Omega_{v}\in Alt$$. Let $$A=\{\Omega_{v}\mid v\in [t] \}$$. Then we have $$A\in$$SuccessorPreBubbles$${(B,a)}$$. We define $$f':[t]\to A$$ as $$f'(v)=\Omega_{v}$$. It follows that $$\mathcal{M},v\vDash f'(v)$$ for each $$v\in [t]$$ and $$f'$$ is surjective. Thus, $$\mathcal{M},t\vDash^{f'}A$$. Moreover, we also have it that $$f'(t)$$ is annotated with $$f(s)$$ and that $$\neg\phi\in f'(t)$$. ■ 3.2 Construction of the initial and final tableau From the construction of the pre-tableau, we know that if the pre-bubble $$A$$ is an $$a$$-successor of a bubble $$B$$ for some $$a\in{\mathbf{\Pi_0}}$$ and the bubble $$B'$$ is a bubble-extension of $$A$$, namely $$B{\stackrel{a}{\rightarrow}}A{\dashrightarrow} B'$$, then it follows by Propositions 7 and 9 that the annotation of each state $$\Delta'\in B'$$ is some state $$\Delta\in B$$. The initial tableau is produced by removing the pre-bubbles and redirecting the arrows in the pre-tableau. For example, if $$B{\stackrel{a}{\rightarrow}}A{\dashrightarrow} B'$$, we then delete the pre-bubble $$A$$ and add an $$a$$-arrow from $$B$$ to $$B'$$, namely $$B{\stackrel{a}{\rightarrow}}B'$$. Since each bubble is a finite set of states and each state is a subset of $$\mathcal{FL}({\phi})$$, the initial tableau is a finite graph. Next, we construct the final tableau based on the initial tableau. Before this, we firstly introduce the notion of a fulfilling chain. Definition 9 (Fulfilling chain) Let $$B$$ be a bubble of a tableau $${\mathcal{T}}$$ and $$\neg[\pi]\phi\in\Delta\in B$$. A fulfilling chain for $$(\neg[\pi]\phi,\Delta,B)$$ is defined as follows. $$(\Delta_0,B_0){\stackrel{a_1}{\rightarrow}}(\Delta_1,B_1)$$ is a fulfilling chain for $$(\neg[a_1]\phi,\Delta_0,B_0)$$ if $$B_0{\stackrel{a_1}{\rightarrow}}B_1$$, $${\neg\phi\in\Delta_1}$$ and the annotation of $$\Delta_1\in B_1$$ is $$\Delta_0\in B_0$$; $$(\Delta_0,B_0)$$ is a fulfilling chain for $$(\neg[?\psi]\phi,\Delta_0,B_0)$$ if $$\neg[?\psi]\phi,{\neg\phi}\in\Delta_0$$; $$(\Delta_0,B_0){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_n,B_n)$$ is a fulfilling chain for $$(\neg[\pi_1;\pi_2]\phi,\Delta_0,B_0)$$ if there exists $$0\leq i\leq n$$ such that $$(\Delta_0,B_0){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_i,B_i)$$ is a fulfilling chain for $$(\neg[\pi_1][\pi_2]\phi,\Delta_0,B_0)$$ and $$(\Delta_i,B_i){\stackrel{a_{i+1}}{\rightarrow}}\cdots(\Delta_n,B_n)$$ is a fulfilling chain for $$(\neg[\pi_2]\phi,\Delta_0,B_0)$$; $$(\Delta_0,B_0){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_n,B_n)$$ is a fulfilling chain for $$(\neg[\pi_1+\pi_2]\phi,\Delta_0,B_0)$$ if it is a fulfilling chain for $$(\neg[\pi_1]\phi,\Delta_0,B_0)$$ or $$(\neg[\pi_2]\phi,\Delta_0,B_0)$$; $$(\Delta_0,B_0){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_n,B_n)$$ is a fulfilling chain for $$(\neg[\pi^*]\phi,\Delta_0,B_0)$$ if $$\neg[\pi^*]\phi\in\Delta_0,{\neg\phi}\in \Delta_n$$, and if $$n>0$$ then there are $$0= i_0\leq \cdots\leq i_{k}= n$$ such that $$(\Delta_{i_j},B_{i_j}){\stackrel{a_{i_j}}{\rightarrow}}\cdots(\Delta_{i_{j+1}},B_{i_{j+1}})$$ is a fulfilling chain for $$(\neg[{\pi}][{\pi^*}]\phi,\Delta_{i_j},B_{i_j})$$ for all $$0\leq j\leq k-1$$. The final tableau $${\mathcal{T}}$$ is obtained by the following procedure. (1) Let $${\mathcal{T}}_0$$ be the initial tableau. (2) Let $$B$$ be a bubble of $${\mathcal{T}}_n$$. If there is $$\neg[\pi]\phi\in\Delta\in B$$ such that there is no fulfilling chain for $$(\neg[\pi]\phi,\Delta,B)$$, then let $${\mathcal{T}}_{n+1}={\mathcal{T}}_n\setminus\{B\}$$. (3) Repeat the step 2 until no bubble is to be deleted, i.e. $${\mathcal{T}}_{n+1}={\mathcal{T}}_n$$. Definition 10 The final tableau $${\mathcal{T}}$$ for $$\phi$$ is open if there is a bubble $$B$$ in $${\mathcal{T}}$$ and $$\Delta\in B$$ such that $$\phi\in\Delta$$. 4 Soundness and completeness In this section, we will show that the tableau constructed in the previous section is proper, i.e. it is sound and complete. 