The Finite Model Property of Quasi-transitive Modal Logic

TL;DR

This paper proves the finite model property for a specific quasi-transitive modal logic and its extension to tense logic, using a Gentzen sequent calculus with finite algebra properties.

Contribution

It establishes the finite model property for $ extsf{K}_2^3$ and $ extsf{Kt}_2^3$ through a novel sequent calculus with finite algebra construction.

Findings

Finite model property proven for $ extsf{K}_2^3$ and $ extsf{Kt}_2^3$

Development of a Gentzen sequent calculus $ extsf{G}$

Sequent calculus has finite algebra property

Abstract

The finite model property of quasi-transitive modal logic $\mathsf{K}_2^3=\mathsf{K}\oplus \Box\Box p\rightarrow \Box\Box\Box p$ is established. This modal logic is conservatively extended to the tense logic $\mathsf{Kt}_2^3$. We present a Gentzen sequent calculus $\mathsf{G}$ for $\mathsf{Kt}_2^3$. The sequent calculus $\mathsf{G}$ has the finite algebra property by a finite syntactic construction. It follows that $\mathsf{Kt}_2^3$ and $\mathsf{K}_2^3$ have the finite model property.

Equations

$${\mathcal{L}}_{◊}\ni \phi ::=p\mid \perp \mid \top \mid \neg \phi \mid ({\phi }_1\wedge {\phi }_2)\mid ({\phi }_1\vee {\phi }_2)\mid ◊\phi ,\mathrm{where}p\in \mathsf{Prop}.$$

$$c(p)$$

$$c(\phi \odot \psi )$$

$$c(\#\phi )$$

$$\frac{s_1\dots s_n}{s_0}(R)$$

$$(\mathrm{Id})\phi \Rightarrow \phi (\mathrm{D})\phi \wedge (\psi \vee \chi )\Rightarrow (\phi \wedge \psi )\vee (\phi \wedge \chi )(\top )\phi \Rightarrow \top (\perp )⟨\perp {⟩}^n\Rightarrow \psi$$

$$(\mathrm{LC})\phi \wedge \neg \phi \Rightarrow \perp (\mathrm{EM})\top \Rightarrow \phi \vee \neg \phi ({◊}_3^2){◊}^3\phi \Rightarrow {◊}^2\phi$$

$$\frac{⟨{\phi }_i{⟩}^n\Rightarrow \psi }{⟨{\phi }_1\wedge {\phi }_2{⟩}^n\Rightarrow \psi }(\wedge \Rightarrow )(i=1,2)\frac{\Gamma \Rightarrow {\psi }_1\Gamma \Rightarrow {\psi }_2}{\Gamma \Rightarrow {\psi }_1\wedge {\psi }_2}(\Rightarrow \wedge )$$

$$\frac{⟨{\phi }_1{⟩}^n\Rightarrow \psi ⟨{\phi }_2{⟩}^n\Rightarrow \psi }{⟨{\phi }_1\vee {\phi }_2{⟩}^n\Rightarrow \psi }(\vee \Rightarrow )\frac{\Gamma \Rightarrow {\psi }_i}{\Gamma \Rightarrow {\psi }_1\vee {\psi }_2}(\Rightarrow \vee )(i=1,2)$$

$$\frac{⟨\phi {⟩}^{n+1}\Rightarrow \psi }{⟨◊\phi {⟩}^n\Rightarrow \psi }(◊\Rightarrow )\frac{\Gamma \Rightarrow \psi }{⟨\Gamma ⟩\Rightarrow ◊\psi }(\Rightarrow ◊)$$

$$\frac{⟨\phi {⟩}^n\Rightarrow \psi }{⟨■\phi {⟩}^{n+1}\Rightarrow \psi }(■\Rightarrow )\frac{⟨\Gamma ⟩\Rightarrow \psi }{\Gamma \Rightarrow ■\psi }(\Rightarrow ■)$$

$$\frac{\Gamma \Rightarrow \phi ⟨\phi {⟩}^n\Rightarrow \psi }{⟨\Gamma {⟩}^n\Rightarrow \psi }(Cut)$$

$${\phi }_1\Rightarrow {\psi }_1$$

$$\phi \Rightarrow \psi$$

$$⟨\phi ⟩\Rightarrow ◊\phi$$

$$\chi {\Rightarrow }_{T^{\circ }}\chi$$

$$\frac{⟨{\phi }_i{⟩}^n{\Rightarrow }_{T^{\circ }}\psi }{⟨{\phi }_1\wedge {\phi }_2{⟩}^n{\Rightarrow }_{T^{\circ }}\psi }(\wedge \Rightarrow )$$

$$\frac{⟨\phi {⟩}^n{\Rightarrow }_{T^{\circ }}{\psi }_1⟨\phi {⟩}^n{\Rightarrow }_{T^{\circ }}{\psi }_2}{⟨\phi {⟩}^n{\Rightarrow }_{T^{\circ }}{\psi }_1\wedge {\psi }_2}(\Rightarrow \wedge )$$

$$\frac{⟨{\phi }_1{⟩}^n{\Rightarrow }_{T^{\circ }}\psi ⟨{\phi }_2{⟩}^n{\Rightarrow }_{T^{\circ }}\psi }{⟨{\phi }_1\vee {\phi }_2{⟩}^n{\Rightarrow }_{T^{\circ }}\psi }(\vee \Rightarrow )$$

$$\frac{⟨\phi {⟩}^n{\Rightarrow }_{T^{\circ }}{\psi }_i}{⟨\phi {⟩}^n{\Rightarrow }_{T^{\circ }}{\psi }_1\vee {\psi }_2}(\Rightarrow \vee )$$

$$\frac{⟨\chi {⟩}^{n+1}{\Rightarrow }_{T^{\circ }}\psi }{⟨◊\chi {⟩}^n{\Rightarrow }_{T^{\circ }}\psi }(◊\Rightarrow )$$

$$\frac{⟨\phi {⟩}^{n-1}\Rightarrow \chi }{⟨\phi {⟩}^n\Rightarrow ◊\chi }(\Rightarrow ◊)$$

$$\frac{⟨\chi {⟩}^{n-1}\Rightarrow \psi }{⟨■\chi {⟩}^n\Rightarrow \psi }(■\Rightarrow )$$

$$\frac{⟨\phi {⟩}^{n+1}\Rightarrow \chi }{⟨\phi {⟩}^n\Rightarrow ■\chi }(\Rightarrow ■)$$

$$\frac{⟨\phi {⟩}^i{\Rightarrow }_{T^{\circ }}\chi ⟨\chi {⟩}^j{\Rightarrow }_{T^{\circ }}\psi }{⟨\phi {⟩}^n{\Rightarrow }_{T^{\circ }}\psi }(Cut)$$

$$G(\phi )=\{⟨\chi {⟩}^n\in FS(T^{◊})\mid \mathsf{G}⊢⟨\chi {⟩}^n{\Rightarrow }_{T^{\circ }}\phi \}\text{and}[[\phi ]]=G(\phi )\cup FS(T^{\bullet }).$$

$${\top }^{\circ }=[[\top ]],{\perp }^{\circ }=[[\perp ]],{\neg }^{\circ }[[\phi ]]=[[\neg \phi ]],[[\phi ]]{\wedge }^{\circ }[[\psi ]]=[[\phi \wedge \psi ]],[[\phi ]]{\vee }^{\circ }[[\psi ]]=[[\phi \vee \psi ]].$$

$$C(X)=\bigcap \{[[\phi ]]\mid X\subseteq [[\phi ]]\in [[T^{\circ }]]\}.$$

$$◊X=\{⟨\Gamma ⟩\mid \Gamma \in X\}\text{and}■X=\{\Gamma \mid ⟨\Gamma ⟩\in X\}.$$

Full text

0.1 Introduction

Modal reduction principles (MRPs) are modal formulas of the form $\mathsf{M}p\rightarrow\mathsf{N}p$ where $\mathsf{M},\mathsf{N}$ are finite (possibly empty) sequences of modal operators $\Box$ or $\Diamond$. Fitch [5] investigated the problem of correspondence between MRPs and first-order properties. Van Benthem [1] proved that every MRP corresponds to a first-order relational property over the class of all transitive frames. Besides the correspondence theory, the finite model property (FMP) of normal modal logics generated by MRPs is also much concerned in the literature.

For any normal modal logic $\Lambda$ and a set $\Sigma$ of modal formulas, let $\mathsf{NExt}(\Lambda)$ be the class of all normal modal logics extending $\Lambda$, and let $\Lambda\oplus\Sigma$ be the normal extension of $\Lambda$ by adding all formulas in $\Sigma$ as axioms. Using the method of canonical formulas, Zakharyashev [10] proved that all logics in $\mathsf{NExt}(\mathsf{K4})$ axiomatized by MRPs have the FMP. However, the FMP of normal modal logics axiomatizable by MRPs over the least normal modal logic $\mathsf{K}$ is a longstanding open problem (cf. [9, p.452]). In particular, it is unknown whether all normal modal logic of the form $\mathsf{K}_{n}^{m}=\mathsf{K}\oplus\Box^{n}p\rightarrow\Box^{m}p$ ($n\neq m\geq 1$) have the FMP. This most intriguing open problem in modal logic was highlighted by Zakharyashev [10] as follows:

Unfortunately, the technical apparatus developed is applicable only to logics with transitive frames, and the situation of extensions of $\mathsf{K}$ by modal reduction principles, even by axioms $\Box^{n}p\rightarrow\Box^{m}p$ still remains unclear. I think at present this is one of the major challenges in completeness theory.

This problem has a long history and was traced back to Krister Segerberg in 1970s by Chagrov and Zakharyashev [4, 11.8 Notes]. It is worth mentioning that Gabbay [6] used a general filtration method to show the FMP of modal logics $\mathsf{K}\oplus\Box p\rightarrow\Box^{m}p$ where $m\geq 0$.

A part of this intriguing open problem is the FMP of all $n$-transitive modal logics of the form $\mathsf{K}\oplus\Box^{n}p\rightarrow\Box^{n+1}p$ (cf. [4, Problem 11.2]). The most well-known example of this open problem is perhaps the FMP of the quasi-transitive modal logic $\mathsf{K}_{2}^{3}=\mathsf{K}\oplus\Box\Box p\rightarrow\Box\Box\Box p$ (cf. e.g. [7]). We suggest the name ‘quasi-transitive modal logic’ for $\mathsf{K}_{2}^{3}$ because frames for it are ‘almost transitive’. The aim of the present work is to show that $\mathsf{K}_{2}^{3}$ has the FMP.

