The fixed point and the Craig interpolation properties for sublogics of $\textbf{IL}$

Abstract

We study the fixed point property and the Craig interpolation property for sublogics of the interpretability logic $\textbf{IL}$. We provide a complete description of these sublogics concerning the uniqueness of fixed points, the fixed point property and the Craig interpolation property.

1 Introduction

De Jongh and Sambin’s fixed point theorem [12] for the modal propositional logic $\textbf{GL}$ is one of notable results of modal logical investigation of formalized provability. For any modal formula A, let v(A) be the set of all propositional variables contained in A. A logic L is said to have the fixed point property (FPP) if for any modal formula A(p) in which the propositional variable p appears only in the scope of $\Box$, there exists a modal formula B such that $v(B) \subseteq v(A) \setminus \{p\}$ and $L \vdash B \leftrightarrow A(B)$. De Jongh and Sambin’s theorem states that $\textbf{GL}$ has FPP, and this is understood as a counterpart of the fixed point theorem in formal arithmetic (see [4]). Bernardi [2] also proved the uniqueness of fixed points (UFP) for $\textbf{GL}$.

A logic L is said to have the Craig interpolation property (CIP) if for any formulas A and B, if $L \vdash A \rightarrow B$, then there exists a formula C such that $v(C) \subseteq v(A) \cap v(B)$, $L \vdash A \rightarrow C$, and $L \vdash C \rightarrow B$. Smoryński [13] and Boolos [3] independently proved that $\textbf{GL}$ has CIP. Smoryński also made an important observation that FPP for $\textbf{GL}$ follows from CIP and UFP.

The interpretability logic $\textbf{IL}$ is an extension of $\textbf{GL}$ in the language of $\textbf{GL}$ equipped with the binary modal operator $\rhd$, where the modal formula $A \rhd B$ is read as “$T + B$ is relatively interpretable in $T + A$”. It is natural to ask whether $\textbf{IL}$ also has the properties that hold for $\textbf{GL}$. Indeed, de Jongh and Visser [8] proved UFP for $\textbf{IL}$ and that $\textbf{IL}$ has FPP. Also Areces, Hoogland, and de Jongh [1] proved that $\textbf{IL}$ has CIP.

Ignatiev [5] introduced the sublogic $\textbf{CL}$ of $\textbf{IL}$ as a base logic of the modal logical investigation of the notion of partial conservativity, and proved that $\textbf{CL}$ is complete with respect to relational semantics (that is, usual Veltman semantics). Kurahashi and Okawa [9] also introduced several sublogics of $\textbf{IL}$, and showed the completeness and the incompleteness of these sublogics with respect to relational semantics.

In this paper, we investigate UFP, FPP, and CIP for sublogics of $\textbf{IL}$ shown in Fig. 1.

Fig. 1

Sublogics of $\textbf{IL}$

Moreover, for technical reasons, we introduce and investigate the notions of $\ell$UFP and $\ell$FPP that are restricted versions of UFP and FPP with respect to some particular forms of formulas, respectively. Table 1 summarizes a complete description of these sublogics concerning $\ell$UFP, UFP, $\ell$FPP, FPP, and CIP.

Table 1 $\ell$UFP, UFP, $\ell$FPP, FPP, and CIP for sublogics of $\textbf{IL}$

The paper is organized as follows. In Sect. 3, we show that UFP holds for extensions of $\textbf{IL}^-(\textbf{J4}_+)$, and that UFP is not the case for sublogics of $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$. We also show that $\ell$UFP holds for extensions of $\textbf{IL}^-$. In Sect. 4, we prove that the logic $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ has CIP by modifying a semantical proof of CIP for $\textbf{IL}$ by Areces, Hoogland, and de Jongh. We also notice that CIP for $\textbf{IL}$ easily follows from CIP for $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$. In Sect. 5, we observe that FPP for $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ immediately follows from our results in the previous sections. Also, we give a syntactical proof of FPP for $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$. Moreover, we prove that $\textbf{IL}^-(\textbf{J4}, \textbf{J5})$ has $\ell$FPP. In Sect. 6, we provide counter models of $\ell$FPP for $\textbf{CL}$ and $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$ and a counter model of FPP for $\textbf{IL}^-(\textbf{J1}, \textbf{J4}_+, \textbf{J5})$. As a consequence, we also show that CIP is not the case for these sublogics except $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ and $\textbf{IL}$.

2 Preliminaries

2.1 $\textbf{IL}$ and its sublogics

The interpretability logic $\textbf{IL}$ is a base logic of modal logical investigations of the notion of relative interpretability (see [15, 16]). The language of $\textbf{IL}$ consists of propositional variables $p, q, \ldots$, the propositional constant $\bot$, the logical connective $\rightarrow$, the unary modal operator $\Box$, and the binary modal operator $\rhd$. Other logical connectives, the propositional constant $\top$, and the modal operator $\Diamond$ are introduced as usual abbreviations. The formulas of $\textbf{IL}$ are generated by the following grammar:

$$\begin{aligned} A {::} = \bot \mid p \mid A \rightarrow A \mid \Box A \mid A \rhd A. \end{aligned}$$

For each formula A, let $\boxdot A \equiv A \wedge \Box A$.

Definition 1

The axioms of the modal propositional logic $\textbf{IL}$ are as follows:

$\textbf{L1}$:

All tautologies in the language of $\textbf{IL}$;

$\textbf{L2}$:

$\Box (A \rightarrow B) \rightarrow (\Box A \rightarrow \Box B)$;

$\textbf{L3}$:

$\Box (\Box A \rightarrow A) \rightarrow \Box A$;

$\textbf{J1}$:

$\Box (A \rightarrow B) \rightarrow A \rhd B$;

$\textbf{J2}$:

$(A \rhd B) \wedge (B \rhd C) \rightarrow A \rhd C$;

$\textbf{J3}$:

$(A \rhd C) \wedge (B \rhd C) \rightarrow (A \vee B) \rhd C$;

$\textbf{J4}$:

$A \rhd B \rightarrow (\Diamond A \rightarrow \Diamond B)$;

$\textbf{J5}$:

$\Diamond A \rhd A$.

The inference rules of $\textbf{IL}$ are Modus Ponens $\dfrac{A\ \ \ A \rightarrow B}{B}$ and Necessitation $\dfrac{A}{\Box A}$.

The conservativity logic $\textbf{CL}$ is obtained from $\textbf{IL}$ by removing the axiom scheme $\textbf{J5}$, that was introduced by Ignatiev [5] as a base logic of modal logical investigations of the notion of partial conservativity. Several other sublogics of $\textbf{IL}$ were introduced in [9]. The basis for these newly introduced logics is the logic $\textbf{IL}^-$.

Definition 2

The language of $\textbf{IL}^-$ is that of $\textbf{IL}$, and the axioms of $\textbf{IL}^-$ are $\textbf{L1}, \textbf{L2}, \textbf{L3}, \textbf{J3}$, and $\textbf{J6}$: $\Box A \leftrightarrow (\lnot A) \rhd \bot$. The inference rules of $\textbf{IL}^-$ are Modus Ponens, Necessitation, $\textbf{R1}$ $\dfrac{A \rightarrow B}{C \rhd A \rightarrow C \rhd B}$, and $\textbf{R2}$ $\dfrac{A \rightarrow B}{B \rhd C \rightarrow A \rhd C}$.

For schemata $\varSigma _1, \ldots , \varSigma _n$, let $\textbf{IL}^-(\varSigma _1, \ldots , \varSigma _n)$ be the logic obtained by adding $\varSigma _1, \ldots , \varSigma _n$ as axiom schemata to $\textbf{IL}^-$. The following schemata $\textbf{J2}_+$ and $\textbf{J4}_+$ were introduced in [9] and [15], respectively:

$\textbf{J2}_+$:

$(A \rhd (B \vee C)) \wedge (B \rhd C) \rightarrow A \rhd C$;

$\textbf{J4}_+$:

$\Box (A \rightarrow B) \rightarrow (C \rhd A \rightarrow C \rhd B)$.

In this paper, we mainly deal with logics consisting of some of the axiom schemata $\textbf{J1}, \textbf{J2}_+, \textbf{J4}_+$, and $\textbf{J5}$ (see Fig. 1 in Sect. 1). Then, we have the following proposition.

Proposition 1

Let A, B, and C be any formulas.

• 1. $\textbf{IL}^- \vdash \Box \lnot A \rightarrow A \rhd B$.

• 2. $\textbf{IL}^- \vdash \Box (A \rightarrow B) \rightarrow (B \rhd C \rightarrow A \rhd C)$.

• 3. $\textbf{IL}^- \vdash (\lnot A \wedge B) \rhd C \rightarrow (A \rhd C \rightarrow B \rhd C)$.

• 4. $\textbf{IL}^-(\textbf{J4}_+) \vdash \textbf{J4}$.

• 5. $\textbf{IL}^-(\textbf{J2}_+) \vdash \textbf{J2} \wedge \textbf{J4}_+$.

• 6. $\textbf{IL}^-(\textbf{J2}_+) \vdash (A \rhd B) \wedge ((B \wedge \lnot C) \rhd C) \rightarrow (A \rhd C)$.

• 7. $\textbf{IL}^-(\textbf{J1}) \vdash A \rhd A$.

• 8. $\textbf{CL}$ is deductively equivalent to $\textbf{IL}^-(\textbf{J1}, \textbf{J2}_+)$.

• 9. $\textbf{IL}$ is deductively equivalent to $\textbf{IL}^-(\textbf{J1}, \textbf{J2}_+, \textbf{J5})$.

Proof

Except clause 3, see [9]. For 3, by $\textbf{J3}$, $\textbf{IL}^- \vdash ((\lnot A \wedge B) \rhd C) \wedge (A \rhd C) \rightarrow ((\lnot A \wedge B) \vee A) \rhd C$. Since $\textbf{IL}^- \vdash B \rightarrow ((\lnot A \wedge B) \vee A)$, we have $\textbf{IL}^- \vdash ((\lnot A \wedge B) \vee A) \rhd C \rightarrow B \rhd C$ by the rule $\textbf{R2}$. Thus, $\textbf{IL}^- \vdash ((\lnot A \wedge B) \rhd C) \wedge (A \rhd C) \rightarrow B \rhd C$. $\square$

The following lemma (Lemma 1) plays an important role in our proofs of CIP and FPP for $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ in Sects. 4 and 5.

