Bisimulation Quotient in Inquisitive Modal Logic

Abstract

Inquisitive modal logic ${\mathrm{InqML}}_{⊞}$ is a natural generalization of basic modal logic, with ⊞ as a primitive modal operator. In this paper, we study the bisimulation quotients in the logic ${\mathrm{InqML}}_{⊞}$. For a given inquisitive modal model $\mathfrak{M}=(W,\Sigma ,V)$, we first show that the bisimilarity relation is an equivalence relation on W and that there is the largest bisimulation on $\mathfrak{M}$. We then define the bisimulation quotient and prove that a model is connected to its bisimulation quotient by a surjective bounded morphism. Finally, we prove that two models are globally bisimilar if and only if their bisimulation quotients are isomorphic.

1. Introduction

Inquisitive logic is a generalization of classical logic that can express both statements and questions. The language of inquisitive logic is obtained by extending the language of classical logic with a connective $\mathbf{\backslash }\vee$, which is called the inquisitive disjunction. With the connective $\backslash \vee$, it is possible to form formulas that represent questions. For example, the formula $\phi \backslash \vee \psi$ can be read as “is $\phi$ or $\psi$ the case?”. Although it is not clear what it means for a question to be true or false, it is possible to say whether the question is settled by the given information or not. The semantics of inquisitive logic is thus defined by a relationship between information states and formulas, where the information states are modeled as sets of worlds1.

The inquisitive modal logic ${\mathrm{InqML}}_{⊞}$ is a generalization of the basic modal logic with a new modal operator ⊞, called window. The logic ${\mathrm{InqML}}_{⊞}$ was first introduced in [1] in an epistemic context. In [1], it is shown that, in this logic, it is possible to express not only the information that agents have but also the information that they are interested in. For example, the formula $⊞?\phi$ can be read as “the agent wonders whether $\phi$”, where $?\phi$ is an abbreviation for $\phi \backslash \vee \neg \phi$.

The inquisitive modal logic ${\mathrm{InqML}}_{⊞}$ was systematically elaborated in [2], while in [3,4,5], various results in the model theory of ${\mathrm{InqML}}_{⊞}$ were presented. In particular, in [3] the notions of bisimulation and bisimulation games were defined, as was the standard translation from ${\mathrm{InqML}}_{⊞}$ to two-sorted first-order logic. Furthermore, analogues of the Ehrenfeucht–Fraïssé theorem and the van Benthem characterization theorem for ${\mathrm{InqML}}_{⊞}$ were proved. In [5], using so-called pseudo-models and an alternative definition of the standard translation, the compactness of ${\mathrm{InqML}}_{⊞}$ was proved in a purely model-theoretic way and a Hennessy–Milner class for ${\mathrm{InqML}}_{⊞}$ was defined. In [4], the finite model property of ${\mathrm{InqML}}_{⊞}$ was proved via filtrations, and as a consequence, the decidability of ${\mathrm{InqML}}_{⊞}$ was obtained.

This paper deals with the bisimulation quotients and thus makes a further contribution to the model theory of ${\mathrm{InqML}}_{⊞}$. In basic modal logic, using the facts that the bisimilarity relation is an equivalence relation and that the largest bisimulation exists, one can define the bisimulation quotient of an observed Kripke model. Furthermore, it is proved that a Kripke model is connected to its bisimulation quotient by a surjective bounded morphism, as well as that two Kripke models are globally bisimilar if and only if their bisimulation quotients are isomorphic (cf. [6]).

In this paper, we generalize these results for the inquisitive modal logic ${\mathrm{InqML}}_{⊞}$. We first show that for a given inquisitive modal model $\mathfrak{M}=(W,\Sigma ,V)$, the bisimilarity relation is an equivalence relation on W and that $\mathfrak{M}$ admits the largest bisimulation. Furthermore, we define the bisimulation quotient and show that a model is connected to its bisimulation quotient by a surjective bounded morphism that guarantees the modal equivalence between the model and its bisimulation quotient. Finally, we prove that two inquisitive modal models are globally bisimilar if and only if their bisimulation quotients are isomorphic.

In Section 2, we give an overview of basic definitions and results in ${\mathrm{InqML}}_{⊞}$. In Section 3, we show that the bisimilarity relation is an equivalence relation and that there is the largest bisimulation between considered models. In Section 4, we define the bisimulation quotient and show that this structure is an inquisitive modal model. Furthermore, we prove that when switching from an inquisitive modal model to its bisimulation quotient, the satisfaction of the formulas is preserved. Finally, we prove the main result of this paper: models are globally bisimilar if and only if their bisimulation quotients are isomorphic.

2. Preliminaries

Definitions and claims from this section are taken from [2].

Let W be a non-empty set of elements called the worlds. An information state (or simply, state) over W is any subset $s\subseteq W$. Furthermore, an inquisitive state over W is a non-empty set of information states $\Pi \subseteq \mathcal{P}\left(W\right)$ that is downward closed, i.e., if $s\in \Pi$ and $t\subseteq s$, then $t\in \Pi$.

Definition 1.

Let $\mathrm{P}$ be a set of propositional variables. An inquisitive modal model is a triple $\mathfrak{M}=(W,\Sigma ,V)$, where W is a set of worlds, $\Sigma :W\to \mathcal{P}\left(\mathcal{P}\right(W\left)\right)$ is a function that to each world w assigns an inquisitive state $\Sigma \left(w\right)$, and $V:\mathrm{P}\to \mathcal{P}\left(W\right)$ is a function that to each propositional variable assigns a set of worlds.

The language ${\mathcal{L}}_{⊞}$ of inquisitive modal logic ${\mathrm{InqML}}_{⊞}$ is given as follows:

$$\begin{array}{c}\hfill \phi ::=p\mid \perp \mid (\phi \wedge \phi )\mid (\phi \to \phi )\mid (\phi \backslash \vee \phi )\mid ⊞\phi .\end{array}$$

We take negation and disjunction as defined connectives in the usual way:

$$\neg \phi :=\phi \to \perp ,\phi \vee \psi :=\neg (\neg \phi \wedge \neg \psi ).$$

Furthermore, we use the following abbreviation: $?\phi :=\phi \backslash \vee \neg \phi .$

Definition 2 (Semantics of ${\mathrm{InqML}}_{⊞}$).

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model. The relation of support between states in $\mathfrak{M}$ and formulas of ${\mathcal{L}}_{⊞}$ is defined as follows:

• $\mathfrak{M},s\vDash p⟺s\subseteq V\left(p\right)$;
• $\mathfrak{M},s\vDash \perp ⟺s=\varnothing$;
• $\mathfrak{M},s\vDash \phi \wedge \psi ⟺\mathfrak{M},s\vDash \phi$ and $\mathfrak{M},s\vDash \psi$;
• $\mathfrak{M},s\vDash \phi \to \psi ⟺\mathit{for}\mathit{all}t\subseteq s$, $\mathfrak{M},t\vDash \phi$ implies $\mathfrak{M},t\vDash \psi$;
• $\mathfrak{M},s\vDash \phi \backslash \vee \psi ⟺\mathfrak{M},s\vDash \phi$ or $\mathfrak{M},s\vDash \psi$;
• $\mathfrak{M},s\vDash ⊞\phi ⟺\mathit{for}\mathit{all}w\in s$, $\mathit{for}\mathit{all}t\in \Sigma \left(w\right)$, $\mathfrak{M},t\vDash \phi$.

A formula $\phi$ is true at a world w in an inquisitive modal model $\mathfrak{M}$, denoted $\mathfrak{M},w\vDash \phi$, if $\mathfrak{M},\left\{w\right\}\vDash \phi$.