4.1 Soundness To show the soundness of the procedure, we need to show that the final tableau for $$\phi$$ is open if $$\phi$$ is satisfiable. We will show that, in the initial tableau, there exists a bubble that contains a state including $$\phi$$ and this bubble is to survive in the final tableau. Proposition 11 If $$\phi$$ is satisfiable, there exists a bubble $$B$$ in the initial tableau and a state $$\Delta\in B$$ such that $$\phi\in\Delta$$ and $$B$$ is satisfiable. Proof. Assume that $$\mathcal{M},s\vDash\phi$$. It follows that $$\mathcal{M},s\vDash\Delta_s$$ for some $$\Delta_s\in{\mathtt{S}}({\phi})$$. For each $$u\in[s]$$ and $$u\neq s$$, we have $$\mathcal{M},u\vDash Epi(\Delta_s)$$. This implies that $$\mathcal{M},u\vDash \Delta_u$$ for some $$\Delta_u\in{\mathtt{S}}({Epi(\Delta_s)})$$. Let $$A=\{\Delta_u\mid u\in[s],u\neq s \}\cup\{\Delta_s \}$$. Then it follows by the step 1 of the construction of the pre-tableau that $$A$$ is a pre-bubble in the pre-tableau. We define $$f:[s]\to A$$ as $$f(s)=\Delta_s$$ and $$f(u)=\Delta_u$$ for each $$u\in[s]$$ and $$u\neq s$$. Therefore we have $$\mathcal{M},s\vDash^f A$$ and $$\phi\in f(s)$$. It follows by Proposition 8 that $$\mathcal{M},s\vDash^{f'}B$$ and $$f(s)\subseteq f'(s)$$ for some function $$f'$$ and some bubble $$B\in$$ExtendToBubbles$${(A)}$$. ■ Definition 11 (Witness chain) Let $$\mathcal{M},s$$ be a pointed model. A witness chain for $$(\mathcal{M},s,\neg[{\pi}]\phi)$$ is defined as follows. $$(s_0,\{\neg[a_1]\phi \}){\stackrel{a_1}{\rightarrow}}(s_1,\{\neg\phi\})$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{a_1}]\phi)$$ if $$s_0{\stackrel{a_1}{\rightarrow}}s_1$$, $$\mathcal{M},s_0\vDash\neg[a_1]\phi$$ and $$\mathcal{M},s_1\vDash\neg\phi$$; $$(s,\{\neg[?\psi]\phi,\neg\phi\})$$ is a witness chain for $$(\mathcal{M},s,\neg[{?\psi}]\phi)$$ if $$\mathcal{M},s\vDash\neg[?\psi]\phi$$; $$(s_0,\Gamma\cup\{\neg[\pi_1;\pi_2]\phi \}){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{\pi_1;\pi_2}]\phi)$$ if there exists $$0\leq i\leq n$$ such that $$(s_0,\Gamma){\stackrel{a_1}{\rightarrow}}\cdots(s_i,\Gamma_i)$$ is a witness chain for $$(\mathcal{M},s_0,\neg[\pi_1][\pi_2]\phi)$$ and $$(s_i,\Gamma_i){\stackrel{a_i}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_i,\neg[\pi_2]\phi)$$. $$(s_0,\Gamma\cup\{\neg[\pi_1+\pi_2]\phi \}){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{\pi_1+\pi_2}]\phi)$$ if $$(s_0,\Gamma){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{\pi_1}]\phi)$$ or $$(\mathcal{M},s_0,\neg[{\pi_2}]\phi)$$. $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n\cup\{\neg\phi\})$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{\pi^*}]\phi)$$ if $$\neg[\pi^*]\phi\in\Gamma_0$$, $$\mathcal{M},s_n\vDash\neg\phi$$, and if $$n>0$$ then there are $$0= i_0\leq \cdots\leq i_{k}= n$$ such that $$(s_{i_j},\Gamma_{i_j}\setminus\{\neg[\pi^*]\phi\}),\cdots,(s_{i_{j+1}},\Gamma_{i_{j+1}})$$ is a witness chain for $$(\mathcal{M},s_{i_j},\neg[{\pi}][{\pi^*}]\phi)$$ for all $$0\leq j\leq k-1$$. By the definition above, it is easy to show the following propositions. Proposition 12 If $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s,\neg[{\pi}]\phi)$$ then $$s_0=s$$, $$\neg[{\pi}]\phi\in\Gamma_0$$, $$\neg\phi\in\Gamma_n$$ and $$\mathcal{M},s_i\vDash\Gamma_i$$ for all $$0\leq i\leq n$$. Proposition 13 If $$\mathcal{M},s\vDash^f B$$, $$\neg[\pi]\phi\in f(s)$$ then there is a witness chain $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ for $$(\mathcal{M},s,\neg[\pi]\phi)$$ such that $$\Gamma_0\subseteq f(s)$$. Proposition 14 Let $$B$$ be a bubble in the initial tableau $${\mathcal{T}}_0$$. If $$\mathcal{M},s\vDash^f B$$, $$\neg[a]\phi\in f(s)$$, and $$\mathcal{M},t\vDash\neg\phi$$ for some $$t\in R^\mathcal{M}_a(s)$$, then there are a bubble $${B'}$$ in $${\mathcal{T}}_0$$ and a function $$f':[t]\to B'$$ such that $$\mathcal{M},t\vDash^{f'} B'$$ and $$(f(s),B){\stackrel{a}{\rightarrow}}(f'(t),B')$$ is a fulfilling chain for $$(\neg[a]\phi,f(s),B)$$. Proof. First, it follows by Proposition 10 that there exists a pre-bubble $$A\in$$SuccessorPreBubbles$${(B,a)}$$ and a function $$f'':[t]\to A$$ such that $$\mathcal{M},t\vDash^{f''} A$$, $$\neg\phi\in f''(t)$$ and $$f''(t)$$ is annotated with $$f(s)$$. It follows by Proposition 8 that there are a bubble $$B'\in$$ExtendToBubbles$${(A)}$$ and a function $$f':[t]\to