Public announcement logic with distributed knowledge: expressivity, completeness and complexity

Abstract

While dynamic epistemic logics with common knowledge have been extensively studied, dynamic epistemic logics with distributed knowledge have so far received far less attention. In this paper we study extensions of public announcement logic ($\mathcal{PAL }$) with distributed knowledge, in particular their expressivity, axiomatisations and complexity. $\mathcal{PAL }$ extended only with distributed knowledge is not more expressive than standard epistemic logic with distributed knowledge. Our focus is therefore on $\mathcal{PACD }$, the result of adding both common and distributed knowledge to $\mathcal{PAL }$, which is more expressive than each of its component logics. We introduce an axiomatisation of $\mathcal{PACD }$, which is not surprising: it is the combination of well-known axioms. The completeness proof, however, is not trivial, and requires novel combinations and extensions of techniques for dealing with $S5$ knowledge, distributed knowledge, common knowledge and public announcements at the same time. We furthermore show that $\mathcal{PACD }$ is decidable, more precisely that it is $\textsc {exptime}$-complete. This result also carries over to $\mathcal{S 5\mathcal CD }$ with common and distributed knowledge operators for all coalitions (and not only the grand coalition). Finally, we propose a notion of a trans-bisimulation to generalise certain results and give deeper insight into the proofs.

Appendices

Appendix A: Supporting definitions and intermediate results

The following until Lemma 43 are used in investigating the expressivity results (Sect. 3).

Definition 39

(Bisimulation) Given two models $\mathfrak{M }\!=\!{(M,{\sim ^{{\textsc {ag}}}},V)}$ and $\mathfrak{N }\!=\!(N,{\approx ^{{\textsc {ag}}}},\nu )$, a non-empty relation $Z\subseteq M\times N$ is called a bisimulation between $\mathfrak{M }$ and $\mathfrak{N }$, denoted by $\mathfrak{M }\rightleftarrows \mathfrak{N }$, if for all $a\in {\textsc {ag}}$, $m\in M$ and $n\in N$ with $mZn$, it holds that: (${\textsc {at}}$) for all $p\in \textsc {prop}$, $m\in V(p)$ iff $n\in \nu (p)$; (zig) for every $m^{\prime }\in M$, if $m\sim _a m^{\prime }$, there is an $n^{\prime }\in N$ such that $n\approx _a n^{\prime }$ and $m^{\prime }Zn^{\prime }$; and (zag) for every $n^{\prime }\in N$, if $n\approx _a n^{\prime }$, there is an $m^{\prime }\in M$ such that $m\sim _a m^{\prime }$ and $m^{\prime }Zn^{\prime }$. We say pointed models $(\mathfrak{M },m)$ and $(\mathfrak{N },n)$ are bisimilar, if there is a bisimulation $Z$ between $\mathfrak{M }$ and $\mathfrak{N }$ linking $m$ and $n$.

Proposition 40

Distributed knowledge does not preserve bisimulation.

Proof

Consider the following models:

It is easy to observe that $(\mathfrak{M },m)$ and $(\mathfrak{N },m)$ are bisimilar, and the formula $D_{ab}p$ is true at $(\mathfrak{M },m)$ but false at $(\mathfrak{N },m)$. $\square$

Definition 41

($\mathcal{S 5\mathcal CD }_{\mathcal{P }}$ game) Fix a set ${\mathcal{P }}$ of propositions. For any two pointed models $(\mathfrak{M },m)=(M,{\sim ^{{\textsc {ag}}}},V,m)$ and $(\mathfrak{N },n)=(N,{\approx ^{{\textsc {ag}}}},\nu ,n)$, the $r$-round $\mathcal{S 5\mathcal CD }_{\mathcal{P }}$ game between spoiler and duplicator on $(\mathfrak{M },m)$ and $(\mathfrak{N },n)$ is the following.

If $r=0$, spoiler wins iff $V(p)\ne \nu (p)$ for some $p\in {\mathcal{P }}$.

If $r\ne 0$, spoiler can initiate one of the following scenarios:

• $C$-forth Spoiler chooses a group $A$ and an $m^{\prime }\in M$ such that $m\sim _{C_A}m^{\prime }$. Duplicator responds by choosing an $n^{\prime }\in N$ such that $n\approx _{C_A}n^{\prime }$.

• $C$-back Spoiler chooses a group $A$ and an $n^{\prime }\in N$ such that $n\approx _{C_A}n^{\prime }$. Duplicator responds by choosing an $m^{\prime }\in M$ such that $m\sim _{C_A}m^{\prime }$.

• $D$-forth Spoiler chooses a group $A$ and an $m^{\prime }\in M$ such that $m\sim _{D_A}m^{\prime }$. Duplicator responds by choosing an $n^{\prime }\in N$ such that $n\approx _{D_A}n^{\prime }$.