Our proof proceeds in the algebraic way. It is well-known that, by the duality between modal algebras and frames, the FMP of a normal modal logic $\Lambda$ is equivalent to the finite algebra property (FAP), i.e., every formula $\varphi$ which is not provable in $\Lambda$ is refuted by a finite $\Lambda$-algebra (cf. [8]). In order to show the FAP of the quasi-transitive modal logic $\mathsf{K}_{2}^{3}$, we shall prove the FAP of the tense logic $\mathsf{Kt}_{2}^{3}$ which is a conservative extension of $\mathsf{K}_{2}^{3}$. The tense logic $\mathsf{Kt}_{2}^{3}$ extends the minimal tense logic $\mathsf{Kt}$ (cf. [2]) by adding the axiom $\Diamond\Diamond\Diamond p\rightarrow\Diamond\Diamond p$. The core of the proof is a syntactic construction of finite algebra based on a Gentzen sequent calculus $\mathsf{G}$ for $\mathsf{Kt}_{2}^{3}$. We show that every sequent which is not derivable in $\mathsf{G}$ is refuted by a finite algebra for $\mathsf{Kt}_{2}^{3}$. There are two main innovative points in the syntactic construction. First, in the definition of sequent calculus $\mathsf{G}$, a structural operator $\langle.\rangle$ for the modal operator $\Diamond$ is introduced (cf. Definition 0.2.3). Second, in order to show that the finite algebra is a $\mathsf{Kt}_{2}^{3}$-algebra, a particular form of interpolation lemma for $\mathsf{G}$ is required (cf. Lemma 0.3.3). This kind of interpolation lemma was used by Buszkowski [3] to show the finite embeddability property of residuated algebras.

0.2 A Gentzen sequent calculus

The language of modal logic consists of a denumerable set of propositional variables $\mathsf{Prop}$, propositional connectives $\bot,\top,\neg,\wedge,\vee$ and a unary modal operator $\Diamond$. The set of all modal formulas $\mathcal{L}_{\Diamond}$ is defined inductively by the following rule:

[TABLE]

The tense language is the extension of the modal language by a unary modal operator $\blacksquare$. The set of all tense formulas is denoted by $\mathscr{L}_{t}$. For any number $k\geq 0$, $\Diamond^{n}\varphi$ is defined by $\Diamond^{0}\varphi:=\varphi$ and $\Diamond^{n+1}\varphi:=\Diamond\Diamond^{k}\varphi$. The complexity $c(\varphi)$ of a tense formula $\varphi$ is defined inductively as follows:

[TABLE]

The tense formula algebra is denoted by $\mathscr{T}$.

Definition 0.2.1**.**

A modal algebra is $\mathfrak{A}=(A,\wedge,\vee,\neg,0,1,\Diamond)$ where $(A,\wedge,\vee,\neg,0,1)$ is a Boolean algebra, and $\Diamond$ is a unary operator on $A$ with $\Diamond 0=0$ and $\Diamond(a\vee b)=\Diamond a\vee\Diamond b$ for all $a,b\in A$. A modal algebra $\mathfrak{A}=(A,\wedge,\vee,\neg,0,1,\Diamond)$ is quasi-transitive, if $\Diamond\Diamond\Diamond a\leq\Diamond\Diamond a$ for all $a\in A$.

A tense algebra is $\mathfrak{A}=(A,\wedge,\vee,\neg,0,1,\Diamond,\blacksquare)$ where $(A,\wedge,\vee,\neg,0,1)$ is a Boolean algebra, and $\Diamond,\blacksquare$ are unary operators on $A$ such that for all $a,b\in A$:

(Adj) $\Diamond a\leq b$ if and only if $a\leq\blacksquare b$.

A tense algebra $\mathfrak{A}=(A,\wedge,\vee,\neg,0,1,\blacksquare,\Diamond)$ is quasi-transitive, if $\Diamond\Diamond\Diamond a\leq\Diamond\Diamond a$ for all $a\in A$. Let $\mathbb{Q}$ be the class of all quasi-transitive algebras. **

Lemma 0.2.2**.**

For any quasi-transitive tense algebra $\mathfrak{A}=(A,\wedge,\vee,\neg,0,1,\Diamond,\blacksquare)$, the following hold for all $a,b,c\in A$:

1. $(1)$* $\Diamond 0=0$ and $\blacksquare 1=1$.*
2. $(2)$* $a\leq\blacksquare\Diamond a$ and $\Diamond\blacksquare a\leq a$.*
3. $(3)$* if $a\leq b$, then $\Diamond a\leq\Diamond b$ and $\blacksquare a\leq\blacksquare b$.*
4. $(4)$* $\Diamond(a\vee b)=\Diamond a\vee\Diamond b$ and $\blacksquare(a\wedge b)=\blacksquare a\wedge\blacksquare b$.*
5. $(5)$* $\blacksquare\blacksquare a\leq\blacksquare\blacksquare\blacksquare a$.*

Proof.

We show only $\Diamond(a\vee b)=\Diamond a\vee\Diamond b$ in (5). Other items can be shown easily. Clearly $a\leq a\vee b$ and $b\leq a\vee b$. By (4), $\Diamond a\leq\Diamond(a\vee b)$ and $\Diamond b\leq\Diamond(a\vee b)$. Then $\Diamond a\vee\Diamond b\leq\Diamond(a\vee b)$. By (2), $a\leq\blacksquare\Diamond a$. By $\Diamond a\leq\Diamond a\vee\Diamond b$ and (3), $\blacksquare\Diamond a\leq\blacksquare(\Diamond a\vee\Diamond b)$. Then $a\leq\blacksquare(\Diamond a\vee\Diamond b)$. Similarly $b\leq\blacksquare(\Diamond a\vee\Diamond b)$. Then $a\vee b\leq\blacksquare(\Diamond a\vee\Diamond b)$. By (Adj), $\Diamond(a\vee b)\leq\Diamond a\vee\Diamond b$. ∎

Definition 0.2.3**.**

For any tense formula $\varphi$, we define $\langle\varphi\rangle^{n}$ by induction on $n\geq 0$ as follows:

$\langle\varphi\rangle^{0}=\varphi$ and $\langle\varphi\rangle^{n+1}=\langle\langle\varphi\rangle^{n}\rangle$.

A formula structure is an expression of the form $\langle\varphi\rangle^{n}$ for some tense formula $\varphi$ and natural number $n\geq 0$. Formula structures are denoted by $\Gamma,\Delta$ etc. For any set of formulas $X$, the set of all formula structures generated by $X$ is defined as $FS(X)=\{\langle\varphi\rangle^{n}\mid\varphi\in X~{}\&~{}n\geq 0\}$.

A sequent is an expression $\Gamma\Rightarrow\psi$ where $\Gamma$ is a formula structure and $\psi$ is a tense formula. Sequents are denoted by $s,t$ etc. with or without subscripts. A sequent rule is a fraction

[TABLE]

where $s_{1},\ldots,s_{n}$ are called the premisses and $s_{0}$ is called the conclusion of $(R)$. **

Definition 0.2.4**.**

The sequent calculus $\mathsf{G}$ for the tense logic $\mathsf{Kt}_{2}^{3}$ consists of the following axiom schemata and rules:

$(1)$ Axiom schemata:

[TABLE]

[TABLE]

$(2)$ Connective rules:

[TABLE]

[TABLE]

$(3)$ Modal rules:

[TABLE]

[TABLE]

$(4)$ Cut rule:

[TABLE]

In the axiom schema $(\bot)$ and sequent rules, the number $n\geq 0$ is arbitrary. A derivation in $\mathsf{G}$ is a finite tree of sequents $\mathcal{D}$ in which each node is either an instance of an axiom schema or derived from child node(s) by a sequent rule. The height of a derivation $\mathcal{D}$, denoted by $|\mathcal{D}|$, is the maximal length of branches in $\mathcal{D}$. In a derivation, we use $(R)^{n}$ to denote $n$ times application of the rule $(R)$. A sequent $\Gamma\Rightarrow\psi$ is derivable in $\mathsf{G}$, notation $\mathsf{G}\vdash\Gamma\Rightarrow\psi$, if there is a derivation $\mathcal{D}$ in $\mathsf{G}$ with root node $\Gamma\Rightarrow\psi$. **

Example 0.2.5**.**

$(1)$ For $\odot\in\{\wedge,\vee\}$, if $\mathsf{G}\vdash\varphi_{1}\Rightarrow\psi_{1}$ and $\mathsf{G}\vdash\varphi_{2}\Rightarrow\psi_{2}$, then $\mathsf{G}\vdash\varphi_{1}\odot\varphi_{2}\Rightarrow\psi_{1}\odot\psi_{2}$. We have the following derivation:

[TABLE]

By a similar derivation, we have $\mathsf{G}\vdash\varphi_{1}\vee\varphi_{2}\Rightarrow\psi_{1}\vee\psi_{2}$.

$(2)$ If $\mathsf{G}\vdash\varphi\Rightarrow\psi$, then $\mathsf{G}\vdash\neg\psi\Rightarrow\neg\varphi$. Suppose $\mathsf{G}\vdash\varphi\Rightarrow\psi$. Clearly $\mathsf{G}\vdash\psi\wedge\neg\psi\Rightarrow\varphi$. Then we have the following derivation:

[TABLE]

By $(\mathrm{D})$, $\mathsf{G}\vdash(\varphi\vee\neg\varphi)\wedge\neg\psi\Rightarrow(\varphi\wedge\neg\psi)\vee(\neg\varphi\wedge\neg\psi)$. By $(Cut)$, $\mathsf{G}\vdash(\varphi\vee\neg\varphi)\wedge\neg\psi\Rightarrow\neg\varphi$. By $(\top)$ and $(1)$, $\mathsf{G}\vdash\top\wedge\neg\psi\Rightarrow(\varphi\vee\neg\varphi)\wedge\neg\psi$. Clearly $\mathsf{G}\vdash\neg\psi\Rightarrow\top\wedge\neg\psi$. By $(Cut)$, $\mathsf{G}\vdash\neg\psi\Rightarrow\neg\varphi$.