Fact 1

(See [17]) For any formula A,

$$\begin{aligned} \textbf{IL}^- \vdash (A \vee \Diamond A) \leftrightarrow ((A \wedge \Box \lnot A) \vee \Diamond (A \wedge \Box \lnot A)). \end{aligned}$$

Lemma 1

Let A and C be any formulas.

• 1. $\textbf{IL}^-(\textbf{J2}, \textbf{J5}) \vdash ((A \wedge \Box \lnot A) \rhd C) \leftrightarrow (A \rhd C)$.

• 2. $\textbf{IL}^-(\textbf{J2}_{+}, \textbf{J5}) \vdash (C \rhd (A \wedge \Box \lnot A)) \leftrightarrow (C \rhd A)$.

Proof

In this proof, let $B \equiv (A \wedge \Box \lnot A)$.

1. $(\leftarrow )$: Since $\textbf{IL}^- \vdash B \rightarrow A$, we have $\textbf{IL}^- \vdash A \rhd C \rightarrow B \rhd C$ by $\textbf{R2}$.

$(\rightarrow )$: Since $\textbf{IL}^-(\textbf{J5}) \vdash \Diamond B \rhd B$, we have $\textbf{IL}^-(\textbf{J2}, \textbf{J5}) \vdash B \rhd C \rightarrow \Diamond B \rhd C$. Hence, by $\textbf{J3}$,

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}, \textbf{J5}) \vdash B \rhd C \rightarrow (B \vee \Diamond B)\rhd C. \end{aligned}$$

By Fact 1 and $\textbf{R2}$, we obtain

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}, \textbf{J5}) \vdash B \rhd C \rightarrow (A \vee \Diamond A) \rhd C. \end{aligned}$$

Since $\textbf{IL}^- \vdash A \rightarrow (A \vee \Diamond A)$, we obtain

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}, \textbf{J5}) \vdash B \rhd C \rightarrow A \rhd C \end{aligned}$$

by $\textbf{R2}$.

2. $(\rightarrow )$: This is immediate from $\textbf{IL}^- \vdash B \rightarrow A$ and $\textbf{R1}$.

$(\leftarrow )$: Since $\textbf{IL}^- \vdash A \rightarrow (A \vee \Diamond A)$, we obtain

$$\begin{aligned} \textbf{IL}^- \vdash C \rhd A \rightarrow C \rhd (A \vee \Diamond A) \end{aligned}$$

by $\textbf{R1}$. Then, by Fact 1 and $\textbf{R1}$,

$$\begin{aligned} \textbf{IL}^- \vdash C \rhd A \rightarrow C \rhd (B \vee \Diamond B). \end{aligned}$$

Since $\textbf{IL}^-(\textbf{J5}) \vdash \Diamond B \rhd B$, we obtain

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}_{+}, \textbf{J5}) \vdash C \rhd A \rightarrow C \rhd B \end{aligned}$$

because $(C \rhd (\Diamond B \vee B)) \wedge (\Diamond B \rhd B) \rightarrow C \rhd B$ is an instance of $\textbf{J2}_+$. $\square$

2.2 $\textbf{IL}^-$-frames and models

Definition 3

We say that a system $\langle W, R, \{S_w\}_{w \in W} \rangle$ is an $\textbf{IL}^-$-frame if it satisfies the following three conditions:

• 1. W is a non-empty set;

• 2. R is a transitive and conversely well-founded binary relation on W;

• 3. For each $w \in W$, $S_w$ is a binary relation on W with

  $$\begin{aligned} \forall x, y \in W (x S_w y \Rightarrow w R x). \end{aligned}$$

A system $\langle W, R, \{S_w\}_{w \in W}, \Vdash \rangle$ is called an $\textbf{IL}^-$-model if $\langle W, R, \{S_w\}_{w \in W} \rangle$ is an $\textbf{IL}^-$-frame and $\Vdash$ is a usual satisfaction relation on the Kripke frame $\langle W, R \rangle$ with the following additional condition:

$$ \begin{aligned} w \Vdash A \rhd B \iff \forall x \in W(w R x \ \& \ x \Vdash A \Rightarrow \exists y \in W (x S_w y\ \& \ y \Vdash B)). \end{aligned}$$

A formula A is said to be valid in an $\textbf{IL}^-$-frame $\langle W, R, \{S_w\}_{w \in W} \rangle$ if $w\Vdash A$ for any satisfaction relation $\Vdash$ on the frame and any $w \in W$.

For each $w \in W$, let $R[w]: = \{x \in W: w R x\}$.

Proposition 2

(See [9] and [15]) Let ${\mathcal {F}} = \langle W, R, \{S_w\}_{w \in W} \rangle$ be any $\textbf{IL}^-$-frame.

• 1. $\textbf{J1}$ is valid in ${\mathcal {F}}$ if and only if for any $w, x \in W$, if wRx, then $x S_w x$.

• 2. $\textbf{J2}_+$ is valid in ${\mathcal {F}}$ if and only if $\textbf{J4}_+$ is valid in ${\mathcal {F}}$ and for any $w \in W$, $S_w$ is transitive.

• 3. $\textbf{J4}_+$ is valid in ${\mathcal {F}}$ if and only if for any $w \in W$, $S_w$ is a binary relation on R[w].

• 4. $\textbf{J5}$ is valid in ${\mathcal {F}}$ if and only if for any $w, x, y \in W$, wRx and xRy imply $x S_w y$.

Theorem 1

(See [5, 7] and [9]) Let L be one of logics shown in Fig. 1 in Sect. 1. Then, for any formula A, the following are equivalent:

• 1. $L \vdash A$.

• 2. A is valid in all (finite) $\textbf{IL}^-$-frames in which all axioms of L are valid.

2.3 The fixed point and the Craig interpolation properties

For each formula A, let v(A) be the set of all propositional variables contained in A.

Definition 4

We say that a formula A is modalized in a propositional variable p if every occurrence of p in A is in the scope of some modal operators $\Box$ or $\rhd$.

Definition 5

A logic L is said to have the fixed point property (FPP) if for any propositional variable p and any formula A(p) which is modalized in p, there exists a formula F such that $v(F) \subseteq v(A) {\setminus } \{p\}$ and $L \vdash F \leftrightarrow A(F)$.

Definition 6

We say that the uniqueness of fixed points (UFP) holds for a logic L if for any propositional variables p, q and any formula A(p) which is modalized in p and does not contain q,

$$\begin{aligned} L \vdash \boxdot (p \leftrightarrow A(p)) \wedge \boxdot (q \leftrightarrow A(q)) \rightarrow (p \leftrightarrow q). \end{aligned}$$

Theorem 2

(De Jongh and Visser [8])

• 1. $\textbf{IL}$ has FPP.

• 2. UFP holds for $\textbf{IL}$.

In particular, de Jongh and Visser showed that a fixed point of a formula $A(p) \rhd B(p)$ is $A(\top ) \rhd B(\Box \lnot A(\top ))$. Then, a fixed point of every formula A(p) which is modalized in p is explicitly calculable by a usual argument.

Definition 7

A logic L is said to have the Craig interpolation property (CIP) if for any formulas A and B, if $L \vdash A \rightarrow B$, then there exists a formula C such that $v(C) \subseteq v(A) \cap v(B)$, $L \vdash A \rightarrow C$, and $L \vdash C \rightarrow B$.

Theorem 3

(Areces, Hoogland, and de Jongh [1]) $\textbf{IL}$ has CIP.

3 Uniqueness of fixed points

In this section, we investigate the uniqueness of fixed points for sublogics. Firstly, we show that UFP holds for extensions of $\textbf{IL}^-(\textbf{J4}_{+})$. Secondly, we prove that UFP is not the case for sublogics of $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$. Then, we investigate the newly introduced notion that a formula A(p) is left-modalized in a propositional variable p. We prove that UFP with respect to formulas which are left-modalized in p ($\ell$UFP) holds for all extensions of $\textbf{IL}^-$. Finally, we discuss Smoryński’s implication “CIP + UFP $\Rightarrow$ FPP” in our framework.

3.1 UFP

By adapting Smoryński’s argument [14], de Jongh and Visser [8] showed that UFP holds for every logic closed under Modus Ponens and Necessitation, and containing $\textbf{L1}$, $\textbf{L2}$, $\textbf{L3}$, $\textbf{E1}$, and $\textbf{E2}$, where

$\textbf{E1}$:

$\Box (A \leftrightarrow B) \rightarrow (A \rhd C \leftrightarrow B \rhd C)$;

$\textbf{E2}$:

$\Box (A \leftrightarrow B) \rightarrow (C \rhd A \leftrightarrow C \rhd B)$.

Since $\textbf{E1}$ and $\textbf{E2}$ are easy consequences of Proposition 1.2 and $\textbf{J4}_+$ respectively, we obtain the following theorem.

Theorem 4

(UFP for $\textbf{IL}^-(\textbf{J4}_+)$) UFP holds for every extension of the logic $\textbf{IL}^-(\textbf{J4}_+)$.

As shown in [8], in the proof of Theorem 4, the use of the following substitution principle is essential.

Proposition 3

(The Substitution Principle) Let A, B, and C(p) be any formulas.

• 1. $\textbf{IL}^-(\textbf{J4}_{+}) \vdash \boxdot (A \leftrightarrow B) \rightarrow (C(A) \leftrightarrow C(B))$.

• 2. If C(p) is modalized in p, then $\textbf{IL}^-(\textbf{J4}_{+}) \vdash \Box (A \leftrightarrow B) \rightarrow (C(A) \leftrightarrow C(B))$.

Proposition 3.2 shows that every extension L of $\textbf{IL}^-(\textbf{J4}_+)$ proves $\Box (A \leftrightarrow B) \rightarrow (C(A) \leftrightarrow C(B))$ for any formula C(p) which is modalized in p. We show that the converse of this statement also holds.