• $D$-back Spoiler chooses a group $A$ and an $n^{\prime }\in N$ such that $n\approx _{D_A}n^{\prime }$. Duplicator responds by choosing an $m^{\prime }\in M$ such that $m\sim _{D_A}m^{\prime }$.

For any of the above scenarios, the rest of the game is the $(r-1)$-round $\mathcal{S 5\mathcal CD }_{\mathcal{P }}$ game on $(\mathfrak{M },m^{\prime })$ and $(\mathfrak{N },n^{\prime })$. If either player cannot perform an action prescribed above, that player loses.

Definition 42

($\mathcal{S 5\mathcal CD }$-degree) The degree of an $\mathcal{S 5\mathcal CD }$-formula is given by the function $d:\mathcal{S 5\mathcal CD }\rightarrow \mathbb{N }$, with $p \in \textsc {prop}$, $a \in {\textsc {ag}}$ and $A\in {\textsc {gr}}$:

$$\begin{aligned} \begin{array}{llll} d(p)=0 \quad d(\lnot \varphi )=d(\varphi ) \quad d(\varphi \wedge \psi )=\text{ max}(d(\varphi ),d(\psi ))\\ d(K_a\varphi ) = d(C_A\varphi )=d(D_A\varphi )=d(\varphi )+1. \end{array}\end{aligned}$$

Lemma 43

Let $(\mathfrak{M },m)$ and $(\mathfrak{N },n)$ be two pointed models and ${\mathcal{P }}$ be a finite set of propositional variables. Let $\mathcal{S 5\mathcal CD }_{\mathcal{P }}$ be the sublanguage of $\mathcal{S 5\mathcal CD }$ which uses only propositional variables in ${\mathcal{P }}$. Then the following are equivalent for all $r\in \mathbb{N }$:

• 1. Duplicator has a winning strategy for the $r$-round $\mathcal{S 5\mathcal CD }_{\mathcal{P }}$ game on $(\mathfrak{M },m)$ and $(\mathfrak{N },n)$.

• 2. $(\mathfrak{M },m)$ and $(\mathfrak{N },n)$ satisfy exactly the same $\mathcal{S 5\mathcal CD }_{\mathcal{P }}$-formulae of degree at most $r$.

Proof

The careful reader should be able to come up with a proof following a similar proof for $\mathcal{S 5\mathcal C }$ in (van Ditmarsch et al. (2007), Chapter 8). Definitions 42 and 43 shall replace corresponding ones in order to deal with the distributed knowledge operators. $\square$

Definition 44 and Proposition 45 are used throughout the completeness proof for $\mathcal{PACD }$ (Sect. 4.2.1).

Definition 44

(Pseudo semantics) Let $\mathfrak{M }={(M,{\sim ^{{\textsc {ag}}}},{\sim ^{{{\textsc {gr}}}}},V)}$ be a pre-model, and $m$ a state in $M$. Satisfaction at $(\mathfrak{M },m)$ is defined as follows:

$$ \begin{aligned}\begin{array}{lll} \mathfrak{M },m\models _\mathsf{{p}}p&{\text{ iff}}\,&m\in V(p)\\ \mathfrak{M },m\models _\mathsf{{p}}\lnot \varphi&{\text{ iff}}\,&\mathfrak{M },m\not \models _\mathsf{{p}}\varphi \\ \mathfrak{M },m\models _\mathsf{{p}}\varphi \wedge \psi&{\text{ iff}}\,&\mathfrak{M },m\models _\mathsf{{p}}\varphi \ \& \ \mathfrak{M },m\models _\mathsf{{p}}\psi \\ \mathfrak{M },m\models _\mathsf{{p}}K_a\varphi&{\text{ iff}}\,&(\forall n\in \mathfrak{M })(m\sim _an\Rightarrow \mathfrak{M },n\models _\mathsf{{p}}\varphi )\\ \mathfrak{M },m\models _\mathsf{{p}}C_A\varphi&{\text{ iff}}\,&(\forall n\in \mathfrak{M })(m\sim _{C_A}n\Rightarrow \mathfrak{M },n\models _\mathsf{{p}}\varphi )\\ \mathfrak{M },m\models _\mathsf{{p}}D_A\varphi&{\text{ iff}}\,&(\forall n\in \mathfrak{M })(m\sim _{D_A}n\Rightarrow \mathfrak{M },n\models _\mathsf{{p}}\varphi )\\ \mathfrak{M },m\models _\mathsf{{p}}[\varphi ]\psi&{\text{ iff}}\,&\mathfrak{M },m\models _\mathsf{{p}}\varphi \Rightarrow \mathfrak{M }|\varphi ,m\models _\mathsf{{p}}\psi ,\\ \end{array}\end{aligned}$$

where $\mathfrak{M }|\varphi =(N,{\approx ^{{\textsc {ag}}}},{\approx ^{{{\textsc {gr}}}}},\nu )$ is a pre-model such that

$$\begin{aligned}\begin{array}{lllllll} N=\{m\in M\ |\ \mathfrak{M },m\models _\mathsf{{p}}\varphi \},&\ \approx _a=\sim _a\cap (N\times N) \text{ for} a\in {\textsc {ag}},\\ \nu (p)=V(p)\cap N \text{ for} p\in \textsc {prop},&\, \approx _{D_A}=\sim _{D_A}\cap (N\times N) \text{ for} A\in {\textsc {gr}}.\\ \end{array}\end{aligned}$$

