Characterizing Frame Definability in Team Semantics via the Universal Modality

Abstract

Let denote the fragment of modal logic extended with the universal modality in which the universal modality occurs only positively. We characterize the definability of in the spirit of the well-known Goldblatt–Thomason theorem. We show that an elementary class ${\mathbb {F}}$ of Kripke frames is definable in if and only if ${\mathbb {F}}$ is closed under taking generated subframes and bounded morphic images, and reflects ultrafilter extensions and finitely generated subframes. In addition, we initiate the study of modal frame definability in team-based logics. We show that, with respect to frame definability, the logics , modal logic with intuitionistic disjunction, and (extended) modal dependence logic all coincide. Thus we obtain Goldblatt–Thomason -style theorems for each of the logics listed above.

The work of the first author was partially supported by JSPS KAKENHI Grant-in-Aid for Young Scientists (B) Grant Numbers 24700146 and 15K21025. The work of the second author was supported by grant 266260 of the Academy of Finland, and by Jenny and Antti Wihuri Foundation.

Appendices

A Modal Logic with Universal Modality

Proposition A.1

Let such that $\psi$ is closed. Then, $(\mathrm {i})$ $\Box (\varphi \vee \psi )\equiv _{K} (\Box \varphi \vee \psi );$ $(\mathrm {ii})$ $\Diamond (\varphi \wedge \psi ) \equiv _{K} (\Diamond \varphi \wedge \psi );$ $(\mathrm {iii})$ .

Proof

$(\mathrm {i})$ and $(\mathrm {iii})$ follow from [10, Proposition 3.6]. $(\mathrm {ii})$ is completely analogous to the item $(\mathrm {i})$. $\square$

Theorem A.1

For each -formula $\varphi$, there exists a -formula $\psi$ in -form such that $\varphi \equiv _K\psi$.

Proof

The proof is done by induction on $\varphi$. The cases for literals and connectives are trivial. As for the case $\varphi = \Box \psi$, we proceed as follows. By induction hypothesis there exists a conjunctive -form $\bigwedge _{i \in I} \psi _{i}$, where each $\psi _i$ is a disjunctive -clause, such that $\bigwedge _{i \in I} \psi _{i} \equiv _{K}\psi$. By the semantics of $\Box$, we have that $(\Box \psi \equiv _{K})$ $\Box \bigwedge _{i \in I} \psi _{i} \equiv _{K} \bigwedge _{i \in I} \Box \psi _{i}$. Now since each $\psi _i$ is a disjunctive -clause, it follows from item $(\mathrm {i})$ of Proposition A.1 that, for each $i \in I$, the formula $\Box \psi _i$ is equivalent to some disjunctive -clause $\psi _i'$. Thus $\bigwedge _{i \in I} \psi _{i}'$ is a conjunctive -form that is equivalent to $\Box \psi$.

The proof for the case of is otherwise the same as the proof for the case $\Box \varphi$, but instead of item $(\mathrm {i})$ of Proposition A.1, item $(\mathrm {iii})$ is used. The proof for the case $\Diamond \varphi$ is likewise analogous to that of $\Box \varphi$. The proof uses a disjunctive -form instead of the conjunctive one and item $(\mathrm {ii})$ of Proposition A.1 instead of item $(\mathrm {i})$. $\square$

Example A.1

Consider the following examples from [10, p.14]: the formula defines the class $\{(W,R)\in {\mathbb {F}}_{{\mathrm {all}}}\mid |W |= 1\}$, whereas the formula defines the class $\{(W,R)\in {\mathbb {F}}_{{\mathrm {all}}}\mid R \ne \emptyset \}$. Clearly, the former is not closed under taking disjoint unions, and the latter is not closed under taking generated subframes. Note that both of the classes above are elementary.

Lemma A.1

For each -formula $\varphi$, there exists a finite set ${\varGamma }$ of closed disjunctive -clauses such that $\mathfrak {M} \Vdash \varphi$ iff $\mathfrak {M} \Vdash {\varGamma }$ for every model $\mathfrak {M}$.

Proof

Let $\varphi$ be an -formula. By Theorem 1, we may assume that $\varphi$ is a conjunctive -form $\bigwedge _{i \in I} \psi _{i}$, where each is a disjunctive -clause. By Proposition A.1 (iii), for each $i\in I$, is equivalent to the closed disjunctive -clause . Thus, for every model ${\mathfrak {M}}$,

Proposition A.2

.