$(3)$ If $\mathsf{G}\vdash\varphi\Rightarrow\psi$, then $\mathsf{G}\vdash\Diamond\varphi\Rightarrow\Diamond\psi$ and $\mathsf{G}\vdash\blacksquare\varphi\Rightarrow\blacksquare\psi$. Suppose $\mathsf{G}\vdash\varphi\Rightarrow\psi$. We have the following derivations:

[TABLE]

$(4)$ For any $n\geq 0$, $\mathsf{G}\vdash\langle\varphi\rangle^{n}\Rightarrow\Diamond^{n}\varphi$. The case $n=0$ is trivial. Let $n>0$. Starting from the axiom $\varphi\Rightarrow\varphi$, by $n$ times application of $({\Rightarrow}\Diamond)$, we have $\mathsf{G}\vdash\langle\varphi\rangle^{n}\Rightarrow\Diamond^{n}\varphi$.

$(5)$ $\mathsf{G}\vdash\Diamond(\varphi\vee\psi)\Rightarrow\Diamond\varphi\vee\Diamond\psi$. We have the following derivation:

[TABLE]

$(6)$ $\mathsf{G}\vdash\Diamond\blacksquare\varphi\Rightarrow\varphi$. Starting from $\varphi\Rightarrow\varphi$, using $(\blacksquare{\Rightarrow})$ and $(\Diamond{\Rightarrow})$, we have $\mathsf{G}\vdash\Diamond\blacksquare\varphi\Rightarrow\varphi$. **

Definition 0.2.6**.**

For any quasi-transitive tense algebra $\mathfrak{A}=(A,\wedge,\vee,\neg,0,1,\Diamond,\blacksquare)$, an assignment in $\mathfrak{A}$ is a function $\theta:\mathsf{Prop}\rightarrow A$. Let $\widehat{\theta}:\mathscr{L}_{t}\rightarrow A$ be the homomorphic extension of $\theta$ to the tense formula algebra $\mathfrak{T}$. For any formula structure $\langle\varphi\rangle^{n}$, we define $\tau(\langle\varphi\rangle^{n})=\Diamond^{n}\varphi$. A sequent $\Gamma\Rightarrow\psi$ is valid in $\mathbb{Q}$, notation $\Gamma\models_{\mathbb{Q}}\psi$, if $\widehat{\theta}(\tau(\Gamma))\leq\widehat{\theta}(\psi)$ for any quasi-transitive tense algebra $\mathfrak{A}$ and assignment $\theta$ in $\mathfrak{A}$. A sequent rule with premisses $s_{1},\ldots,s_{n}$ and conclusion $s_{0}$ preserves validity in $\mathbb{Q}$, if $s_{0}$ is valid in $\mathbb{Q}$ whenever $s_{i}$ for all $1\leq i\leq n$ are valid in $\mathbb{Q}$. **

A formula $\varphi$ is equivalent to $\psi$ with respect to $\mathsf{G}$, notation $\varphi\sim_{\mathsf{G}}\psi$, if $\mathsf{G}\vdash\varphi\Rightarrow\psi$ and $\mathsf{G}\vdash\psi\Rightarrow\varphi$. Let $|\varphi|_{\mathsf{G}}=\{\chi\in\mathscr{L}\mid\varphi\sim_{\mathsf{G}}\chi\}$. For any set of formulas $T$, let $|T|_{\mathsf{G}}=\{|\varphi|_{\mathsf{G}}\mid\varphi\in T\}$.

Lemma 0.2.7**.**

The relation $\sim_{t}$ is a congruence relation on the tense formula algebra $\mathscr{T}$.

Proof.

It is trivial that $\sim_{t}$ is an equivalence relation on $\mathscr{L}_{t}$. Suppose $\varphi_{1}\sim_{t}\psi_{1}$ and $\varphi_{2}\sim_{t}\psi_{2}$. By Example 0.2.5 $(1)$, $\varphi_{1}\odot\varphi_{2}\sim_{t}\psi_{1}\odot\psi_{2}$ for $\odot\in\{\wedge,\vee\}$. Suppose $\varphi\sim_{t}\psi$. By Example 0.2.5 $(2)$ and $(3)$, $\#\varphi\sim_{t}\#\psi$ for $\#\in\{\neg,\Diamond,\blacksquare\}$. ∎

Let $\mathfrak{L}_{t}$ be the quotient algebra of the tense formula algebra $\mathscr{T}$ under $\sim_{t}$. One can easily show that $\mathfrak{L}_{t}$ is quasi-transitive. Moreover, for any $\varphi,\psi\in\mathscr{L}_{t}$, $|\varphi|_{\mathsf{G}}\leq|\psi|_{\mathsf{G}}$ if and only if $\mathsf{G}\vdash\varphi\Rightarrow\psi$.

Theorem 0.2.8**.**

For any sequent $\Gamma\Rightarrow\psi$, $\mathsf{G}\vdash\Gamma\Rightarrow\psi$ if and only if $\mathbb{Q}\models\Gamma\Rightarrow\psi$.

Proof.

Suppose $\mathsf{G}\vdash\Gamma\Rightarrow\psi$. All axioms are obviously valid in $\mathbb{Q}$. One can show that all sequent rules of $\mathsf{G}$ preserve validity in $\mathbb{Q}$. For $(Cut)$, assume $\Gamma\models_{\mathbb{Q}}\varphi$ and $\langle\varphi\rangle^{n}\models_{\mathbb{Q}}\psi$. Let $\mathfrak{A}$ be any algebra in $\mathbb{Q}$ and $\theta$ be an assignment in $\mathfrak{A}$. Then $\widehat{\theta}(\tau(\Gamma))\leq\widehat{\theta}(\varphi)$ and $\widehat{\theta}(\Diamond^{n}\varphi)\leq\widehat{\theta}(\psi)$. Then $\Diamond^{n}\widehat{\theta}(\tau(\Gamma))\leq\Diamond^{n}\widehat{\theta}(\varphi)$ and $\Diamond^{n}\widehat{\theta}(\varphi)\leq\widehat{\theta}(\psi)$. Hence $\Diamond^{n}\widehat{\theta}(\tau(\Gamma))\leq\widehat{\theta}(\psi)$. Then $\langle\Gamma\rangle^{n}\models_{\mathbb{Q}}\psi$. The other cases can be shown easily. Hence $\mathbb{Q}\models\Gamma\Rightarrow\psi$. Suppose $\mathsf{G}\not\vdash\Gamma\Rightarrow\psi$. By $\mathsf{G}\vdash\Gamma\Rightarrow\tau(\Gamma)$ and $(Cut)$, $\mathsf{G}\not\vdash\tau(\Gamma)\Rightarrow\psi$. Hence $|\tau(\Gamma)|\not\leq|\psi|$. Let $\theta$ be the assignment in $\mathfrak{L}_{t}$ with ${\theta}(p)=|p|$ for each $p\in\mathsf{Prop}$. One can easily show by induction on the complexity of $\varphi$ that $\widehat{\theta}(\varphi)=|\varphi|$. Then $\widehat{\theta}(\tau(\Gamma))\not\leq\widehat{\theta}(\psi)$. Hence $\mathbb{Q}\not\models\Gamma\Rightarrow\psi$. ∎

0.3 Finite Model Property

In this section, we prove the finite algebra property (FAP) of the sequent calculus $\mathsf{G}$, i.e., if $\mathsf{G}\not\vdash\Gamma\Rightarrow\psi$, there is a finite pretransitive tense algebra that refutes $\Gamma\Rightarrow\psi$. The FMP of $\mathsf{Kt}_{2}^{3}$ and $\mathsf{K}_{2}^{3}$ is derived from the FAP.

Definition 0.3.1**.**

For any set of tense formulas $X$ with $\top,\bot\in X$, the sets $X^{b}$ and $X^{\Diamond}$ are defined as follows:

• •

$X^{b}$ is the smallest set of tense formulas such that $X\subseteq X^{b}$ and $X^{b}$ is closed under the operations $\neg$, $\wedge$ and $\vee$.

• •

$X^{\Diamond}=\{\Diamond^{k}\varphi\mid\varphi\in X~{}\&~{}0\leq k\leq 3\}$.

For any finite set of tense formulas $T$ with $\top,\bot\in T$, let $T^{\circ}=(T^{\Diamond})^{b}$ and $T^{\bullet}=T^{\circ}\setminus T^{\Diamond}$. **

In this section, we stipulate that $T$ is a finite set of tense formulas with $\top,\bot\in T$. Obviously $T\subseteq T^{\Diamond}$. A sequent $\Gamma\Rightarrow\psi$ is $T^{\circ}$-derivable in $\mathsf{G}$, notation $\mathsf{G}\vdash\Gamma\Rightarrow_{T^{\circ}}\psi$, if there is a derivation $\mathcal{D}$ of $\Gamma\Rightarrow\psi$ in $\mathsf{G}$ such that all formulas in $\mathcal{D}$ belong to $T^{\circ}$.

Lemma 0.3.2**.**

For any $n\geq 0$, if $\mathsf{G}\vdash\langle\varphi\rangle^{n+2}\Rightarrow_{T^{\circ}}\psi$ and $\varphi\in T^{\Diamond}$, then $\mathsf{G}\vdash\langle\varphi\rangle^{n+3}\Rightarrow_{T^{\circ}}\psi$.

Proof.

Assume $\mathsf{G}\vdash\langle\varphi\rangle^{n+2}\Rightarrow_{T^{\circ}}\psi$. Let $\varphi=\Diamond^{k}\chi$ for some formula $\chi\in T$ where $k\geq 0$. Clearly $\mathsf{G}\vdash\langle\chi\rangle^{k}\Rightarrow_{T^{\circ}}\Diamond^{k}\chi$. We have the following derivation:

[TABLE]

Note that $\Diamond^{3}\chi,\Diamond^{2}\chi\in T^{\circ}$ since $\chi\in T$. Hence $\mathsf{G}\vdash\langle\varphi\rangle^{n+3}\Rightarrow_{T^{\circ}}\psi$. ∎

Lemma 0.3.3** (Interpolation).**

For any set of tense formulas $T$ and $n\geq m\geq 0$, if $\mathsf{G}\vdash\langle\varphi\rangle^{n}\Rightarrow_{T^{\circ}}\psi$, there exists a formula $\gamma\in T^{\circ}$ with $\mathsf{G}\vdash\langle\varphi\rangle^{m}\Rightarrow_{T^{\circ}}\gamma$ and $\mathsf{G}\vdash\langle\gamma\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi$.