Satisfaction in a pre-model $\mathfrak{M }$ (denoted by $\mathfrak{M }\models _\mathsf{{p}}\varphi$) is defined as usual. We use $\models _\mathsf{{p}}\varphi$ to denote validity with respect to models, i.e., $\mathfrak{M },m\models _\mathsf{{p}}\varphi$ for any pointed model $(\mathfrak{M },m)$. We write $\models$ instead of $\models _\mathsf{{p}}$ when there is no confusion.

Proposition 45

If $\mathfrak{M }$ is a pseudo model, and $\varphi$ is a formula such that $\mathfrak{M },m\models \varphi$ for some $m\in \mathfrak{M }$, then $\mathfrak{M }|\varphi$ is also a pseudo model.

Theorem 46

(Pseudo soundness) Any PACD-theorem is valid in the class of all pseudo models.

Theorem 47

(Closure). Given a formula $\varphi$, the closure of $\varphi$ is given by the function $cl:\mathcal{PACD }\rightarrow \wp (\mathcal{PACD })$ which is defined as follows:

• 1. $\varphi \in cl(\varphi )$, and if $\psi \in cl(\varphi )$, so are all of its subformulae;

• 2. If $\varphi$ is not a negation, then $\varphi \in cl(\varphi )\Rightarrow \lnot \varphi \in cl(\varphi )$;

• 3. $K_a\psi \in cl(\varphi )$ iff $D_a\psi \in cl(\varphi )$;

• 4. $C_A\psi \in cl(\varphi )\Rightarrow \{K_aC_A\psi \ |\ a\in A\}\subseteq cl(\varphi )$;

• 5. $[\psi ]C_A\chi \in cl(\varphi )\Rightarrow (([\psi ]\chi )\in cl(\varphi ) \& \{K_a[\psi ]C_A\chi \ |\ a\in A\}\subseteq cl(\varphi ))$;

• 6. For all of the six reduction axioms, such as R$_{[]p}$and R$_{[]\lnot }$, if the left-hand side of the equivalence is in $cl(\varphi )$, so is the right-hand side. That is, for each reduction axiom of the form $\alpha \,{\leftrightarrow }\,\beta : \alpha \in cl(\varphi )\Rightarrow \beta \in cl(\varphi )$.

It is easy to observe that the closure of a formula is finite, and a stronger result is shown in Proposition 30.

Lemmas 48 and 49, Definition 50 and Proposition 51 are mainly introduced for the proof of the Pseudo Truth Lemma (Lemma 15).

Lemma 48

Let ${\mathcal{S }}=\{\Gamma \ |\ \Gamma \text{ is} \text{ maximal} \text{ consistent} \text{ in} cl(\alpha )\}$ with $\alpha$ a formula. It holds that $\vdash \bigvee _{\Gamma \in {\mathcal{S }}}\underline{\Gamma }$  and $\vdash \varphi {\leftrightarrow }\bigvee _{\varphi \in \Gamma \in {\mathcal{S }}}\underline{\Gamma }$  for any $\varphi \in cl(\alpha )$.

Lemma 49

If $\Gamma$ and $\Delta$ are maximal consistent in $cl(\alpha )$, then

• 1. $\Gamma$ is deductively closed in $cl(\alpha )$, i.e., $\Gamma \vdash \varphi {\Leftrightarrow }\varphi \in \Gamma$ for any $\varphi \in cl(\alpha )$;

• 2. If $\lnot \varphi \in cl(\alpha )$, then $\varphi \in \Gamma {\Leftrightarrow }\lnot \varphi \notin \Gamma$;

• 3. If $\varphi \wedge \psi \in cl(\alpha )$, then $\varphi \wedge \psi \in \Gamma {\Leftrightarrow }\varphi \in \Gamma \& \psi \in \Gamma$;

• 4. If $\underline{\Gamma }\wedge \hat{K}_a\underline{\Delta }$ is consistent, $\Gamma \sim _a^c\Delta$; if $\underline{\Gamma }\wedge \hat{D}_A\underline{\Delta }$ is consistent, $\Gamma \sim _{D_A}^c\Delta$;

• 5. If $K_a\psi \in cl(\alpha )$, then $K_a\Gamma \vdash \psi {\Leftrightarrow }K_a\Gamma \vdash K_a\psi$;

• 6. If $D_A\psi \in cl(\alpha )$, then $D_A\Gamma \vdash \psi {\Leftrightarrow }D_A\Gamma \vdash D_A\psi$;

• 7. If $C_A\varphi \in cl(\alpha )$, then $C_A\varphi \in \Gamma {\Leftrightarrow }\forall \Delta .(\Gamma \sim ^c_{C_A}\Delta \Rightarrow \varphi \in \Delta )$;

• 8. If $[\varphi ]C_A\psi \in cl(\alpha )$, then $[\varphi ]C_A\psi \in \Gamma$ iff every $A$-$\varphi$-path from $\Gamma$ is a $[\varphi ]\psi$-path.