Proposition 4

Let L be any extension of $\textbf{IL}^-$. Suppose that for any formula C(p) which is modalized in p, $L \vdash \Box (A \leftrightarrow B) \rightarrow (C(A) \leftrightarrow C(B))$. Then, $L \vdash \textbf{J4}_+$.

Proof

Let A, B, and C be any formulas and assume $p \notin v(C)$. Then, the formula $C \rhd p$ is modalized in p. By the supposition, we have

$$\begin{aligned} L \vdash \Box (A \leftrightarrow A \wedge B) \rightarrow (C \rhd A \leftrightarrow C \rhd (A \wedge B)). \end{aligned}$$

Since $\textbf{IL}^- \vdash \Box (A \rightarrow B) \rightarrow \Box (A \leftrightarrow A \wedge B)$ and $\textbf{IL}^- \vdash C \rhd (A \wedge B) \rightarrow C \rhd B$, we obtain $L \vdash \Box (A \rightarrow B) \rightarrow (C \rhd A \rightarrow C \rhd B)$. $\square$

On the other hand, we show that UFP does not hold for sublogics of $\textbf{IL}^{-}(\textbf{J1}, \textbf{J5})$ in general.

Proposition 5

Let p, q be distinct propositional variables. Then,

$$\begin{aligned} \textbf{IL}^{-}(\textbf{J1}, \textbf{J5}) \nvdash \boxdot (p \leftrightarrow (\top \rhd \lnot p)) \wedge \boxdot (q \leftrightarrow (\top \rhd \lnot q)) \rightarrow (p \leftrightarrow q). \end{aligned}$$

Proof

We define an $\textbf{IL}^-$-frame ${\mathcal {F}} = \langle W, R, \{S_{w}\}_{w \in W}\rangle$ as follows:

• $W:= \{w, x, y\}$;

• $R:= \{\langle w, x\rangle \}$;

• $S_{w}:=\{\langle x, x\rangle , \langle x, y\rangle \}$, $S_{x}:= \emptyset$, $S_{y}:= \emptyset$.

Obviously, by Proposition 2, $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$ is valid in ${\mathcal {F}}$. Let $\Vdash$ be a satisfaction relation on ${\mathcal {F}}$ satisfying the following conditions:

• $w \Vdash p$ and $w \nVdash q$;

• $x \Vdash p$ and $x \Vdash q$;

• $y \nVdash p$ and $y \Vdash q$.

Fig. 2

A counter model of UFP for $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$

We prove $w \Vdash \boxdot (p \leftrightarrow (\top \rhd \lnot p)) \wedge \boxdot (q \leftrightarrow (\top \rhd \lnot q)) \wedge \lnot (p \leftrightarrow q)$. Since $w \Vdash p$ and $w \nVdash q$, $w \Vdash \lnot (p \leftrightarrow q)$ is obvious. We show $w \Vdash (p \leftrightarrow (\top \rhd \lnot p)) \wedge (q \leftrightarrow (\top \rhd \lnot q))$. Since $w \Vdash p$ and $w \nVdash q$, it suffices to prove $w \Vdash \top \rhd \lnot p$ and $w \Vdash \lnot (\top \rhd \lnot q)$.

• $w \Vdash \top \rhd \lnot p$: Let $z \in W$ be any element with wRz. Then, $z = x$. Since $x S_{w} y$ and $y \Vdash \lnot p$, we obtain $w \Vdash \top \rhd \lnot p$.

• $w \Vdash \lnot (\top \rhd \lnot q)$: Let $z \in W$ be any element with $x S_{w} z$. Then, $z = x$ or $z = y$. In either case, we obtain $z \Vdash q$. Since xRy, we conclude $w \Vdash \lnot (\top \rhd \lnot q)$.

At last, we show $w \Vdash \Box (p \leftrightarrow (\top \rhd \lnot p)) \wedge \Box (q \leftrightarrow (\top \rhd \lnot q))$. Let $z \in W$ be such that wRz. Then, $z = x$. Since there is no $z' \in W$ such that $x R z'$, $x \Vdash (\top \rhd \lnot p) \wedge (\top \rhd \lnot q)$. Since $x \Vdash p$ and $x \Vdash q$, we have $x \Vdash (p \leftrightarrow (\top \rhd \lnot p)) \wedge (q \leftrightarrow (\top \rhd \lnot q))$. Hence, we obtain $w \Vdash \Box (p \leftrightarrow (\top \rhd \lnot p)) \wedge \Box (q \leftrightarrow (\top \rhd \lnot q))$.

Therefore, $w \Vdash \boxdot (p \leftrightarrow (\top \rhd \lnot p)) \wedge \boxdot (q \leftrightarrow (\top \rhd \lnot q)) \wedge \lnot (p \leftrightarrow q)$. $\square$

3.2 $\ell$UFP

Even for extensions of $\textbf{IL}^-$, Proposition 1.2 suggests that the uniqueness of fixed points may hold with respect to formulas in some particular forms. From this perspective, we introduce the notion that formulas are left-modalized in p.

Definition 8

We say that a formula A is left-modalized in a propositional variable p if A is modalized in p and $p \notin v(C)$ for any subformula $B \rhd C$ of A.

Then, we obtain the following version of the substitution principle.

Proposition 6

Let A, B, and C(p) be any formulas such that $p \notin v(E)$ for any subformula $D \rhd E$ of C.

• 1. $\textbf{IL}^- \vdash \boxdot (A \leftrightarrow B) \rightarrow (C(A) \leftrightarrow C(B))$.

• 2. If C(p) is left-modalized in p, then $\textbf{IL}^- \vdash \Box (A \leftrightarrow B) \rightarrow (C(A) \leftrightarrow C(B))$.

Proof

1. This proposition is proved by induction on the construction of C(p). We only prove the case $C(p) \equiv D(p) \rhd E$ (By our supposition, $p \notin v(E)$). For any subformula $D' \rhd E'$ of D, it is also a subformula of C, and hence $p \notin v(E')$. Then, by the induction hypothesis, we obtain

$$\begin{aligned} \textbf{IL}^- \vdash \boxdot (A \leftrightarrow B) \rightarrow (D(A) \leftrightarrow D(B)). \end{aligned}$$

Then, $\textbf{IL}^- \vdash \Box (A \leftrightarrow B) \rightarrow \Box (D(A) \leftrightarrow D(B))$. Therefore, by Proposition 1.2,

$$\begin{aligned} \textbf{IL}^- \vdash \Box (A \leftrightarrow B) \rightarrow (D(A) \rhd E \leftrightarrow D(B) \rhd E). \end{aligned}$$

Since $p \notin v(E)$, we find $C(A) \equiv (D(A) \rhd E)$ and $C(B) \equiv (D(B) \rhd E)$. Therefore,

$$\begin{aligned} \textbf{IL}^- \vdash \Box (A \leftrightarrow B) \rightarrow (C(A) \leftrightarrow C(B)). \end{aligned}$$

2. This follows from our proof of clause 1. $\square$

We introduce our restricted versions of UFP and FPP.

Definition 9

We say that $\ell$UFP holds for a logic L if for any formula A(p) which is left-modalized in p, $L \vdash \boxdot (p \leftrightarrow A(p)) \wedge \boxdot (q \leftrightarrow A(q)) \rightarrow (p \leftrightarrow q)$.

Definition 10

We say that a logic L has $\ell$FPP if for any formula A(p) which is left-modalized in p, there exists a formula F such that $v(F) \subseteq v(A) {\setminus } \{p\}$ and $L \vdash F \leftrightarrow A(F)$.

Then, $\ell$UFP holds for every our sublogic of $\textbf{IL}$.

Theorem 5

$(\ell$UFP for $\textbf{IL}^-)$ $\ell$UFP holds for all extensions of $\textbf{IL}^-$.

Proof

Let A(p) be any formula which is left-modalized in p. Then, by Proposition 6.2, $\textbf{IL}^- \vdash \Box (p \leftrightarrow q) \rightarrow (A(p) \leftrightarrow A(q))$. Therefore,

$$\begin{aligned} \textbf{IL}^- \vdash \boxdot (p \leftrightarrow A(p)) \wedge \boxdot (q \leftrightarrow A(q))&\rightarrow (\Box (p \leftrightarrow q) \rightarrow (A(p) \leftrightarrow A(q))) \\&\rightarrow (\Box (p \leftrightarrow q) \rightarrow (p \leftrightarrow q)) \\&\rightarrow (\Box (\Box (p \leftrightarrow q) \rightarrow (p \leftrightarrow q))) \\&\rightarrow \Box (p \leftrightarrow q)\\&\rightarrow (p \leftrightarrow q). \end{aligned}$$

$\square$

3.3 Applications of Smoryński’s argument

We have shown that UFP and the substitution principle hold for extensions of $\textbf{IL}^-(\textbf{J4}_+)$ (Theorem 4 and Proposition 3). Then, by applying Smoryński’s argument [13], we prove that for any appropriate extension of $\textbf{IL}^-(\textbf{J4}_+)$, CIP implies FPP.

Lemma 2

Let L be any extension of $\textbf{IL}^{-}(\textbf{J4}_{+})$ that is closed under substituting a formula for a propositional variable. If L has CIP, then L also has FPP.