Proof.

The required formula $\gamma\in T^{\circ}$ is called an interpolant. If $m=n$, we choose $\psi\in T^{\circ}$ as a required interpolant. If $m=0$, we choose $\varphi\in T^{\circ}$ as a required interpolant. Let $n>m>0$. Assume $\mathsf{G}\vdash\langle\varphi\rangle^{n}\Rightarrow_{T^{\circ}}\psi$. There is a derivation $\mathcal{D}$ of $\langle\varphi\rangle^{n}\Rightarrow_{T^{\circ}}\psi$ in $\mathsf{G}$. The proof proceeds by induction on the height $|\mathcal{D}|$. Suppose $|\mathcal{D}|=0$. Then $\varphi=\bot$. We choose $\bot$ as a required interpolant. Suppose $|\mathcal{D}|>0$. Let $\langle\varphi\rangle^{n}\Rightarrow_{T^{\circ}}\psi$ be obtained by a rule $(R)$.

(1) $(R)$ is a connective rule. One can get the required interpolant by induction hypothesis and the rule $(R)$. We have the following cases:

(1.1) Let $(R)$ be $(\wedge{\Rightarrow})$ and the derivation end with

[TABLE]

where $\varphi=\varphi_{1}\wedge\varphi_{2}$ and $i=1,2$. By induction hypothesis, there is a formula $\gamma\in T^{\circ}$ with (i) $\mathsf{G}\vdash\langle\varphi_{i}\rangle^{m}\Rightarrow_{T^{\circ}}\gamma$ and (ii) $\mathsf{G}\vdash\langle\gamma\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi$. By (i) and $(\wedge{\Rightarrow})$, $\mathsf{G}\vdash\langle\varphi_{1}\wedge\varphi_{2}\rangle^{m}\Rightarrow_{T^{\circ}}\gamma$. Hence $\gamma$ is a required interpolant.

(1.2) Let $(R)$ be $({\Rightarrow}\wedge)$ and the derivation end with

[TABLE]

By induction hypothesis, there are formulas $\chi_{1},\chi_{2}\in T^{\circ}$ with (i) $\mathsf{G}\vdash\langle\varphi\rangle^{m}\Rightarrow_{T^{\circ}}\chi_{1}$; (ii) $\mathsf{G}\vdash\langle\chi_{1}\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi_{1}$; (iii) $\mathsf{G}\vdash\langle\varphi\rangle^{m}\Rightarrow_{T^{\circ}}\chi_{2}$; (iv) $\mathsf{G}\vdash\langle\chi_{2}\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi_{2}$. By (i) and (iii), using $({\Rightarrow}\wedge)$, we have $\mathsf{G}\vdash\langle\varphi\rangle^{m}\Rightarrow_{T^{\circ}}\chi_{1}\wedge\chi_{2}$. Since $\mathsf{G}\vdash\chi_{1}\wedge\chi_{2}\Rightarrow_{T^{\circ}}\chi_{1}$ and $\mathsf{G}\vdash\chi_{1}\wedge\chi_{2}\Rightarrow_{T^{\circ}}\chi_{2}$, by (ii) and (iv), using $(Cut)$, we have $\mathsf{G}\vdash\langle\chi_{1}\wedge\chi_{2}\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi_{1}$ and $\mathsf{G}\vdash\langle\chi_{1}\wedge\chi_{2}\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi_{2}$. By $({\Rightarrow}\wedge)$, $\mathsf{G}\vdash\langle\chi_{1}\wedge\chi_{2}\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi_{1}\wedge\psi_{2}$. Note that $\chi_{1}\wedge\chi_{2}\in T^{\circ}$ since $\chi_{1},\chi_{2}\in T^{\circ}$. Hence $\chi_{1}\wedge\chi_{2}$ is a required interpolant.

(1.3) Let $(R)$ be $(\vee{\Rightarrow})$ and the derivation end with

[TABLE]

where $\varphi=\varphi_{1}\vee\varphi_{2}$. By induction hypothesis, there are formulas $\gamma_{1},\gamma_{2}\in T^{\circ}$ with (i) $\mathsf{G}\vdash\langle\varphi_{1}\rangle^{m}\Rightarrow_{T^{\circ}}\gamma_{1}$; (ii) $\mathsf{G}\vdash\langle\gamma_{1}\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi$; (iii) $\mathsf{G}\vdash\langle\varphi_{2}\rangle^{m}\Rightarrow_{T^{\circ}}\gamma_{2}$; (iv) $\mathsf{G}\vdash\langle\gamma_{2}\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi$. By (i) and (iii), applying $({\Rightarrow}\vee)$ and $(\vee{\Rightarrow})$, we have $\mathsf{G}\vdash\langle\varphi_{1}\vee\varphi_{2}\rangle^{m}\Rightarrow_{T^{\circ}}\gamma_{1}\vee\gamma_{2}$. By (ii) and (iv), using $({\Rightarrow}\vee)$, $\mathsf{G}\vdash\langle\gamma_{1}\vee\gamma_{2}\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi$. Note that $\gamma_{1}\vee\gamma_{2}\in T^{\circ}$ since $\gamma_{1},\gamma_{2}\in T^{\circ}$. Hence $\chi_{1}\vee\chi_{2}$ is a required interpolant.

(1.4) Let $(R)$ be $({\Rightarrow}\vee)$ and the derivation end with

[TABLE]

where $\psi=\psi_{1}\vee\psi_{2}$ and $i=1,2$. By induction hypothesis, there is a formula $\chi\in T^{\circ}$ with (i) $\mathsf{G}\vdash\langle\varphi\rangle^{m}\Rightarrow_{T^{\circ}}\chi$ and (ii) $\mathsf{G}\vdash\langle\chi\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi_{i}$. By (ii) and $({\Rightarrow}\vee)$, we have $\mathsf{G}\vdash\langle\chi\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi_{1}\vee\psi_{2}$. Hence $\chi$ is a required interpolant.

(2) $(R)$ is a modal rule. We have the following cases:

(2.1) $(R)$ is $(\Diamond{\Rightarrow})$. Let the derivation end with

[TABLE]

where $\varphi=\Diamond\chi$. By induction hypothesis, (i) $\mathsf{G}\vdash\langle\chi\rangle^{m+1}\Rightarrow_{T^{\circ}}\gamma$ and (ii) $\mathsf{G}\vdash\langle\gamma\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi$ for some $\gamma\in T^{\circ}$. By (i) and $(\Diamond{\Rightarrow})$, $\mathsf{G}\vdash\langle\Diamond\chi\rangle^{m}\Rightarrow_{T^{\circ}}\gamma$. Then $\gamma$ is a required interpolant.

(2.2) $(R)$ is $({\Rightarrow}\Diamond)$. Let the derivation end with

[TABLE]

where $\psi=\Diamond\chi$. By induction hypothesis, (i) $\mathsf{G}\vdash\langle\varphi\rangle^{m}\Rightarrow_{T^{\circ}}\gamma$ and (ii) $\mathsf{G}\vdash\langle\gamma\rangle^{n-m-1}\Rightarrow_{T^{\circ}}\chi$ for some $\gamma\in T^{\circ}$. By (ii) and $({\Rightarrow}\Diamond)$, $\mathsf{G}\vdash\langle\gamma\rangle^{n-m}\Rightarrow_{T^{\circ}}\Diamond\chi$. Then $\gamma$ is a required interpolant.

(2.3) $(R)$ is $(\blacksquare{\Rightarrow})$. Let the derivation end with

[TABLE]

where $\varphi=\blacksquare\chi$. By induction hypothesis, (i) $\mathsf{G}\vdash\langle\chi\rangle^{m-1}\Rightarrow_{T^{\circ}}\gamma$ and (ii) $\mathsf{G}\vdash\langle\gamma\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi$ for some $\gamma\in T^{\circ}$. By (i) and $(\blacksquare{\Rightarrow})$, $\mathsf{G}\vdash\langle\blacksquare\chi\rangle^{m}\Rightarrow_{T^{\circ}}\gamma$. Then $\gamma$ is a required interpolant.

(2.4) $(R)$ is $({\Rightarrow}\blacksquare)$. Let the derivation end with

[TABLE]

where $\psi=\blacksquare\chi$. By induction hypothesis, (i) $\mathsf{G}\vdash\langle\varphi\rangle^{m}\Rightarrow_{T^{\circ}}\gamma$ and (ii) $\mathsf{G}\vdash\langle\gamma\rangle^{n-m+1}\Rightarrow_{T^{\circ}}\chi$ for some $\gamma\in T^{\circ}$. By (ii) and $({\Rightarrow}\blacksquare)$, $\mathsf{G}\vdash\langle\gamma\rangle^{n-m}\Rightarrow_{T^{\circ}}\blacksquare\chi$. Then $\gamma$ is a required interpolant.

(3) $(R)$ is $(Cut)$. Let the derivation end with

[TABLE]

where $i+j=n$. Suppose $m\leq i$. By induction hypothesis, (i) $\mathsf{G}\vdash\langle\varphi\rangle^{m}\Rightarrow_{T^{\circ}}\gamma$ and (ii) $\mathsf{G}\vdash\langle\gamma\rangle^{i-m}\Rightarrow_{T^{\circ}}\chi$ for some $\gamma\in T^{\circ}$. By the right premiss of $(Cut)$ and (ii), using $(Cut)$, $\mathsf{G}\vdash\langle\gamma\rangle^{n-m}\Rightarrow_{T^{\circ}}\psi$. Then $\gamma$ is a required interpolant. Suppose $m>i$. By induction hypothesis, (iii) $\mathsf{G}\vdash\langle\chi\rangle^{m-i}\Rightarrow_{T^{\circ}}\gamma^{\prime}$ and (iv) $\mathsf{G}\vdash\langle\gamma^{\prime}\rangle^{n-m}\Rightarrow_{T^{\circ}}\chi$ for some $\gamma^{\prime}\in T^{\circ}$. By the left premiss of $(Cut)$ and (iii), using $(Cut)$, $\mathsf{G}\vdash\langle\varphi\rangle^{m}\Rightarrow_{T^{\circ}}\gamma^{\prime}$. Then $\gamma^{\prime}$ is a required interpolant. ∎

Definition 0.3.4**.**

Let $FS(T^{\Diamond}),FS(T^{\circ})$ and $FS(T^{\bullet})$ be sets of all formula structures generated by $T^{\Diamond},T^{\circ}$ and $T^{\bullet}$ respectively. For any $\varphi\in T^{\circ}$, we define

[TABLE]

Let $\llbracket{T^{\circ}}\rrbracket=\{\llbracket{\varphi}\rrbracket\mid\varphi\in T^{\circ}\}$. We define operations $\bot^{\circ},\top^{\circ},\neg^{\circ},\wedge^{\circ}$ and $\vee^{\circ}$ on $\llbracket{T^{\circ}}\rrbracket$ as follows:

[TABLE]

Let $\mathfrak{B}(T^{\circ})=(\llbracket{T^{\circ}}\rrbracket,\wedge^{\circ},\vee^{\circ},\neg^{\circ},\top^{\circ},\bot^{\circ})$. The binary relation $\leq^{\circ}$ on $\llbracket{T^{\circ}}\rrbracket$ is defined as follows: $\llbracket{\varphi}\rrbracket\leq^{\circ}\llbracket{\psi}\rrbracket$ if and only if $\llbracket{\varphi}\rrbracket\wedge^{\circ}\llbracket{\psi}\rrbracket=\llbracket{\varphi}\rrbracket$.

