Neighborhood semantics for logic of knowing how

Abstract

In this paper, we give an alternative semantics to the non-normal logic of knowing how proposed by Fervari et al. (Proc IJCAI 2017:1031–1038, 2017), based on a class of Kripke neighborhood models with both the epistemic relations and neighborhood structures. This alternative semantics is inspired by the same quantifier alternation pattern of $\exists \forall$ in the semantics of the know-how modality and the (monotonic) neighborhood semantics for the standard modality. We show that this new semantics is equivalent to the original Kripke semantics in terms of the validities. A key result is a representation theorem showing that the more abstract Kripke neighborhood models can be represented by the concrete Kripke models with action transitions modulo the valid formulas. We prove the completeness of the logic for the neighborhood semantics. The neighborhood semantics can be adapted to other variants of logics of knowing how. It provides us a powerful technical tool to study these logics while preserving the basic semantic intuition.

A Appendix

1.1 A.1 Proof of Proposition 7.

Proof

If $s\not \in \mathtt {Dom}(\sigma )$, then $\mathtt {CELeaf}(\sigma ,[s]^i)=[s]^i$. The proposition holds immediately. Next we focus on the case of $s\in \mathtt {Dom}(\sigma )$.

According to the well-ordering theorem (equivalent to Axiom of Choice), we assume $\mathtt {CELeaf}(\sigma ,[s]^i)=\{s_\xi \mid \xi < \gamma \}$ where $\gamma$ is an ordinal number and $\gamma \ge 1$.

This implies that for each $s_\xi$ there exists a strategy $\sigma _\xi$ of agent i such that all $\sigma _\xi$’s complete executions from states in $[s_\xi ]^i$ are finite $(\pmb \ddagger )$. Next, we will firstly define a strategy $\tau$ of agent i and then show that all complete $\tau$-executions from $s'\in [s]^i$ are finite and $\mathtt {CELeaf}(\tau ,[s]^i)\subseteq \bigcup _{\xi <\gamma }\mathtt {CELeaf}(\sigma _\xi ,[s_\xi ]^i)$. The definition of $\tau$ consists of the following steps.

Step I. By induction on $\xi$, we define a set of strategies $\tau _\xi$ where $0\le \xi < \gamma$ as follows:

• for $\xi =0$, $\tau _0=\sigma _0|_{\{[w]^i\mid w\in \mathtt {CEInner}(\sigma _0,[s_0]^i)\}}$;

• for each $\xi >0$, $\tau _\xi =f_\xi \cup (\sigma _\xi |_{D_\xi })$ where $f_\xi =\bigcup _{\beta < \xi }\tau _\beta$ and $D_\xi =\{[w]^i\mid w\in \mathtt {CEInner}(\sigma _{\xi },[s_{\xi }]^i)\}\setminus (\mathtt {dom}(f_\xi )\cup \{[v]^i\mid v\in \mathtt {CELeaf}(f_\xi ,\mathtt {Dom}(f_\xi )) \})$.

$\square$

Claim 1

We have the following results:

• (1.) For each $0\le \xi < \gamma$, $\tau _j\subseteq \tau _\xi$ if $j< \xi$;

• (2.) For each $0\le \xi < \gamma$, $\tau _\xi$ is a strategy of agent i;

• (3.) For each $0\le \xi < \gamma$, if $j< \xi$ then $\mathtt {Dom}(\tau _\xi )\cap \mathtt {CELeaf}(\tau _j,\mathtt {Dom}(\tau _j))=\emptyset$;

• (4.) For each $0\le \xi < \gamma$, if $\delta =t_0t_1\cdots$ is a complete $\tau _\xi$-execution from a state $t\in \mathtt {Dom}(\tau _\xi )$, we have that $\delta =t_0\cdots t_n$ for some $n\in \mathbb {N}$ and $t_n\in \bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$;

• (5.) For each $0{\le } \xi < \gamma$, either $[s_\xi ]^i{\in }\mathtt {dom}(\tau _\xi )$ or $[s_\xi ]^i\subseteq (\bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$.

Proof of claim

(1.) It is obvious.

(2.) We prove it by induction on $\xi$. For the case of $\xi =0$, we only need to show that $\{[w]^i\mid w\in \mathtt {CEInner}(\sigma _0,[s_0]^i)\}$ is a subset of $\mathtt {dom}(\sigma _0)$. For each $w\in \mathtt {CEInner}(\sigma _0,s_0)$, this implies that there is a non-complete $\sigma _0$-execution $t_0\cdots t_n$ such that $t_n=w$. Since this execution is non-complete, this implies that $[w]^i\in \mathtt {dom}(\sigma _0)$.

For the case of $\xi >0$, it follows by IH that, for each $\beta <\xi$, $\tau _\beta$ is a strategy of agent i. It follows by (1) that $\tau _{\beta '}\subseteq \tau _{\beta }$ if $\beta '<\beta <\xi$. Thus, we have $f_\xi =\bigcup _{\beta <\xi }\tau _\beta$ is a strategy of agent i. Furthermore, since $\sigma _\xi$ is a strategy of agent i and $D_\xi \subseteq \mathtt {dom}(\sigma _\xi )$, in order to show $\tau _\xi$ is also a strategy of agent i, we only need to show that $\mathtt {dom}(f_\xi )\cap D_\xi =\emptyset$. Since $D_\xi =\{[w]^i\mid w\in \mathtt {CEInner}(\sigma _{\xi },[s_{\xi }]^i)\}\setminus \mathtt {dom}(f_\xi )\setminus \{[v]^i\mid v\in \mathtt {CELeaf}(f_\xi ,\mathtt {Dom}(f_\xi )) \}$, it is obvious that $\mathtt {dom}(f_\xi )\cap D_\xi =\emptyset$.