Definition 50

(Complexity of formulae). The complexity of a formula is defined by the function $c:\mathcal{PACD }\rightarrow \mathbb{N }$:

$$\begin{aligned} \begin{array}{l} c(p)=1,\qquad c(\lnot \varphi )=c(K_a\varphi )=c(C_A\varphi )=c(D_A\varphi )=1+c(\varphi ),\\ c([\varphi ]\psi )=(4+c(\varphi ))\cdot c(\psi ),\qquad c(\varphi \wedge \psi )=1+\text{ max}(c(\varphi ),c(\psi )).\\ \end{array}\end{aligned}$$

Proposition 51

For any formulae $\varphi$, $\psi$ and $\chi$,

• 1. $c(\varphi )\ge c(\psi )$, if $\psi$ is a subformula of $\varphi$

• 2. $c([\varphi ]p)>c(\varphi \rightarrow p)$

• 3. $c([\varphi ]\lnot \psi )>c(\varphi \rightarrow \lnot [\varphi ]\psi )$

• 4. $c([\varphi ](\psi \wedge \chi ))>c([\varphi ]\psi \wedge [\varphi ]\chi )$

• 5. $c([\varphi ]K_a\psi )>c(\varphi \rightarrow K_a[\varphi ]\psi )$

• 6. $c([\varphi ]C_A\psi )>c([\varphi ]\psi )$

• 7. $c([\varphi ]D_A\psi )>c(\varphi \rightarrow D_A[\varphi ]\psi )$

• 8. $c([\varphi ][\psi ]\chi )>c([\varphi \wedge [\varphi ]\psi ]\chi )$.

Appendix B: Proofs of Theorems

Proof of Lemma 15

We show this lemma by induction on $c(\varphi )$ (see Definition 50). Despite the slight difference that we are using pseudo models here, the cases for Boolean operators, individual knowledge operators, common knowledge operators, and public announcement operators are like those cases in the proof of the Truth Lemma for $\mathcal{PAC }$ ((van Ditmarsch et al. 2007, p. 193)) (extensions and adaptions of supporting definitions and lemmas can be found in the “Appendix A”; the only missing subcase is $[\psi ]D_A\chi$ which can be easily shown). Thus, we only discuss the case when $\varphi$ is of the form $D_A\psi$.

From left to right. Suppose $D_A\psi \in \Gamma$. Take an arbitrary $\Delta \in M^c$. Suppose that $\Gamma \sim _{D_A}^c\Delta$. So $D_A\psi \in \Delta$ by the definition of $\sim ^c_{D_A}$. Since $\vdash D_A\psi \rightarrow \psi$ (Axiom T$_D$), and $\Delta$ is deductively closed, it must be the case that $\psi \in \Delta$. This is equivalent to $\mathfrak{M }^c,\Delta \models \psi$ by IH. Since all the $\Delta$s “$\sim ^c_{D_A}$-reachable” from $\Gamma$ forces $\psi$, therefore $\mathfrak{M }^c,\Gamma \models D_A\psi$ by the semantics.

From right to left. Suppose $\mathfrak{M }^c,\Gamma \models D_A\psi$. Then $\mathfrak{M }^c,\Delta \models \psi$ for any $\Delta$ such that $\Gamma \sim _{D_A}^c\Delta$. We must show $D_A\psi \in \Gamma$. Suppose not, then $\lnot D_A\psi \in \Gamma$, i.e., $\hat{D}_A\lnot \psi \in \Gamma$. Hence $\underline{\Gamma }\wedge \hat{D}_A\lnot \psi$ is consistent, and so $\underline{\Gamma }\wedge \hat{D}_A\bigvee _{\lnot \psi \in \Theta \in {M^c}}\underline{\Theta }$ is consistent by Lemma 48. This is equivalent to that $\bigvee _{\lnot \psi \in \Theta \in {M^c}}(\underline{\Gamma }\wedge \hat{D}_A\underline{\Theta })$ is consistent. It entails that there is a $\Psi \in M^c$ such that $\lnot \psi \in \Psi$ and $\underline{\Gamma }\wedge \hat{D}_A\underline{\Psi }$ is consistent. From $\lnot \psi \in \Psi$, we have $\psi \notin \Psi$, and therefore $\mathfrak{M }^c,\Psi \not \models \psi$ by induction hypothesis. Hence $\mathfrak{M }^c,\Psi \models \lnot \psi$. From the fact that $\underline{\Gamma }\wedge \hat{D}_A\underline{\Psi }$ is consistent, we have $\Gamma \sim _{D_A}^c\Psi$ by Lemma 49(4). Therefore $\mathfrak{M }^c,\Gamma \models \hat{D}_A\lnot \psi$, which contradicts the assumption that $\mathfrak{M }^c,\Gamma \models D_A\psi$.