Proof

Suppose $L \supseteq \textbf{IL}^-(\textbf{J4}_+)$ and L has CIP. Let A(p) be any formula modalized in p. Then, by Theorem 4,

$$\begin{aligned} L \vdash \boxdot (p \leftrightarrow A(p)) \wedge \boxdot (q \leftrightarrow A(q)) \rightarrow (p \leftrightarrow q). \end{aligned}$$

We have

$$\begin{aligned} L \vdash \boxdot (p \leftrightarrow A(p)) \wedge p \rightarrow (\boxdot (q \leftrightarrow A(q)) \rightarrow q). \end{aligned}$$

Since L has CIP, there exists a formula F such that $v(F) \subseteq v(A) {\setminus } \{p\}$, $L \vdash \boxdot (p \leftrightarrow A(p)) \wedge p \rightarrow F$ and $L \vdash F \rightarrow (\boxdot (q \leftrightarrow A(q)) \rightarrow q)$. Since $q \notin v(F)$, we have $L \vdash F \rightarrow (\boxdot (p \leftrightarrow A(p)) \rightarrow p)$ by substituting p for q. Then,

$$\begin{aligned} L \vdash \boxdot (p \leftrightarrow A(p)) \rightarrow (F \leftrightarrow p). \end{aligned}$$

By substituting A(F) for p, we get

$$\begin{aligned} L \vdash \boxdot (A(F) \leftrightarrow A(A(F))) \rightarrow (F \leftrightarrow A(F)). \end{aligned}$$

(1)

Then

$$\begin{aligned} L \vdash \Box (A(F) \leftrightarrow A(A(F))) \rightarrow \Box (F \leftrightarrow A(F)). \end{aligned}$$

Since A(p) is modalized in p, by Proposition 3.2,

$$\begin{aligned} L \vdash \Box (A(F) \leftrightarrow A(A(F))) \rightarrow (A(F) \leftrightarrow A(A(F))). \end{aligned}$$

Then, by applying the axiom scheme $\textbf{L3}$, we obtain $L \vdash A(F) \leftrightarrow A(A(F))$. From this with (1), we conclude $L \vdash F \leftrightarrow A(F)$. Therefore, F is a fixed point of A(p) in L. $\square$

Also, we have shown that $\ell$UFP and the substitution principle with respect to left-modalized formulas hold for extensions of $\textbf{IL}^-$ (Theorem 5 and Proposition 6). Thus, our proof of Lemma 2 also works for proving the following lemma.

Lemma 3

Let L be any extension of $\textbf{IL}^{-}$ that is closed under substituting a formula for a propositional variable. If L has CIP, then L also has $\ell$FPP.

4 The Craig interpolation property

In this section, we prove the following theorem.

Theorem 6

(CIP for $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5}))$ The logic $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ has CIP.

Our proof of Theorem 6 is based on a semantical proof of CIP for $\textbf{IL}$ due to Areces, Hoogland, and de Jongh [1].

4.1 Preparations for our proof of Theorem 6

In this subsection, we prepare several definitions and prove some lemmas that are used in our proof of Theorem 6. Only in this section, we write $\vdash A$ instead of $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5}) \vdash A$ if there is no confusion. Notice that by Proposition 1, $\vdash \textbf{J2} \wedge \textbf{J4} \wedge \textbf{J4}_+$.

For a formula A, we define the formula ${\sim } A$ as follows:

$$\begin{aligned} {\sim } A:\equiv \left\{ \begin{array}{ll} B &{} \text {if}\ A \equiv \lnot B\ \text {for some formula}\ B, \\ \lnot A &{} \text {otherwise.} \end{array} \right. \end{aligned}$$

For a set X of formulas, by ${\mathcal {L}}_X$ we denote the set of all formulas built up from $\bot$ and propositional variables occurring in formulas in X. We simply write ${\mathcal {L}}_A$ instead of ${\mathcal {L}}_{\{ A \}}$. For a finite set X of formulas, let $\bigwedge X$ be a conjunction of all elements of X. For the sake of simplicity, only in this section, $\vdash \bigwedge X \rightarrow A$ will be written as $\vdash X \rightarrow A$.

For a set $\varPhi$ of formulas, we define

$$\begin{aligned} \varPhi _\rhd := \{A: \text {there exists a formula}\ B \ \text {such that}\ A \rhd B \in \varPhi \ \text {or}\ B \rhd A \in \varPhi \}. \end{aligned}$$

Definition 11

A set $\varPhi$ of formulas is said to be adequate if it satisfies the following conditions:

• 1. $\varPhi$ is closed under taking subformulas and the ${\sim }$-operation;

• 2. $\bot \in \varPhi _\rhd$;

• 3. If $A, B \in \varPhi _\rhd$, then $A \rhd B \in \varPhi$;

• 4. If $A \in \varPhi _\rhd$, then $\Box {\sim } A \in \varPhi$.

Our notion of adequate sets is essentially the same as that introduced by de Jongh and Veltman [7].

Note that for any finite set X of formulas, there exists the smallest finite adequate set $\varPhi$ containing X. We denote this set by $\varPhi _X$.

Definition 12

• 1. A pair $(\varGamma _1, \varGamma _2)$ of finite sets of formulas is said to be separable if for some formula $I \in {\mathcal {L}}_{\varGamma _1} \cap {\mathcal {L}}_{\varGamma _2}$, $\vdash \varGamma _1 \rightarrow I$, and $\vdash \varGamma _2 \rightarrow \lnot I$. A pair is said to be inseparable if it is not separable.

• 2. A pair $(\varGamma _1, \varGamma _2)$ of finite sets of formulas is said to be complete if it is inseparable and

  • For each $F \in \varPhi _{\varGamma _1}$, either $F \in \varGamma _1$ or ${\sim } F \in \varGamma _1$;

  • For each $F \in \varPhi _{\varGamma _2}$, either $F \in \varGamma _2$ or ${\sim } F \in \varGamma _2$.

We say a finite set X of formulas is consistent if $\nvdash X \rightarrow \bot$. If a pair $(\varGamma _1, \varGamma _2)$ is inseparable, then it can be shown that both of $\varGamma _1$ and $\varGamma _2$ are consistent.

In the rest of this subsection, we fix some sets X and Y of formulas. Put $\varPhi ^1:= \varPhi _X$ (resp. $\varPhi ^2:= \varPhi _Y$) and ${\mathcal {L}}_1:= {\mathcal {L}}_X$ (resp. ${\mathcal {L}}_2:= {\mathcal {L}}_Y$). Let $X' \subseteq \varPhi ^1$ and $Y' \subseteq \varPhi ^2$. It is easily proved that if $(X', Y')$ is inseparable, then for any formula $A \in \varPhi ^1$, at least one of $(X' \cup \{A\}, Y')$ and $(X' \cup \{{\sim } A\}, Y')$ is inseparable. Also, a similar statement holds for $\varPhi ^2$ and $Y'$. Then, we obtain the following proposition.

Proposition 7

If (X, Y) is inseparable, then there exists some complete pair $\varGamma ' = (\varGamma _1, \varGamma _2)$ such that $X \subseteq \varGamma _1 \subseteq \varPhi ^1$ and $Y \subseteq \varGamma _2 \subseteq \varPhi ^2$.

Let $K(\varPhi ^1, \varPhi ^2)$ be the set of all complete pairs $(\varGamma _1, \varGamma _2)$ satisfying $\varGamma _1 \subseteq \varPhi ^1$ and $\varGamma _2 \subseteq \varPhi ^2$. Note that the set $K(\varPhi ^1, \varPhi ^2)$ is finite. For each $\varGamma \in K(\varPhi ^1, \varPhi ^2)$, let $\varGamma _1$ and $\varGamma _2$ be the first and the second components of $\varGamma$, respectively.

Definition 13

We define a binary relation $\prec$ on $K(\varPhi ^1, \varPhi ^2)$ as follows: For $\varGamma , \varDelta \in K(\varPhi ^1, \varPhi ^2)$,

$$\begin{aligned} \varGamma \prec \varDelta :\Leftrightarrow \begin{array}{l} \text {For}\ i = \{ 1, 2 \},\ \text {if}\ \Box A \in \varGamma _i, \ \text {then}\ \Box A, A \in \varDelta _i, \ \text {and} \\ \text {there exists some}\ \Box B\ \text {such that}\ \Box B \in \varDelta _1 \cup \varDelta _2 \ \text {and} \ \Box B \not \in \varGamma _1 \cup \varGamma _2. \end{array} \end{aligned}$$

Then, $\prec$ is a transitive and conversely well-founded binary relation on $K(\varPhi ^1, \varPhi ^2)$.

Definition 14

Let $\varGamma , \varDelta \in K(\varPhi ^1, \varPhi ^2)$ and $A \in \varPhi ^1_\rhd \cup \varPhi ^2_\rhd$. We say that $\varDelta$ is an A-critical successor of $\varGamma$ (write $\varGamma \prec _A \varDelta$) if the following conditions are met:

• 1. $\varGamma \prec \varDelta$;

• 2. If $A \in \varPhi ^1_\rhd$, then

  $$ \begin{aligned} \varGamma _1^A:=&\{ \Box {\sim } B, {\sim } B : B \rhd A \in \varGamma _1 \} \subseteq \varDelta _1; \\ \varGamma _2^A:=&\{ \Box {\sim } C, {\sim } C : C \in \varPhi ^2_\rhd \ \text {and for some}\ I \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2, \\&\quad \vdash \varGamma _1 \rightarrow (I \wedge \lnot A) \rhd A \ \& \ \vdash \varGamma _2 \rightarrow C \rhd I \} \subseteq \varDelta _2. \end{aligned}$$

• 3. If $A \in \varPhi ^2_\rhd$, then

  $$ \begin{aligned} \varGamma _1^A :=&\{ \Box {\sim } B, {\sim } B : B \in \varPhi ^1_\rhd \ \text {and for some}\ I \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2, \\&\quad \vdash \varGamma _1 \rightarrow B \rhd I\ \& \ \vdash \varGamma _2 \rightarrow (I \wedge \lnot A) \rhd A \} \subseteq \varDelta _1; \\ \varGamma _2^A :=&\{ \Box {\sim } C, {\sim } C : C \rhd A \in \varGamma _2 \} \subseteq \varDelta _2. \end{aligned}$$

The notion of A-critical successors is originally introduced by de Jongh and Veltman [7]. Our definition is based on a modification due to Areces, Hoogland, and de Jongh [1].

From the following claim, Definition 14 makes sense.

Proposition 8

If $A \in \varPhi ^1_\rhd \cap \varPhi ^2_\rhd$, then the sets $\varGamma _1^A$ in clauses 2 and 3 of Definition 14 coincide. This is also the case for $\varGamma _2^A$.

Proof

We prove only for $\varGamma _1^A$. It suffices to show that for any formula B, the following are equivalent:

• 1. $B \rhd A \in \varGamma _1$.

• 2. $B \in \varPhi ^1_\rhd$ and for some $I \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2$, $\vdash \varGamma _1 \rightarrow B \rhd I$, and $\vdash \varGamma _2 \rightarrow (I \wedge \lnot A) \rhd A$.