(3.) We prove it by induction on $\xi$. It is obvious for the case of $\xi =0$. For the case of $\xi >0$, we need to show that $\mathtt {Dom}(\tau _\xi )\cap \mathtt {CELeaf}(\tau _j,\mathtt {Dom}(\tau _j))=\emptyset$ where $j<\xi$. Supposing $v\in \mathtt {CELeaf}(\tau _j,\mathtt {Dom}(\tau _j))$, we will show that $[v]^i\not \in \mathtt {dom}(\tau _\xi )$, namely $[v]^i\not \in \mathtt {dom}(f_\xi )\cup D_\xi$. Firstly, we will show that $[v]^i\not \in \mathtt {dom}(f_\xi )$. Assuming $[v]^i\in \mathtt {dom}(f_\xi )$, it follows that $[v]^i\in \mathtt {dom}(\tau _\beta )$ for some $\beta <\xi$. There are two cases: $j<\beta$ or $j\ge \beta$. If $j<\beta$, it follows by IH that $\mathtt {Dom}(\tau _\beta )\cap \mathtt {CELeaf}(\tau _j,\mathtt {Dom}(\tau _j))=\emptyset$. Contradiction. If $j\ge \beta$, it follows by (1) that $\tau _\beta \subseteq \tau _j$. Due to $[v]^i\in \mathtt {dom}(\tau _\beta )$, it follows $[v]^i\in \mathtt {dom}(\tau _j)$. It is contradictory with $v\in \mathtt {CELeaf}(\tau _j,\mathtt {Dom}(\tau _j))$.

Next, we continue to show that $[v]^i\not \in D_\xi$. Since we know that

$$\begin{aligned} D_\xi =\{[w]^i\mid w\in \mathtt {CEInner}(\sigma _{\xi },[s_{\xi }]^i)\}\setminus (\mathtt {dom}(f_\xi )\cup \{[w]^i\mid w\in \mathtt {CELeaf}(f_\xi ,\mathtt {Dom}(f_\xi ))\}), \end{aligned}$$

we only need to show that $v\in \mathtt {CELeaf}(f_\xi ,\mathtt {Dom}(f_\xi ))$. Since $j<\xi$ and $f_\xi =\bigcup _{\beta <\xi }\tau _\xi$, it follows that $\tau _j\subseteq f_\xi$. Thus, each execution of $\tau _j$ is also an execution of $f_\xi$. Since $v\in \mathtt {CELeaf}(\tau _j,\mathtt {Dom}(\tau _j))$, we have that there is an execution $\delta$ of $\tau _j$ such that $\delta$ ends with v. Thus, $\delta$ is also an execution of $f_\xi$. We have shown that $[v]^i\not \in \mathtt {dom}(f_\xi )$, thus we have that $\delta$ is a complete execution of $f_\xi$. Thus, we have $v\in \mathtt {CELeaf}(f_\xi ,\mathtt {Dom}(f_\xi ))$. This implies $[v]^i\not \in D_\xi$. Thus, we have $[v]^i\not \in \mathtt {dom}(\tau _\xi )$.

(4.) We prove it by induction on $\xi$. For the case of $\xi =0$, we have that $\delta =t_0t_1\cdots$ is a complete $\tau _0$-execution from $t\in \mathtt {Dom}(\tau _0)$. We need to show that $\delta =t_0\cdots t_n$ for some $n\in \mathbb {N}$ and $t_n\in \mathtt {CELeaf}(\sigma _0,[s_0]^i)$. Due to $\mathtt {dom}(\tau _0)=\{[w]^i\mid w\in \mathtt {CEInner}(\sigma _0,[s_0]^i)\}$, it follows from $t\in \mathtt {Dom}(\tau _0)$ that there exists $w\in \mathtt {CEInner}(\sigma _0,[s_0]^i)$ such that $t\in [w]^i$. Since $w\in \mathtt {CEInner}(\sigma _0,[s_0]^i)$, this implies that there is a non-complete $\sigma _0$-execution $w_0\cdots w_m$ such that $w_0\in [s_0]^i$ and $w_m=w$. Moreover, since $t\in [w_m]^i$ and the ETS $\mathcal {M}$ has the property of perfect recall, this implies by Proposition 4, that there is a non-complete $\sigma _0$-execution $u_0\cdots u_m$ such that $u_k\in [w_k]^i$ for all $0\le k\le m$ and $u_m=t$. Since $w_0\in [s_0]^i$, this implies that $u_0\in [s_0]^i$. This implies that $u_k\in \mathtt {CEInner}(\sigma _0,[s_0]^i)$ for all $0\le k\le m$. Thus, $u_0\cdots u_m$ is also a possible execution of $\tau _0$. Let $\mu =u_0\cdots u_{m-1}\circ \delta$. (If $m=0$ then $\mu =\delta$). Since $\delta$ is a $\tau _0$’s complete execution from t, this implies that $\mu$ is a $\tau _0$’s complete execution from $u_0$. Since $u_0\in [s_0]^i$, this implies by ($\pmb \ddagger$) that the length of $\mu$ is finite. Thus, $\delta =t_0\cdots t_n$ for some $n\in \mathbb {N}$. We continue to show that $t_n\in \mathtt {CELeaf}(\sigma _0,[s]^i)$. Since $u_0\in [s_0]^i$ and $\mu =u_0\cdots u_{m-1}\circ t_0\cdots t_n$ is a complete $\tau _0$-execution from $u_0$, $\mu$ must also be complete given $\sigma _0$. If it is not, we then have that $t_n\in \mathtt {CEInner}(\sigma _0,[s_0]^i)$. By the definition of $\tau _0$, this implies that $[t_n]^i\in \mathtt {dom}(\tau _0)$. This is contradictory with the fact that $\mu$ is complete given $\tau _0$. Therefore, $\mu$ is complete given $\sigma _0$. We then have that $t_n\in \mathtt {CELeaf}(\sigma _0,[s]^i)$.