Proof

The direction ${\le _F}$ is trivial. We show that . Consider any -definable class of frames $\mathbb {F}$. Let ${\varGamma }$ be a set of formulae that defines $\mathbb {F}$. By Lemma A.1, for each $\varphi \in {\varGamma }$, there is a finite set ${\varDelta }_{\varphi }$ of closed disjunctive -clauses such that $\mathfrak {M} \Vdash \varphi$ iff $\mathfrak {M} \Vdash {\varDelta }_{\varphi }$ for every Kripke model $\mathfrak {M}$. It follows that $\mathfrak {M} \Vdash {\varGamma }$ iff $\mathfrak {M} \Vdash \bigcup _{\varphi \in {\varGamma }} {\varDelta }_{\varphi }$ for every Kripke model $\mathfrak {M}$. Therefore, $\bigcup _{\varphi \in {\varGamma }} {\varDelta }_{\varphi }$ also defines $\mathbb {F}$, as desired. $\square$

Proposition A.3

Let ${\mathfrak {F}}$ be a frame and $\varphi$ a closed disjunctive -clause. If ${\mathfrak {F}}\Vdash \varphi$, then ${\mathfrak {G}}\Vdash \varphi$ for all generated subframes ${\mathfrak {G}}$ of ${\mathfrak {F}}$.

Proof

Fix any generated subframe ${\mathfrak {G}}$ of a frame ${\mathfrak {F}}$ and put . Suppose that ${\mathfrak {F}}\Vdash \varphi$. To show ${\mathfrak {G}}\Vdash \varphi$, fix any valuation V and any state w in ${\mathfrak {G}}$. We show that for some $i \in I$. Since we can regard V as a valuation on ${\mathfrak {F}}$, . Thus there is some $i \in I$ such that $({\mathfrak {F}},V), u \Vdash \psi _{i}$, for every $u\in |{\mathfrak {F}}|$. Fix such $i \in I$. Since $\psi _{i}$ is in ${\mathcal {ML}}$ and the satisfaction of ${\mathcal {ML}}$ is invariant under taking generated submodels (cf. [3, Proposition2.6]), $({\mathfrak {G}},V), u \Vdash \psi _{i}$ for every $u\in {\mathfrak {G}}$. Therefore, , as desired. $\square$

Proposition A.4

Let ${\mathfrak {F}}$ be a frame and $\varphi$ a closed disjunctive -clause. If ${\mathfrak {G}}\Vdash \varphi$ for all finitely generated subframes ${\mathfrak {G}}$ of ${\mathfrak {F}}$, then ${\mathfrak {F}}\Vdash \varphi$.

Proof

We show the contrapositive implication. Let $\varphi$ be and suppose that . Now, we can find a valuation V and a state w such that for all $i \in I$. Thus, for each $i \in I$, there is a state $w_{i}$ such that $({\mathfrak {F}},V),w_{i} \not \Vdash \psi _{i}$. Define $X := \left\{ \, {w_{i}} \,|\, {i \in I} \,\right\}$ and note that X is finite. Consider the submodel $({\mathfrak {F}}_{X},V_{X})$ of ${\mathfrak {F}}$ generated by X. Since for each $i\in I$, $({\mathfrak {F}},V),w_{i} \not \Vdash \psi _{i}$ and $\psi _{i}\in {\mathcal {ML}}$, and since the satisfaction of ${\mathcal {ML}}$ is invariant under generated submodels (cf. [3, Proposition2.6]), it follows that $({\mathfrak {F}}_{X},V_{X}),w_{i} \not \Vdash \psi _{i}$ for each $i\in I$. Thus for each $i \in I$. Hence , which implies our goal . $\square$

B Goldblatt-Thomason Theorem

Definition B.1

(Satisfiability). Let ${\varGamma }$ be a set of formulae, ${\mathfrak {M}}$ a model and ${\mathbb {F}}$ a class of frames. We say that ${\varGamma }$ is satisfiable in ${\mathfrak {M}}$ if there exists a point w of ${\mathfrak {M}}$ such that ${\mathfrak {M}},w\Vdash \gamma$ for all $\gamma \in {\varGamma }$. We say that ${\varGamma }$ is finitely satisfiable in ${\mathfrak {M}}$ if each finite subset of ${\varGamma }$ is satisfiable in ${\mathfrak {M}}$. We say that ${\varGamma }$ is satisfiable in ${\mathbb {F}}$ if there exists a frame ${\mathfrak {F}}\in {\mathbb {F}}$ and a valuation V on ${\mathfrak {F}}$ such that ${\varGamma }$ is satisfiable in $({\mathfrak {F}}, V)$. Finally, we say that ${\varGamma }$ is finitely satisfiable in ${\mathbb {F}}$ if each finite subset of ${\varGamma }$ is satisfiable in ${\mathbb {F}}$.