$(1 \Rightarrow 2)$: Suppose $B \rhd A \in \varGamma _1$, then $B \in \varPhi ^1_\rhd$. By Proposition 1.1, we have $\textbf{IL}^- \vdash (A \wedge \lnot A) \rhd A$ because $\textbf{IL}^- \vdash \Box \lnot (A \wedge \lnot A)$. Since $A \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2$, the clause 2 holds by letting $I \equiv A$.

$(2 \Rightarrow 1)$: Assume that clause 2 holds. Then, $A \rhd B \in \varPhi ^1$ because $A, B \in \varPhi ^1_\rhd$. Suppose, towards a contradiction, that $\lnot (B \rhd A) \in \varGamma _1$. By Proposition 1.6, $\vdash (B \rhd I) \wedge ((I \wedge \lnot A) \rhd A) \rightarrow B \rhd A$. Then, we obtain $\vdash \varGamma _1 \rightarrow \lnot ((I \wedge \lnot A) \rhd A)$. This contradicts the inseparability of $\varGamma$ because $(I \wedge \lnot A) \rhd A \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2$. Hence, $\lnot (B \rhd A) \notin \varGamma _1$. Since $\varGamma$ is complete, $B \rhd A \in \varGamma _1$. $\square$

Lemma 4

For $\varGamma , \varDelta \in K(\varPhi ^1, \varPhi ^2)$, if $\varGamma \prec \varDelta$, then $\varGamma \prec _\bot \varDelta$.

Proof

Notice that $\bot \in \varPhi ^1_\rhd \cap \varPhi ^2_\rhd$. By Proposition 8, it suffices to show that if $C \rhd \bot \in \varGamma _1$ (resp. $\varGamma _2$), then $\Box {\sim } C, {\sim } C \in \varDelta _1$ (resp. $\varDelta _2$). Suppose $C \rhd \bot \in \varGamma _1$. Then, by $(\textbf{J6})$, $\vdash \varGamma _1 \rightarrow \Box {\sim } C$. Note that $\Box {\sim } C \in \varPhi ^1$, and hence $\Box {\sim } C \in \varGamma _1$. By $\varGamma \prec \varDelta$, $\Box {\sim } C, {\sim } C \in \varDelta _1$. The case $C \rhd \bot \in \varGamma _2$ is proved similarly. Therefore, $\varGamma \prec _\bot \varDelta$. $\square$

Lemma 5

For $\varGamma , \varDelta , \varTheta \in K(\varPhi ^1, \varPhi ^2)$ and $A \in \varPhi ^1_\rhd \cup \varPhi ^2_\rhd$, if $\varGamma \prec _A \varDelta$ and $\varDelta \prec \varTheta$, then $\varGamma \prec _A \varTheta$.

Proof

We only prove the case $A \in \varPhi ^1_\rhd$. Let $\varGamma _1^A$ and $\varGamma _2^A$ be the sets as in Definition 14. If $\Box {\sim } B, {\sim } B \in \varGamma _1^A$, then $\Box {\sim } B \in \varDelta _1$ because $\varGamma \prec _A \varDelta$. Thus, $\Box {\sim } B, {\sim } B \in \varTheta _1$ because $\varDelta \prec \varTheta$. Similarly, if $\Box {\sim } C, {\sim } C \in \varGamma _2^A$, then $\varTheta _2$ contains $\Box {\sim } C$ and ${\sim } C$. This means $\varGamma \prec _A \varTheta$. $\square$

In order to prove the Truth Lemma (Lemma 4.2), we show the following two lemmas.

Lemma 6

Let $\varGamma \in K(\varPhi ^1, \varPhi ^2)$. If $\lnot (G \rhd F) \in \varGamma _1 \cup \varGamma _2$, then there exists a pair $\varDelta \in K(\varPhi ^1, \varPhi ^2)$ such that:

• 1. $\varGamma \prec _F \varDelta$;

• 2. $G, \Box {\sim } F \in \varDelta _1 \cup \varDelta _2$.

Proof

Suppose $\lnot (G \rhd F) \in \varGamma _1$. Let

$$ \begin{aligned} X':=&\boxdot \varGamma _1 \cup \{ G, \Box {\sim } G, \Box {\sim } F \} \cup \{ \Box {\sim } A, {\sim } A : A \rhd F \in \varGamma _1 \}; \\ Y':=&\boxdot \varGamma _2 \cup \{ \Box {\sim } B, {\sim } B : B \in \varPhi ^2_\rhd \ \text {and for some}\ I \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2, \\&\quad \vdash \varGamma _1 \rightarrow (I \wedge \lnot F) \rhd F\ \& \ \vdash \varGamma _2 \rightarrow B \rhd I \}, \end{aligned}$$

where $\boxdot \varGamma _i$ ($i =1, 2$) denotes the set $\{ \Box C, C: \Box C \in \varGamma _i \}$.

We claim $\Box {\sim } G \not \in \varGamma _1 \cup \varGamma _2$. Assume $\Box \lnot G \in \varGamma _1$. Then, $\vdash \varGamma _1 \rightarrow \Box \lnot G$. By Proposition 1.1, $\vdash \Box \lnot G \rightarrow G \rhd F$. Hence, $\vdash \varGamma _1 \rightarrow G \rhd F$. This implies that $\varGamma _1$ is inconsistent, a contradiction. Thus, $\Box {\sim } G \not \in \varGamma _1$. Moreover, if $\Box {\sim } G \in \varGamma _2$, then $\Diamond G$ separates $(\varGamma _1, \varGamma _2)$ because $\Diamond G \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2$. This contradicts the inseparability of $\varGamma$. Hence, $\Box {\sim } G \not \in \varGamma _2$.

We show that $( X', Y' )$ is inseparable. Suppose, for a contradiction, that $J \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2$ separates $(X', Y')$. From $\vdash Y' \rightarrow \lnot J$,

$$\begin{aligned} \vdash \boxdot \varGamma _2 \rightarrow \left( J \rightarrow \bigvee _{j \in \kappa }{(\Diamond B_j \vee B_j)} \right) , \end{aligned}$$

where $\kappa$ is an appropriate index set for $Y'$ such that for each $j \in \kappa$, $B_j \in \varPhi ^2_\rhd$ and there exists a formula $I_j \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2$ such that

$$\begin{aligned}{} & {} \vdash \varGamma _1 \rightarrow (I_j \wedge \lnot F) \rhd F, \ \text {and} \end{aligned}$$

(2)

$$\begin{aligned}{} & {} \vdash \varGamma _2 \rightarrow B_j \rhd I_j. \end{aligned}$$

(3)

Then

$$\begin{aligned} \vdash \varGamma _2 \rightarrow \Box \left( J \rightarrow \bigvee _{j \in \kappa }{(\Diamond B_j \vee B_j)} \right) . \end{aligned}$$

By Proposition 1.2,

$$\begin{aligned} \vdash \varGamma _2 \rightarrow \left( \left( \bigvee _{j \in \kappa }{(\Diamond B_j \vee B_j)} \right) \rhd \bigvee _{j \in \kappa } I_j \rightarrow J \rhd \bigvee _{j \in \kappa } I_j \right) . \end{aligned}$$

By (3), $\textbf{J2}$, $\textbf{J3}$, and $\textbf{J5}$, we have $\vdash \varGamma _2 \rightarrow \left( \bigvee _{j \in \kappa }{(\Diamond B_j \vee B_j)}\right) \rhd \bigvee _{j \in \kappa } I_j$. Hence

$$\begin{aligned} \vdash \varGamma _2 \rightarrow J \rhd \bigvee _{j \in \kappa }I_j. \end{aligned}$$

(4)

On the other hand, from $\vdash X' \rightarrow J$,

By $\textbf{J2}$, $\textbf{J3}$, and $\textbf{J5}$, we have $\vdash \varGamma _1 \rightarrow \left( \bigvee \nolimits _{A \rhd F \in \varGamma _1}{(\Diamond A \vee A)} \vee \Diamond F\right) \rhd F$. Hence, we obtain $\vdash \varGamma _1 \rightarrow (\lnot J \wedge G \wedge \Box \lnot G) \rhd F$. By Proposition 1.3, $\vdash \varGamma _1 \rightarrow \left( J \rhd F \rightarrow (G \wedge \Box \lnot G ) \rhd F \right)$. By Lemma 1.1, $\vdash \varGamma _1 \rightarrow \left( J \rhd F \rightarrow G \rhd F \right)$. Since $\vdash \varGamma _1 \rightarrow \lnot ( G \rhd F )$, we get $\vdash \varGamma _1 \rightarrow \lnot ( J \rhd F)$. From (2) and $\textbf{J3}$, we obtain $\vdash \varGamma _1 \rightarrow \left( \bigvee \nolimits _{j \in \kappa }I_j \wedge \lnot F \right) \rhd F$. By Proposition 1.6, $\vdash \varGamma _1 \rightarrow \left( J \rhd \bigvee \nolimits _{j \in \kappa } I_j \rightarrow J \rhd F \right)$. Hence,

$$\begin{aligned} \vdash \varGamma _1 \rightarrow \lnot \left( J \rhd \bigvee _{j \in \kappa } I_j \right) . \end{aligned}$$

From this and (4), we conclude that $\lnot (J \rhd \bigvee _{j \in \kappa } I_j)$ separates $(\varGamma _1, \varGamma _2)$, a contradiction. Therefore, $(X', Y')$ is inseparable.

Now let $\varDelta \in K(\varPhi ^1, \varPhi ^2)$ be a complete pair extending $(X', Y')$. We have $\varGamma \prec _F \varDelta$ and $G, \Box {\sim } F \in \varDelta _1$. The other case where $\lnot (G \rhd F) \in \varGamma _2$ is proved in a similar way. $\square$

Lemma 7

Let $\varGamma , \varDelta \in K(\varPhi ^1, \varPhi ^2)$. Suppose that $\varGamma \prec _A \varDelta$, $G \rhd F \in \varGamma _1 \cup \varGamma _2$ and $G \in \varDelta _1 \cup \varDelta _2$. Then, there exists a pair $\varTheta \in K(\varPhi ^1, \varPhi ^2)$ such that:

• $\varGamma \prec _A \varTheta$;

• $F \in \varTheta _1 \cup \varTheta _2$;

• $\Box {\sim } A, {\sim } A \in \varTheta _1 \cup \varTheta _2$.