For the case of $\xi >0$, there are two situations: $[t]^i\in \mathtt {dom}(f_\xi )$ or $[t]^i\in D_\xi$. If $[t]^i\in \mathtt {dom}(f_\xi )$, it follows that $[t]^i\in \mathtt {dom}(\tau _\beta )$ for some $\beta <\xi$. By (3), we have $\mathtt {Dom}(\tau _\xi )\cap \mathtt {CELeaf}(\tau _\beta ,t)=\emptyset$. Since $\delta$ is a $\tau _\xi$’s complete execution, it follows by Proposition 6 that $\delta$ is also a $\tau _\beta$’s complete execution from t. It follows by IH that $\delta =t_0\cdots t_n$ for some $n\in \mathbb {N}$ and that $t_n\in \bigcup _{j\le \beta }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$. Since $\beta <\xi$, this implies that $t_n\in \bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$.

If $[t]^i\in D_\xi$, we will firstly show that $\delta$ is finite. There are two cases: there exist $k<|\delta |$ ($|\delta |$ means the length of $\delta$, similarly in later proofs) such that $[t_k]^i\in \mathtt {dom}(f_\xi )$, or not. (Please note that $|\delta |>1$ due to the fact that $\delta$ is $\tau _\xi$’s complete execution from t and $[t]^i\in \mathtt {dom}(\tau _\xi )$.)

• If there exist $k<|\delta |$ such that $[t_k]^i\in \mathtt {dom}(f_\xi )$, this implies that $[t_k]^i\in \mathtt {dom}(\tau _\beta )$ for some $\beta <\xi$. Please note that $\mu =t_k t_{k+1}\cdots$ is a $\tau _{\xi }$’s complete execution from $t_k$. By (3) and Proposition 6, $\mu$ is a $\tau _\beta$’s complete execution from $t_k$. By IH, $\mu =t_k\cdots t_{k+n}$ for some $n\in \mathbb {N}$. Therefore, $\delta =t_0\cdots t_{k+n}$ is finite.

• If there does not exist $k<|\delta |$ such that $[t_k]^i\in \mathtt {dom}(f_\xi )$, this implies by $[t]^i\in \mathtt {dom}(\tau _\xi )$ that $[t]^i\in D_\xi$ and $\delta =t_0\cdots$ is a $\sigma _{\xi }$’s possible execution from t. Since $D_\xi \subseteq \{[w]^i\mid w\in \mathtt {CEInner}(\sigma _{\xi },[s_{\xi }]^i)\}$, this implies that there is $w\in \mathtt {CEInner}(\sigma _\xi ,[s_\xi ]^i)$ such that $[t]^i=[w]^i$, namely $t\in [w]^i$. Since $w\in \mathtt {CEInner}(\sigma _\xi ,[s_\xi ]^i)$, this implies that there is a non-complete $\sigma _\xi$-execution $w_0\cdots w_m$ such that $w_0\in [s_\xi ]^i$ and $w_m=w$. Moreover, since $t\in [w_m]^i$ and the ETS $\mathcal {M}$ has the property of perfect recall, this implies by Proposition 4, that there is a non-complete $\sigma _\xi$-execution $u_0\cdots u_m$ such that $u_k\in [w_k]^i$ for all $0\le k\le m$ and $u_m=t$. Since $w_0\in [s_\xi ]^i$, this implies that $u_0\in [s_\xi ]^i$. Let $\mu =u_0\cdots u_{m-1}\circ \delta$. (If $m=0$ then $\mu =\delta$.) This implies that $\mu$ is a $\sigma _\xi$’s possible execution from $u_0\in [s_\xi ]^i$. By $(\pmb \ddagger )$, all $\sigma _\xi$’s complete executions from states in $[s_\xi ]^i$ are finite. Thus, $\mu$ is finite. Therefore, $\delta =t_0\cdots t_n$ for some $n\in \mathbb {N}$.