Theorem B.1

Given any elementary frame class ${\mathbb {F}}$, the following are equivalent:

• (i) ${\mathbb {F}}$ is -definable.

• (ii) ${\mathbb {F}}$ is closed under taking generated subframes and bounded morphic images, and reflects ultrafilter extensions and finitely generated subframes.

Proof

The direction from $(\mathrm {i})$ to $(\mathrm {ii})$ follows directly by Propositions 1 and 3, and Theorem 2. In the proof of the converse direction, we use some notions from first-order model theory such as elementary extensions and $\omega$-saturation. The reader unfamiliar with them is referred to [5]. Assume $(\mathrm {ii})$ and define . We show that, for any frame ${\mathfrak {F}}$, ${\mathfrak {F}}\in {\mathbb {F}}$ iff ${\mathfrak {F}}\Vdash \mathrm {Log}({\mathbb {F}})$.

Consider any ${\mathfrak {F}}=(W,R)$. It is trivial to show the Only-If-direction, and so we show the If-direction. Assume that ${\mathfrak {F}}\Vdash \mathrm {Log}({\mathbb {F}})$. To show ${\mathfrak {F}}\in {\mathbb {F}}$, we may assume, without loss of generality, that ${\mathfrak {F}}$ is finitely generated. This is because: otherwise, it would suffice to show, since ${\mathbb {F}}$ reflects finitely generated subframes, that ${\mathfrak {G}}\in {\mathbb {F}}$ for all finitely generated subframes ${\mathfrak {G}}$ of ${\mathfrak {F}}$. Let U be a finite generator of ${\mathfrak {F}}$. Let us expand our syntax with a (possibly uncountable) set $\left\{ \, {p_{A}} \,|\, {A \subseteq W} \,\right\}$ of new propositional variables and define ${\varDelta }$ to be the set containing exactly:

$$\begin{aligned} p_{A \cap B}\leftrightarrow p_{A} \wedge p_{B}, \quad p_{W \setminus A}\leftrightarrow \lnot p_{A}, \quad p_{m_{R}(A)}\leftrightarrow \Diamond p_{A}, \quad p_{W}, \end{aligned}$$

where $A, B \subseteq W$ and $m_{R}(A) := \left\{ \, {x \in W} \,|\, {xRy\text { for some }y \in A} \,\right\}$. Define

$$\begin{aligned} {\varDelta }_{{\mathfrak {F}},u} :=\{p_{\{u\}} \wedge \Box ^n \varphi \mid n\in \omega \text { and } \varphi \in {\varDelta }\}, \end{aligned}$$

for each $u\in U$. Recall that ${\mathfrak {F}}$ is finitely generated by U. The intuition here is that $({\varDelta }_{{\mathfrak {F}},u})_{u \in U}$ provides a “complete enough description” of $\mathfrak {F}$.