$\square$

Proof of Lemma 21

Let $\mathfrak{X }\!=\!\langle {{\overline{x_{0}}},R_{\tau _0},\ldots ,R_{\tau _{s-1}},{\overline{x_{s}}}}\rangle$ and $\mathfrak{Y }\!=\!\langle {\overline{y_{0}}},$ $R_{\theta _0},\ldots ,R_{\theta _{t-1}},{\overline{y_{t}}}\rangle$ be two reduced paths of $\mathfrak{T }^\mathfrak{M }$ from ${\overline{m}}$ to ${\overline{n}}$. It suffices to show that $\mathfrak{X }$ and $\mathfrak{Y }$ are identical (including states and relations).

First, we show that if the states at the same position of the paths match, then $\mathfrak{X }$ and $\mathfrak{Y }$ are identical. In other words, if $\mathfrak{X }\ne \mathfrak{Y }$, then there must be some $i\ (1\le i< s)$ such that ${\overline{x_{i}}}\ne {\overline{y_{i}}}$. Suppose not, then ${\overline{x_{i}}}={\overline{y_{i}}}$ for every $i\ (0\le i\le s)$. It suffices to show that $R_{\tau _i}=R_{\theta _i}$ for every $i\ (0\le i\le s-1)$, as this will lead to a contradiction that $\mathfrak{X }=\mathfrak{Y }$. Fix an $i\ (0\le i\le s-1)$. Since ${\overline{x_{i}}}R_{\tau _i}{\overline{x_{i+1}}}$, we consider the following four cases:

• (1) ${\overline{x_{i}}}={\overline{x_{i+1}}}$. This is not a possible situation, as $\mathfrak{X }$ is a reduced path.

• (2) ${\overline{x_{i}}}$ $\tau _i$-grafts ${\overline{x_{i+1}}}$. Since ${\overline{x_{i}}}={\overline{y_{i}}}$ and ${\overline{x_{i+1}}}={\overline{y_{i+1}}}$, ${\overline{y_{i}}}$ must also $\tau _i$-graft ${\overline{y_{i+1}}}$. Therefore $R_{\tau _i}=R_{\theta _i}$.

• (3) ${\overline{x_{i}}}$ $\tau _i$-extends ${\overline{x_{i+1}}}$. ${\overline{y_{i}}}$ must also $\tau _i$-extend ${\overline{y_{i+1}}}$. Therefore $R_{\tau _i}=R_{\theta _i}$.

• (4) ${\overline{x_{i+1}}}$ $\tau _i$-extends ${\overline{x_{i}}}$. Similarly $R_{\tau _i}=R_{\theta _i}$ holds.

Therefore, it suffices to show that the states at the same position of the paths match. We first show that ${\overline{x_{n}}}={\overline{y_{n}}}$ for $0\le n\le \min (s,t)$. Suppose ${\overline{x_{i}}}={\overline{y_{i}}}\ (0\le i<\min (s,t))$, and we explore the different cases of ${\overline{x_{i+1}}}$:

• (1) ${\overline{x_{i}}}={\overline{x_{i+1}}}$. Impossible.

• (2) ${\overline{x_{i+1}}}$ $\tau _i$-extends ${\overline{x_{i}}}$. By Lemma 20(1), every node ${\overline{x_{j}}}\ (i<j\le s)$ extends its previous node, and therefore contains ${\overline{x_{i}}}$ as an initial segment. In particular, ${\overline{x_{s}}}$ contains ${\overline{x_{i}}}$ as an initial segment. We show that ${\overline{y_{i+1}}}$ must $\theta _i$-extend ${\overline{y_{i}}}$ by excluding other cases:

  • (2a) It is impossible to have ${\overline{y_{i+1}}}={\overline{y_{i}}}$ since $\mathfrak{Y }$ is a reduced path.

  • (2b) If ${\overline{y_{i+1}}}$ $\theta _i$-grafts ${\overline{y_{i}}}$, then by Lemma 20(2), all the nodes that follow ${\overline{y_{i+1}}}$ does not contain ${\overline{y_{i}}}$ as an initial segment. In particular, ${\overline{y_{t}}}$ does not contain ${\overline{y_{i}}}$ as an initial segment, which leads to a contradiction that ${\overline{x_{s}}}\ne {\overline{y_{t}}}$. (This says that “being growing does not match with being grafted”.)