We have shown that $\delta$ is finite if $[t]^i\in D_\xi$. We continue to show that if $[t]^i\in D_\xi$ and $\delta =t_0\cdots t_n$ is a complete $\tau _\xi$-execution from $t\in \mathtt {Dom}(\tau _\xi )$ then $t_n\in \bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$. If there is $0<k<n$ such that $[t_k]^i\in \mathtt {dom}(f_\xi )$, since $t_k\cdots t_n$ is a complete $\tau _\xi$-execution, by what we have just shown, we then have that $t_n\in \bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$. If there is no $0<k<n$ such that $[t_k]^i\in \mathtt {dom}(f_\xi )$, this implies that $\delta$ is a $\sigma _\xi$’s possible execution from t. There then are two cases: either $\delta$ is already complete given $\sigma _\xi$, or $t_n\in \mathtt {CELeaf}(f_\xi ,\mathtt {Dom}(f_\xi ))$. If $\delta$ is complete given $\sigma _\xi$, this implies that $t_n\in \mathtt {CELeaf}(\sigma _\xi ,[s_\xi ]^i)$, and then $t_n\in \bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$. If $t_n\in \mathtt {CELeaf}(f_\xi ,\mathtt {Dom}(f_\xi ))$, there is a complete $f_\xi$-execution $\delta '$ such that $\delta '$ ends with $t_n$. Since $\delta '$ is complete, this implies that $|\delta '|$ is finite. Since $f_\xi =\bigcup _{\beta < \xi }\tau _\beta$, this implies that there is some $\beta <\xi$ such that $\delta '$ is a possible execution of $\tau _\beta$. Since $\delta '$ is complete given $f_\xi$ and $\tau _\beta \subseteq f_\xi$, this implies that $\delta '$ is complete given $\tau _\beta$. Since $\beta <\xi$, by IH, we have that $t_n\in \bigcup _{j\le \beta }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$, and then $t_n\in \bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$.

(5.) Firstly, there are two cases: $[s_\xi ]^i\not \in \mathtt {dom}(\sigma _\xi )$ or $[s_\xi ]^i\in \mathtt {dom}(\sigma _\xi )$. If $[s_\xi ]^i\not \in \mathtt {dom}(\sigma _\xi )$, this implies that $[s_\xi ]=\mathtt {CELeaf}(\sigma _\xi ,[s_\xi ]^i)$. Thus, if $[s_\xi ]^i\not \in \mathtt {dom}(\sigma _\xi )$, then $[s_\xi ]^i\subseteq (\bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$. If $[s_\xi ]^i\in \mathtt {dom}(\sigma _\xi )$, there are two cases: $\xi =0$ or $\xi >0$. If $\xi =0$, since $[s_0]^i\in \mathtt {dom}(\sigma _0)$, this implies that $s_0\in \mathtt {CEInner}(\sigma _0,[s_0]^i)$. Thus, we have $[s_0]^i\in \mathtt {dom}(\tau _0)$. Therefore, we are only left to show that if $[s_\xi ]^i\in \mathtt {dom}(\sigma _\xi )$ and $\xi >0$ then either $[s_\xi ]^i\in \mathtt {dom}(\tau _\xi )$ or $[s_\xi ]^i\subseteq (\bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$.

If $[s_\xi ]^i\in \mathtt {dom}(\sigma _\xi )$ and $\xi >0$, we show that if $[s_\xi ]^i\not \in \mathtt {dom}(\tau _\xi )$ then $[s_\xi ]^i\subseteq (\bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$. Firstly, we have that $s_\xi \in \mathtt {CEInner}(\sigma _\xi ,[s_\xi ]^i)$ due to $[s_\xi ]^i\in \mathtt {dom}(\sigma _\xi )$. Since $[s_\xi ]^i\not \in \mathtt {dom}(\tau _\xi )$ and $\mathtt {dom}(f_\xi )\subseteq \mathtt {dom}(\tau _\xi )$, this implies that $[s_\xi ]^i\in \{[w]^i\mid w\in \mathtt {CELeaf}(f_\xi ,\mathtt {Dom}(f_\xi ))\}$. Thus, there exists $u\in \mathtt {CELeaf}(f_\xi ,\mathtt {Dom}(f_\xi ))$ such that $[u]^i=[s_\xi ]^i$. We then only need to show that $[u]^i\subseteq (\bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$.

Since $u\in \mathtt {CELeaf}(f_\xi ,\mathtt {Dom}(f_\xi ))$, this implies that there is a complete $f_\xi$-execution $u_0\cdots u_k$ such that $u_0\in \mathtt {Dom}(f_\xi )$ and $u_k=u$. Since $f_\xi =\bigcup _{\beta < \xi }$, this implies that there exists some $\tau _\beta$ where $\beta <\xi$ such that $u_0\cdots u_k$ is a complete $\tau _\beta$-execution from $u_0\in \mathtt {Dom}(\tau _\beta )$. Let t be a state in $[t]^i=[t_k]^i$. Since $t\in [u_k]^i$ and the ETS $\mathcal {M}$ has the property of perfect recall, by Proposition 4, there are $t_h\in [u_h]^i$ for all $0\le h\le k$ such that $t_0\cdots t_k$ is a complete $\tau _\beta$-execution and $t_k=t$. Since $\beta <\xi$, by (3) of this claim, we have that $t_0\cdots t_k$ is a complete $\tau _\xi$-execution. By (4) of this claim, we have that $t_k\in (\bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$. Since $t_k=t$ and $t\in [u]^i$, thus we have shown that $[u]^i\subseteq (\bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$. $\square$

Step II. We define $\tau _\gamma =\bigcup _{\xi <\gamma }\tau _\xi$. It follows by (1) and (2) of Claim 1 that $\tau _\gamma$ is a strategy of agent i. Then we prove the following claim.