Let us introduce a finite set $\{x_{u} | u \in U\}$ of variables in first-order syntax and let $ST_{x_{u}}$ be the standard translation from to the corresponding first-order logic via the variable $x_{u}$. We will show that $\bigcup _{u \in U} \{ST_{x_{u}} (\varphi ) \,|\, \varphi \in {\varDelta }_{{\mathfrak {F}},u}\}$ is satisfiable in ${\mathbb {F}}$ in the sense of the satisfaction in first-order model theory. Since ${\mathbb {F}}$ is elementary, it follows from the compactness of first-order logic that it suffices to show that $\bigcup _{u \in U} \{ ST_{x_{u}} (\varphi ) \,|\, \varphi \in {\varDelta }_{{\mathfrak {F}},u}\}$ is finitely satisfiable in ${\mathbb {F}}$. Let ${\varGamma }$ be a finite subset of this set. Then, we may write ${\varGamma }= \bigcup _{1 \le k \le n} ST_{x_{u_{k}}} [{\varGamma }_{u_{k}}]$ for some $u_{1}$, ..., $u_{n} \in U$ and some finite ${\varGamma }_{u_{k}} \subseteq {\varDelta }_{{\mathfrak {F}},u_{k}}$ ($1 \le k \le n$). Assume, for the sake of a contradiction, that ${\varGamma }$ is not satisfiable in ${\mathbb {F}}$. It follows that ${\mathbb {F}}\Vdash \vartheta$ in the sense of modal logic, where . Since $\vartheta$ is an -formula, it belongs to $\mathrm {Log}({\mathbb {F}})$. Thus by the assumption ${\mathfrak {F}}\Vdash \mathrm {Log}({\mathbb {F}})$, we conclude that ${\mathfrak {F}}\Vdash \vartheta$; and therefore ${\varGamma }$ is not satisfiable in $\mathfrak {F}$ in the sense of first-order model theory. However, ${\varGamma }$ is clearly satisfiable in ${\mathfrak {F}}$ under the natural structure interpreting $p_{A}$ as A and the natural assignment sending $x_{u}$ to u. This is a contradiction. Therefore, $\bigcup _{u \in U} \{ ST_{x_{u}} (\varphi ) \,|\, \varphi \in {\varDelta }_{{\mathfrak {F}},u}\}$ is satisfiable in ${\mathbb {F}}$.

Let ${\mathfrak {G}}\in {\mathbb {F}}$ be such that $\bigcup _{u \in U} \{ST_{x_{u}} (\varphi ) \,|\, \varphi \in {\varDelta }_{{\mathfrak {F}},u}\}$ is satisfiable in ${\mathfrak {G}}$. Let us fix a valuation V and a finite set $Z := \{w_{u} |u \in U\}$ of points such that $\bigcup _{u \in U} \{ST_{x_{u}} (\varphi ) \,|\, \varphi \in {\varDelta }_{{\mathfrak {F}},u}\}$ is satisfied in $({\mathfrak {G}},V)$ under an assignment sending each $x_{u}$ to $w_{u}$. Then, $({\mathfrak {G}},V), w_{u} \Vdash {\varDelta }_{{\mathfrak {F}},u}$. Now let $({\mathfrak {G}}^*_Z,V^*_Z)$ denote some $\omega$-saturated elementary extension of the Z generated submodel of $({\mathfrak {G}},V)$. It is easy to check that $({\mathfrak {G}}^*_Z,V^*_Z), w_{u}^{*} \Vdash {\varDelta }_{{\mathfrak {F}},u}$ where $w_{u}^{*}$ is the corresponding element in ${\mathfrak {G}}^*_Z$ to $w_{u}$ of ${\mathfrak {G}}_Z$ and that $({\mathfrak {G}}^*_Z,V^*_Z)\Vdash {\varDelta }$. Since ${\mathbb {F}}$ is elementary and closed under taking generated subframes, we conclude first that ${\mathfrak {G}}_{Z} \in {\mathbb {F}}$ and then that ${\mathfrak {G}}^*_Z\in {\mathbb {F}}$. We can now prove the following claim.

Claim. The ultrafilter extension $\mathfrak {ue F}$ is a bounded morphic image of ${\mathfrak {G}}_{Z}^{*}$.

By closure of ${\mathbb {F}}$ under bounded morphic images, we oftain $\mathfrak {ueF} \in {\mathbb {F}}$. Finally, since ${\mathbb {F}}$ reflects ultrafilter extensions, ${\mathfrak {F}}\in {\mathbb {F}}$, as required. $\square$

(Proof of Claim ) Define a mapping $f: |{\mathfrak {G}}_{Z}^{*}| \rightarrow \mathrm {Uf}(W)$ (where $\mathrm {Uf}(W)$ is the set of all ultrafilters on W) by

$$\begin{aligned} f(s) := \left\{ \, {A \subseteq W} \,|\, { ({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), s \Vdash p_{A}} \,\right\} . \end{aligned}$$

We will show that (a) f(s) is an ultrafilter on W; (b)f is a bounded morphism; (c) f is surjective. Below, we denote by S the underlying binary relation of ${\mathfrak {G}}_{Z}^{*}$.

• (a) f(u) is an ultrafilter: Follows immediately from the fact that $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}) \Vdash {\varDelta }$.