The modality ⊞ can be seen as a natural counterpart to the standard modal operator □. To illustrate this, consider a Kripke model $\mathfrak{M}=(W,R,V)$ for basic modal logic. The accessibility relation R can equivalently be viewed as a function $\sigma :W\to \mathcal{P}\left(W\right)$ that assigns to each world w the set of worlds accessible from w. Thus, we have $\mathfrak{M},w\vDash \square \phi$ if and only if $\sigma \left(w\right)\subseteq \left|\phi \right|$, where $\left|\phi \right|\subseteq W$ denotes the set of worlds where $\phi$ is true.

In inquisitive modal logic, the function $\sigma$ is replaced by a function $\Sigma :W\to \mathcal{P}\left(\mathcal{P}\right(W\left)\right)$, which assigns to each world w a set of sets of worlds. If $\left[\phi \right]$ denotes the set of sets of worlds where $\phi$ is supported, then we have $\mathfrak{M},w\vDash ⊞\phi$ if and only if $\Sigma \left(w\right)\subseteq \left[\phi \right]$. This shows that, from the perspective of mathematical logic, inquisitive modal logic is a natural generalization of basic modal logic.

For further comments on the modality □, including a discussion of the differences between □ and ⊞, see Remark 1.

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model. A state $s\subseteq W$ is compatible with a formula $\phi$ if there is $t\subseteq s,t\ne \varnothing$, such that $\mathfrak{M},t\vDash \phi$. Using this notion, the support conditions for the connectives ¬ and ∨ can be expressed as follows:

• $\mathfrak{M},s\vDash \neg \phi ⟺s$ is not compatible with $\phi$;
• $\mathfrak{M},s\vDash \phi \vee \psi ⟺$ for all $t\subseteq s$, $t\ne \varnothing$, so we prove that t is compatible with $\phi$ or t is compatible with $\psi$.

Let us illustrate the difference between the connectives $\backslash \vee$ and ∨ with the below example.

Example 1.

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model, with $s\subseteq W$ and $p,q$ propositional variables. Then, for the formulas $p \backslash \vee q$ and $p\vee q$, the following holds:

• $\mathfrak{M},s\vDash p \backslash \vee q\iff \mathfrak{M},s\vDash p$ or $\mathfrak{M},s\vDash q\iff s\subseteq V\left(p\right)$ or $s\subseteq V\left(q\right)\iff$ for all $w\in s$, so we have $w\in V\left(p\right)$ or for all $w\in s$, we have $w\in V\left(q\right)\iff$ for all $w\in s$, we have $\mathfrak{M},w\vDash p$ or for all $w\in s$, and we have $\mathfrak{M},w\vDash q$;
• $\mathfrak{M},s\vDash p\vee q\iff$ for all $t\subseteq s,t\ne \varnothing$, so we have t is compatible with p or t is compatible with q⇔ for all $w\in s$, we have $\left\{w\right\}$ is compatible with p or $\left\{w\right\}$ is compatible with q⇔ for all $w\in s$, and we have $\mathfrak{M},w\vDash p$ or $\mathfrak{M},w\vDash q$.

It is easy to show that the following properties hold:

• Persistence: $If \mathfrak{M},s\vDash \phi \mathrm{and} t\subseteq s,\mathrm{then} \mathfrak{M},t\vDash \phi$;
• Empty state property: $\mathfrak{M},\varnothing \vDash \phi$.

If the support condition of a formula is completely determined by its truth condition, we refer to such a formula as truth-conditional.

Definition 3.

A formula $\phi \in {\mathcal{L}}_{⊞}$ is truth-conditional if, for all $\mathfrak{M},s$, the following holds:

$$\mathfrak{M},s\vDash \phi \mathit{if}\mathit{and}\mathit{only}\mathit{if}\mathit{for}\mathit{all}w\in s\mathit{we}\mathit{have}\mathfrak{M},w\vDash \phi .$$

We regard truth-conditional formulas as statements, and non-truth-conditional formulas as questions.

Example 2.

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model, where $W=\{w,u\}$, $\Sigma \left(w\right)=\Sigma \left(u\right)=\{\varnothing \}$, and $V\left(p\right)=\left\{w\right\}$ for the propositional variable p. Consider the formula $?p$. Although $\mathfrak{M},w\vDash ?p$ and $\mathfrak{M},u\vDash ?p$, we have $\mathfrak{M},\{w,u\}/\vDash ?p$. Hence, the formula $?p$ is a question.

The set of declarative formulas is defined as follows:

$$\begin{array}{c}\hfill \alpha ::=p\mid \perp \mid (\alpha \wedge \alpha )\mid (\alpha \to \alpha )\mid ⊞\phi ,\end{array}$$

where $\phi$ is an arbitrary formula.

The set of declarative formulas corresponds exactly to the set of truth-conditional formulas up to logical equivalence (cf. [2]).

Remark 1.

In some of the literature, the modal operator □ is included in the alphabet with the following semantics:

$$\mathfrak{M},s\vDash \square \phi ⟺\mathit{for}\mathit{all}w\in s\mathit{we}\mathit{have}\mathfrak{M},\bigcup \Sigma \left(w\right)\vDash \phi .$$

However, it is shown that every formula of the form $\square \phi$ is equivalent to a disjunction of formulas of the form $⊞\psi$; thus, □ can be eliminated from the alphabet (see cf. [2] for more details).

Furthermore, both formulas $\square \phi$ and $⊞\phi$ are statements, but using only the operator □, it is not possible to obtain a statement that is not equivalent to some formula of basic modal logic. For example, the formula $\square ?p$ is equivalent to the formula $\square p\vee \square \neg p$. On the other hand, using the operator $⊞$, it is possible to obtain a statement that has no counterpart in basic modal logic (e.g., $⊞?p$). By embedding questions under the modality $⊞$, we can therefore formulate new statements that cannot be expressed in basic modal logic (for more details, see cf. [2]).

In an epistemic context, the operator □ is used to express only the information that agents have. In contrast, the operator $⊞$ also allows us to express the information that agents are interested in, thus capturing not only what they know but also what they want to resolve (see cf. [1,2] for more details on inquisitive epistemic logic).

For more basic properties of ${\mathrm{InqML}}_{⊞}$, see [2].

3. The Bisimilarity Relation

As previously mentioned, bisimulation in inquisitive modal logic ${\mathrm{InqML}}_{⊞}$ was first defined and studied in [3]. The goal of bisimulation is to establish a relation between two inquisitive modal models in which related worlds agree on the propositional variables and have corresponding inquisitive states.

Let $\mathfrak{M}=(W,R,V)$ be a Kripke model for basic modal logic. In this model, there is only one type of transition: from a world to a world. Therefore, bisimulation in basic modal logic is defined as a binary relation on the set of worlds.

In an inquisitive modal model, the function $\Sigma$ assigns to each world a set of states, which leads to two types of transitions: from a world to a state and from a state to a world. Namely, a world w can be associated with a state $s\in \Sigma \left(w\right)$, while a state s can be associated with a world $u\in s$. Therefore, the definition of bisimulation in ${\mathrm{InqML}}_{⊞}$ must be such that it takes both types of transitions into account.