Claim 2

If $\delta =t_0\cdots$ is a $\tau _\gamma$’s complete execution from some $t\in \mathtt {Dom}(\tau _\gamma )$ then $\delta =t_0\cdots t_n$ for some $n\in \mathbb {N}$ and $t_n\in (\bigcup _{j < \gamma }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$.

Proof of claim

Since $t\in \mathtt {Dom}(\tau _\gamma )$, it follows that $t\in \mathtt {Dom}(\tau _\xi )$ for some $\xi <\gamma$. It follows by (4) of Claim 1 that all $\tau _\xi$’s complete executions are finite. Moreover, since $\tau _\xi \subseteq \tau _\gamma$, this implies that, for some $n\in \mathbb {N}$, the initial segment $t_0\cdots t_n$ of $\delta$ is complete given $\tau _\xi$. Since $t_0=t\in \mathtt {Dom}(\tau _\xi )$, this implies by (4) of Claim 1 that $t_n\in (\bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$. Since $\xi <\gamma$, it follows that $t_n\in (\bigcup _{j < \gamma }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$.

Next, we only need to show $\delta =t_0\cdots t_n$. If not, then $\delta =t_0\cdots t_n t_{n+1}\cdots$. We then have that there exists $j<\gamma$ such that $t_k\in \mathtt {Dom}(\tau _j)$ for each $0\le k\le n$. Since $t_n\not \in \mathtt {Dom}(\tau _\xi )$, this implies by (1) of Claim 1 that $j>\xi$.

By (3) of Claim 1, we then have that $\mathtt {Dom}(\tau _j)\cap \mathtt {CELeaf}(\tau _\xi ,t)=\emptyset$. This is in contradiction with $t_n\in \mathtt {Dom}(\tau _j)$ and $t_n\in \mathtt {CELeaf}(\tau _\xi ,t)$. Therefore, we have $\delta =t_0\cdots t_n$. $\square$

Step III. We define $\tau$ as $\tau =\tau _\gamma \cup (\sigma |_C)$ where $C=\{[w]^i\mid w\in \mathtt {CEInner}(\sigma ,[s]^i)\}\setminus (\mathtt {dom}(\tau _\gamma )\cup \{[v]^i\mid v\in \mathtt {CELeaf}(\tau _\gamma ,\mathtt {Dom}(\tau _\gamma ))\})$.

Since both $\tau _\gamma$ and $\sigma$ are strategies of agent i and $\mathtt {dom}(\tau _\gamma )\cap C=\emptyset$, this implies that $\tau$ is also a strategy of agent i. We then prove the following claim.

Claim 3

If $\delta =t_0\cdots$ is a $\tau$’s complete execution from some state $t\in \mathtt {Dom}(\tau )$ then $\delta =t_0\cdots t_n$ for some $n\in \mathbb {N}$ and $t_n\in (\bigcup _{j< \gamma }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$.

Proof of claim

Since $\mathtt {dom}(\tau )=\mathtt {dom}(\tau _\gamma )\cup C$, there are two cases: $[t]^i\in \mathtt {dom}(\tau _\gamma )$ or $[t]^i\in C$. We firstly show that the claim 3 holds if $[t]^i\in \mathtt {dom}(\tau _\gamma )$. Please note that $\{[v]^i\mid v\in \mathtt {CELeaf}(\tau _\gamma ,t)\}\cap C=\emptyset$ due to the definition of C. Moreover, we have $\{[v]^i\mid v\in \mathtt {CELeaf}(\tau _\gamma ,t)\}\cap \mathtt {dom}(\tau _\gamma )=\emptyset$. Thus, we have $\mathtt {CELeaf}(\tau _\gamma ,t)\cap \mathtt {Dom}(\tau )=\emptyset$. If $t\in \mathtt {Dom}(\tau _\gamma )$, it follows by Proposition 6 that $\delta$ is a $\tau _\gamma$’s complete execution from t. It follows by Claim 2 that $\delta =t_0\cdots t_n$ for some $n\in \mathbb {N}$ and $t_n\in (\bigcup _{j <\gamma }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$.

If $[t]^i\in C$, next we will show that the claim 3 also holds. There are two cases: there exists $k<|\delta |$ such that $t_k\in \mathtt {Dom}(\tau _\gamma )$, or there does not exist such k. (Please note that $|\delta |>1$ due to the fact that $\delta =t_0\cdots$ is $\tau$’s complete execution from t and $t\in \mathtt {Dom}(\tau )$.)