• (b1) f satisfies (Forth): We show that $sSs'$ implies $f(s)R^{\mathfrak {ue}}f(s')$. Assume that $sSs'$. By the definition of $R^{\mathfrak {ue}}$, it suffices to show that $A \in f(s')$ implies $m_{R}(A) \in f(s)$. Suppose $A \in f(s')$. Thus $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), s' \Vdash p_{A}$. Since $sSs'$, we obtain $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), s \Vdash \Diamond p_{A}$. Since $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}) \Vdash {\varDelta }$, $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}) \Vdash \Diamond p_{A} \leftrightarrow p_{m_{R}(A)}$. Therefore $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), s \Vdash p_{m_{R}(A)}$, and hence $m_{R}(A) \in f(s)$, as desired.

• (b2) f satisfies (Back): We show that $f(s)R^{\mathfrak {ue}}\mathcal {U}$ implies $sSs'$ and $f(s')$ = $\mathcal {U}$ for some $s' \in |{\mathfrak {G}}_{Z}^{*}|$. Assume that $f(s)R^{\mathfrak {ue}}\mathcal {U}$. We will find a state $s'$ such that $sSs'$ and $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), s' \Vdash p_{A}$ for all $A \in \mathcal {U}$. By $\omega$-saturation, it suffices to show that $\left\{ \, {p_{A}} \,|\, {A \in \mathcal {U}} \,\right\}$ is finitely satisfiable in the set $\left\{ \, {t \in |{\mathfrak {G}}_{Z}^{*}|} \,|\, {sSt} \,\right\}$ of the successors of s. Take any $A_{1}$, $\ldots$, $A_{n} \in \mathcal {U}$. Then, $\bigcap _{1 \le i \le n} A_{i} \in \mathcal {U}$. Now since $f(s)R^{\mathfrak {ue}}\mathcal {U}$, $m_{R}(\bigcap _{1 \le i \le n} A_{i}) \in f(s)$. Hence $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), s \Vdash p_{m_{R}(\bigcap _{1 \le i \le n} A_{i})}$. Since $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}) \Vdash {\varDelta }$, $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}) \Vdash p_{m_{R}(\bigcap _{1 \le i \le n} A_{i})} \leftrightarrow \Diamond p_{\bigcap _{1 \le i \le n} A_{i}}$. Therefore $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), s \Vdash \Diamond p_{\bigcap _{1 \le i \le n} A_{i}}$. Thus there is a state $s' \in |{\mathfrak {G}}_{Z}^{*}|$ such that $sSs'$ and $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), s' \Vdash p_{\bigcap _{1 \le i \le n} A_{i}}$. Therefore and since $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}) \Vdash {\varDelta }$, it follows that $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), s' \Vdash p_{A_{i}}$ for all $1 \le i \le n$.

• (c) f is surjective: Let us take any ultrafilter $\mathcal {U} \in |\mathfrak {ueF}|$. To prove surjectiveness, we show that the set $\left\{ \, {p_{A}} \,|\, {A \in \mathcal {U}} \,\right\}$ is satisfiable in $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*})$. By $\omega$-saturatedness of $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*})$, it suffices to show finite satisfiability. Fix any $A_{1}, \ldots , A_{n} \in \mathcal {U}$. It follows that $\bigcap _{1 \le k \le n} A_{k} \in \mathcal {U}$, and hence $\bigcap _{1 \le k \le n} A_{k} \ne \emptyset$. Pick $w \in \bigcap _{1 \le k \le n} A_{k}$. Since ${\mathfrak {F}}$ is finitely generated by U, w is reachable (in ${\mathfrak {F}}$) from some point $u \in U$ in a finite number of steps. But then there is some $l \in \omega$ such that $({\mathfrak {F}}, V_{0}), u \Vdash p_{ (m_{R})^{l} ( \bigcap _{1 \le k \le n} A_{k} )}$, where $V_{0}$ is the natural valuation on ${\mathfrak {F}}$ sending $p_{X}$ to X. Since $V_0$ is the natural valuation, we also obtain that $u\in (m_{R})^{l}( \bigcap _{1 \le k \le n} A_{k})$, and thus ${\varDelta }$ contains $p_{\left\{ \,{u}\,\right\} } \leftrightarrow p_{\left\{ \,{u}\,\right\} } \wedge p_{(m_{R})^{l} ( \bigcap _{1 \le k \le n} A_{k})}$. It now follows from $({\mathfrak {G}}^*_Z,V^*_Z), w_{u}^{*} \Vdash {\varDelta }_{{\mathfrak {F}},u}$ that $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), w_{u}^{*} \Vdash p_{\left\{ \,{u}\,\right\} }$. Since $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}) \Vdash {\varDelta }$, we obtain $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), w_{u}^{*} \Vdash p_{ (m_{R})^{l} ( \bigcap _{1 \le k \le n} A_{k})}$, and hence also that $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*}), w_{u}^{*} \Vdash \Diamond ^{l} p_{\bigcap _{1 \le k \le n} A_{k}}$. Therefore, $\left\{ \,{p_{A_{1}},\ldots ,p_{A_{n}}}\,\right\}$ is satisfiable in $({\mathfrak {G}}_{Z}^{*},V_{Z}^{*})$. $\dashv$