Lemma 0.3.5**.**

For any $\varphi,\psi\in T^{\circ}$, the following hold:

$(1)$* $\llbracket{\varphi}\rrbracket\subseteq\llbracket{\psi}\rrbracket$ if and only if $G(\varphi)\subseteq G(\psi)$.*

$(2)$* if $\mathsf{G}\vdash\varphi\Rightarrow_{T^{\circ}}\psi$, then $\llbracket{\varphi}\rrbracket\subseteq\llbracket{\psi}\rrbracket$.*

$(3)$* $\llbracket{T^{\circ}}\rrbracket$ is finite.*

$(4)$* $\llbracket{\varphi}\rrbracket\leq^{\circ}\llbracket{\psi}\rrbracket$ if and only if $\llbracket{\varphi}\rrbracket\subseteq\llbracket{\psi}\rrbracket$.*

$(5)$* $\llbracket{\varphi\wedge\psi}\rrbracket=\llbracket{\varphi}\rrbracket\cap\llbracket{\psi}\rrbracket$.*

Proof.

$(1)$ Assume $\llbracket{\varphi}\rrbracket\subseteq\llbracket{\psi}\rrbracket$. Let $\langle\chi\rangle^{n}\in G(\varphi)\subseteq\llbracket{\varphi}\rrbracket$. Then $\chi\in T^{\Diamond}$ and $\langle\chi\rangle^{n}\in\llbracket{\psi}\rrbracket$. Then $\langle\chi\rangle^{n}\in G(\psi)$. Hence $G(\varphi)\subseteq G(\psi)$. Assume $G(\varphi)\subseteq G(\psi)$. Then $G(\varphi)\cup FS(T^{\bullet})\subseteq G(\psi)\cup FS(T^{\bullet})$, i.e., $\llbracket{\varphi}\rrbracket\subseteq\llbracket{\psi}\rrbracket$.

$(2)$ Assume $\mathsf{G}\vdash\varphi\Rightarrow_{T^{\circ}}\psi$. Suppose $\langle\chi\rangle^{n}\in G(\varphi)$. Then $\chi\in T^{\Diamond}$ and $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\varphi$. By $(Cut)$, $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\psi$. Then $\langle\chi\rangle^{n}\in G(\psi)$. Then $G(\varphi)\subseteq G(\psi)$. By $(1)$, $\llbracket{\varphi}\rrbracket\subseteq\llbracket{\psi}\rrbracket$.

$(3)$ Since $T$ is finite, $T^{\Diamond}$ is also finite. There are only finitely many non-equivalent formulas in $(T^{\Diamond})^{b}$. Then $|T^{\circ}|_{\mathsf{G}}$ is finite. By $(2)$, if $\varphi\sim_{\mathsf{G}}\psi$, then $\llbracket{\varphi}\rrbracket=\llbracket{\psi}\rrbracket$. Hence $\llbracket{T^{\circ}}\rrbracket$ is finite.

$(4)$ Assume $\llbracket{\varphi}\rrbracket\leq^{\circ}\llbracket{\psi}\rrbracket$. Then $\llbracket{\varphi}\rrbracket\wedge^{\circ}\llbracket{\psi}\rrbracket=\llbracket{\varphi\wedge\psi}\rrbracket=\llbracket{\varphi}\rrbracket$. Suppose $\langle\chi\rangle^{n}\in G(\varphi)$. Then $\chi\in T^{\Diamond}$ and $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\varphi$. Since $\llbracket{\varphi\wedge\psi}\rrbracket=\llbracket{\varphi}\rrbracket$, we have $\langle\chi\rangle^{n}\in G(\varphi\wedge\psi)$. Then $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\varphi\wedge\psi$. Clearly $\mathsf{G}\vdash\varphi\wedge\psi\Rightarrow_{T^{\circ}}\psi$. By $(Cut)$, $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\psi$. Then $\langle\chi\rangle^{n}\in G(\psi)$. Hence $G(\varphi)\subseteq G(\psi)$. By $(1)$, we have $\llbracket{\varphi}\rrbracket\subseteq\llbracket{\psi}\rrbracket$.

Assume $\llbracket{\varphi}\rrbracket\subseteq\llbracket{\psi}\rrbracket$. By $(1)$, we have $G(\varphi)\subseteq G(\psi)$. Suppose $\langle\chi\rangle^{n}\in G(\varphi)$. Then $\langle\chi\rangle^{n}\in G(\psi)$. Then $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\varphi$ and $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\psi$. By $({\Rightarrow}\wedge)$, $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\varphi\wedge\psi$. Then $\langle\chi\rangle^{n}\in G(\varphi\wedge\psi)$. Hence $G(\varphi)\subseteq G(\varphi\wedge\psi)$. By $(1)$, $\llbracket{\varphi}\rrbracket\subseteq\llbracket{\varphi\wedge\psi}\rrbracket$. Suppose $\langle\chi\rangle^{n}\in G(\varphi\wedge\psi)$. Then $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\varphi\wedge\psi$. Clearly $\mathsf{G}\vdash\varphi\wedge\psi\Rightarrow_{T^{\circ}}\varphi$. By $(Cut)$, $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\varphi$. Then $\langle\chi\rangle^{n}\in G(\varphi)$. By $(1)$, $\llbracket{\varphi\wedge\psi}\rrbracket\subseteq\llbracket{\varphi}\rrbracket$. Hence $\llbracket{\varphi\wedge\psi}\rrbracket=\llbracket{\varphi}\rrbracket$, i.e., $\llbracket{\varphi}\rrbracket\leq^{\circ}\llbracket{\psi}\rrbracket$.

$(5)$ Clearly $\mathsf{G}\vdash\varphi\wedge\psi\Rightarrow_{T^{\circ}}\varphi$ and $\mathsf{G}\vdash\varphi\wedge\psi\Rightarrow_{T^{\circ}}\psi$. By (2), $\llbracket{\varphi\wedge\psi}\rrbracket\subseteq\llbracket{\varphi}\rrbracket$ and $\llbracket{\varphi\wedge\psi}\rrbracket\subseteq\llbracket{\psi}\rrbracket$. Then $\llbracket{\varphi\wedge\psi}\rrbracket\subseteq\llbracket{\varphi}\rrbracket\cap\llbracket{\psi}\rrbracket$. Clearly $\llbracket{\varphi}\rrbracket\cap\llbracket{\psi}\rrbracket=(G(\varphi)\cap G(\psi))\cup FS(T^{\bullet})$. Suppose $\langle\chi\rangle^{n}\in G(\varphi)\cap G(\psi)$. Then $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\varphi$ and $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\psi$. By $({\Rightarrow}\wedge)$, $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\varphi\wedge\psi$.Then $\langle\chi\rangle^{n}\in G(\varphi\wedge\psi)$. Then $G(\varphi)\cap G(\psi)\subseteq G(\varphi\wedge\psi)$. Then $\llbracket{\varphi}\rrbracket\cap\llbracket{\psi}\rrbracket\subseteq\llbracket{\varphi\wedge\psi}\rrbracket$. ∎

Lemma 0.3.6**.**

$\mathfrak{B}(T^{\circ})$* is a finite Boolean algebra.*

Proof.

By Lemma 0.3.5 $(3)$, $\llbracket{T^{\circ}}\rrbracket$ is finite. It is easy to show that $(\llbracket{T^{\circ}}\rrbracket,\wedge^{\circ},\vee^{\circ})$ is a distributive lattice. Here we show only the law of distributivity. Suppose $\varphi,\psi,\chi\in T^{\circ}$. Then $\llbracket{\varphi}\rrbracket\wedge^{\circ}(\llbracket{\psi}\rrbracket\vee^{\circ}\llbracket{\chi}\rrbracket)=\llbracket{\varphi\wedge(\psi\vee\chi)}\rrbracket$ and $(\llbracket{\varphi}\rrbracket\wedge^{\circ}\llbracket{\psi}\rrbracket)\vee^{\circ}(\llbracket{\varphi}\rrbracket\wedge^{\circ}\llbracket{\chi}\rrbracket)=\llbracket{(\varphi\wedge\psi)\vee(\varphi\wedge\chi)}\rrbracket$. Clearly $\mathsf{G}\vdash\varphi\wedge(\psi\vee\chi)\Rightarrow_{T^{\circ}}(\varphi\wedge\psi)\vee(\varphi\wedge\chi)$. Then $G(\varphi\wedge(\psi\vee\chi))\subseteq G((\varphi\wedge\psi)\vee(\varphi\wedge\chi))$. By Lemma 0.3.5 $(1)$ and $(4)$, $\llbracket{\varphi}\rrbracket\wedge^{\circ}(\llbracket{\psi}\rrbracket\vee^{\circ}\llbracket{\chi}\rrbracket)\leq^{\circ}(\llbracket{\varphi}\rrbracket\wedge^{\circ}\llbracket{\psi}\rrbracket)\vee^{\circ}(\llbracket{\varphi}\rrbracket\wedge^{\circ}\llbracket{\chi}\rrbracket)$. By Lemma 0.3.5 $(4)$, the order $\leq^{\circ}$ is equal to $\subseteq$. By $\mathsf{G}\vdash\varphi\Rightarrow_{T^{\circ}}\top$ and $\mathsf{G}\vdash\bot\Rightarrow_{T^{\circ}}\varphi$, using Lemma 0.3.5 $(2)$ and $(4)$, $\llbracket{\varphi}\rrbracket\leq^{\circ}\llbracket{\top}\rrbracket$ and $\llbracket{\bot}\rrbracket\leq^{\circ}\llbracket{\varphi}\rrbracket$. Hence the $(\wedge^{\circ},\vee^{\circ},\top^{\circ},\bot^{\circ})$-reduct of $\mathfrak{B}(T^{\circ})$ is a bounded distributive lattice. By $\mathsf{G}\vdash\varphi\wedge\neg\varphi\Rightarrow_{T^{\circ}}\bot$ and $\mathsf{G}\vdash\top\Rightarrow_{T^{\circ}}\varphi\vee\neg\varphi$, using Lemma 0.3.5 $(2)$ and $(4)$, $\llbracket{\varphi}\rrbracket\wedge^{\circ}\llbracket{\neg\varphi}\rrbracket=\llbracket{\varphi\wedge\neg\varphi}\rrbracket\leq^{\circ}\llbracket{\bot}\rrbracket$ and $\llbracket{\top}\rrbracket\leq^{\circ}\llbracket{\varphi\vee\neg\varphi}\rrbracket=\llbracket{\varphi}\rrbracket\vee^{\circ}\llbracket{\neg\varphi}\rrbracket$. Hence $\mathfrak{B}(T^{\circ})$ is a Boolean algebra. ∎