  • (2c) If ${\overline{y_{i}}}$ $\theta _i$-extends ${\overline{y_{i+1}}}$, then by Lemma 20(3), all the nodes that follow ${\overline{y_{i+1}}}$ (including itself) does not contain ${\overline{y_{i}}}$ as an initial segment. In particular, ${\overline{y_{t}}}$ does not contain ${\overline{y_{i}}}$ as an initial segment, which leads to a contradiction that ${\overline{x_{s}}}\ne {\overline{y_{t}}}$. (This says that “being growing does not match with being withering”.)

  So it can only happen that ${\overline{y_{i+1}}}$ $\theta _i$-extends ${\overline{y_{i}}}$. Furthermore, it must be the case that $x_{i+1}=y_{i+1}$, otherwise ${\overline{y_{j}}}\ (j>i)$ will always contain different initial segments, leading to a contradiction that ${\overline{x_{s}}}\ne {\overline{y_{t}}}$.

• (3) ${\overline{x_{i+1}}}$ $\tau _i$-grafts ${\overline{x_{i}}}$. By Lemma 20(2), every node ${\overline{x_{j}}}\ (i+1<j\le s)$ extends its previous node, but does not contain ${\overline{x_{i}}}$ as an initial segment. In particular, ${\overline{x_{s}}}$ contains ${\overline{x_{i+1}}}$ but not ${\overline{x_{i}}}$ as an initial segment. We can show that ${\overline{y_{i+1}}}$ must $\theta _i$-graft ${\overline{y_{i}}}$ also by excluding other cases; the details are omitted here. Furthermore, it must be the case that $x_{i+1}=y_{i+1}$, otherwise ${\overline{y_{j}}}\ (j>i+1)$ will always contain different initial segments, leading to a contradiction that ${\overline{x_{s}}}\ne {\overline{y_{t}}}$.

• (4) ${\overline{x_{i}}}$ $\tau _i$-extends ${\overline{x_{i+1}}}$. By Lemma 20(3), every node ${\overline{x_{j}}}\ (i<j\le s)$ does not contain ${\overline{x_{i}}}$ as an initial segment. In particular, ${\overline{x_{s}}}$ does not contain ${\overline{x_{i}}}$ as an initial segment. We can similarly show that ${\overline{y_{i+1}}}$ $\theta _i$-graft ${\overline{y_{i}}}$ by excluding other cases, details of which are omitted. Thus, $x_{i+1}=y_{i+1}$.

The above shows that ${\overline{x_{n}}}={\overline{y_{n}}}$ for $0\le n\le \min (s,t)$. If $s=t$, then we have $\mathfrak{X }=\mathfrak{Y }$, as we want. Finally, it suffices to show that $s\ne t$ leads to a contradiction. Suppose without loss of generality that $s<t$. From the above we get ${\overline{x_{s}}}={\overline{y_{s}}}$, and we show ${\overline{y_{s}}}\ne {\overline{y_{t}}}$ by exploring all the cases of ${\overline{y_{s+1}}}$:

• (1) It is impossible to have ${\overline{y_{s+1}}}={\overline{y_{s}}}$.

• (2) If ${\overline{y_{s+1}}}\ \theta _s$-extends ${\overline{y_{s}}}$, then every ${\overline{y_{x}}}\ (s<x\le t)$ extends its previous node by Lemma 20(1). Therefore ${\overline{y_{t}}}\ne {\overline{y_{s}}}$.

• (3) If ${\overline{y_{s+1}}}\ \theta _s$-grafts ${\overline{y_{s}}}$, then every ${\overline{y_{x}}}\ (s+1<x\le t)$ does not contain ${\overline{y_{s}}}$ as an initial segment by Lemma 20(2). Therefore ${\overline{y_{t}}}\ne {\overline{y_{s}}}$.

• (4) If ${\overline{y_{s}}}\ \theta _s$-extends ${\overline{y_{s+1}}}$, then every ${\overline{y_{x}}}\ (s<x\le t)$ does not contain ${\overline{y_{s}}}$ as an initial segment, by Lemma 20(3). Therefore ${\overline{y_{t}}}\ne {\overline{y_{s}}}$.

$\square$

Proof of Lemma 27

Let $Z=\{(m,{\overline{m}})\ |\ m\in \mathfrak{M }\}$ and let $\tau$ be of the type $a$ or $D_A$. We show that $Z:(\mathfrak{M },m)\,{\rightleftarrows }\,(\mathfrak{T }^\mathfrak{M },{\overline{m}})$.

• (AT) For every proposition $p$, $\mathfrak{M },m\models p$ iff $\mathfrak{T }^\mathfrak{M },{\overline{m}}\models p$ by definition.

• (zig) If $m\sim _\tau n$ for some $n\in \mathfrak{M }$, then ${\overline{m}}R_\tau {\overline{n}}$ and $nZ{\overline{n}}$ also by definition.