B'$$ such that $$\mathcal{M},t\vDash^{f'} B'$$, $$f''(t)\subseteq f'(t)$$, and the annotation of $$f'(t)$$ (which is the same as $$f''(t)$$) is $$f(s)$$. Since $$B{\stackrel{a}{\rightarrow}}A{\dashrightarrow} B'$$, it follows that $$B{\stackrel{a}{\rightarrow}}B'$$ in the tableau $${\mathcal{T}}_0$$. Since the annotation of $$f'(t)\in B'$$ is $$f(s)\in B$$ and $$\neg\phi\in f''(t)\subseteq f'(t)$$, we have $$\neg\phi\in f'(t)$$. Therefore, $$(f(s),B){\stackrel{a}{\rightarrow}}(f'(t),B')$$ is a fulfilling chain for $$(\neg[a]\phi,f(s),B)$$. ■ Proposition 15 Let $$B$$ be a bubble of the initial tableau $${\mathcal{T}}_0$$. If $$\mathcal{M},s\vDash^f B$$, $$\neg[\pi]\phi\in f(s)$$, and $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s,\neg[\pi]\phi)$$ such that $$\Gamma_0\subseteq f(s)$$, then there are a sequence of bubbles $${B_1}\cdots B_n$$ of $${\mathcal{T}}_0$$ and a sequence of functions $$f_1\cdots f_n$$ such that $$f_i:[s_i]\to B_i$$, $$\mathcal{M},s_i\vDash^{f_i} B_i$$, $$\Gamma_i\subseteq f_i(s_i)$$ for all $$1\leq i\leq n$$, and $$(f(s),{B}){\stackrel{a_1}{\rightarrow}}\cdots (f_n(s_n),B_n)$$ is a fulfilling chain for $$(\neg[\pi]\phi,f(s),B)$$. Proof. We prove it by induction on $$\pi$$. In the case of $$\neg[a]\phi$$, it is obvious by Proposition 14. Case of $$\neg[?\psi]\phi$$: It follows by $$\neg[?\psi]\phi\in f(s)$$ that both $$\psi$$ and $$\neg \phi$$ are in $$f(s)$$. Therefore, $$(f(s),B)$$ is a fulfilling chain for $$(\neg[?\psi]\phi,f(s),B)$$. Case of $$\neg [\pi_1;\pi_2]\phi$$: Since we have that $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s,\neg[\pi_1;\pi_2]\phi)$$, it follows that there exists $$i$$ such that $$(s_0,\Gamma_0\setminus\{\neg[\pi_1;\pi_2]\phi\}){\stackrel{a_1}{\rightarrow}}\cdots(s_i,\Gamma_i)$$ is a witness chain for $$(\mathcal{M},s,\neg[\pi_1][\pi_2]\phi)$$ and $$(s_i,\Gamma_i){\stackrel{a_i}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_i,\neg[\pi_2]\phi)$$. It follows by $$\neg [\pi_1;\pi_2]\phi\in f(s)$$ that $$\neg{[\pi_1][\pi_2]}\phi\in f(s)$$. By induction on $$\pi_1$$, there are a bubble sequence $${B_1}\cdots B_i$$ and a function sequence $$f_1\cdots f_i$$ such that $$f_j:[s_j]\to B_j$$, $$\mathcal{M},s_j\vDash^{f_j} B_j$$, $$\Gamma_j\subseteq f_j(s_j)$$ for all $$1\leq j\leq i$$, and $$(f(s),{B}){\stackrel{a_1}{\rightarrow}}\cdots(f_i(s_i),B_i)$$ is a fulfilling chain for $$(\neg[\pi_1][\pi_2]\phi,f(s),B)$$. It follows by Proposition 12 that $$\neg[\pi_2]\phi\in f_i(s_i)$$. Then, by induction on $$\pi_2$$, there are a bubble sequence $${B_{i+1}}\cdots B_n$$ and a function sequence $$f_{i+1}\cdots f_n$$ such that $$f_j:[s_j]\to B_j$$, $$\mathcal{M},s_j\vDash^{f_j} B_j$$, $$\Gamma_j\subseteq f_j(s_j)$$ for all $$i+1\leq j\leq i$$, and $$(f(s_i),{B_i}){\stackrel{a_i}{\rightarrow}}\cdots(f_n(s_n),B_n)$$ is a fulfilling chain for $$(\neg[\pi_2]\phi,f(s),B)$$. Therefore, $$(f(s),{B}){\stackrel{a_i}{\rightarrow}}\cdots(f_n(s_n),B_n)$$ is a fulfilling chain for $$(\neg[\pi_1;\pi_2]\phi,f(s),B)$$. Case of $$\neg [\pi_1+\pi_2]\phi$$: Since $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s,\neg[\pi_1+\pi_2]\phi)$$, it follows that $$(s_0,\Gamma_0\setminus\{\neg[\pi_1+\pi_2]\phi\}){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s,\neg[\pi_1]\phi)$$ or $$(\mathcal{M},s,\neg[\pi_1]\phi)$$. Assuming it is a witness chain for $$(\mathcal{M},s,\neg[\pi_1]\phi)$$, it follows by Proposition 12 that $$\neg[\pi_1]\phi\in \Gamma_0$$. Because of $$\Gamma_0\subseteq f(s)$$, it follows that $$\neg[\pi_1]\phi\in f(s)$$. By induction on $$\pi_1$$, there are a bubble sequence $${B_{1}}\cdots B_n$$ and a function sequence $$f_{1}\cdots f_n$$ such that $$f_j:[s_j]\to B_j$$, $$\mathcal{M},s_j\vDash^{f_j} B_j$$ and $$\Gamma_j\subseteq f_j(s_j)$$ for all $$1\leq j\leq n$$, and $$(f(s),{B}){\stackrel{a_1}{\rightarrow}}\cdots(f_n(s_n),B_n)$$ is a fulfilling chain for $$(\neg[\pi_1]\phi,f(s),B)$$. Therefore, it is a fulfilling chain for $$(\neg[\pi_1+\pi_2]\phi,f(s),B)$$. Case of $$\neg [\pi^*]\phi$$: If $$n=0$$, then $$(f(s),B)$$ is a fulfilling chain for $$(\neg[\pi^*]\phi,f(s),B)$$. If $$n>0$$, it follows that there are $$0= i_0\leq \cdots\leq i_{k}= n$$ such that $$(s_{i_j},\Gamma_{i_j}\setminus\{\neg[\pi^*]\phi\}){\stackrel{a_{i_j}}{\rightarrow}}\cdots(s_{i_{j+1}},\Gamma_{i_{j+1}})$$ is a witness chain for $$(\mathcal{M},s_{i_j},\neg[{\pi}][{\pi^*}]\phi)$$ for all $$0\leq j\leq k-1$$. Therefore, $$(s_{i_0},\Gamma_{i_0}\setminus\{\neg[\pi^*]\phi\}),\cdots,(s_{i_{1}},\Gamma_{i_{1}})$$ is a witness chain for $$(\mathcal{M},s_{i_0},\neg[{\pi}][{\pi^*}]\phi)$$ (please note that $$i_0=0$$). It follows by Proposition 12 that $$\neg[\pi][\pi^*]\phi\in \Gamma_{i_j}$$ for all $$0\leq j\leq k-1$$. Because of $$\Gamma_{i_0}\subseteq f(s)$$, it follows that $$\neg[\pi][\pi^*]\phi\in f(s)$$. By induction on $$\pi$$, there are a bubble sequence $$B_1\cdots{B_{i_1}}$$ and a function sequence $$f_1\cdots f_{i_1}$$ such that $$f_{j}:[s_{j}]\to B_{j}$$, $$\mathcal{M},s_{j}\vDash^{f_{j}} B_{j}$$, $$\Gamma_{j}\subseteq f_{j}(s_{j})$$ for all $$1\leq j\leq i_1$$, and $$(f(s),{B}){\stackrel{a_{1}}{\rightarrow}}\cdots(f_{i_1}(s_{i_1})_{B_{i_1}})$$. Similarly, we can do this for all $$0\leq j\leq k-1$$. Therefore, $$\neg[\pi^*]\phi\in\Gamma_{i_{j+1}}\subseteq f_{i_{j+1}}(s_{i_{j+1}})$$, and $$(f_{i_j}(s_{i_j}),{B_{i_j}}){\stackrel{a_{i_j+1}}{\rightarrow}}\cdots (f_{i_{j+1}}(s_{i_{j+1}}),{B_{i_{j+1}}})$$ is a fulfilling chain for $$(\neg[\pi][\pi^*]\phi,f_{i_j}(s_{i_j}),{B_{i_j}})$$ for all $$0\leq j\leq k-1$$. Moreover, since $$(s_0,\Gamma_0){\stackrel{a_1}{\rightarrow}}\cdots(s_n,\Gamma_n)$$ is a witness chain for $$(\mathcal{M},s_0,\neg[{\pi^*}]\phi)$$, it follows that $$\mathcal{M},s_n\vDash\neg\phi$$. Because of $$\mathcal{M},s_n\vDash^{f_n}B_n$$ and $$\neg[\pi^*]\phi\in f_n(s_n)$$, it follows that $$\neg\phi\in f_n(s_n)$$. Otherwise, by Table 2, we will have $$\phi\in f_n(s_n)$$. This is contradictory with $$\mathcal{M},s_n\vDash\neg\phi$$ and $$\mathcal{M},s\vDash f_n(s_n)$$. Therefore, $$(f(s),B){\stackrel{a_1}{\rightarrow}}\cdots(f_n(s_n),B_n)$$ is a fulfilling chain for $$(\neg[\pi^*]\phi,f(s),B)$$. ■ Lemma 1 For each bubble $$B$$ in the initial tableau, if $$B$$ is satisfiable, then $$B$$ is to survive in the final tableau. Proof. We only need to show that if $$B$$ is satisfiable then $$B$$ is to survive in $${\mathcal{T}}_{i}$$ for each $$i\in\mathbb{N}$$. We prove it by induction on $$i$$. It is obvious if $$i=0$$. Next we only need to show that if $$B$$ survives in $${\mathcal{T}}_i$$ and is satisfiable, then $$B$$ will also survive in $${\mathcal{T}}_{i+1}$$. Therefore, we need to show that for each $$\neg[\pi]\phi\in\Delta\in B$$ there is a fulfilling chain of $${\mathcal{T}}_i$$ for $$(\neg[\pi]\phi,\Delta,B)$$. If $$\neg[\pi]\phi\in\Delta\in B$$ and $$B$$ is satisfiable, it follows by Propositions 13 and 15 that there is a fulfilling chain $$(\Delta,B){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_n,B_n)$$ of $${\mathcal{T}}_0$$ for $$(\neg[\pi]\phi,\Delta,B)$$ such that each $$B_i$$ is satisfiable for all $$1\leq j\leq n$$. By induction on $$i$$, each $$B_j (1\leq j\leq n)$$ is to survive in $${\mathcal{T}}_i$$. Therefore, $$(\Delta,B){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_n,B_n)$$ is also a chain of $${\mathcal{T}}_i$$, and then $$B$$ survives in $${\mathcal{T}}_{i+1}$$. Thus, we have shown that if $$B$$ is satisfiable, then $$B$$ is to survive in $${\mathcal{T}}_{i}$$ for each $$i\in\mathbb{N}$$, namely $$B$$ is to survive in the final tableau. ■ Theorem 1 (Soundness) If $$\phi$$ is satisfiable, the final tableau $${\mathcal{T}}$$ for $$\phi$$ is open. Proof. It follows from Proposition 11 and Lemma 1. ■ 4.2 Completeness To show the completeness of the procedure, we need to show that there is a model for the input formula $$\phi$$ if the final tableau for $$\phi$$ is open. First, we construct a deterministic tree of bubbles based on the open final tableau such that each $$\neg[\pi]\psi$$ from each bubble is fulfilled on this tree. Next, this deterministic bubble tree can be turned into a model with $$\texttt{PR}$$ and $$\texttt{NM}$$ on which $$\phi$$ is satisfied. Please note that in this section we always assume the final tableau is open. Definition 12 (Bubble tree) Let $${\langle {S,\{R_a\mid a\in {\mathbf{\Pi_0}}\}} \rangle}$$ be a tree in which $$S\subseteq{\mathbb{N}}$$. $${L}$$ is a function on $$S$$ which labels each $$n\in S$$ a bubble from the final tableau $${\mathcal{T}}$$. A labelled bubble tree (or just bubble tree) $${\mathfrak{T}}$$ based on $${\mathcal{T}}$$ is a tuple $${\mathfrak{T}}={\langle {S,\{R_a\mid a\in {\mathbf{\Pi_0}}\},{L}} \rangle}$$ such that $$n{\stackrel{a}{\rightarrow}}k$$ implies $$l(n){\stackrel{a}{\rightarrow}}l(k)$$ in $${\mathcal{T}}$$ for each $$a\in{\mathbf{\Pi_0}}$$. We refer to the components of $${\mathfrak{T}}$$ as $$S^{\mathfrak{T}}$$, $$R^{\mathfrak{T}}_a$$ and $${L}^{\mathfrak{T}}$$. A bubble tree $${\mathfrak{T}}$$ is deterministic if $$n{\stackrel{a}{\rightarrow}}k$$ and $$n{\stackrel{a}{\rightarrow}}l$$ imply $$k=l$$. Definition 13 (Expanding a bubble tree) Given a bubble tree $${\mathfrak{T}}$$ based on $${\mathcal{T}}$$, let $$n$$ be a node on the tree and $$\sigma=B_0{\stackrel{a_1}{\rightarrow}}\cdots B_k$$ be a chain from $${\mathcal{T}}$$ such that $${L}(n)=B_0$$. We expand $${\mathfrak{T}}$$ at $$n$$ by $$\sigma$$ as below. If there is $$i\in S^{\mathfrak{T}}$$ such that $$n{\stackrel{a_1}{\rightarrow}}i$$ and $${L}^{\mathfrak{T}}(i)=B_1$$, we continue expanding $${\mathfrak{T}}$$ at $$i$$ by $$B_1{\stackrel{a_2}{\rightarrow}}\cdots B_k$$. If not, we first add a new number $$j$$, add an arrow $$n{\stackrel{a_1}{\rightarrow}}j$$, and label $$j$$ as $${B_1}$$. Next, we continue expanding it at the node $$j$$ by $$B_1{\stackrel{a_2}{\rightarrow}}\cdots B_k$$. Proposition 16 let $$k$$ be a node on a finite deterministic bubble tree $${\mathfrak{T}}$$ and $${L}(k)=B$$. For each $$\neg[\pi]\phi\in\Delta\in B$$, there is a fulfilling chain $$(\Delta,B){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_n,B_n)$$ for $$(\neg[\pi]\phi,\Delta,B)$$ such that expanding $${\mathfrak{T}}$$ at $$k$$ by $$B_0{\stackrel{a_1}{\rightarrow}}\cdots B_n$$ is still a finite deterministic bubble tree. Proof. We prove it by induction on $$\pi$$. We only focus on the cases of $$\neg[a]\phi$$ and $$\neg[\pi^*]\phi$$; the others are obvious. In the case of $$\neg[a]\phi$$, if there is a node $$j$$ on the tree such that $$k{\stackrel{a}{\rightarrow}}j$$, then it follows that $${L}(k){\stackrel{a}{\rightarrow}}{L}(j)$$ in the final tableau. It follows by Propositions 9 and 7 that there is some $$\Delta'\in{L}(j)$$ such that $$\Delta'$$’s notation is $$\Delta\in B={L}(k)$$. Therefore, $$(\Delta,{L}(k)){\stackrel{a}{\rightarrow}}(\Delta',{L}(j))$$ is a fulfilling chain for $$(\neg[a]\phi,\Delta,{L}(k))$$. If there is no such $$j$$ on the tree, it follows that there is a fulfilling chain $$(\Delta,{L}(k)){\stackrel{a}{\rightarrow}}(\Delta',B')$$ in the final tableu for $$(\neg[a]\phi,\Delta,{L}(k))$$. Then we expand $${\mathfrak{T}}$$ at $$k$$ by $${L}(k){\stackrel{a}{\rightarrow}}B'$$. It is obvious that it is still a finite deterministic tree after the expansion. In the case of $$\neg[\pi^*]\phi\in\Delta\in B$$, there are two situations: $$\neg\phi\in\Delta$$ or $$\neg[\pi][\pi^*]\phi\in\Delta$$. If $$\neg\phi\in\Delta$$, it is obvious. If $$\neg[\pi][\pi^*]\phi\in\Delta$$, it follows by induction on $$\pi$$ that there is a fulfilling chain $$(\Delta,B){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_n,B_n)$$ for $$(\neg[\pi][\pi^*]\phi,\Delta,B)$$ such that expanding $${\mathfrak{T}}$$ at $$k$$ by $$B_0{\stackrel{a_1}{\rightarrow}}\cdots B_n$$ is still a finite deterministic bubble tree. If $$\neg\phi$$ is not in $$\Delta_n\in B_n$$, it follows by $$\neg[\pi^*]\phi\in\Delta_n$$ that $$\neg[\pi][\pi^*]\phi\in\Delta_n$$. Therefore, we can repeat this procedure until $$\neg\phi\in\Delta_n\in B_n$$ or $$B_n$$ is labelled on a leaf node. If $$\neg[\pi^*]\phi$$ is still not fulfilled and $$B_n$$ is labelled on a leaf node $$l$$, since the final tableau is open, it follows that there is a fulfilling chain $$(\Delta_n,{L}(l)){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_h,B_h)$$ for $$(\neg[\pi^*]\phi,\Delta_n,{L}(l))$$, which is a chain of the final tableau. We can than continue expanding the expanded tree with $${L}(l){\stackrel{a_1}{\rightarrow}}\cdots B_h$$. It is still a finite deterministic