Proof

Suppose $G \rhd F \in \varGamma _1$. From $G \in \varDelta _1 \cup \varDelta _2$ and $G \in \varPhi _1$, we obtain $G \in \varDelta _1$ by the inseparability of $\varDelta$. We distinguish the following two cases:

(Case 1): Assume $A \in \varPhi ^1_\rhd$. Then, $G \rhd A \in \varPhi ^1$. If $G \rhd A \in \varGamma _1$, then ${\sim } G \in \varDelta _1$ because $\varGamma \prec _A \varDelta$. This contradicts the consistency of $\varDelta _1$. Therefore, $G \rhd A \notin \varGamma _1$. Since $\varGamma$ is complete, we have $\lnot (G \rhd A) \in \varGamma _1$.

Let:

$$ \begin{aligned} X':=&\boxdot \varGamma _1 \cup \{ \Box {\sim } F, F, \Box {\sim } A, {\sim } A \} \cup \{ \Box {\sim } B, {\sim } B : B \rhd A \in \varGamma _1 \};\\ Y':=&\boxdot \varGamma _2 \cup \{ \Box {\sim } C, {\sim } C : C \in \varPhi ^2_\rhd \ \text {and for some}\ I \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2, \\&\quad \vdash \varGamma _1 \rightarrow (I \wedge \lnot A) \rhd A\ \& \ \vdash \varGamma _2 \rightarrow C \rhd I \}. \end{aligned}$$

We show $\Box {\sim } F \not \in \varGamma _1 \cup \varGamma _2$. If $\Box {\sim } G \in \varGamma _1$, then ${\sim } G \in \varDelta _1$ because $\varGamma \prec \varDelta$. This contradicts the consistency of $\varDelta _1$. Hence, $\Box {\sim } G \notin \varGamma _1$. Since $\vdash \varGamma _1 \rightarrow (G \rhd F) \wedge \Diamond G$, we have $\vdash \varGamma _1 \rightarrow \Diamond F$ by $\textbf{J4}$. Therefore, $\Box {\sim } F \notin \varGamma _1$. Moreover, if $\Box {\sim } F \in \varGamma _2$, then $\Diamond F$ would separate $(\varGamma _1, \varGamma _2)$, a contradiction. Thus, $\Box {\sim } F \notin \varGamma _2$.

We show that $(X', Y')$ is inseparable. Suppose, for a contradiction, that for some $J \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2$, $\vdash X' \rightarrow J$ and $\vdash Y' \rightarrow \lnot J$.

From $\vdash Y' \rightarrow \lnot J$,

$$\begin{aligned} \vdash \boxdot \varGamma _2 \rightarrow \left( J \rightarrow \bigvee _{j \in \kappa }{(\Diamond C_j \vee C_j)} \right) , \end{aligned}$$

where $\kappa$ is an appropriate index set such that for each $j \in \kappa$, $C_j \in \varPhi ^2_\rhd$ and there exists a formula $I_j \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2$ such that $\vdash \varGamma _1 \rightarrow (I_j \wedge \lnot A) \rhd A$ and $\vdash \varGamma _2 \rightarrow C_j \rhd I_j$. Then,

$$\begin{aligned} \vdash \varGamma _2 \rightarrow \Box \left( J \rightarrow \bigvee _{j \in \kappa }{(\Diamond C_j \vee C_j)} \right) . \end{aligned}$$

Since $\vdash \varGamma _2 \rightarrow \left( \bigvee _{j \in \kappa }{(\Diamond C_j \vee C_j)}\right) \rhd \bigvee I_j$, by Proposition 1.2, we obtain

$$\begin{aligned} \vdash \varGamma _2 \rightarrow J \rhd \bigvee I_j. \end{aligned}$$

(5)

On the other hand, from $\vdash X' \rightarrow J$,

$$\begin{aligned}&\vdash \boxdot \varGamma _1 \rightarrow \left( \lnot J \wedge \Box \lnot F \wedge F \wedge \lnot A \rightarrow \Diamond A \vee \bigvee _{B \rhd A \in \varGamma _1}{(\Diamond B \vee B)} \right) , \\&\vdash \varGamma _1 \rightarrow \Box \left( \lnot J \wedge \Box \lnot F \wedge F \wedge \lnot A \rightarrow \Diamond A \vee \bigvee _{B \rhd A \in \varGamma _1}{(\Diamond B \vee B)} \right) . \end{aligned}$$

Then, by Proposition 1.2, we obtain $\vdash \varGamma _1 \rightarrow (\lnot J \wedge \Box \lnot F \wedge F \wedge \lnot A) \rhd A$ because $\vdash \varGamma _1 \rightarrow \left( \Diamond A \vee \bigvee _{B \rhd A \in \varGamma _1}{(\Diamond B \vee B)} \right) \rhd A$. By Proposition 1.3, $\vdash \varGamma _1 \rightarrow \left( J \rhd A \rightarrow (\Box \lnot F \wedge F \wedge \lnot A) \rhd A \right)$. By Lemma 1.2, we have $\vdash \varGamma _1 \rightarrow G \rhd (\Box \lnot F \wedge F)$. Then, by Proposition 1.6, we obtain $\vdash \varGamma _1 \rightarrow ((\Box \lnot F \wedge F \wedge \lnot A) \rhd A \rightarrow G \rhd A)$. Thus, $\vdash \varGamma _1 \rightarrow (J \rhd A \rightarrow G \rhd A)$. Since $\lnot (G \rhd A) \in \varGamma _1$, we get $\vdash \varGamma _1 \rightarrow \lnot (J \rhd A)$. Since $\vdash \varGamma _1 \rightarrow \left( \bigvee _{j \in \kappa } I_j \wedge \lnot A \right) \rhd A$, we have $\vdash \varGamma _1 \rightarrow \left( J \rhd \bigvee _{j \in \kappa } I_j \rightarrow J \rhd A \right)$ by Proposition 1.6. Therefore,

$$\begin{aligned} \vdash \varGamma _1 \rightarrow \lnot \left( J \rhd \bigvee _{j \in \kappa } I_j \right) . \end{aligned}$$

From this and (5), we conclude that $\lnot (J \rhd \bigvee _{j \in \kappa } I_j)$ separates $(\varGamma _1, \varGamma _2)$, a contradiction.

(Case 2): Assume $A \in \varPhi ^2_\rhd$. Let:

$$ \begin{aligned} X':=&\boxdot \varGamma _1 \cup \{ \Box {\sim } F, F \}\\&\cup \{ \Box {\sim } B, {\sim } B : B \in \varPhi ^1_\rhd \ \text {and for some}\ I \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2, \\&\quad \vdash \varGamma _1 \rightarrow B \rhd I \ \& \ \vdash \varGamma _2 \rightarrow (I \wedge \lnot A) \rhd A \}; \\ Y':=&\boxdot \varGamma _2 \cup \{ \Box {\sim } A, {\sim } A \} \cup \{ \Box {\sim } C, {\sim } C : C \rhd A \in \varGamma _2 \}. \end{aligned}$$

As in Case 1, it can be shown $\Box {\sim } F \not \in \varGamma _1 \cup \varGamma _2$. We prove that $(X', Y')$ is inseparable. Suppose, for a contradiction, that for some $J \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2$, $\vdash X' \rightarrow J$ and $\vdash Y' \rightarrow \lnot J$. From $\vdash X' \rightarrow J$,

$$\begin{aligned} \vdash \boxdot \varGamma _1 \rightarrow \left( \Box \lnot F \wedge F \wedge \lnot J \rightarrow \bigvee _{j\in \kappa }{(\Diamond B_j \vee B_j)} \right) , \end{aligned}$$

where $\kappa$ is an appropriate index set such that for each $j \in \kappa$, $B_j \in \varPhi ^1_\rhd$ and there exists a formula $I_j \in {\mathcal {L}}_1 \cap {\mathcal {L}}_2$ such that $\vdash \varGamma _1 \rightarrow B_j \rhd I_j$ and $\vdash \varGamma _2 \rightarrow (I_j \wedge \lnot A) \rhd A$. Then,

$$\begin{aligned} \vdash \varGamma _1 \rightarrow \Box \left( \Box \lnot F \wedge F \wedge \lnot J \rightarrow \bigvee _{j\in \kappa }{(\Diamond B_j \vee B_j)} \right) . \end{aligned}$$

Since $\vdash \varGamma _1 \rightarrow \left( \bigvee _{j \in \kappa }{(\Diamond B_j \vee B_j)} \right) \rhd \bigvee I_j$, we have

$$\begin{aligned} \vdash \varGamma _1 \rightarrow (\Box \lnot F \wedge F \wedge \lnot J) \rhd \bigvee _{j \in \kappa } I_j \end{aligned}$$

by Proposition 1.2. Then,

$$\begin{aligned}&\vdash \varGamma _1 \rightarrow \left( \Box \lnot F \wedge F \wedge \bigwedge _{j \in \kappa } \lnot I_j \wedge \lnot J \right) \rhd \left( \bigvee _{j \in \kappa } I_j \vee J \right) ,\\&\vdash \varGamma _1 \rightarrow \left( \Box \lnot F \wedge F \wedge \lnot \left( \bigvee _{j \in \kappa } I_j \vee J \right) \right) \rhd \left( \bigvee _{j \in \kappa } I_j \vee J \right) . \end{aligned}$$

Since $G \rhd F \in \varGamma _1$, by Lemma 1.2, we obtain $\vdash \varGamma _1 \rightarrow G \rhd (\Box \lnot F \wedge F)$. Therefore, by Proposition 1.6, we obtain

$$\begin{aligned} \vdash \varGamma _1 \rightarrow G \rhd \left( \bigvee _{j \in \kappa } I_j \vee J \right) . \end{aligned}$$

(6)

On the other hand, from $\vdash Y' \rightarrow \lnot J$,

$$\begin{aligned}&\vdash \boxdot \varGamma _2 \rightarrow \left( J \wedge \lnot A \rightarrow \Diamond A \vee \bigvee _{C \rhd A \in \varGamma _2}{(\Diamond C \vee C)} \right) ,\\&\vdash \varGamma _2 \rightarrow \Box \left( J \wedge \lnot A \rightarrow \Diamond A \vee \bigvee _{C \rhd A \in \varGamma _2}{(\Diamond C \vee C)}\right) . \end{aligned}$$

Since $\vdash \varGamma _2 \rightarrow \left( \Diamond A \vee \bigvee _{C \rhd A \in \varGamma _2}{(\Diamond C \vee C)} \right) \rhd A$, we obtain $\vdash \varGamma _2 \rightarrow (J \wedge \lnot A) \rhd A$ by Proposition 1.2. Since $\vdash \varGamma _2 \rightarrow \left( \bigvee _{j \in \kappa } I_j \wedge \lnot A \right) \rhd A$, we have

$$\begin{aligned} \vdash \varGamma _2 \rightarrow \left( \left( \bigvee _{j \in \kappa } I_j \vee J \right) \wedge \lnot A \right) \rhd A. \end{aligned}$$

From this and (6), we conclude ${\sim } G \in \varDelta _1$ because $\varGamma \prec _A \varDelta$. This contradicts the consistency of $\varDelta _1$.