• (zag) If ${\overline{m}}R_\tau {\overline{n}}$ for some ${\overline{n}}\in \mathfrak{T }^\mathfrak{M }$, we explore the following possible cases:

  • ${\overline{m}}={\overline{n}}$. It follows that $m=n$, which entails $m\sim _\tau n$.

  • ${\overline{m}}$ $\tau$-grafts ${\overline{n}}$. It follows that ${\overline{m}}=\langle {{\overline{x}}\sim _\tau m}\rangle$ and ${\overline{n}}=\langle {{\overline{x}}\sim _\tau n}\rangle$ for some ${\overline{x}}$. Then $m\sim _\tau x\sim _\tau n$, and so $m\sim _\tau n$ by transitivity of $\sim _\tau$.

  • ${\overline{m}}$ $\tau$-extends ${\overline{n}}$. It follows that ${\overline{m}}=\langle {{\overline{n}}\sim _\tau m}\rangle$, and so $n\sim _\tau m$. We get $m\sim _\tau n$ by symmetry of $\sim _\tau$.

  • ${\overline{n}}$ $\tau$-extends ${\overline{m}}$. It follows that ${\overline{n}}=\langle {{\overline{m}}\sim _\tau n}\rangle$, and so $m\sim _\tau n$.

  In all cases we have $m\sim _\tau n$. Besides, $nZ{\overline{n}}$ holds by definition. $\square$

Proof of Lemma 36

For the left to right direction, if $\varphi$ is satisfiable, then it must be consistent, and thus belong to a maximal consistent set $\Phi$ in $cl(\alpha )$, where $\Phi$ is a state of $\mathfrak{M }^h$ by Lemma 35. For the right to left direction, we show a stronger result that

$$\begin{aligned} \text{ for} \text{ any} \text{ state} \Phi \text{ of} \mathfrak{M }^h, \varphi \in \Phi \text{ iff} \mathfrak{M }^h,\Phi \models \varphi . \end{aligned}$$

(1)

This looks similar to the Pseudo Truth Lemma, and its proof is more or less related. Let $\mathfrak{M }^h={(M,{\sim ^{{\textsc {ag}}}},{\sim ^{{{\textsc {gr}}}}},V)}$. We show (1) by induction on $c(\varphi )$. The base case holds by the definition of the evaluation function in $\mathfrak{M }^h$, and the Boolean cases are guaranteed by Boolean consistency of Hintikka sets.

Consider the case for $K_a\psi$. For the left to right direction, suppose $K_a\psi \in \Phi$. For all $\Psi \in M$ such that $\Phi \sim _a\Psi$, we have $\psi \in \Psi$ by Proposition 34(1). By the induction hypothesis (IH), $\mathfrak{M }^h,\Psi \models \psi$, and hence $\mathfrak{M }^h,\Phi \models K_a\psi$. For the right to left direction, suppose $\mathfrak{M }^h,\Phi \models K_a\psi$. It follows that $\mathfrak{M }^h,\Psi \models \psi$ holds for any state $\Psi$ of $\mathfrak{M }^h$ such that $\Phi \sim _a\Psi$. By IH, $\psi \in \Psi$. From the algorithm of eliminating Hintikka sets, we observe that $K_a\psi \in \Phi$. Hence, $\psi \in \Psi$ holds by IH. The case for $D_A\psi$ is very similar.

Now we consider the case for $C_A\psi$. It is similar to the above for the right to left direction. As for the left to right direction, suppose $C_A\psi \in \Phi$. It suffices to show that $C_A\psi \in \Psi _j(0\le j\le n)$ for any path $\Psi _0\sim _{a_0}\cdots \sim _{a_{n-1}}\Psi _n$ of $\mathfrak{M }^h$ such that $\Psi _0=\Phi$ and $\{a_0,\ldots ,a_{n-1}\}\subseteq A$, as this entails $\psi \in \Psi _j$ and thus $\mathfrak{M }^h,\Phi \models C_A\psi$ by IH. Since $\Psi _0$ respects CK2, $K_{a_0}C_A\psi \in \Psi _0$, Therefore, $C_A\psi \in \Psi _1$ by Proposition 34(1). In such a way, we can show $C_A\psi \in \Psi _j$ for all $0\le j\le n$.