Let $\mathfrak{M}=(W,\Sigma ,V)$ and ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$ be inquisitive modal models. The conditions under which a relation Z can be considered a bisimulation are as follows: First, if $w\in W$ and $w^{\prime }\in W^{\prime }$ are worlds such that $wZ{w}^{\prime }$, then the same propositional variables must be true at both worlds. Second, if $wZ{w}^{\prime }$, then for each state $s\in \Sigma \left(w\right)$, there must exist a state $s^{\prime }\in {\Sigma }^{\prime }\left(w^{\prime }\right)$ such that $sZ{s}^{\prime }$, and vice versa. Third, if $sZ{s}^{\prime }$, then for each $w\in s$, there must exist $w^{\prime }\in s^{\prime }$ such that $wZ{w}^{\prime }$, and vice versa. A bisimulation Z should therefore be a relation $Z\subseteq W\times W^{\prime }\cup \mathcal{P}\left(W\right)\times \mathcal{P}\left(W^{\prime }\right)$ that satisfies the above conditions (cf. [3]).

However, as shown in [3], the definition of bisimulation can also be formulated as a relation defined exclusively on the set of worlds of the observed models, i.e., as a relation $Z\subseteq W\times W^{\prime }$. In this paper, we will use this formulation. For this purpose, we first define the below relation.

The lifting of a relation $Y\subseteq W\times W^{\prime }$ is the relation $\overline{Y}\subseteq \mathcal{P}\left(W\right)\times \mathcal{P}\left(W^{\prime }\right)$ defined with $s\overline{Y}{s}^{\prime }$ if and only if the following conditions are satisfied:

• For all $w\in s$, there is $w^{\prime }\in s^{\prime }$ such that $wY{w}^{\prime }$;
• For all $w^{\prime }\in s^{\prime }$, there is $w\in s$ such that $wY{w}^{\prime }$.

The purpose of the relation $\overline{Y}$ is to define how subsets of two sets relate to each other based on a binary relation between their elements. Furthermore, the relation $\overline{Y}$ is clearly uniquely determined by Y.

Remark 2.

Notice that for a relation $Y\subseteq W\times W^{\prime }$, we have ${\overline{Y}}^{-1}=\overline{Y^{-1}}$. Furthermore, if $Y_1\subseteq W\times W^{\prime }$ and $Y_2\subseteq W^{\prime }\times W^{″}$ are relations, then $\overline{Y_1}\circ \overline{Y_2}=\overline{Y_1\circ Y_2}$.

Now, we can define the notion of bisimulation in ${\mathrm{InqML}}_{⊞}$ (cf. [3]).

Definition 4.

Let $\mathfrak{M}=(W,\Sigma ,V)$ and ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$ be inquisitive modal models. A non-empty relation $Z\subseteq W\times W^{\prime }$ is called a bisimulation if the following conditions are satisfied:

(at) 

If $wZ{w}^{\prime }$, then $w\in V\left(p\right)$ if and only if $w^{\prime }\in V^{\prime }\left(p\right)$, for all $p\in \mathrm{P}$;

(forth) 

If $wZ{w}^{\prime }$ and $s\in \Sigma \left(w\right)$, then there is $s^{\prime }\in {\Sigma }^{\prime }\left(w^{\prime }\right)$ such that $s\overline{Z}{s}^{\prime }$;

(back) 

If $wZ{w}^{\prime }$ and $s^{\prime }\in {\Sigma }^{\prime }\left(w^{\prime }\right)$, then there is $s\in \Sigma \left(w\right)$ such that $s\overline{Z}{s}^{\prime }$.

We say that the worlds w and $w^{\prime }$ are bisimilar, denoted by $\mathfrak{M},w\sim {\mathfrak{M}}^{\prime },w^{\prime }$, if there exists a bisimulation Z such that $wZ{w}^{\prime }$.

We say that the states s and $s^{\prime }$ are bisimilar, denoted by $\mathfrak{M},s\sim {\mathfrak{M}}^{\prime },s^{\prime }$, if there exists a bisimulation Z such that $s\overline{Z}{s}^{\prime }$.

The following remarks point out connections between the notion of bisimulation used here and related approaches used in the literature.

Remark 3.

The definition of bisimilarity between states in inquisitive modal logic is similar to that in modal team logic, where formulas are also evaluated over sets of worlds (cf. [7]). However, modal team logic uses Kripke models so every world is directly connected to other worlds, whereas in ${\mathrm{InqML}}_{⊞}$, each world is connected to information states.

Therefore, the notions of bisimilarity between worlds and bisimilarity between states in ${\mathrm{InqML}}_{⊞}$ are closely related, unlike in modal team logic, where the latter is defined on the basis of the former.

Remark 4.

Inquisitive modal models can be seen as downward closed neighborhood models2, i.e., models where the set of neighborhoods is closed under subsets. In neighborhood semantics for modal logic (cf. [8]), a formula $\square \phi$ is true at a world w if the set of worlds where φ is true is one of the neighborhoods of w.

Although ${\mathrm{InqML}}_{⊞}$ and neighborhood semantics are based on similar models, they lead to different logics, since the neighborhood function is used differently to interpret formulas. Furthermore, as shown in [3], ${\mathrm{InqML}}_{⊞}$ and neighborhood semantics are incomparable in terms of expressive power, which results in different notions of bisimilarity (cf. [9]).

On the other hand, the definition of bisimilarity in ${\mathrm{InqML}}_{⊞}$ is quite similar to that in monotonic modal logic, where neighborhood models are closed under supersets, in contrast to the subset closure condition for inquisitive modal models (cf. [10]).

Let us now consider the connection between bisimilarity and modal equivalence.

Definition 5.

Let $\mathfrak{M}=(W,\Sigma ,V)$ and ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$ be inquisitive modal models. The states s and $s^{\prime }$ are modally equivalent, denoted by $\mathfrak{M},s\equiv {\mathfrak{M}}^{\prime },s^{\prime }$, if for every formula $\phi \in {\mathcal{L}}_{⊞}$ the following holds:

$$\mathfrak{M},s\vDash \phi \mathit{if}\mathit{and}\mathit{only}\mathit{if}{\mathfrak{M}}^{\prime },s^{\prime }\vDash \phi .$$

As expected, it is shown in [3] that if the states are bisimilar, then they are modally equivalent. The converse, i.e., the claim that modal equivalence implies bisimilarity, does not hold. However, in [3], an Ehrenfeucht–Fraïssé theorem for ${\mathrm{InqML}}_{⊞}$ is proved, showing that, assuming a finite set of propositional variables, the finite levels ${\sim }_n$ of bisimulation correspond to the levels of modal equivalence up to a modal depth of at most n.

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model. Consider the bisimilarity relation between worlds in W, i.e., the relation defined as follows:

$$\begin{array}{c}\hfill w\sim u\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{bisimulation}Z\subseteq W\times W\mathrm{such}\mathrm{that}wZu.\end{array}$$

The first goal is to prove that the bisimilarity relation ∼ is an equivalence relation, which leads to the below proposition.

Proposition 1.

Let $\mathfrak{M}=(W,\Sigma ,V)$, ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$, and ${\mathfrak{M}}^{″}=(W^{″},{\Sigma }^{″},V^{″})$ be inquisitive modal models. Then, the following hold:

$\left(a\right)$

The relation $Z=\left\{\right(w,w):w\in W\}\subseteq W\times W$ is a bisimulation;

$\left(b\right)$

If $Z\subseteq W\times W^{\prime }$ is a bisimulation, then $Z^{-1}=\{(w^{\prime },w):wZ{w}^{\prime }\}\subseteq W^{\prime }\times W$ is a bisimulation;

$\left(c\right)$

If $Z^{\prime }\subseteq W\times W^{\prime }$ and $Z^{″}\subseteq W^{\prime }\times W^{″}$ are bisimulations, then $Z=Z^{\prime }\circ Z^{″}\subseteq W\times W^{″}$ is a bisimulation.

Proof. 