Definition 0.3.7**.**

The operation $C:\mathcal{P}(FS(T^{\circ}))\rightarrow\mathcal{P}(FS(T^{\circ}))$ is defined as follows:

[TABLE]

The unary operations $\Diamond$ and $\blacksquare$ on $\mathcal{P}(FS(T^{\circ}))$ are defined as follows:

[TABLE]

The unary operation $\Diamond_{c}$ on $\mathcal{P}(FS(T^{\circ}))$ is defined by $\Diamond_{c}X=C(\Diamond X)$.

Lemma 0.3.8**.**

For any $X,Y\in\mathcal{P}(FS(T^{\circ}))$, the following hold:

$(1)$* $\llbracket{\top}\rrbracket=FS(T^{\circ})$.*

$(2)$* $C(X)=\llbracket{\varphi}\rrbracket$ for some formula $\varphi\in T^{\circ}$.*

$(3)$* $X\subseteq C(X)$.*

$(4)$* if $X\subseteq Y$, then $C(X)\subseteq C(Y)$.*

$(5)$* $C(C(X))\subseteq C(X)$.*

$(6)$* $C(\llbracket{\varphi}\rrbracket)=\llbracket{\varphi}\rrbracket$.*

Proof.

$(1)$ Clearly $FS(T^{\circ})=FS(T^{\Diamond})\cup FS(T^{\bullet})$ and $\llbracket{\top}\rrbracket\subseteq FS(T^{\circ})$. It is obvious that $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow\top$ for any $n\geq 0$ and $\chi\in T^{\Diamond}$. Then $FS(T^{\Diamond})\subseteq\llbracket{\top}\rrbracket$. Clearly $FS(T^{\bullet})\subseteq\llbracket{\top}\rrbracket$. Then $FS(T^{\circ})\subseteq\llbracket{\top}\rrbracket$. Hence $\llbracket{\top}\rrbracket=FS(T^{\circ})$.

$(2)$ Let $\mathfrak{X}=\{\llbracket{\psi}\rrbracket\mid X\subseteq\llbracket{\psi}\rrbracket\in\llbracket{T^{\circ}}\rrbracket\}$. By Lemma 0.3.5 $(3)$, $\llbracket{T^{\circ}}\rrbracket$ is finite. Then $\mathfrak{X}$ is finite. By (1), $\mathfrak{X}\neq\varnothing$. Let $\mathfrak{X}=\{\llbracket{\psi_{0}}\rrbracket,\ldots,\llbracket{\psi_{n}}\rrbracket\}$. Then $C(X)=\llbracket{\psi_{0}}\rrbracket\cap\ldots\cap\llbracket{\psi_{n}}\rrbracket$. By Lemma 0.3.5 $(5)$, $C(X)=\llbracket{\bigwedge_{i\leq n}\psi_{i}}\rrbracket$. Clearly $\varphi=\bigwedge_{i\leq n}\psi_{i}\in T^{\circ}$.

$(3)$ It follows from the definition of $C(X)$.

$(4)$ Assume $X\subseteq Y$. By $(2)$, let $C(Y)=\llbracket{\psi}\rrbracket$ for some $\psi\in T^{\circ}$. By $(3)$, $Y\subseteq\llbracket{\psi}\rrbracket$. Then $X\subseteq\llbracket{\psi}\rrbracket$. By the definition of $C$, we have $C(X)\subseteq\llbracket{\psi}\rrbracket=C(Y)$.

$(5)$ By $(2)$, let $C(X)=\llbracket{\chi}\rrbracket$ for some $\chi\in T^{\circ}$. Then $C(C(X))\subseteq\llbracket{\chi}\rrbracket=C(X)$.

$(6)$ By $(3)$, $\llbracket{\varphi}\rrbracket\subseteq C(\llbracket{\varphi}\rrbracket)$. By the definition of $C$, $C(\llbracket{\varphi}\rrbracket)\subseteq\llbracket{\varphi}\rrbracket$. ∎

Lemma 0.3.9**.**

For any $X,Y\in\mathcal{P}(FS(T^{\circ}))$, the following hold:

$(1)$* if $X\subseteq Y$, then $\Diamond X\subseteq\Diamond Y$.*

$(2)$* $\Diamond X\subseteq Y$ if and only if $X\subseteq\blacksquare Y$.*

$(3)$* $\Diamond C(X)\subseteq C(\Diamond X)$.*

$(4)$* $C(\Diamond^{3}X)\subseteq C(\Diamond^{2}X)$.*

Proof.

$(1)$ Assume $X\subseteq Y$. Suppose $\langle\Gamma\rangle\in\Diamond X$ with $\Gamma\in X$. Then $\Gamma\in Y$. Then $\langle\Gamma\rangle\in\Diamond Y$.

$(2)$ Assume $\Diamond X\subseteq Y$. Suppose $\Gamma\in X$. Then $\langle\Gamma\rangle\in\Diamond X$. Then $\langle\Gamma\rangle\in Y$. Then $\Gamma\in\blacksquare Y$. Hence $X\subseteq\blacksquare Y$. Assume $X\subseteq\blacksquare Y$. Suppose $\langle\Gamma\rangle\in\Diamond X$ with $\Gamma\in X$. Then $\Gamma\in\blacksquare Y$. Then $\langle\Gamma\rangle\in Y$. Hence $\Diamond X\subseteq Y$.

$(3)$ By Lemma 0.3.8 $(2)$, let $C(\Diamond X)=\llbracket{\varphi}\rrbracket$ with $\varphi\in T^{\circ}$. Then $\Diamond X\subseteq\llbracket{\varphi}\rrbracket$. Let $\mathfrak{X}=\{\langle\chi\rangle^{n}\in X\mid\chi\in T^{\Diamond}~{}\&~{}n\geq 0\}$. Suppose $\mathfrak{X}=\varnothing$. Then $X\subseteq FS(T^{\bullet})\subseteq\llbracket{\bot}\rrbracket$. Hence $C(X)\subseteq\llbracket{\bot}\rrbracket$. By $(1)$, $\Diamond C(X)\subseteq\Diamond\llbracket{\bot}\rrbracket$. It is easy to show that $\Diamond\llbracket{\bot}\rrbracket\subseteq\llbracket{\bot}\rrbracket$. Clearly $\llbracket{\bot}\rrbracket\subseteq\llbracket{\varphi}\rrbracket$. Hence $\Diamond C(X)\subseteq C(\Diamond X)$. Suppose $\mathfrak{X}\neq\varnothing$. Take any $\langle\chi_{i}\rangle^{n_{i}}\in X$ with $\chi_{i}\in T^{\Diamond}$. Since $\Diamond X\subseteq\llbracket{\varphi}\rrbracket$, we have $\langle\chi_{i}\rangle^{n_{i}+1}\in\llbracket{\varphi}\rrbracket$. Then $\mathsf{G}\vdash\langle\chi_{i}\rangle^{n_{i}+1}\Rightarrow_{T^{\circ}}\varphi$. By Lemma 0.3.3, there exists $\gamma_{i}\in T^{\circ}$ with $\mathsf{G}\vdash\langle\chi_{i}\rangle^{n_{i}}\Rightarrow\gamma_{i}$ and $\mathsf{G}\vdash\langle\gamma_{i}\rangle\Rightarrow\varphi$. Let $Y\subseteq T^{\circ}$ be the set of all such interpolants. Since $|T^{\circ}|_{\mathsf{G}}$ is finite, $|Y|_{\mathsf{G}}$ is finite. Let $\{\gamma_{0},\ldots,\gamma_{k}\}$ be the set of all representatives selected from equivalence classes in $|Y|_{\mathsf{G}}$. Let $\xi=\gamma_{0}\vee\ldots\vee\gamma_{k}$. By $({\Rightarrow}\vee)$ and $(\vee{\Rightarrow})$, (i) $\mathsf{G}\vdash\langle\chi_{i}\rangle^{n_{i}}\Rightarrow\xi$ and (ii) $\mathsf{G}\vdash\langle\xi\rangle\Rightarrow\varphi$. Then $\mathfrak{X}\subseteq\llbracket{\xi}\rrbracket$. Clearly $FS(T^{\bullet})\subseteq\llbracket{\xi}\rrbracket$. Hence $X\subseteq\llbracket{\xi}\rrbracket$. Then $C(X)\subseteq\llbracket{\xi}\rrbracket$. By $(1)$, $\Diamond C(X)\subseteq\Diamond\llbracket{\xi}\rrbracket$. It suffices to show that $\Diamond\llbracket{\xi}\rrbracket\subseteq\llbracket{\varphi}\rrbracket$. Take any $\langle\langle\alpha\rangle^{j}\rangle\in\Diamond\llbracket{\xi}\rrbracket$ with $\langle\alpha\rangle^{j}\in\llbracket{\xi}\rrbracket$. If $\alpha\in T^{\bullet}$, then $\langle\langle\alpha\rangle^{j}\rangle\in\llbracket{\varphi}\rrbracket$. Suppose $\alpha\in T^{\Diamond}$. Then (iii) $\mathsf{G}\vdash\langle\alpha\rangle^{j}\Rightarrow_{T^{\circ}}\xi$. By applying $(Cut)$ to (iii) and (ii), $\mathsf{G}\vdash\langle\langle\alpha\rangle^{j}\rangle\Rightarrow_{T^{\circ}}\varphi$. Then $\langle\langle\alpha\rangle^{j}\rangle\in\llbracket{\varphi}\rrbracket$. Hence $\Diamond\llbracket{\xi}\rrbracket\subseteq\llbracket{\varphi}\rrbracket$. Therefore $\Diamond C(X)\subseteq C(\Diamond X)$.