As to the seven cases for $[\chi ]\psi$, we make use of the fact that Hintikka sets respect the reduction axioms, and it can be shown in the same way as in the proof of the Pseudo Truth Lemma. (In the case for $[\chi ]C_A\psi$, a counterpart of Lemma 49(8) is ensured by the construction of $\mathfrak{M }^h$.)$\square$

Proof of Theorem 38

Let $\mathfrak{T }^\mathfrak{M }=(T,{R^{{{\textsc {ag}}}}},{R^{{{\textsc {gr}}}}},V)$ be the unravelled treelike pre-model of $\mathfrak{M }$. By Lemma 27 and the invariance of bisimulation, we have for any $\varphi$ and ${\overline{m}}$, $\mathfrak{M },m\models \varphi$ iff $\mathfrak{T }^\mathfrak{M },{\overline{m}}\models \varphi$. Thus, we get a natural equivalence relation, , on $T$ such that iff $m=n$. We write m for the equivalence class of the state ${\overline{m}}$. Let $\mathtt{T }=(U,Q^{\textsc {ag}},\upsilon )$ be the model such that

$$\begin{aligned} \begin{array}{rcl} U&= \{{\mathtt{m }}\ |\ {\overline{m}}\in T\}\\ Q_a&= \{(\mathtt{m },\mathtt n )\ |\ \exists {\overline{m}}\exists {\overline{n}}.\ ({\overline{m}},{\overline{n}})\in (R_a^\cup )^*\}\\ \upsilon (p)&= \{{\mathtt{m }}\ |\ {\overline{m}}\in V(p)\}.\\ \end{array} \end{aligned}$$

We show that $Z=\{(\mathtt m ,{\overline{m}})\ |\ {\overline{m}}\in T\}$ is a trans-bisimulation between $\mathtt{T }$ and $\mathfrak{T }^\mathfrak{M }$. The condition (${\textsc {at}}$) holds trivially by definition.

For (zig$_a$) , suppose $\mathtt{m }Q_a\mathtt{n }$ or $\mathtt{m }Q_{D_a}\mathtt{n }$. Then, there are ${\overline{m}}$ and ${\overline{n}}$ such that $({\overline{m}},{\overline{n}})\in (R_a^\cup )^*$, i.e., there is a sequence ${\overline{m}}R_a^\cup \cdots R_a^\cup {\overline{n}}$. It follows that there is a sequence ${\overline{m}}R_{\tau _0}\cdots R_{\tau _x}{\overline{n}}$ with each $\tau _i$ being $a$ or $D_A$ such that $a\in A$.

For (zig$_A$), suppose $\mathtt{m }Q_{D_A}\mathtt{n }$ with $|A|\ge 2$. It follows that $(\mathtt{m },\mathtt{n })\in (Q_{a_1}\cap \cdots \cap Q_{a_j})$ with $A=\{a_1,\ldots ,a_j\}$. Therefore, there are ${\overline{m}}_i$ and ${\overline{n}}_i$ such that $({\overline{m}}_i,{\overline{n}}_i)\in (R_{a_i}^\cup )^*$ for each $1\le i\le j$. It follows that there are $j$ reduced paths of $\mathfrak{T }^\mathfrak{M }$ such that:

$$\begin{aligned} \begin{array}{c} {\overline{m}}_1R_{\tau _{1,1}}\cdots R_{\tau _{1,x_1}}{\overline{n}}_1\\ \vdots \\ {\overline{m}}_jR_{\tau _{j,1}}\cdots R_{\tau _{j,x_j}}{\overline{n}}_j \end{array} \end{aligned}$$

where each $\tau _{i,k}$ ($1\le i\le j$ and $1\le k\le x_i$) is $a_i$ or $D_{A_i}$ such that $a_i\in A_i$. Since $\mathfrak{T }^\mathfrak{M }$ is treelike, there can only be a unique reduced path from ${\overline{m}}$ to ${\overline{n}}$. Moreover, each $\sim _{\tau _{i,k}}$ cannot be $a_i$ and has to be some $D_{A_i}$ such that $A\subseteq A_i$, for this is the only way to have a unique reduced path. Hence, there is a path ${\overline{m}}R_{D_{A_0}}\cdots R_{D_{A_x}}{\overline{m}}$ such that $A\subseteq A_0\cap \cdots \cap A_x$, for some ${\overline{m}}$ and ${\overline{n}}$.

For (zag), suppose ${\overline{m}}R_\tau {\overline{n}}$. If $\tau$ is some $a\in {\textsc {ag}}$, it follows by definition that $\mathtt{m }Q_a\mathtt{n }$. Otherwise, if $\tau$ is some $D_A$, it follows by definition that ${\overline{m}}R_a{\overline{n}}$, and also $\mathtt{m }Q_a\mathtt{n }$, for all $a\in A$; and thus $\mathtt{m }Q_{D_A}\mathtt{n }$.

Now, since $(\mathtt{T },\mathtt m )\,{\rightleftarrows ^{\textsc {t}}}\,(\mathfrak{T }^\mathfrak{M },{\overline{m}})\,{\rightleftarrows }\,(\mathfrak{M },m)$, we have $(\mathtt{T },\mathtt m )\,{\equiv ^{\textsc {t}}}\,(\mathfrak{T }^\mathfrak{M },{\overline{m}})$ by Lemma 26. Therefore $(\mathfrak{M },m)\,{\equiv ^{\textsc {t}}}\,(\mathtt{T },\mathtt m )$ by Lemma 27. $\square$