Claims $\left(a\right)$ and $\left(b\right)$ are straightforward, so we focus on proving claim $\left(c\right)$. We show that Z satisfies the conditions $\left(at\right),\left(forth\right)$, and $\left(back\right)$:

$\left(at\right)$

Let $wZ{w}^{″}$. Then, there is $w^{\prime }$ such that $w{Z}^{\prime }{w}^{\prime }$ and $w^{\prime }{Z}^{″}{w}^{″}$. Since $Z^{\prime }$ and $Z^{″}$ are bisimulations, it follows that $w\in V\left(p\right)$ if and only if $w^{\prime }\in V^{\prime }\left(p\right)$ and if and only if $w^{″}\in V^{″}\left(p\right)$.

$\left(forth\right)$

Let $wZ{w}^{″}$ and $s\in \Sigma \left(w\right)$. Then, there is $w^{\prime }$ such that $w{Z}^{\prime }{w}^{\prime }$ and $w^{\prime }{Z}^{″}{w}^{″}$. Since $Z^{\prime }$ is a bisimulation, there is $s^{\prime }\in {\Sigma }^{\prime }\left(w^{\prime }\right)$ such that $s\overline{Z^{\prime }}{s}^{\prime }$. Since $Z^{″}$ is a bisimulation, there is $s^{″}\in {\Sigma }^{″}\left(w^{″}\right)$ such that $s^{\prime }\overline{Z^{″}}{s}^{″}$.

We claim that $s\overline{Z}{s}^{″}$. Suppose $u\in s$. Since $s\overline{Z^{\prime }}{s}^{\prime }$, there is $u^{\prime }\in s^{\prime }$ such that $u{Z}^{\prime }{u}^{\prime }$, and since $s^{\prime }\overline{Z^{″}}{s}^{″}$, there is $u^{″}\in s^{″}$ such that $u^{\prime }{Z}^{″}{u}^{″}$. From this, it follows that $uZ{u}^{″}$. The second condition can be proved in a similar way.

$\left(back\right)$

Similar to condition $\left(forth\right)$.

Hence, $Z=Z^{\prime }\circ Z^{\prime \prime }$ is a bisimulation. □

From the previous proposition, we obtain the following corollary:

Corollary 1.

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model. The bisimilarity relation ∼ is an equivalence relation on W.

In the following proposition, we show that the union of bisimulations is a bisimulation.

Proposition 2.

Let $\mathfrak{M}=(W,\Sigma ,V)$ and ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$ be inquisitive modal models. If $\{Z_i:i\in I\}$ is a set of bisimulations $Z_i\subseteq W\times W^{\prime }$, then ${\displaystyle \mathcal{Z}=\bigcup _{i\in I}{Z}_i}$ is a bisimulation.

Proof. 

$\left(at\right)$

Let $w\mathcal{Z}{w}^{\prime }$. Then, there is $i\in I$ such that $w{Z}_i{w}^{\prime }$. Since $Z_i$ is a bisimulation, it follows that $w\in V\left(p\right)$ if and only if $w^{\prime }\in V^{\prime }\left(p\right)$.

$\left(forth\right)$

Let $w\mathcal{Z}{w}^{\prime }$ and $s\in \Sigma \left(w\right)$. Then, there is $i\in I$ such that $w{Z}_i{w}^{\prime }$. Since $Z_i$ is a bisimulation, there is $s^{\prime }\in {\Sigma }^{\prime }\left(w^{\prime }\right)$ such that $s\overline{Z_i}{s}^{\prime }$.

We claim that $s\overline{\mathcal{Z}}{s}^{\prime }$. Suppose $u\in s$. Since $s\overline{Z_i}{s}^{\prime }$, there is $u^{\prime }\in s^{\prime }$ such that $u{Z}_i{u}^{\prime }$. This implies $u\mathcal{Z}{u}^{\prime }$. The second condition can be proved in a similar way.

$\left(back\right)$

Similar to condition $\left(forth\right)$.

Hence, ${\displaystyle \mathcal{Z}=\bigcup _{i\in I}{Z}_i}$ is a bisimulation. □

The union of all bisimulations between inquisitive modal models $\mathfrak{M}$ and ${\mathfrak{M}}^{\prime }$ is clearly the largest bisimulation between these models.

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model and let Z be the largest bisimulation on $\mathfrak{M}$. Since $wZu$ if and only if $w\sim u$, it follows that Z is an equivalence relation. For $w\in W$, denote by $\left[w\right]$ the equivalence class of an element w under the relation Z. Furthermore, for $s\subseteq W$, let

$$\left[s\right]=\left\{\right[w]:w\in s\}.$$

Observe that if $\left[w\right]\in \left[s\right]$, then there is $u\in s$ such that $wZu$.

For the worlds, it holds that $wZu$ if and only if $\left[w\right]=\left[u\right]$. A similar result holds for the states, as expressed in the below proposition.

Proposition 3.

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model and let Z be the largest bisimulation on $\mathfrak{M}$. Then, for all states s and $s^{\prime }$, we have

$$s\overline{Z}{s}^{\prime }\mathit{if}\mathit{and}\mathit{only}\mathit{if}\left[s\right]=\left[s^{\prime }\right].$$

Proof. 

First, suppose that $s\overline{Z}{s}^{\prime }$. Let $\left[w\right]\in \left[s\right]$. Then, there is $u\in s$ such that $wZu$. Since $s\overline{Z}{s}^{\prime }$, there is $u^{\prime }\in s^{\prime }$ such that $uZ{u}^{\prime }$, so we have $wZ{u}^{\prime }$, i.e., $\left[w\right]=\left[u^{\prime }\right]$. Since $u^{\prime }\in s^{\prime }$, it follows that $\left[u^{\prime }\right]\in \left[s^{\prime }\right]$, so we have $\left[w\right]\in \left[s^{\prime }\right]$. Hence, $\left[s\right]\subseteq \left[s^{\prime }\right]$. We can prove $\left[s^{\prime }\right]\subseteq \left[s\right]$ in a similar way.

Conversely, suppose that $\left[s\right]=\left[s^{\prime }\right]$. Let $w\in s$. Then, we have $\left[w\right]\in \left[s\right]$. This implies that $\left[w\right]\in \left[s^{\prime }\right]$. Then, there is $u^{\prime }\in s^{\prime }$ such that $wZ{u}^{\prime }$. The second condition can be proved in a similar way, so we obtain $s\overline{Z}{s}^{\prime }$. □

4. Bisimulation Quotient

Having shown that the bisimilarity relation is an equivalence relation on the set of worlds W of an inquisitive modal model $\mathfrak{M}=(W,\Sigma ,V)$, we define the quotient of the model $\mathfrak{M}$ with respect to the largest bisimulation.

Definition 6.

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model and Z be the largest bisimulation on $\mathfrak{M}$. The model ${\mathfrak{M}}_Z=(W_Z,{\Sigma }_Z,V_Z)$ is called the bisimulation quotient of $\mathfrak{M}$ if the following conditions are satisfied:

• $W_Z=\{\left[w\right]:w\in W\}$;
• $\left[s\right]\in {\Sigma }_Z\left(\left[w\right]\right)$ if and only if there are $w^{\prime }\in \left[w\right]$ and $s^{\prime }$ with $\left[s^{\prime }\right]=\left[s\right]$ such that $s^{\prime }\in \Sigma \left(w^{\prime }\right)$;
• $\left[w\right]\in V_Z\left(p\right)$ if and only if $w\in V\left(p\right)$ for all $p\in \mathrm{P}$.

It is easy to check whether the definition does not depend on the choice of representatives.

Now, we show that the bisimulation quotient is an inquisitive modal model. First, we have the following lemma:

Lemma 1.