• $t_k\in \mathtt {Dom}(\tau _\gamma )$ for some $k<|\delta |$: Please note that $\mu =t_k\cdots$ is a $\tau$’s complete execution from $t_k$. Since $\mathtt {dom}(\tau )\cap \mathtt {CELeaf}(\tau _\gamma ,t_k)=\emptyset$, it follows by Proposition 6 that $\mu$ is also a $\tau _\gamma$’s complete execution from $t_k$. It follows by Claim 2 that $\mu =t_k\cdots t_{k+m}$ for some $m\in \mathbb {N}$ and $t_{k+m}\in (\bigcup _{j< \gamma }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$.

• If there does not exist $k<|\delta |$ such that $t_k\in \mathtt {Dom}(\tau _\gamma )$, then $\delta =t_0\cdots$ is a $\sigma$’s possible execution from t. Firstly, we will show that $\delta$ is finite.

  Since $[t]^i\in C\subseteq \{[w]^i\mid w\in \mathtt {CEInner}(\sigma ,[s]^i)\}$, this implies that there exists $w\in \mathtt {CEInner}(\sigma ,[s]^i)$ such that $t\in [w]^i$. Since $w\in \mathtt {CEInner}(\sigma ,[s]^i)$, this implies that there is a $\sigma$-execution $w_0\cdots w_m$ such that $w_0\in [s]^i$ and $w_m=w$. Since $t\in [w]^i=[w_m]^i$ and the ETS $\mathcal {M}$ has the property of perfect recall, we then have that there is a $\sigma$-execution $u_0\cdots u_m$ such that $u_h\in [w_h]^i$ for all $0\le h\le m$ and $u_m=t$. Let $\mu =u_0\cdots u_{m-1}\circ \delta$. (If $m=0$ then $\mu =\delta$.) This implies that $\mu$ is a $\sigma$-execution from $u_0\in [s]^i$. By $(\pmb *)$, all $\sigma$’s complete executions from s are finite. Thus, the length of $\mu$ is finite. Therefore, $\delta =t_0\cdots t_n$ for some $n\in \mathbb {N}$. We continue to show that $t_n\in (\bigcup _{j< \gamma }\mathtt {CELeaf}(\sigma _j,[s_j]^i))$. Since $\delta =t_0\cdots t_n$ is a $\tau$’s complete execution from t and it is also a $\sigma$’s possible execution from t, there are two cases: either $t_n\in \mathtt {CELeaf}(\tau _\gamma ,\mathtt {Dom}(\tau _\gamma ))$, or $\delta$ is a $\sigma$’s complete execution from t. If $t_n\in \mathtt {CELeaf}(\tau _\gamma ,\mathtt {Dom}(\tau _\gamma ))$, it follows by Claim 2 that $t_n\in \bigcup _{j<\gamma }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$. If $\delta$ is already complete given $\sigma$, this implies that $t_n=s_\xi$ for some $0\le \xi <\gamma$. Since $\delta =t_0\cdots t_n$ is a complete $\tau$-execution, this implies that $t_n\not \in \mathtt {Dom}(\tau )$. Since $\tau _\gamma \subseteq \tau$, this implies $t_n\not \in \mathtt {Dom}(\tau _\gamma )$. Moreover, since $\tau _\gamma =\bigcup _{\beta < \gamma }\tau _\beta$, this implies that $t_n\not \in \mathtt {Dom}(\tau _\xi )$, namely $s_\xi \not \in \mathtt {Dom}(\tau _\xi )$. By (5) of Claim 1, we then have that $t_n\in \bigcup _{j\le \xi }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$. Since $\xi <\gamma$, it follows that $t_n\in \bigcup _{j<\gamma }\mathtt {CELeaf}(\sigma _j,[s_j]^i)$.

$\square$

Finally, we will show that the i-strategy $\tau$ defined through Step I to Step III is what we want, that is, we will show that all complete $\tau$-executions from $s'\in [s]^i$ are finite and $\mathtt {CELeaf}(\tau ,[s]^i)\subseteq (\bigcup _{\xi <\gamma }\mathtt {CELeaf}(\sigma _\xi ,[s_\xi ]^i))$. There are two cases: $s\in \mathtt {Dom}(\tau )$ or not. If $s\in \mathtt {Dom}(\tau )$, this implies that $s'\in \mathtt {Dom}(\tau )$ for all $s'\in [s]^i$. By Claim 3, we then have that all complete $\tau$-executions from $s'\in [s]^i$ are finite and $\mathtt {CELeaf}(\tau ,[s]^i)\subseteq (\bigcup _{\xi <\gamma }\mathtt {CELeaf}(\sigma _\xi ,[s_\xi ]^i))$.