C Modal Logics with Team Semantics

Proposition C.1

Let ${\varPhi }$ be an infinite set of proposition symbols. For every formula $\varphi \in {\mathcal {EMDL}} ({\varPhi })$ there exists a formula $\varphi ^*\in {\mathcal {MDL}} ({\varPhi })$ such that ${\mathfrak {F}}\models \varphi$ iff ${\mathfrak {F}}\models \varphi ^*$ for every frame ${\mathfrak {F}}$.

Proof

We give a sketch of the proof here. An analogous proof is given in the extended version of [19] (to appear). The translation $\varphi \mapsto \varphi ^*$ is defined inductively in the following way. For (negated) proposition symbols the translation is the identity. For propositional connectives and modalities we define

$$\begin{aligned} (\psi _1 \oplus \psi _2) \mapsto (\psi _1^* \oplus \psi _2^*), \quad \text {and}\quad \nabla \psi \mapsto \nabla \psi ^*, \end{aligned}$$

where $\oplus \in \{\wedge ,\vee \}$ and $\nabla \in \{\Diamond , \Box \}$. The only nontrivial case is the case for the dependence atoms. Let $\varphi$ be the dependence atom ${\mathrm {dep}} \!\left( \psi _1,\dots ,\psi _n\right)$, let k be the modal depth of $\varphi$, and let $p_1,\dots ,p_n$ be distinct fresh proposition symbols. Define

$$\begin{aligned} \varphi ^* :=\big (\bigwedge _{0\le i\le k} \Box ^i \bigwedge _{1\le j\le n} (p_j\leftrightarrow \psi _j) \big ) \rightarrow {\mathrm {dep}} \!\left( p_0,\dots ,p_n\right) . \end{aligned}$$

It is now straightforward to show that the claim follows.

Definition C.1

We say that an -formula $\varphi$ is in -normal form if for some $n\in \omega$ and $\psi _1,\psi _2,\dots ,\psi _n\in {\mathcal {ML}} ({\varPhi })$.

Proposition C.2

( -normal form, [19, 21]). For every -formula $\varphi$ there exists an equivalent formula in -normal form.

Lemma C.1

For every ${\mathcal {ML}}$-formula $\varphi$ and model ${\mathfrak {M}}$: iff ${\mathfrak {M}},W \models \varphi$.

Proof

By the semantics of iff ${\mathfrak {M}}, w \Vdash \varphi$ for every $w \in W$. Furthermore by Proposition 4, ${\mathfrak {M}}, w \Vdash \varphi$ for every $w \in W$ iff ${\mathfrak {M}}, W \models \varphi$. $\square$

Lemma C.2

For every -formula $\varphi$ there exists a closed disjunctive -clause $\varphi ^-$ such that ${\mathfrak {M}}\models \varphi$ iff ${\mathfrak {M}}\Vdash \varphi ^-$ for every Kripke model ${\mathfrak {M}}$.

Proof

Let $\varphi$ be an arbitrary -formula. By Proposition C.2, we may assume that , for some $n\in \omega$ and $\psi _1,\dots ,\psi _n\in {\mathcal {ML}}$. Let ${\mathfrak {M}}=(W,R,V)$ be an arbitrary model. It suffices to show . This is shown as follows.

Lemma C.3

For every closed disjunctive -clause there exists an -formula $\varphi ^{*}$ such that ${\mathfrak {M}}\Vdash \varphi$ iff ${\mathfrak {M}}\models \varphi ^*$ for every Kripke model ${\mathfrak {M}}$.

Proof

Let $\varphi$ be an arbitrary closed disjunctive -clause, i.e., for some $n\in \omega$ and $\psi _1,\dots ,\psi _n\in {\mathcal {ML}}$. Let ${\mathfrak {M}}=(W,R,V)$ be an arbitrary Kripke model. It suffices to show . We proceed as follows.