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model and Z be the largest bisimulation on $\mathfrak{M}$ and $t,s\subseteq W$. If $\left[t\right]\subseteq \left[s\right]$, then there is $t^{\prime }\subseteq s$ such that $\left[t\right]=\left[t^{\prime }\right]$.

Proof. 

Let $t^{\prime }=\{w\in s:\mathrm{there}\mathrm{is}u\in t\mathrm{such}\mathrm{that}uZw\}$. It is obvious that $t^{\prime }\subseteq s$. We claim that $\left[t\right]=\left[t^{\prime }\right]$. It suffices to show $t\overline{Z}{t}^{\prime }$. Suppose $u\in t$. Since $\left[t\right]\subseteq \left[s\right]$, there is $w\in s$ such that $uZw$. Now, we have $w\in t^{\prime }$. Thus, for every $u\in t$, there is $w\in t^{\prime }$ such that $uZw$. Conversely, suppose $w\in t^{\prime }$. By the definition of the set $t^{\prime }$, there is $u\in t$ such that $uZw$. Hence, $t\overline{Z}{t}^{\prime }$. □

Proposition 4.

Let $\mathfrak{M}=(W,\Sigma ,V)$ be an inquisitive modal model and Z be the largest bisimulation on $\mathfrak{M}$. The bisimulation quotient ${\mathfrak{M}}_Z=(W_Z,{\Sigma }_Z,V_Z)$ is an inquisitive modal model.

Proof. 

We need to show that, for every $w\in W$, the sets ${\Sigma }_Z\left(\left[w\right]\right)$ are downward closed. So, let $\left[s\right]\in {\Sigma }_Z\left(\left[w\right]\right)$ and $\left[t\right]\subseteq \left[s\right]$. From $\left[s\right]\in {\Sigma }_Z\left(\left[w\right]\right)$, it follows that there are $w^{\prime }\in \left[w\right]$ and $s^{\prime }$ with $\left[s^{\prime }\right]=\left[s\right]$ such that $s^{\prime }\in \Sigma \left(w^{\prime }\right)$. By the previous lemma, $\left[t\right]\subseteq \left[s\right]=\left[s^{\prime }\right]$ implies that there is $t^{\prime }\subseteq s^{\prime }$ such that $\left[t\right]=\left[t^{\prime }\right]$. Since $\mathfrak{M}$ is an inquisitive modal model, $t^{\prime }\subseteq s^{\prime }$ and $s^{\prime }\in \Sigma \left(w^{\prime }\right)$ imply that $t^{\prime }\in \Sigma \left(w^{\prime }\right)$. Now, we obtain $\left[t^{\prime }\right]\in {\Sigma }_Z\left(\left[w^{\prime }\right]\right)$, i.e., $\left[t\right]\in {\Sigma }_Z\left(\left[w\right]\right)$. Hence, the set ${\Sigma }_Z\left(\left[w\right]\right)$ is downward closed. □

Naturally, we expect a model to be connected to its bisimulation quotient by a surjective bounded morphism and thus by modal equivalence (cf. [6]). Here, we adapt this for the inquisitive modal logic ${\mathrm{InqML}}_{⊞}$.

For $s\subseteq W$, denote $f\left(s\right)=\left\{f\right(w):w\in s\}$.

Definition 7.

Let $\mathfrak{M}=(W,\Sigma ,V)$ and ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$ be inquisitive modal models. A function $f:W\to W^{\prime }$ is called a bounded morphism of the models $\mathfrak{M}$ and ${\mathfrak{M}}^{\prime }$ if the following conditions are satisfied:

$\left(a{t}_{bm}\right)$

$w\in V\left(p\right)$ if and only if $f\left(w\right)\in V^{\prime }\left(p\right)$ for all $p\in \mathrm{P}$;

$\left(fort{h}_{bm}\right)$

If $s\in \Sigma \left(w\right)$, then $f\left(s\right)\in {\Sigma }^{\prime }\left(f\left(w\right)\right)$;

$\left(bac{k}_{bm}\right)$

If $s^{\prime }\in {\Sigma }^{\prime }\left(f\left(w\right)\right)$, then there is $s\in \Sigma \left(w\right)$ such that $f\left(s\right)=s^{\prime }$.

We now show that modal satisfaction is invariant under a bounded morphism, i.e., the below proposition holds.

Proposition 5.

Let $\mathfrak{M}=(W,\Sigma ,V)$ and ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$ be inquisitive modal models and let $f:W\to W^{\prime }$ be a bounded morphism of the models $\mathfrak{M}$ and ${\mathfrak{M}}^{\prime }$. Then, for every $s\subseteq W$ and every formula $\phi \in {\mathcal{L}}_{⊞}$, we have

$$\mathfrak{M},s\vDash \phi \mathit{if\; and\; only\; if} {\mathfrak{M}}^{\prime },f\left(s\right)\vDash \phi .$$

Proof. 

Since bisimilar states are modally equivalent, it suffices to show that there is a bisimulation Z such that $s\overline{Z}f\left(s\right)$. Let $Z=\left\{\right(w,f\left(w\right)\left)\right):w\in W\}$.

Suppose $wZf\left(w\right)$. Then, we have $w\in V\left(p\right)$ if and only if $f\left(w\right)\in V^{\prime }\left(p\right)$, so condition $\left(at\right)$ holds.

Suppose $wZf\left(w\right)$ and $s\in \Sigma \left(w\right)$. From condition $\left(fort{h}_{bm}\right)$, it follows that $f\left(s\right)\in {\Sigma }^{\prime }\left(f\left(w\right)\right)$. We need to show that $s\overline{Z}f\left(s\right)$. If $u\in s$, then for $f\left(u\right)\in f\left(s\right)$, we have $uZf\left(u\right)$. Conversely, if $u^{\prime }\in f\left(s\right)$, then there is $u\in s$ such that $f\left(u\right)=u^{\prime }$. This implies $uZf\left(u\right)$, i.e., $uZ{u}^{\prime }$. Hence, condition $\left(forth\right)$ holds.

Suppose $wZf\left(w\right)$ and $s^{\prime }\in {\Sigma }^{\prime }\left(f\left(w\right)\right)$. From condition $\left(bac{k}_{bm}\right)$, it follows that there is $s\in \Sigma \left(w\right)$ such that $f\left(s\right)=s^{\prime }$. As in the previous case, it can be proved that $s\overline{Z}{s}^{\prime }$, so condition $\left(back\right)$ also holds.

Hence, Z is a bisimulation.

Let $s\subseteq W$ be an arbitrary state. Suppose $u\in s$. Then, for $f\left(u\right)\in f\left(s\right)$, we have $uZf\left(u\right)$. Conversely, suppose $u^{\prime }\in f\left(s\right)$. Then, there is $u\in s$ such that $f\left(u\right)=u^{\prime }$. Now, we have $uZf\left(u\right)$, i.e., $uZ{u}^{\prime }$. Hence, $s\overline{Z}f\left(s\right)$. □

Proposition 6.

The projection $\pi :\mathfrak{M}\to {\mathfrak{M}}_Z$, $\pi \left(w\right)=\left[w\right]$, is a surjective bounded morphism.

Proof. 

It is obvious that $\pi$ is surjective. Let us show that $\pi$ is a bounded morphism:

$\left(a{t}_{bm}\right)$

We have $w\in V\left(p\right)$ if and only if $\left[w\right]\in V_Z\left(p\right)$ and if and only if $\pi \left(w\right)\in V_Z\left(p\right)$.

$\left(fort{h}_{bm}\right)$

Let $s\in \Sigma \left(w\right)$. Then, $\left[s\right]\in {\Sigma }_Z\left(\left[w\right]\right)$, i.e., $\pi \left(s\right)\in {\Sigma }_Z\left(\pi \left(w\right)\right)$, by the definition of the bisimulation quotient.