If $s\not \in \mathtt {Dom}(\tau )$, it is obvious that all complete $\tau$-executions from $s'\in [s]^i$ are finite. Since $\mathtt {CELeaf}(\tau ,[s]^i)=[s]^i$ because of $s\not \in \mathtt {Dom}(\tau )$, we only left to show that $[s]^i\subseteq (\bigcup _{\xi <\gamma }\mathtt {CELeaf}(\sigma _\xi ,[s_\xi ]^i))$. Since $s\in \mathtt {Dom}(\sigma )$, we have $s\in \mathtt {CEInner}(\sigma ,[s]^i)$. If $s\not \in \mathtt {Dom}(\tau )$, due to $s\in \mathtt {CEInner}(\sigma ,{[s]^i})$, by the definition of $\tau$ in Step III, this implies that there exists $w\in \mathtt {CELeaf}(\tau _\gamma ,\mathtt {Dom}(\tau _\gamma ))$ such that $[s]^i= [w]^i$. Since $\mathcal {M}$ has the property of perfect recall and $\tau _\gamma$ is an i-strategy, by Proposition 4, we know that $\mathtt {CELeaf}(\tau _\gamma ,\mathtt {Dom}(\tau _\gamma ))$ is closed under $\sim _i$. This implies that $[s]^i\subseteq \mathtt {CELeaf}(\tau _\gamma ,\mathtt {Dom}(\tau _\gamma ))$. By Claim 2, we know that $\mathtt {CELeaf}(\tau _\gamma ,\mathtt {Dom}(\tau _\gamma )) \subseteq (\bigcup _{\xi <\gamma }\mathtt {CELeaf}(\sigma _\xi ,[s_\xi ]^i))$. This implies that $[s]^i\subseteq (\bigcup _{\xi <\gamma }\mathtt {CELeaf}(\sigma _\xi ,[s_\xi ]^i))$. $\square$

1.2 Proof of Theorem 3.

The soundness is based on the following propositions.

Proposition 17

The rule $\mathtt {MONOKh}$ preserves validity on KN-frames with $\mathrm {MonoKh}$, that is, if $\varphi \rightarrow \psi$ is valid on KN-frames with $\mathrm {MonoKh}$, so is the formula ${\mathcal {K}h}_i\varphi \rightarrow {\mathcal {K}h}_i\psi$, where $i\in I$.

Proof

Let $\mathcal {N}$ be an KN-model, and its frame satisfies the condition $\mathrm {MonoKh}$. We only need to show that, for each $s\in W^\mathcal {N}$, $\mathcal {N},s\Vdash {\mathcal {K}h}_i\varphi \rightarrow {\mathcal {K}h}_i\psi$. If $\mathcal {N},s\Vdash {\mathcal {K}h}_i\varphi$, we will show that $\mathcal {N},s\Vdash {\mathcal {K}h}_i\psi$. Since $\mathcal {N},s\Vdash {\mathcal {K}h}_i\varphi$, this implies that $\llbracket \varphi \rrbracket ^\mathcal {N}\in N_i(s)$. Moreover, since $\varphi \rightarrow \psi$ is valid on KN-frames with $\mathrm {MonoKh}$, this implies that, for each $w\in W$, if $\mathcal {N},w\Vdash \varphi$ then $\mathcal {N},w\Vdash \psi$. Thus, we have that $\llbracket \varphi \rrbracket ^\mathcal {N}\subseteq \llbracket \psi \rrbracket ^\mathcal {N}$. Since the frame of $\mathcal {N}$ satisfies $\mathrm {MonoKh}$ and $\llbracket \varphi \rrbracket ^\mathcal {N}\in N_i(s)$, this implies that $\llbracket \psi \rrbracket ^\mathcal {N}\in N_i(s)$. Therefore, $\mathcal {N},s\Vdash {\mathcal {K}h}_i\psi$. $\square$

Proposition 18

The axiom $\mathtt {AxKhbot}$ is valid on KN-frames with $\mathrm {Khbot}$.

Proof

Let $\mathcal {N}$ be an KN-model, and its frame satisfies the condition $\mathrm {Khbot}$. We only need to show that, for each $s\in W^\mathcal {N}$, $\mathcal {N},s\Vdash \lnot {\mathcal {K}h}_i\bot$. If there is $s\in W^\mathcal {N}$ such that $\mathcal {N},s\Vdash {\mathcal {K}h}_i\bot$, this implies that $\llbracket \bot \rrbracket ^\mathcal {N}\in N_i(s)$. Since $\llbracket \bot \rrbracket ^\mathcal {N}=\emptyset$, this implies that $\emptyset \in N_i(s)$. This is contradictory with the assumption that the frame of $\mathcal {N}$ satisfies $\mathrm {Khbot}$. Therefore, there is no $s\in W^\mathcal {N}$ such that $\mathcal {N},s\Vdash {\mathcal {K}h}_i\bot$. Thus, $\mathcal {N},s\Vdash \lnot {\mathcal {K}h}_i\bot$ for each $s\in W^\mathcal {N}$. $\square$

Proposition 19

The axiom $\mathtt {AxKhtoKKh}$ is valid on KN-frames with $\mathrm {KhtoKKh}$.

Proof

Let $\mathcal {N}$ be an KN-model, and its frame satisfies the condition $\mathrm {KhtoKKh}$. We only need to show that, for each $s\in W^\mathcal {N}$, $\mathcal {N},s\Vdash {\mathcal {K}h}_i\varphi \rightarrow {\mathcal {K}}_i{\mathcal {K}h}_i\varphi$. If $\mathcal {N},s\Vdash {\mathcal {K}h}_i\varphi$, this implies that $\llbracket \varphi \rrbracket ^\mathcal {N}\in N_i(s)$. Since the frame of $\mathcal {N}$ satisfies $\mathrm {KhtoKKh}$, this implies that $N_i(s)=N_i(t)$ for each $t\in [s]^i$. Thus, $\llbracket \varphi \rrbracket ^\mathcal {N}\in N_i(t)$ for each $t\in [s]^i$. This implies that $\mathcal {N},t\Vdash {\mathcal {K}h}_i\varphi$ for all $t\in [s]^i$. Thus, we have that $\mathcal {N},s\Vdash {\mathcal {K}}_i{\mathcal {K}h}_i\varphi$. $\square$

Proposition 20

The axiom $\mathtt {AxKtoKh}$ is valid on KN-frames with $\mathrm {KtoKh}$ and $\mathrm {MonoKh}$.