In both cases, $(X', Y')$ is inseparable, and hence we can obtain a complete pair $\varTheta \in K(\varPhi ^1, \varPhi ^2)$ which extends $(X', Y')$ and satisfies the desired conditions. $\square$

4.2 Proof of Theorem 6

We are ready to prove Theorem 6.

Proof of Theorem 6

Suppose that the implication $A_0 \rightarrow B_0$ has no interpolant, and we would like to show $\nvdash A_0 \rightarrow B_0$. It follows that $(\{A_0\}, \{ \lnot B_0\} )$ is inseparable. Let $\varPhi ^1$ (resp. $\varPhi ^2$) be the smallest finite adequate set containing $A_0$ (resp. $\lnot B_0$), and put $K:= K(\varPhi ^1, \varPhi ^2)$. There exists $\varGamma ' \in K(\varPhi ^1, \varPhi ^2)$ such that $A_0 \in \varGamma _1'$ and $\lnot B_0 \in \varGamma _2'$. For $\varGamma \in K$, we define inductively the rank of $\varGamma$ (write $\textrm{rank}{(\varGamma )}$) as $\textrm{rank}{(\varGamma )}:= \sup \{ \textrm{rank}{(\varDelta )} +1: \varGamma \prec \varDelta \}$, where $\sup \emptyset = 0$. This notion is well-defined because $\prec$ is conversely well-founded.

For finite sequences $\tau$ and $\sigma$ of formulas, let $\tau \subseteq \sigma$ mean that $\sigma$ is an end-extension of $\tau$. Let $\tau *\langle A \rangle$ be the sequence obtained from $\tau$ by adding A as the last element.

We define an $\textbf{IL}^-$-model $M = \langle W, R, \{S_w\}_{w \in W}, \Vdash \rangle$ as follows:

$$\begin{aligned}{} & {} W:= \{ \langle \varGamma , \tau \rangle : \varGamma \in K \ \text {and} \ \tau \ \text {is a finite sequence of elements of}\\{} & {} \quad \varPhi _\rhd ^1 \cup \varPhi _\rhd ^2 \ \text {with}\ \textrm{rank}(\varGamma ) + |\tau | \le \textrm{rank}(\varGamma ') \}; \\{} & {} \langle \varGamma , \tau \rangle R \langle \varDelta , \sigma \rangle :\Leftrightarrow \varGamma \prec \varDelta \ \text {and}\ \tau \subsetneq \sigma ;\\{} & {} \langle \varDelta , \sigma \rangle S_{\langle \varGamma , \tau \rangle } \langle \varTheta , \rho \rangle \\{} & {} \quad :\Leftrightarrow \left\{ \begin{array}{l} \langle \varGamma , \tau \rangle R \langle \varDelta , \sigma \rangle , \langle \varGamma , \tau \rangle R \langle \varTheta , \rho \rangle \ \text {and}\ \\ \text {if}\ \tau *\langle A \rangle \subseteq \sigma , \varGamma \prec _A \varDelta \ \text {and}\ \Box {\sim } A \in \varDelta _1 \cup \varDelta _2, \\ \text {then}\ \tau *\langle A \rangle \subseteq \rho , \varGamma \prec _A \varTheta \ \text {and}\ \Box {\sim } A, {\sim } A \in \varTheta _1 \cup \varTheta _2; \end{array} \right. \\{} & {} \langle \varGamma , \tau \rangle \Vdash p:\iff p \in \varGamma _1\cup \varGamma _2. \end{aligned}$$

Claim

$\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ is valid in the frame of M.

Proof

It is clear that R is transitive and conversely well-founded.

• Suppose $\langle \varDelta , \sigma \rangle S_{\langle \varGamma , \tau \rangle } \langle \varTheta , \rho \rangle$. Then, we have $\langle \varGamma , \tau \rangle R \langle \varTheta , \rho \rangle$ by the definition of $S_{\langle \varGamma , \tau \rangle }$. Therefore, $\textbf{J4}_+$ is valid in the frame of M.

• Suppose $\langle \varDelta , \sigma \rangle S_{\langle \varGamma , \tau \rangle } \langle \varTheta , \rho \rangle S_{\langle \varGamma , \tau \rangle } \langle \varLambda , \pi \rangle$. Then, we have $\langle \varGamma , \tau \rangle R \langle \varDelta , \sigma \rangle$ and $\langle \varGamma , \tau \rangle R \langle \varLambda , \pi \rangle$.

  Assume $\tau *\langle A \rangle \subseteq \sigma$, $\varGamma \prec _A \varDelta$, and $\Box {\sim } A \in \varDelta _1 \cup \varDelta _2$. By $\langle \varDelta , \sigma \rangle S_{\langle \varGamma , \tau \rangle } \langle \varTheta , \rho \rangle$, we obtain $\tau *\langle A \rangle \subseteq \rho$, $\varGamma \prec _A \varTheta$, and $\Box {\sim } A \in \varTheta _1 \cup \varTheta _2$. By $\langle \varTheta , \rho \rangle S_{\langle \varGamma , \tau \rangle } \langle \varLambda , \pi \rangle$, we conclude $\tau *\langle A \rangle \subseteq \pi$, $\varGamma \prec _A \varLambda$, and $\Box {\sim } A, {\sim } A \in \varLambda _1 \cup \varLambda _2$.

  Thus, $\langle \varDelta , \sigma \rangle S_{\langle \varGamma , \tau \rangle } \langle \varLambda , \pi \rangle$. We obtain that $\textbf{J2}_+$ is valid in the frame of M.

• Suppose $\langle \varGamma , \tau \rangle R \langle \varDelta , \sigma \rangle R \langle \varTheta , \rho \rangle$. Then, $\langle \varGamma , \tau \rangle R \langle \varDelta , \sigma \rangle$, and hence $\langle \varGamma , \tau \rangle R \langle \varTheta , \rho \rangle$ by the transitivity of R.

  Assume $\tau *\langle A \rangle \subseteq \sigma$, $\varGamma \prec _A \varDelta$, and $\Box {\sim } A \in \varDelta _1 \cup \varDelta _2$. Since $\sigma \subseteq \rho$, we have $\tau *\langle A \rangle \subseteq \rho$. Since $\varDelta \prec \varTheta$, we have $\Box {\sim } A, {\sim } A \in \varTheta _1 \cup \varTheta _2$. Also, by Lemma 5, $\varGamma \prec _A \varTheta$.

  Thus, $\langle \varDelta , \sigma \rangle S_{\langle \varGamma , \tau \rangle } \langle \varTheta , \rho \rangle$. We conclude that $\textbf{J5}$ is valid in the frame of M.$\square$

Claim

(The Truth Lemma) For $B \in \varPhi ^1 \cup \varPhi ^2$ and $\langle \varGamma , \tau \rangle \in W$, the following are equivalent:

• 1. $B \in \varGamma _1 \cup \varGamma _2$.

• 2. $\langle \varGamma , \tau \rangle \Vdash B$.

Proof

The lemma is proved by induction on the construction of B. We only prove for $B \equiv G \rhd F$.

($1 \Rightarrow 2$): Assume $G \rhd F \in \varGamma _1 \cup \varGamma _2$. Let $\langle \varDelta , \sigma \rangle \in W$ be any element such that $\langle \varGamma , \tau \rangle R \langle \varDelta , \sigma \rangle$ and $\langle \varDelta , \sigma \rangle \Vdash G$. By the induction hypothesis, $G \in \varDelta _1 \cup \varDelta _2$. We distinguish the following two cases:

(Case 1): Assume that $\tau *\langle A \rangle \subseteq \sigma$, $\varGamma \prec _A \varDelta$, and $\Box {\sim } A \in \varDelta _1 \cup \varDelta _2$. By Lemma 7, there exists a pair $\varTheta \in K$ such that $\varGamma \prec _A \varTheta$, $F \in \varTheta _1 \cup \varTheta _2$, and $\Box {\sim } A, {\sim } A \in \varTheta _1 \cup \varTheta _2$.

Take $\rho := \tau *\langle A \rangle$. By $\varGamma \prec \varTheta$, $\textrm{rank}{(\varTheta )}+1 \le \textrm{rank}{(\varGamma )}$. We have

$$\begin{aligned} \textrm{rank}{(\varTheta )} + |\rho | = \textrm{rank}{(\varTheta )} + 1 + |\tau | \le \textrm{rank}{(\varGamma )} + |\tau | \le \textrm{rank}{(\varGamma ')}. \end{aligned}$$

It follows that $\langle \varTheta , \rho \rangle \in W$, and we have $\langle \varDelta , \sigma \rangle S_{\langle \varGamma , \tau \rangle } \langle \varTheta , \rho \rangle$. By the induction hypothesis, $\langle \varTheta , \rho \rangle \Vdash F$. Therefore, $\langle \varGamma , \tau \rangle \Vdash G \rhd F$.