$\left(bac{k}_{bm}\right)$

Let $\left[s\right]\in {\Sigma }_Z\left(\pi \left(w\right)\right)$, i.e., $\left[s\right]\in {\Sigma }_Z\left(\left[w\right]\right)$. By the definition of the bisimulation quotient, there are $w^{\prime }\in \left[w\right]$ and $s^{\prime }$ with $\left[s^{\prime }\right]=\left[s\right]$ such that $s^{\prime }\in \Sigma \left(w^{\prime }\right)$. By condition $\left(back\right)$, $wZ{w}^{\prime }$ and $s^{\prime }\in \Sigma \left(w^{\prime }\right)$ imply that there is t with $t\overline{Z}{s}^{\prime }$ such that $t\in \Sigma \left(w\right)$. Now, $\left[s^{\prime }\right]=\left[s\right]$ and $\left[t\right]=\left[s^{\prime }\right]$ imply $\left[t\right]=\left[s\right]$. Hence, $\pi \left(t\right)=\left[s\right]$.

Thus, $\pi$ is a surjective bounded morphism. □

The next corollary follows from the previous propositions.

Corollary 2.

Let ${\mathfrak{M}}_Z$ be the bisimulation quotient of an inquisitive modal model $\mathfrak{M}$. Then, for every state s, we have $\mathfrak{M},s\equiv {\mathfrak{M}}_Z,\left[s\right]$, i.e., the states s and $\left[s\right]$ are modally equivalent.

For the rest of this section, we turn to the main result of the paper, which is also an analogue of a well-known result in basic modal logic (cf. [6]): two models are globally bisimilar if and only if their bisimulation quotients are isomorphic.

First, we define the notions of global bisimulation and isomorphism.

Definition 8.

Inquisitive modal models $\mathfrak{M}=(W,\Sigma ,V)$ and ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$ are globally bisimilar if there is a bisimulation X such that for all $w\in W$, there is $w^{\prime }\in W^{\prime }$ with $wX{w}^{\prime }$, and vice versa. In this case, X is called a global bisimulation.

Notice that of all models that are globally bisimilar to $\mathfrak{M}$, the bisimulation quotient ${\mathfrak{M}}_Z$ is in some sense minimal, since every other model that is globally bisimilar to $\mathfrak{M}$ must contain at least one world for each class obtained by the relation Z.

Definition 9.

Let $\mathfrak{M}=(W,\Sigma ,V)$ and ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$ be inquisitive modal models. The function $f:W\to W^{\prime }$ is called an isomorphism of the models $\mathfrak{M}$ and ${\mathfrak{M}}^{\prime }$ if f is a bijection and the following conditions are satisfied:

$\left(i\right)$

$w\in V\left(p\right)$ if and only if $f\left(w\right)\in V^{\prime }\left(p\right)$ for all $p\in \mathrm{P}$;

$\left(ii\right)$

$s\in \Sigma \left(w\right)$ if and only if $f\left(s\right)\in {\Sigma }^{\prime }\left(f\left(w\right)\right)$.

Before proving the main theorem, we need the following lemma.

Lemma 2.

Let $\mathfrak{M}=(W,\Sigma ,V)$ and ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$ be inquisitive modal models, and let Z and $Z^{\prime }$ be the largest bisimulations on $\mathfrak{M}$ and ${\mathfrak{M}}^{\prime }$, respectively. Let $f:W_Z\to W_{Z^{\prime }}^{\prime }$ be a function defined by

$$f\left(\left[w\right]\right)=\left[w^{\prime }\right] \mathit{if\; there\; exist} x\in \left[w\right]\mathit{and}{x}^{\prime }\in \left[w^{\prime }\right] \mathit{such\; that} xX{x}^{\prime },$$

where X is a global bisimulation between $\mathfrak{M}$ and ${\mathfrak{M}}^{\prime }$.

Then the function f is well-defined, and for all states s and $s^{\prime }$, we have

$$f\left(\left[s\right]\right)=\left[s^{\prime }\right] \mathit{if\; and\; only\; if\; there\; exist} t\mathit{with}\left[t\right]=\left[s\right]\mathit{and}{t}^{\prime }\mathit{with}\left[t^{\prime }\right]=\left[s^{\prime }\right] \mathit{such\; that} t\overline{X}{t}^{\prime }.$$

Proof. 

First, we show that f is well-defined. Suppose $\left[u\right]=\left[v\right]$, i.e., $uZv$. We need to prove that $f\left(\right[u\left]\right)=f\left(\right[v\left]\right)$. Since X is a global bisimulation, there are $u^{\prime }$ and $v^{\prime }$ such that $uX{u}^{\prime }$ and $vX{v}^{\prime }$. Then, we have $f\left(\left[u\right]\right)=\left[u^{\prime }\right]$ and $f\left(\left[v\right]\right)=\left[v^{\prime }\right]$, so it suffices to show that $\left[u^{\prime }\right]=\left[v^{\prime }\right]$, i.e., $u^{\prime }{Z}^{\prime }{v}^{\prime }$. We have $u^{\prime }{X}^{-1}uZvX{v}^{\prime }$, so $u^{\prime }(X^{-1}\circ Z\circ X)v^{\prime }$. Thus, $u^{\prime }{Z}^{\prime }{v}^{\prime }$ since $Z^{\prime }$ is the largest bisimulation on ${\mathfrak{M}}^{\prime }$.

Now, we prove the second claim of the lemma. Let $f\left(\left[s\right]\right)=\left[s^{\prime }\right]$. Suppose, for a contradiction, that for all t with $\left[t\right]=\left[s\right]$ and $t^{\prime }$ with $\left[t^{\prime }\right]=\left[s^{\prime }\right]$, it is not the case that $t\overline{X}{t}^{\prime }.$ In particular, for $t=s$ and $t^{\prime }=s^{\prime }$, it is not the case that $s\overline{X}{s}^{\prime }$. This means that there is either $v\in s$ such that, for every $v^{\prime }\in s^{\prime }$, it is not the case that $vX{v}^{\prime }$, i.e., $f\left(\left[v\right]\right)\ne \left[v^{\prime }\right]$, or there is $v^{\prime }\in s^{\prime }$ such that, for all $v\in s$, it is not the case that $vX{v}^{\prime }$, i.e., $f\left(\left[v\right]\right)\ne \left[v^{\prime }\right]$. Since $\left[v^{\prime }\right]\in \left[s^{\prime }\right]$ for every $v^{\prime }\in s^{\prime }$ and $\left[v\right]\in \left[s\right]$ for every $v\in s$, we obtain a contradiction with $f\left(\left[s\right]\right)=\left[s^{\prime }\right]$ in both cases.

Conversely, suppose that there are t with $\left[t\right]=\left[s\right]$ and $t^{\prime }$ with $\left[t^{\prime }\right]=\left[s^{\prime }\right]$ such that $t\overline{X}{t}^{\prime }$. This means that for every $v\in t$ there is $v^{\prime }\in t^{\prime }$ such that $vX{v}^{\prime }$, i.e., $f\left(\left[v\right]\right)=\left[v^{\prime }\right]$, and vice versa. Suppose, for a contradiction, that $f\left(\left[s\right]\right)/\subseteq \left[s^{\prime }\right]$ or $\left[s^{\prime }\right]/\subseteq f\left(\left[s\right]\right)$.