Proof

Let $\mathcal {N}$ be an KN-model, and its frame satisfies the conditions $\mathrm {KtoKh}$ and $\mathrm {MonoKh}$. We only need to show that, for each $s\in W^\mathcal {N}$, $\mathcal {N},s\Vdash {\mathcal {K}}_i\varphi \rightarrow {\mathcal {K}h}_i\varphi$. If $\mathcal {N},s\Vdash {\mathcal {K}}_i\varphi$, this implies that $\mathcal {N},t\Vdash \varphi$ for all $t\in [s]^i$. Thus, we have that $[s]^i\subseteq \llbracket \varphi \rrbracket ^\mathcal {N}$. Since the frame of $\mathcal {N}$ satisfies $\mathrm {KtoKh}$, this implies that $[s]^i\in N_i(s)$. Since the frame of $\mathcal {N}$ satisfies $\mathrm {MonoKh}$ and $[s]^i\subseteq \llbracket \varphi \rrbracket ^\mathcal {N}$, this implies that $\llbracket \varphi \rrbracket ^\mathcal {N}\in N_i(s)$. Therefore, we have that $\mathcal {N},s\Vdash {\mathcal {K}h}_i\varphi$. $\square$

Proposition 21

The axiom $\mathtt {AxKhtoKhK}$ is valid on KN-frames with $\mathrm {KhtoKhK}$ and $\mathrm {MonoKh}$.

Proof

Let $\mathcal {N}$ be an KN-model, and its frame satisfies the conditions $\mathrm {KhtoKhK}$ and $\mathrm {MonoKh}$. We only need to show that, for each $s\in W^\mathcal {N}$, $\mathcal {N},s\Vdash {\mathcal {K}h}_i\varphi \rightarrow {\mathcal {K}h}_i{\mathcal {K}}_i\varphi$. If $\mathcal {N},s\Vdash {\mathcal {K}h}_i\varphi$, this implies that $\llbracket \varphi \rrbracket ^\mathcal {N}\in N_i(s)$. Next we will show that $\llbracket {\mathcal {K}}_i\varphi \rrbracket ^\mathcal {N}\in N_i(s)$. Since the frame of $\mathcal {N}$ satisfies $\mathrm {KhtoKhK}$ and $\llbracket \varphi \rrbracket ^\mathcal {N}\in N_i(s)$, this implies that $Y=\{t\mid [t]^i\subseteq \llbracket \varphi \rrbracket ^\mathcal {N}\}\in N_i(s)$. For each $t\in Y$, we have that $\mathcal {N},t\Vdash {\mathcal {K}}_i\varphi$ because of $[t]^i\subseteq \llbracket \varphi \rrbracket ^\mathcal {N}$. This implies that $Y\subseteq \llbracket {\mathcal {K}}_i\varphi \rrbracket ^\mathcal {N}$. Since $Y\in N_i(s)$ and the frame of $\mathcal {N}$ satisfies $\mathrm {MonoKh}$, this implies that $\llbracket {\mathcal {K}}_i\varphi \rrbracket ^\mathcal {N}\in N_i(s)$. Therefore, we have that $\mathcal {N},s\Vdash {\mathcal {K}h}_i{\mathcal {K}}_i\varphi$. $\square$

Proposition 22

The axiom $\mathtt {AxKhKh}$ is valid on KN-frames with $\mathrm {KhKh}$.

Proof

Let $\mathcal {N}$ be an KN-model, and its frame satisfies the condition $\mathrm {KhKh}$. We only need to show that, for each $s\in W^\mathcal {N}$, $\mathcal {N},s\Vdash {\mathcal {K}h}_i{\mathcal {K}h}_i\varphi \rightarrow {\mathcal {K}h}_i\varphi$. If $\mathcal {N},s\Vdash {\mathcal {K}h}_i{\mathcal {K}h}_i\varphi$, this implies that $\llbracket {\mathcal {K}h}_i\varphi \rrbracket ^\mathcal {N}\in N_i(s)$. For each $t\in \llbracket {\mathcal {K}h}_i\varphi \rrbracket ^\mathcal {N}$, namely $\mathcal {N},t\Vdash {\mathcal {K}h}_i\varphi$, we then have that $\llbracket \varphi \rrbracket ^\mathcal {N}\in N_i(t)$. Since the frame of $\mathcal {N}$ satisfies $\mathrm {KhKh}$, this implies that $\llbracket \varphi \rrbracket ^\mathcal {N}\in N_i(s)$. Therefore, we have that $\mathcal {N},s\Vdash {\mathcal {K}h}_i\varphi$. $\square$