$(4)$ By Lemma 0.3.8 $(2)$, let $C(\Diamond^{2}X)=\llbracket{\varphi}\rrbracket$ for some $\varphi\in T^{\circ}$. By Lemma 0.3.8 $(3)$, $\Diamond^{2}X\subseteq\llbracket{\varphi}\rrbracket$. Assume $\langle\psi\rangle^{n+3}\in\Diamond^{3}X$ with $\langle\psi\rangle^{n}\in X$. If $\psi\in T^{\bullet}$, then $\langle\psi\rangle^{n+3}\in\llbracket{\varphi}\rrbracket$. Suppose $\psi\in T^{\Diamond}$. Clearly $\langle\psi\rangle^{n+2}\in\Diamond^{2}X$. Then $\langle\psi\rangle^{n+2}\in\llbracket{\varphi}\rrbracket$. Hence $\mathsf{G}\vdash\langle\psi\rangle^{n+2}\Rightarrow_{T^{\circ}}\varphi$. By Lemma 0.3.2, $\mathsf{G}\vdash\langle\psi\rangle^{n+3}\Rightarrow_{T^{\circ}}\varphi$. Then $\langle\psi\rangle^{n+3}\in\llbracket{\varphi}\rrbracket$. Then $\Diamond^{3}X\subseteq\llbracket{\varphi}\rrbracket=C(\Diamond^{2}X)$. Hence $C(\Diamond^{3}X)\subseteq C(\Diamond^{2}X)$. ∎

Lemma 0.3.10**.**

For any $X\in\mathcal{P}(FS(T^{\circ}))$, $\Diamond_{c}^{3}X\subseteq\Diamond_{c}^{2}X$.

Proof.

By Lemma 0.3.9 $(1)$ and $(3)$, $\Diamond_{c}^{3}X=C(\Diamond C(\Diamond C(\Diamond X)))\subseteq C(C(C(\Diamond^{3}X)))$. By Lemma 0.3.8 $(3)$ and $(5)$, $C(C(C(\Diamond^{3}X)))=C(\Diamond^{3}X)$. By Lemma 0.3.9 $(4)$, $C(\Diamond^{3}X)\subseteq C(\Diamond^{2}X)$. By Lemma 0.3.8 $(3)$, $\Diamond X\subseteq C(\Diamond X)$. By Lemma 0.3.9 $(2)$, $\Diamond^{2}X\subseteq\Diamond C(\Diamond X)$. By Lemma 0.3.8 $(4)$, $C(\Diamond^{2}X)\subseteq C(\Diamond C(\Diamond X))=\Diamond_{c}^{2}X$. Hence $\Diamond_{c}^{3}X\subseteq\Diamond_{c}^{2}X$. ∎

Lemma 0.3.11**.**

For any $X,Y\in\mathcal{P}(FS(T^{\circ}))$, $\Diamond_{c}C(X)\subseteq C(Y)$ if and only if $C(X)\subseteq\blacksquare C(Y)$.

Proof.

Assume $\Diamond_{c}C(X)\subseteq C(Y)$. Then $C(\Diamond C(X))\subseteq C(Y)$. Clearly $\Diamond C(X)\subseteq\Diamond C(C(X))$. By Lemma 0.3.9 $(3)$, $\Diamond C(X)\subseteq\Diamond C(C(X))\subseteq C(\Diamond C(X))$. Then $\Diamond C(X)\subseteq C(Y)$. By Lemma 0.3.9 $(2)$, $C(X)\subseteq\blacksquare C(Y)$. Assume $C(X)\subseteq\blacksquare C(Y)$. By Lemma 0.3.9 $(2)$, $\Diamond C(X)\subseteq C(Y)$. By Lemma 0.3.8 $(4)$ and $(5)$, $C(\Diamond C(X))\subseteq C(C(Y))\subseteq C(Y)$. Hence $\Diamond_{c}C(X)\subseteq C(Y)$. ∎

Corollary 0.3.12**.**

For any $\llbracket{\varphi}\rrbracket,\llbracket{\psi}\rrbracket\in\llbracket{T^{\circ}}\rrbracket$, $\Diamond_{c}\llbracket{\varphi}\rrbracket\leq^{\circ}\llbracket{\psi}\rrbracket$ if and only if $\llbracket{\varphi}\rrbracket\leq^{\circ}\blacksquare\llbracket{\psi}\rrbracket$

Proof.

By Lemma 0.3.11 and Lemma 0.3.8 $(6)$. ∎

Lemma 0.3.13**.**

For any $\varphi,\Diamond\varphi,\blacksquare\varphi\in T^{\circ}$, $(1)$ $\Diamond_{c}\llbracket{\varphi}\rrbracket=\llbracket{\Diamond\varphi}\rrbracket$ and $(2)$ $\blacksquare\llbracket{\varphi}\rrbracket=\llbracket{\blacksquare\varphi}\rrbracket$.

Proof.

$(1)$ Assume $\langle\psi\rangle^{n}\in\llbracket{\varphi}\rrbracket$. Then $\langle\psi\rangle^{n+1}\in\Diamond\llbracket{\varphi}\rrbracket$. If $\psi\in T^{\bullet}$, then $\langle\psi\rangle^{n+1}\in\llbracket{\Diamond\varphi}\rrbracket$. Suppose $\psi\in T^{\Diamond}$. Then $\mathsf{G}\vdash\langle\psi\rangle^{n}\Rightarrow_{T^{\circ}}\varphi$. By $({\Rightarrow}\Diamond)$, $\mathsf{G}\vdash\langle\psi\rangle^{n+1}\Rightarrow_{T^{\circ}}\Diamond\varphi$. Then $\langle\psi\rangle^{n+1}\in\llbracket{\Diamond\varphi}\rrbracket$. Then $\Diamond\llbracket{\varphi}\rrbracket\subseteq\llbracket{\Diamond\varphi}\rrbracket$. Hence $C(\Diamond\llbracket{\varphi}\rrbracket)\subseteq\llbracket{\Diamond\varphi}\rrbracket$, i.e., $\Diamond_{c}\llbracket{\varphi}\rrbracket\subseteq\llbracket{\Diamond\varphi}\rrbracket$. Assume $\langle\chi\rangle^{n}\in\llbracket{\Diamond\varphi}\rrbracket$. By Lemma 0.3.8 $(2)$, let $\Diamond_{c}\llbracket{\varphi}\rrbracket=C(\Diamond\llbracket{\varphi}\rrbracket)=\llbracket{\delta}\rrbracket$ for some formula $\delta\in T^{\circ}$. If $\chi\in T^{\bullet}$, then $\langle\chi\rangle^{n}\in\llbracket{\delta}\rrbracket$. Suppose $\chi\in T^{\Diamond}$. Then (i) $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\Diamond\varphi$. Clearly $\varphi\in\llbracket{\varphi}\rrbracket$ and $\varphi,\Diamond\varphi\in T^{\Diamond}$. Then $\langle\varphi\rangle\in\Diamond\llbracket{\varphi}\rrbracket$. By Lemma 0.3.8 $(3)$, $\Diamond\llbracket{\varphi}\rrbracket\subseteq C(\Diamond\llbracket{\varphi}\rrbracket)=\llbracket{\delta}\rrbracket$. Then $\langle\varphi\rangle\in\llbracket{\delta}\rrbracket$. Then $\mathsf{G}\vdash\langle\varphi\rangle\Rightarrow_{T^{\circ}}\delta$. By $(\Diamond{\Rightarrow})$, (ii) $\mathsf{G}\vdash\Diamond\varphi\Rightarrow_{T^{\circ}}\delta$. By applying $(Cut)$ to (i) and (ii), $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\delta$. Then $\langle\chi\rangle^{n}\in\llbracket{\delta}\rrbracket$. Hence $\llbracket{\Diamond\varphi}\rrbracket\subseteq\llbracket{\delta}\rrbracket=\Diamond_{c}\llbracket{\varphi}\rrbracket$.

$(2)$ Assume $\langle\psi\rangle^{n}\in\blacksquare\llbracket{\varphi}\rrbracket$. Then $\langle\psi\rangle^{n+1}\in\llbracket{\varphi}\rrbracket$. If $\psi\in T^{\bullet}$, then $\langle\psi\rangle^{n}\in\llbracket{\blacksquare\varphi}\rrbracket$. Suppose $\psi\in T^{\Diamond}$. Then $\mathsf{G}\vdash\langle\psi\rangle^{n+1}\Rightarrow_{T^{\circ}}\varphi$. By $({\Rightarrow}\blacksquare)$, $\mathsf{G}\vdash\langle\psi\rangle^{n}\Rightarrow_{T^{\circ}}\blacksquare\varphi$. Then $\langle\psi\rangle^{n}\in\llbracket{\blacksquare\varphi}\rrbracket$. Hence $\blacksquare\llbracket{\varphi}\rrbracket\subseteq\llbracket{\blacksquare\varphi}\rrbracket$. Assume $\langle\chi\rangle^{n}\in\llbracket{\blacksquare\varphi}\rrbracket$. If $\chi\in T^{\bullet}$, then $\langle\chi\rangle^{n+1}\in\llbracket{\varphi}\rrbracket$. Suppose $\chi\in T^{\Diamond}$. Then $\mathsf{G}\vdash\langle\chi\rangle^{n}\Rightarrow_{T^{\circ}}\blacksquare\varphi$. Clearly $\mathsf{G}\vdash\langle\blacksquare\varphi\rangle\Rightarrow_{T^{\circ}}\varphi$. By $(Cut)$, $\mathsf{G}\vdash\langle\chi\rangle^{n+1}\Rightarrow_{T^{\circ}}\varphi$. Then $\langle\chi\rangle^{n+1}\in\llbracket{\varphi}\rrbracket$. Then $\langle\chi\rangle^{n}\in\blacksquare\llbracket{\varphi}\rrbracket$. Hence $\llbracket{\blacksquare\varphi}\rrbracket\subseteq\blacksquare\llbracket{\varphi}\rrbracket$. ∎

Lemma 0.3.14**.**

For any $\psi\in T^{\circ}$, there exists a formula $\varphi\in T^{\circ}$ with $\blacksquare\llbracket{\psi}\rrbracket=\llbracket{\varphi}\rrbracket$.