Let $f\left(\left[s\right]\right)/\subseteq \left[s^{\prime }\right]$. This means that there is $\left[v^{\prime }\right]\in f\left(\left[s\right]\right)$ such that $\left[v^{\prime }\right]\notin \left[s^{\prime }\right]$. Then, there is $\left[v\right]\in \left[s\right]$ such that $\left[v^{\prime }\right]=f\left(\left[v\right]\right)$. Since $\left[v\right]\in \left[s\right]=\left[t\right]$, there is $u\in t$ such that $\left[u\right]=\left[v\right]$, so we have $\left[v^{\prime }\right]=f\left(\left[u\right]\right)$. Since $\left[v^{\prime }\right]\notin \left[s^{\prime }\right]=\left[t^{\prime }\right],$ it follows that $f\left(\left[u\right]\right)\notin \left[t^{\prime }\right]$, so for every $u^{\prime }\in t^{\prime }$, we obtain $\left[u^{\prime }\right]\ne f\left(\left[u\right]\right)$, which contradicts $t\overline{X}{t}^{\prime }$.

Let $\left[s^{\prime }\right]/\subseteq f\left(\left[s\right]\right)$. This means that there is $\left[v^{\prime }\right]\in \left[s^{\prime }\right]$ such that $\left[v^{\prime }\right]\notin f\left(\left[s\right]\right)$, i.e., $\left[v^{\prime }\right]\ne f\left(\left[v\right]\right)$ for every $\left[v\right]\in \left[s\right]$. Since $\left[v^{\prime }\right]\in \left[s^{\prime }\right]=\left[t^{\prime }\right]$, there is $u^{\prime }\in t^{\prime }$ such that $\left[u^{\prime }\right]=\left[v^{\prime }\right]$, so we have $\left[u^{\prime }\right]\ne f\left(\left[v\right]\right)$ for every $\left[v\right]\in \left[s\right]=\left[t\right]$. Then, for every $u\in t$, we obtain $\left[u^{\prime }\right]\ne f\left(\left[u\right]\right)$, which contradicts $t\overline{X}{t}^{\prime }$. □

Now, we can prove that global bisimilarity between inquisitive modal models is characterized by isomorphism between their minimal versions.

Theorem 1.

Let $\mathfrak{M}=(W,\Sigma ,V)$ and ${\mathfrak{M}}^{\prime }=(W^{\prime },{\Sigma }^{\prime },V^{\prime })$ be inquisitive modal models, and let Z and $Z^{\prime }$ be the largest bisimulations on $\mathfrak{M}$ and ${\mathfrak{M}}^{\prime }$, respectively. The models $\mathfrak{M}$ and ${\mathfrak{M}}^{\prime }$ are globally bisimilar if and only if the bisimulation quotients ${\mathfrak{M}}_Z$ and ${\mathfrak{M}}_{Z^{\prime }}^{\prime }$ are isomorphic.

Proof. 

Let X be a global bisimulation between the models $\mathfrak{M}$ and ${\mathfrak{M}}^{\prime }$. Let $f:W_Z\to W_{Z^{\prime }}^{\prime }$ be the function defined by

$$f\left(\left[w\right]\right)=\left[w^{\prime }\right]\mathrm{if} \mathrm{there} \mathrm{exist} x\in \left[w\right]\mathrm{and}{x}^{\prime }\in \left[w^{\prime }\right]\mathrm{such} \mathrm{that} xX{x}^{\prime }.$$

By the previous lemma, the function f is well-defined.

We first show the surjectivity of f. Let $\left[w^{\prime }\right]\in W_{Z^{\prime }}^{\prime }$. Take any $x^{\prime }\in \left[w^{\prime }\right]$. Since X is a global bisimulation, there is $w\in W$ such that $wX{x}^{\prime }$. Since $w\in \left[w\right]$, it follows that $f\left(\left[w\right]\right)=\left[w^{\prime }\right]$.

Now, we show the injectivity of f. Let $f\left(\right[u\left]\right)=f\left(\right[v\left]\right)$. Then, there are $x\in \left[u\right]$ and $x^{\prime }\in f\left(\left[v\right]\right)$ such that $xX{x}^{\prime }$. Since $x\in \left[u\right]$, it follows that $uZx$, and since $x^{\prime }\in f\left(\left[v\right]\right)$, i.e., $\left[x^{\prime }\right]=f\left(\left[v\right]\right)$, there are $y\in \left[v\right]$ and $y^{\prime }\in \left[x^{\prime }\right]$ such that $yX{y}^{\prime }$. From $y\in \left[v\right]$, it follows that $yZv$, and since $y^{\prime }\in \left[x^{\prime }\right]$, we obtain $y^{\prime }{Z}^{\prime }{x}^{\prime }$. Now, we have $uZxX{x}^{\prime }{Z}^{\prime -1}{y}^{\prime }{X}^{-1}yZv$, so $u(Z\circ X\circ Z^{\prime -1}\circ X^{-1}\circ Z)v$. Thus, $uZv$, since Z is the largest bisimulation on $\mathfrak{M}$. Hence, $\left[u\right]=\left[v\right]$.

It remains to prove $\left(i\right)$ and $\left(ii\right)$ to show that f is an isomorphism:

$\left(i\right)$

Let $\left[w\right]\in V_Z\left(p\right)$. Then, $w\in V\left(p\right)$. Since X is a global bisimulation, there is $w^{\prime }\in W^{\prime }$ such that $wX{w}^{\prime }$. Now, we have $f\left(\left[w\right]\right)=\left[w^{\prime }\right]$ and $w^{\prime }\in V^{\prime }\left(p\right)$, so $f\left(\left[w\right]\right)\in V_{Z^{\prime }}^{\prime }\left(p\right)$. The converse is proved in a similar way.

$\left(ii\right)$

Let $\left[s\right]\in {\Sigma }_Z\left(\left[w\right]\right)$. Then, there are $x\in \left[w\right]$ and t with $\left[t\right]=\left[s\right]$ such that $t\in \Sigma \left(x\right)$. Since X is a global bisimulation, there is $x^{\prime }\in W^{\prime }$ such that $xX{x}^{\prime }$, so by condition $\left(forth\right)$ of the bisimulation X, there is $t^{\prime }\in {\Sigma }^{\prime }\left(x^{\prime }\right)$ such that $t\overline{X}{t}^{\prime }$.

Now, $x\in \left[w\right]$ and $xX{x}^{\prime }$ imply $\left[x\right]=\left[w\right]$ and $f\left(\left[x\right]\right)=\left[x^{\prime }\right]$, so we obtain $f\left(\left[w\right]\right)=\left[x^{\prime }\right]$. Similarly, since $t\overline{X}{t}^{\prime }$, by the previous lemma, we obtain $f\left(\left[t\right]\right)=\left[t^{\prime }\right]$, so from $\left[t\right]=\left[s\right]$, it follows that $f\left(\left[s\right]\right)=\left[t^{\prime }\right]$. Now, $x^{\prime }\in f\left(\left[w\right]\right)$, $\left[t^{\prime }\right]=f\left(\left[s\right]\right)$ and $t^{\prime }\in {\Sigma }^{\prime }\left(x^{\prime }\right)$ imply $f\left(\left[s\right]\right)\in {\Sigma }_{Z^{\prime }}^{\prime }\left(f\left(\left[w\right]\right)\right)$.

Conversely, suppose $f\left(\left[s\right]\right)\in {\Sigma }_{Z^{\prime }}^{\prime }\left(f\left(\left[w\right]\right)\right)$. Then, there are $x^{\prime }\in f\left(\left[w\right]\right)$ and $t^{\prime }$ with $\left[t^{\prime }\right]=f\left(\left[s\right]\right)$ such that $t^{\prime }\in {\Sigma }^{\prime }\left(x^{\prime }\right)$. Since X is a global bisimulation, there is $x\in W$ such that $xX{x}^{\prime }$, so by condition $\left(back\right)$ of the bisimulation X, there is $t\in \Sigma \left(x\right)$ such that $t\overline{X}{t}^{\prime }$.