tree after all these expansion, and $$(\Delta,{L}(k)){\stackrel{a_1}{\rightarrow}}\cdots(\Delta_h,B_h)$$ is a fulfilling chain for $$(\neg[\pi^*],\Delta,{L}(k))$$. ■ Let the final tableau $${\mathcal{T}}$$ for $$\phi$$ be open, and $$\phi\in\Delta$$ for some state $$\Delta\in B$$ and some bubble $$B$$ of $${\mathcal{T}}$$. Next we will construct a deterministic bubble tree $${\mathfrak{T}}^\phi$$ based on the the final tableau $${\mathcal{T}}$$ for $$\phi$$ and then construct a model $$\mathcal{M}^\phi$$ based on $${\mathfrak{T}}^\phi$$. $${\mathfrak{T}}^\phi$$ is constructed as follows. (1) Let $$0$$ be the root and $${L}(0)=B$$, and mark the root as unfulfilled. (2) Let $$n$$ be the first unfulfilled node with the minimal level of the tree. We expand the tree as follows: – For each $$\neg[\pi]\phi\in \Delta\in {L}(n)$$, let $$\sigma$$ be a fulfilling chain for $$(\neg[\pi]\phi,\Delta,{L}(n))$$ such that expanding the tree at $$n$$ by $$\sigma$$ will keep the tree deterministic and finite. Proposition 16 makes sure that there exists such a fulfilling chain. We expand the tree at $$n$$ by $$\sigma$$ and mark the new nodes as unfulfilled. – After fulfilling all the $$\neg[\pi]\phi\in\Delta$$ for all $$\Delta\in {L}(n)$$, mark $$n$$ as fulfilled. (3) Repeat the step 2 until there are no unfulfilled nodes. Since each step of the construction keeps the tree deterministic, we have the following result. Proposition 17 $${\mathfrak{T}}^\phi$$ is a deterministic bubble tree. Proposition 18 For each node $$n$$ of $${\mathfrak{T}}^\phi$$ and each $$\neg[\pi]\psi\in\Delta\in {L}(n)$$, there exists a fulfilling chain $$(\Delta_0,{L}(n_0)){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_n,{L}(n_k))$$ for $$(\neg[\pi]\psi,\Delta,{L}(n))$$ such that $$n_0{\stackrel{a_1}{\rightarrow}}\cdots n_k$$. Proof. Let the level of node $$n$$ be $$m$$. Since there are only finite states in each bubble and finite formulas in each state, each node of $${\mathfrak{T}}^\phi$$ has a finite number of children. This means that $${\mathfrak{T}}^\phi$$ is finitely branching. Therefore, we will deal with $$\neg[\pi]\phi\in\Delta\in {L}(n)$$ in some step of the construction. It follows by the construction and Definition 13 that there exists a fulfilling chain $$(\Delta_0,{L}(n_0)){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_n,{L}(n_k))$$ for $$(\neg[\pi]\psi,\Delta,{L}(n))$$ such that $$n_0{\stackrel{a_1}{\rightarrow}}\cdots n_k$$. ■ Next, we define the model $$\mathcal{M}^\phi$$ based on $${\mathfrak{T}}^\phi$$. Definition 14 The model $$\mathcal{M}^\phi$$ is defined as follows. $$S=\{(\Delta,n)\mid n$$ is a node of $${\mathfrak{T}}^\phi$$ and $$\Delta\in {L}(n) \}$$ $${\stackrel{a}{\rightarrow}}=\{((\Delta,n),(\Delta',n'))\mid n{\stackrel{a}{\rightarrow}}n'$$ and $$\Delta'$$’s annotation is $$\Delta. \}$$ $$\sim=\{((\Delta,n),(\Delta',n'))\mid n=n' \}$$ $$V(p)=\{(\Delta,n)\mid p\in\Delta \}$$ for each $$p\in\mathcal{FL}({\phi})$$ Proposition 19 The model $$\mathcal{M}^\phi$$ has the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. Proof. First we will show $$\mathcal{M}^\phi$$ has the property of $$\texttt{PR}$$. If $$(\Delta,n){\stackrel{a}{\rightarrow}}(\Delta',n')$$ and $$(\Delta',n')\sim(\Theta',k')$$, it follows that $$k'=n'$$, $$n{\stackrel{a}{\rightarrow}}n'$$ and $$\Theta'\in {L}(n')$$. It follows by Propositions 9 and 7 that $$B{\stackrel{a}{\rightarrow}}B'$$ implies the annotation of each state in $$B'$$ is a state in $$B$$. Therefore, the annotation of $$\Theta'$$ is a state $$\Theta\in {L}(n)$$. It follows that $$(\Delta,n)\sim(\Theta,n)$$ and $$(\Theta,n){\stackrel{a}{\rightarrow}}(\Delta',k')$$. Secondly if $$(\Delta,n){\stackrel{a}{\rightarrow}}(\Delta',n')$$, $$(\Delta,n)\sim(\Theta,n)$$, and $$(\Theta,n){\stackrel{a}{\rightarrow}}(\Theta',k)$$, we need to show that $$(\Delta',n')\sim(\Theta',k)$$. We then have $$n{\stackrel{a}{\rightarrow}}n'$$ and $$n{\stackrel{a}{\rightarrow}}k$$. It follows by Proposition 17 that $$n'=k$$. Thus we have $$(\Delta',n')\sim(\Theta',k)$$. This means the model $$\mathcal{M}^\phi$$ has the property of $$\texttt{NM}$$. ■ Lemma 2 Given a formula $$\phi$$ and the model $$\mathcal{M}^\phi$$, we have the following results. (1) For each $$[\pi]\psi\in\mathcal{FL}({\phi})$$, if $$[\pi]\phi\in\Delta$$ and $$(\Delta,n){\stackrel{\pi}{\rightarrow}}(\Delta',n')$$, then $$\psi\in\Delta'$$; (2) For each $$\neg[\pi]\psi\in\mathcal{FL}({\phi})$$, if $$(\Delta_0,{L}(n_0)){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_k,{L}(n_k))$$ is a fulfilling chain for $$(\neg[\pi]\psi,\Delta_0,{L}(n_0))$$ such that $$n_0{\stackrel{a_1}{\rightarrow}}\cdots n_k$$, then $$(\Delta_0,n_0){\stackrel{\pi}{\rightarrow}}(\Delta_k,n_k)$$; (3) For each $$\psi\in\mathcal{FL}({\phi})$$, if $$\psi\in \Delta\in{L}(n)$$, then $$\mathcal{M}^\phi, (\Delta,n)\vDash\psi$$. Proof. We prove it by simultaneous induction on 1, 2 and 3. For 1, we only focus on the cases of $$[a]\psi$$, $$[?\chi]\psi$$ and $$[\pi^*]\psi$$; the other cases are obvious. In the case of $$[a]\psi\in\Delta$$, it follows by $$(\Delta,n){\stackrel{a}{\rightarrow}}(\Delta',n')$$ that $${L}(n){\stackrel{a}{\rightarrow}}{L}(n')$$ and that $$\Delta'$$’s annotation is $$\Delta$$. Since each state only has one annotation in the tableau, it follows by Propositions 9 and 7 that $$[a]^-(\Delta)\subseteq \Delta'$$. Thus, we have $$\psi\in\Delta'$$. In the case of $$[?\chi]\psi\in\Delta$$, it follows that either $$\neg\chi\in\Delta$$ or $$\psi\in\Delta$$. It follows by $$(\Delta,n){\stackrel{\chi}{\rightarrow}}(\Delta',n')$$ that $$\mathcal{M}^\phi,(\Delta,n)\vDash\chi$$. Assume $$\neg\chi\in\Delta$$. Then it follows by IH of 3 that $$\mathcal{M}^\phi,(\Delta,n)\vDash\neg\chi$$. This results in a contradiction. Therefore, $$\neg\chi\not\in\Delta$$, and consequently $$\psi\in\Delta$$. In the case of $$[\pi^*]\psi\in\Delta$$, it follows by $$(\Delta,n){\stackrel{\pi^*}{\rightarrow}}(\Delta',n')$$ that $$(\Delta,n){\stackrel{\pi^k}{\rightarrow}}(\Delta',n')$$ for some $$k\in\mathbb{N}$$. Since $$[\pi^*]\psi\in\Delta_i$$ implies that $$\psi,[\pi][\pi^*]\psi\in \Delta_i$$ for each state $$\Delta_i$$, where $$0\leq i\leq k$$. By induction on $$\pi$$, it follows that $$\psi\in\Delta'$$. For 2, we only focus on the cases of $$\neg[?\chi]\psi$$ and $$\neg[\pi^*]\psi$$; the other cases are obvious. In the case of $$\neg[?\chi]\psi$$, the fulfilling chain for $$(\neg[?\chi]\psi,\Delta_0,{L}(n_0))$$ is $$(\Delta_0,{L}(n_0))$$. It follows that $$\chi\in\Delta_0\in{L}(n_0)$$. By IH of 3, we have $$\mathcal{M}^\phi,(\Delta_0,n_0)\vDash\chi$$. Thus, we have $$(\Delta_0,n_0){\stackrel{\chi}{\rightarrow}}(\Delta_0,n_0)$$. In the case of $$\neg[\pi^*]\psi$$, it is obvious if $$k=0$$. If $$k>0$$, there are $$0= i_0\leq \cdots\leq i_{h}= k$$ such that $$(\Delta_{i_j},{L}(n_{i_j})){\stackrel{a_{i_j}}{\rightarrow}}\cdots(\Delta_{i_{j+1}},{L}(n_{i_{j+1}}))$$ is a fulfilling chain for $$(\neg[{\pi}][{\pi^*}]\psi,\Delta_{i_j},{L}(n_{i_j}))$$ for all $$0\leq j\leq h-1$$. Moreover, we have $$n_{i_j}{\stackrel{a_{i_j}}{\rightarrow}}\cdots n_{i_{j+1}}$$ for all $$0\leq j\leq h-1$$. By induction on $$\pi$$, it follows that $$(\Delta_{i_j},n_{i_j}){\stackrel{\pi}{\rightarrow}}(\Delta_{i_{j+1}},n_{i_{j+1}})$$ for all $$0\leq j\leq h-1$$. Since $$i_0=0$$ and $$i_h=k$$, we have $$(\Delta_0,n_0){\stackrel{\pi^*}{\rightarrow}}(\Delta_k,n_k)$$. For 3, we only focus on the cases of $$[\pi]\psi$$, $$\neg[\pi]\psi$$, $$K\psi$$ and $$\neg K\psi$$; it is obvious in the cases of $$p$$, $$\neg p$$, $$\psi\land\psi'$$ and $$\neg(\psi\land\psi')$$. In the case of $$[\pi]\psi\in \Delta\in{L}(n)$$, it follows by IH of 1 that $$\psi\in\Delta'\in{L}(n')$$ for each $$(\Delta',n')$$ such that $$(\Delta,n){\stackrel{\pi}{\rightarrow}}(\Delta',n')$$. By induction on $$\psi$$, we have $$\mathcal{M}^\phi,(\Delta',n')\vDash\psi$$. It follows that $$\mathcal{M}^\phi,(\Delta,n)\vDash[\pi]\psi$$. In the case of $$\neg[\pi]\psi\in\Delta\in{L}(n)$$, it follows by Proposition 18 that there exists a fulfilling chain $$(\Delta_0,{L}(n_0)){\stackrel{a_1}{\rightarrow}}\cdots (\Delta_n,{L}(n_k))$$ for $$(\neg[\pi]\psi,\Delta,{L}(n))$$ such that $$n_0{\stackrel{a_1}{\rightarrow}}\cdots n_k$$. By IH of 2, it follows that $$(\Delta,n){\stackrel{\pi}{\rightarrow}}(\Delta_k,n_k)$$. We also have that $$\neg\psi\in \Delta_k\in{L}(n_k)$$ because it is a fulfilling chain. By induction on $$\neg\psi$$, it follows that $$\mathcal{M}^\phi,(\Delta_k,n_k)\vDash\neg\psi$$. Thus, we have $$\mathcal{M}^\phi,(\Delta,n)\vDash\neg[\pi]\psi$$. In the case of $$K\psi\in\Delta\in{L}(n)$$, since $${L}(n)$$ is a bubble, it follows that $$K\psi,\psi\in \Delta'$$ for each $$\Delta'\in {L}(n)$$. It follows that $$\psi\in\Delta'$$ for each $$(\Delta',n)$$ such that $$(\Delta,n)\sim(\Delta',n)$$. By induction on $$\psi$$, it follows that $$\mathcal{M}^\phi,(\Delta',n)\vDash\psi$$, and then we have $$\mathcal{M}^\phi,(\Delta,n)\vDash K\psi$$. In the case of $$\neg K\psi\in\Delta\in{L}(n)$$, since $${L}(n)$$ is a bubble, it follows that $$\neg\psi\in\Delta'$$ for some $$\Delta'\in {L}(n)$$. By induction on $$\psi$$, it follows that $$\mathcal{M}^\phi,(\Delta',n)\vDash\neg\psi$$, and then we have $$\mathcal{M}^\phi,(\Delta,n)\vDash \neg K\psi$$. ■ Theorem 2 (Completeness) If the final tableau $${\mathcal{T}}$$ for $$\phi$$ is open, $$\phi$$ is satisfiable. Proof. If the final tableau $${\mathcal{T}}$$ for $$\phi$$ is open, let $$\phi\in\Delta\in B$$ for some state $$\Delta\in B$$ and some bubble $$B$$. We can construct the deterministic tree $${\mathfrak{T}}^\phi$$ with the root $$B$$ and then construct the model $$\mathcal{M}^\phi$$. It follows by Proposition 19 that $$\mathcal{M}^\phi$$ has the properties of $$\texttt{PR}$$ and $$\texttt{NM}$$. Moreover, it follows by Lemma 2 that $$\mathcal{M}^\phi,(\Delta,0)\vDash\phi$$. ■ 5 Conclusion This article presented a tableau-based procedure for EPDL with the properties of perfect recall and no miracles. Perfect recall and no miracles are important properties for capturing the interactions between knowledge and actions. In [1, 2], a tableau-based decision procedure for TEL with no forgetting and no learning was developed. The property of perfect recall in EPDL makes no great difference to the property of no forgetting in TEL. However, the property of no miracles in EPDL is more complex than the property of no learning in TEL. Compared to TEL, EPDL has branching actions, and accessibility relations are usually not serial. This makes the tableau procedure for EPDL with no miracles more complex than the tableau procedure for TEL with no learning. There are a number of issues to explore in future research. The decision procedure presented in this article is not guaranteed to have an optimal complexity. Relevantly, it is shown in [10] that single-agent TEL with perfect recall and no learning is EXPSPACE. What is more, it is shown in [7] that EPDL with perfect recall and no learning is decidable in coN2EXPTIME. We can improve our procedure and compare the complexity to see whether no miracles makes a real difference with no learning on complexity. As we mentioned at the beginning, the axiomatization and decidability of EPDL extended with various combinations of perfect recall, no learning and Church-Rosser properties are well studied in [18]. Therefore, another interesting direction is to investigate these issues of EPDL extended with no miracles and the other properties. Moreover, we can also extend and adapt this tableau construction for EPDL extended with the other combination of perfect recall, no miracles and no learning and Church-Rosser properties. Acknowledgements The author thanks Barteld Kooi and the two anonymous reviewers of this journal for their helpful comments on the earlier versions of the article. The author is grateful to Max Bialek who proofread the article. Footnotes *A preliminary version of this article appeared in the proceedings of LORI-V [14]. 1The sequences $$b$$, $$ba$$ and $$aaa$$ are not solutions since they are not executable at $$s_2$$. References [1] Ajspur M.. Tableau-based Decision Procedures for Epistemic and Temporal Epistemic Logics . PhD Thesis, Doctoral School of Communication, Business and Information Technologies, Roskilde University, Denmark, 10 2013. [2] Ajspur M. and Goranko V.. 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Journal of Logic and ComputationOxford University Press

Published: Mar 1, 2018

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