Proof.

Let $\mathfrak{X}=\{\langle\chi\rangle^{n}\in\blacksquare\llbracket{\psi}\rrbracket\mid\chi\in T^{\Diamond}~{}\&~{}n\geq 0\}$. Clearly $\bot\in T^{\Diamond}$ and $\langle\bot\rangle\in\llbracket{\psi}\rrbracket$. Then $\bot\in\blacksquare\llbracket{\psi}\rrbracket$. Hence $\mathfrak{X}\neq\varnothing$. Take any $\langle\chi_{i}\rangle^{n_{i}}\in\mathfrak{X}$. Then $\langle\chi_{i}\rangle^{n_{i}+1}\in\llbracket{\psi}\rrbracket$. Hence $\mathsf{G}\vdash\langle\chi_{i}\rangle^{n_{i}+1}\Rightarrow_{T^{\circ}}\psi$. By Lemma 0.3.3, there is $\gamma_{i}\in T^{\circ}$ with $\mathsf{G}\vdash\langle\chi_{i}\rangle^{n_{i}}\Rightarrow\gamma_{i}$ and $\mathsf{G}\vdash\langle\gamma_{i}\rangle\Rightarrow\psi$. Let $Y\subseteq T^{\circ}$ be the set of all such interpolants. Since $|T^{\circ}|_{\mathsf{G}}$ is finite, $|Y|_{\mathsf{G}}$ is finite. Let $\{\gamma_{0},\ldots,\gamma_{k}\}$ be the set of all representatives selected from equivalence classes in $|Y|_{\mathsf{G}}$. Let $\xi=\gamma_{0}\vee\ldots\vee\gamma_{k}$. By $({\Rightarrow}\vee)$ and $(\vee{\Rightarrow})$, (i) $\mathsf{G}\vdash\langle\chi_{i}\rangle^{n_{i}}\Rightarrow\xi$ and (ii) $\mathsf{G}\vdash\langle\xi\rangle\Rightarrow\psi$. Then $\mathfrak{X}\subseteq\llbracket{\xi}\rrbracket$. Clearly $FS(T^{\bullet})\subseteq\llbracket{\xi}\rrbracket$. Hence $\blacksquare\llbracket{\psi}\rrbracket\subseteq\llbracket{\xi}\rrbracket$. Take any $\langle\alpha\rangle^{k}\in\llbracket{\xi}\rrbracket$. Then $\langle\alpha\rangle^{k+1}\in\Diamond\llbracket{\xi}\rrbracket$. Suppose $\alpha\in T^{\bullet}$. Then $\langle\alpha\rangle^{k+1}\in\llbracket{\psi}\rrbracket$. Hence $\langle\alpha\rangle^{k}\in\blacksquare\llbracket{\psi}\rrbracket$. Suppose $\alpha\in T^{\Diamond}$. Then (iii) $\mathsf{G}\vdash\langle\alpha\rangle^{k}\Rightarrow_{T^{\circ}}\xi$. By applying $(Cut)$ to (ii) and (iii), $\mathsf{G}\vdash\langle\alpha\rangle^{k+1}\Rightarrow_{T^{\circ}}\psi$. Then $\langle\alpha\rangle^{k+1}\in\llbracket{\psi}\rrbracket$. Then $\langle\alpha\rangle^{k}\in\blacksquare\llbracket{\psi}\rrbracket$. Hence $\llbracket{\xi}\rrbracket\subseteq\blacksquare\llbracket{\psi}\rrbracket$. Therefore $\blacksquare\llbracket{\psi}\rrbracket=\llbracket{\xi}\rrbracket$. ∎

By Lemma 0.3.8 $(2)$ and Lemma 0.3.14, the operations $\Diamond_{c}$ and $\blacksquare$ are unary operations on $\llbracket{T^{\circ}}\rrbracket$. Now we get the algebra $\mathfrak{A}_{T}=(\llbracket{T^{\circ}}\rrbracket,\wedge^{\circ},\vee^{\circ},\neg^{\circ},\bot^{\circ},\top^{\circ},\Diamond_{c},\blacksquare)$.

Lemma 0.3.15**.**

$\mathfrak{A}_{T}$* is a finite quasi-transitive tense algebra.*

Proof.

By Lemma 0.3.6, the $(\wedge^{\circ},\vee^{\circ},\neg^{\circ},\bot^{\circ},\top^{\circ})$-reduct of $\mathfrak{A}_{T}$ is a finite Boolean algebra. By Corollary 0.3.12, $\mathfrak{A}_{T}$ is a tense algebra. By Lemma 0.3.10, $\Diamond_{c}^{3}\llbracket{\varphi}\rrbracket\leq^{\circ}\Diamond_{c}^{2}\llbracket{\varphi}\rrbracket$ for any $\llbracket{\varphi}\rrbracket\in\llbracket{T^{\circ}}\rrbracket$. Then $\mathfrak{A}_{T}$ is quasi-transitive. ∎

Theorem 0.3.16**.**

$\mathsf{G}$* has the FAP.*

Proof.

Assume $\mathsf{G}\not\vdash\langle\varphi\rangle^{n}\Rightarrow\psi$. Clearly $\mathsf{G}\vdash\langle\varphi\rangle^{n}\Rightarrow\Diamond^{n}\varphi$. Then $\mathsf{G}\not\vdash\Diamond^{n}\varphi\Rightarrow\psi$. Let $\chi=\Diamond^{n}\varphi$ and $T=Sub(\chi)\cup Sub(\psi)\cup\{\top,\bot\}$. Then $\mathsf{G}\not\vdash\chi\Rightarrow_{T^{\circ}}\psi$. Let $\theta_{T}$ be the assignment in $\mathfrak{A}_{T}$ with $\theta_{T}(p)=\llbracket{p}\rrbracket$ for every propositional variable $p$. By the definition of $\mathfrak{A}_{T}$ and Lemma 0.3.13, one can easily show by induction on the complexity of a formula $\xi\in T^{\circ}$ that $\widehat{\theta_{T}}(\xi)=\llbracket{\xi}\rrbracket$. Since $\chi\in T$, we have $\chi\in T^{\Diamond}$. By $\mathsf{G}\vdash\chi\Rightarrow_{T^{\circ}}\chi$, we have $\chi\in\llbracket{\chi}\rrbracket$. By $\mathsf{G}\not\vdash\chi\Rightarrow_{T^{\circ}}\psi$, we have $\chi\not\in\llbracket{\psi}\rrbracket$. Then $\llbracket{\chi}\rrbracket\not\subseteq\llbracket{\psi}\rrbracket$. Hence $\mathfrak{A}_{T}\not\models\chi\Rightarrow\psi$, i.e., $\mathfrak{A}_{T}\not\models\langle\varphi\rangle^{n}\Rightarrow\psi$. By Lemma 0.3.15, $\mathfrak{A}_{T}$ is a finite quasi-transitive tense algebra. ∎

Finally, by the algebraic completeness of tense logic $\mathsf{Kt}_{2}^{3}=\mathsf{Kt}\oplus\Diamond^{3}p\rightarrow\Diamond^{2}p$ (cf. [2, 8]), for any tense formula $\varphi$, $\mathsf{Kt}_{2}^{3}\vdash\varphi$ if and only if $\mathsf{G}\vdash\top\Rightarrow\varphi$. Using duality between tense algebras and bidirectional frames (cf. [2, 8]), by Theorem 0.3.16, one can get the FMP of $\mathsf{Kt}_{2}^{3}$, i.e., if $\mathsf{Kt}_{2}^{3}\not\vdash\varphi$, then $\varphi$ is refuted by the dual frame of the algebra $\mathfrak{A}_{T}$ where $T$ is the set of tense formulas $Sub(\varphi)\cup\{\top,\bot\}$. Finally, since $\mathsf{Kt}_{2}^{3}$ is a conservative extension of $\mathsf{K}_{2}^{3}=\mathsf{K}\oplus\Diamond^{3}p\rightarrow\Diamond^{2}p$, we obtain the FMP of $\mathsf{K}_{2}^{3}$.

Corollary 0.3.17**.**

$\mathsf{Kt}_{2}^{3}$* and $\mathsf{K}_{2}^{3}$ have the FMP and hence are decidable.*

0.4 Concluding remarks

We established the finite model property of the quasi-transitive modal logic $\mathsf{K}\oplus\Diamond\Diamond\Diamond p\rightarrow\Diamond\Diamond p$ by showing the finite model property of its conservative tense extension $\mathsf{Kt}_{2}^{3}$. In the sequent calculus $\mathsf{G}$ for the tense logic $\mathsf{Kt}_{2}^{3}$, for each sequent which is not derivable in $\mathsf{G}$ there exists a finite syntactic algebraic model that refutes the sequent. We can extend the method in the present work to show the FMP of logics $\mathsf{K}\oplus\Box^{n}p\rightarrow\Box^{m}p$ for $n\neq m\geq 0$. Furthermore, we may extend the method to show the FMP of non-classical modal logics. For example, the finite model property of some intuitionistic modal logics and lattice-based modal logics can be proved.

Acknowledgements

The first author was supported by Chinese National Funding of Social Sciences (No. 17CZX048). The second author was supported by Guangdong Province Higher Vocational Colleges $\&$ Schools Pearl River Scholar Funded Scheme (2017-2019). Thanks are given to the reviewers’ insightful and helpful comments on the revision of this paper. In particular, the first reviewer mentioned some proof-theoretic points of the sequent calculus in the first version. The second reviewer pointed out the possibility of a shorter proof of the main result.