Since $x^{\prime }\in f\left(\left[w\right]\right)$, i.e., $\left[x^{\prime }\right]=f\left(\left[w\right]\right)$, by the definition of f, there are $y\in \left[w\right]$ and $y^{\prime }\in \left[x^{\prime }\right]$ such that $yX{y}^{\prime }$. Now, we have $xX{x}^{\prime }{Z}^{\prime }{y}^{\prime }{X}^{-1}yZw$, so $x(X\circ Z^{\prime }\circ X^{-1}\circ Z)w$ and thus $xZw$ since Z is the largest bisimulation on $\mathfrak{M}$.

Similarly, from $\left[t^{\prime }\right]=f\left(\left[s\right]\right)$, it follows by the previous lemma that there are r with $\left[r\right]=\left[s\right]$ and $r^{\prime }$ with $\left[r^{\prime }\right]=\left[t^{\prime }\right]$ such that $r\overline{X}{r}^{\prime }$. Then, we have $t\overline{X}{t}^{\prime }\overline{Z^{\prime }}{r}^{\prime }\overline{X^{-1}}r\overline{Z}s$, so $t(\overline{X}\circ \overline{Z^{\prime }}\circ \overline{X^{-1}}\circ \overline{Z})s$ and thus $t\overline{Z}s$ since Z is the largest bisimulation on $\mathfrak{M}$.

Now, from $xZw$ and $t\overline{Z}s$, it follows that $x\in \left[w\right]$ and $\left[t\right]=\left[s\right]$, so, since $t\in \Sigma \left(x\right)$, we obtain $\left[s\right]\in {\Sigma }_Z\left(\left[w\right]\right)$.

Hence, f is an isomorphism.

For the converse, let $f:W_Z\to W_{Z^{\prime }}^{\prime }$ be an isomorphism of the models ${\mathfrak{M}}_Z$ and ${\mathfrak{M}}_{Z^{\prime }}^{\prime }$. We define a relation $X\subseteq W\times W^{\prime }$ by

$$wX{w}^{\prime }\mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if} f\left(\left[w\right]\right)=\left[w^{\prime }\right].$$

We claim that X is a global bisimulation.

Suppose $w\in W$. Then, there is $w^{\prime }\in W^{\prime }$ such that $f\left(\left[w\right]\right)=\left[w^{\prime }\right]$, so we obtain $wX{w}^{\prime }$. Conversely, suppose $w^{\prime }\in W^{\prime }$. By the surjectivity of f, there is $\left[w\right]\in W_Z$ such that $f\left(\left[w\right]\right)=\left[w^{\prime }\right]$, so we obtain $wX{w}^{\prime }$. This shows the globality of the relation X.

It remains to show that X is a bisimulation:

$\left(at\right)$

Let $wX{w}^{\prime }$. From the definitions of bisimulation quotient and isomorphism and from the claim $f\left(\left[w\right]\right)=\left[w^{\prime }\right]$, it follows that $w\in V\left(p\right)$ if and only if $\left[w\right]\in V_Z\left(p\right)$ if and only if $f\left(\left[w\right]\right)\in V_{Z^{\prime }}^{\prime }\left(p\right)$ if and only if $\left[w^{\prime }\right]\in V_{Z^{\prime }}^{\prime }\left(p\right)$ if and only if $w^{\prime }\in V^{\prime }\left(p\right)$.

$\left(forth\right)$

Let $wX{w}^{\prime }$ and $s\in \Sigma \left(w\right)$. We need to find $s^{\prime }\in {\Sigma }^{\prime }\left(w^{\prime }\right)$ such that $s\overline{X}{s}^{\prime }$.

By the definition of the bisimulation quotient, we have $\left[s\right]\in {\Sigma }_Z\left(\left[w\right]\right)$ and, since f is an isomorphism, it follows that $f\left(\left[s\right]\right)\in {\Sigma }_{Z^{\prime }}^{\prime }\left(f\left(\left[w\right]\right)\right)$, i.e., $f\left(\left[s\right]\right)\in {\Sigma }_{Z^{\prime }}^{\prime }\left(\left[w^{\prime }\right]\right))$. This means that there are $x\in \left[w^{\prime }\right]$ and $t\subseteq W^{\prime }$ with $\left[t\right]=f\left(\right[s\left]\right)$ such that $t\in {\Sigma }^{\prime }\left(x\right)$. Since $x\in \left[w^{\prime }\right]$, we have $x{Z}^{\prime }{w}^{\prime }$, so, since $t\in {\Sigma }^{\prime }\left(x\right)$, by the condition $\left(forth\right)$ of the bisimulation $Z^{\prime }$, there is $s^{\prime }\in {\Sigma }^{\prime }\left(w^{\prime }\right)$ such that $t\overline{Z^{\prime }}{s}^{\prime }$. Now, $t\overline{Z^{\prime }}{s}^{\prime }$ implies $\left[t\right]=\left[s^{\prime }\right]$, so by $\left[t\right]=f\left(\right[s\left]\right)$, we obtain $\left[s^{\prime }\right]=f\left(\left[s\right]\right)$.

It remains to show $s\overline{X}{s}^{\prime }$. Suppose $v\in s$. Then, $\left[v\right]\in \left[s\right]$, so we have $f\left(\right[v\left]\right)\in f\left(\right[s\left]\right)$. Now, we have $f\left(\left[v\right]\right)\in \left[s^{\prime }\right]$, so there is $\left[u^{\prime }\right]\in \left[s^{\prime }\right]$ such that $f\left(\left[v\right]\right)=\left[u^{\prime }\right]$. Since $\left[u^{\prime }\right]\in \left[s^{\prime }\right]$, there is $v^{\prime }\in s^{\prime }$ such that $\left[u^{\prime }\right]=\left[v^{\prime }\right]$, so we obtain $f\left(\left[v\right]\right)=\left[v^{\prime }\right]$. Hence, $vX{v}^{\prime }$. The converse can be proved in a similar way, so we have $s\overline{X}{s}^{\prime }$.

$\left(back\right)$

Similar to condition $\left(forth\right)$.

Hence, X is a global bisimulation. □

5. Conclusions and Future Work

The field of inquisitive modal logic has grown in recent years, and it continues to develop further. In this paper, we make a new contribution to the model theory of inquisitive modal logic. We investigate the notion of bisimulation in ${\mathrm{InqML}}_{⊞}$ and show some of its basic properties. Moreover, we define the bisimulation quotient and show that an inquisitive modal model is related to its bisimulation quotient by modal equivalence. We also prove that two models are globally bisimilar if and only if their bisimulation quotients are isomorphic. These results are well-known in basic modal logic (as well as in several other logics). However, since inquisitive modal logic has a more complex semantics in which not only sets of worlds but also sets of sets of worlds are considered, it is important to verify whether these results hold here and whether the standard proof techniques can be adapted to this logic.

The logic ${\mathrm{InqML}}_{⊞}$ was generalized to the inquisitive modal logic ${\mathrm{InqML}}_{\Rrightarrow }$ in [11]. The language of ${\mathrm{InqML}}_{\Rrightarrow }$ is based on a binary modal operator ⇛, and the formulas of ${\mathrm{InqML}}_{\Rrightarrow }$ are interpreted over models that do not require a downward closure condition. It is expected that the results of this work can be generalized to the inquisitive modal logic ${\mathrm{InqML}}_{\Rrightarrow }$.