(Case 2): Otherwise, by Lemma 4, we have $\varGamma \prec _\bot \varDelta$. By Lemma 7, there exists a pair $\varTheta \in K$ such that $\varGamma \prec _\bot \varTheta$ and $F \in \varTheta _1 \cup \varTheta _2$.

Take $\rho := \tau *\langle \bot \rangle$. Then, we have $\langle \varTheta , \rho \rangle \in W$ by a similar argument as in Case 1. By the definition of $S_{\langle \varGamma , \tau \rangle }$ and induction hypothesis, $\langle \varDelta , \sigma \rangle S_{\langle \varGamma , \tau \rangle } \langle \varTheta , \rho \rangle$ and $\langle \varTheta , \rho \rangle \Vdash F$. Therefore, $\langle \varGamma , \tau \rangle \Vdash G \rhd F$.

($2 \Rightarrow 1$): Assume $G \rhd F \not \in \varGamma _1 \cup \varGamma _2$. Then, $\lnot (G \rhd F) \in \varGamma _1 \cup \varGamma _2$ because $\varGamma$ is complete. By Lemma 6, there exists a pair $\varDelta \in K$ such that $\varGamma \prec _F \varDelta$ and $G, \Box {\sim } F \in \varDelta _1 \cup \varDelta _2$. Let $\sigma := \tau *\langle F \rangle$. We have $\langle \varDelta , \sigma \rangle \in W$. By the induction hypothesis, $\langle \varDelta , \sigma \rangle \Vdash G$. It suffices to show that for any $\langle \varTheta , \rho \rangle \in W$, if $\langle \varDelta , \sigma \rangle S_{\langle \varGamma , \tau \rangle } \langle \varTheta , \rho \rangle$, then $\langle \varTheta , \rho \rangle \nVdash F$. Suppose $\langle \varDelta , \sigma \rangle S_{\langle \varGamma , \tau \rangle } \langle \varTheta , \rho \rangle$. Since $\tau *\langle F \rangle \subseteq \sigma$, $\varGamma \prec _F \varDelta$, and $\Box {\sim } F \in \varDelta _1 \cup \varDelta _2$, we have ${\sim } F \in \varTheta _1 \cup \varTheta _2$ (and hence $F \not \in \varTheta _1 \cup \varTheta _2$). By the induction hypothesis, $\langle \varTheta , \rho \rangle \nVdash F$. $\square$

Let $\epsilon$ be the empty sequence, then $\langle \varGamma ', \epsilon \rangle \in W$ because $\textrm{rank}(\varGamma ') + |\epsilon | \le \textrm{rank}(\varGamma ')$. By the Truth Lemma (Lemma 4.2), $\langle \varGamma ', \epsilon \rangle \Vdash A_0 \wedge \lnot B_0$, and therefore $A_0 \rightarrow B_0$ is not valid in M. It follows that $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ does not prove $A_0 \rightarrow B_0$. $\square$

4.3 Consequences of Theorem 6

In this subsection, we prove some consequences of Theorem 6 on interpolation properties. Firstly, we prove that $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ has a version of the $\rhd$-interpolation property. Secondly, we notice that CIP for $\textbf{IL}$ easily follows from Theorem 6.

Before them, we show the so-called generated submodel lemma. Let $M = \langle W, R, \{S_w\}_{w \in W}, \Vdash \rangle$ be any $\textbf{IL}^-$-model such that $\textbf{J4}_+$ is valid in the frame of M. For each $r \in W$, we define an $\textbf{IL}^-$-model $M^* = \langle W^*, R^*, \{S_w^*\}_{w \in W^*}, \Vdash ^* \rangle$ as follows:

• $W^*: = R[r] \cup \{r\}$;

• $x R^* y: \iff x R y$;

• $y S_x^* z: \iff y S_x z$;

• $x \Vdash ^* p: \iff x \Vdash p$.

We call $M^*$ the submodel of M generated by r. It is easy to show that if $\textbf{J1}$ is valid in the frame of M, then it is also valid in the frame of $M^*$. This is also the case for $\textbf{J2}_+$ and $\textbf{J5}$. Also, the following lemma is easily obtained.

Lemma 8

(The Generated Submodel Lemma) Suppose that $\textbf{J4}_+$ is valid in the frame of an $\textbf{IL}^-$-model $M = \langle W, R, \{S_w\}_{w \in W}, \Vdash \rangle$. For any $r \in W$, let $M^* = \langle W^*, R^*, \{S_w^*\}_{w \in W^*}, \Vdash ^* \rangle$ be the submodel of M generated by r. Then, for any $x \in W^*$ and formula A, $x \Vdash A$ if and only if $x \Vdash ^* A$.

Proof

This is proved by induction on the construction of A. We only prove the case $A \equiv (B \rhd C)$.

$(\Rightarrow )$: Suppose $x \Vdash B \rhd C$. Let $y \in W^*$ be any element such that $x R^* y$ and $y \Vdash ^* B$. Then, xRy, and by the induction hypothesis, $y \Vdash B$. Hence, there exists $z \in W$ such that $y S_x z$ and $z \Vdash C$. Since $\textbf{J4}_+$ is valid in the frame of M, xRz. Since rRx, we have rRz. Thus, $z \in W^*$. It follows $y S_x^* z$. By the induction hypothesis, $z \Vdash ^* C$. Therefore, $x \Vdash ^* B \rhd C$.

$(\Leftarrow )$: Suppose $x \Vdash ^* B \rhd C$. Let $y \in W$ be any element with xRy and $y \Vdash B$. Since $x \in W^*$, we have $y \in W^*$, and hence $x R^* y$. By the induction hypothesis, $y \Vdash ^* B$. Then, for some $z \in W^*$, $y S_x^* z$ and $z \Vdash ^* C$. We have $y S_x z$. By the induction hypothesis, $z \Vdash C$. Thus, we conclude $x \Vdash B \rhd C$. $\square$

Proposition 9

For any formulas A and B, the following are equivalent:

• 1. $\vdash A \rhd B$.

• 2. $\vdash A \rightarrow \Diamond B$.

Proof

$(1 \Rightarrow 2)$: Suppose $\nvdash A \rightarrow \Diamond B$. Then, by Theorem 1, there exist an $\textbf{IL}^-$-model $M = \langle W, R, \{S_w\}_{w \in W}, \Vdash \rangle$ and $r \in W$ such that $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ is valid in the frame of M and $r \Vdash A \wedge \Box \lnot B$. By the Generated Submodel Lemma, we may assume that r is the root of M, that is, for all $w \in W {\setminus } \{r\}$, rRw.

We define a new $\textbf{IL}^-$-model $M' = \langle W', R', \{S_w'\}_{w \in W'}, \Vdash ' \rangle$ as follows:

• $W': = W \cup \{r_0\}$, where $r_0$ is a new element;

• $x R' y: \iff {\left\{ \begin{array}{ll} x R y &{} \text {if}\ x \ne r_0, \\ y \in W &{} \text {if}\ x = r_0; \end{array}\right. }$

• $y S_x' z: \iff {\left\{ \begin{array}{ll} y S_x z &{} \text {if}\ x \ne r_0, \\ y R z &{} \text {if}\ x = r_0; \end{array}\right. }$

• $x \Vdash ' p: \iff x \ne r_0$ and $x \Vdash p$.

Then, $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ is also valid in the frame of $M'$. Also, it is easily shown that for any $x \in W$ and any formula C, $x \Vdash C$ if and only if $x \Vdash ' C$.

Then, $r \Vdash ' A \wedge \Box \lnot B$. Let $x \in W'$ be any element such that $r S_{r_0}' x$. Then, rRx, and hence $r R' x$. We have $x \nVdash ' B$. Therefore, we obtain $r_0 \nVdash ' A \rhd B$. It follows $\nvdash A \rhd B$.

$(2 \Rightarrow 1)$: Suppose $\vdash A \rightarrow \Diamond B$, then $\vdash \Diamond B \rhd B \rightarrow A \rhd B$ by $\textbf{R2}$. Thus, $\vdash A \rhd B$. $\square$

Areces, Hoogland, and de Jongh [1] proved that $\textbf{IL}$ has the $\rhd$-interpolation property. Namely, for every formulas A and B with $\textbf{IL}\vdash A \rhd B$, there exists a formula C such that $v(C) \subseteq v(A) \cap v(B)$, $\textbf{IL}\vdash A \rhd C$, and $\textbf{IL}\vdash C \rhd B$. We prove that $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ has a version of the $\rhd$-interpolation property.

Corollary 1

(A version of the $\rhd$-interpolation property) Let A and B be any formulas. If $\vdash A \rhd B$, then there exists a formula C such that $v(C) \subseteq v(A) \cap v(B)$, $\vdash A \rightarrow C$, and $\vdash C \rhd B$.

Proof

Suppose $\vdash A \rhd B$. Then, by Proposition 9, $\vdash A \rightarrow \Diamond B$. By Theorem 6, there exists a formula C such that $v(C) \subseteq v(A) \cap v(B)$, $\vdash A \rightarrow C$, and $\vdash C \rightarrow \Diamond B$. By Proposition 9 again, we obtain $\vdash C \rhd B$. $\square$

Problem 1

Does the logic $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ have the original version of the $\rhd$-interpolation property?

For each formula A, let $\textrm{Sub}(A)$ be the set of all subformulas of A. Also, let $\textrm{PSub}(A): = \textrm{Sub}(A) \setminus \{A\}$. We prove that $\textbf{IL}$ is embeddable into $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ in some sense.

Proposition 10

For any formula A, the following are equivalent:

• 1. $\textbf{IL}\vdash A$.

• 2. A is valid in all finite $\textbf{IL}^-$-frames in which all axioms of $\textbf{IL}$ are valid.

• 3. $\vdash \boxdot \bigwedge \{B \rhd B: B \in \textrm{PSub}(A)\} \rightarrow A$.

Proof

$(1 \Rightarrow 2)$ is obvious.

$(3 \Rightarrow 1)$ follows from Proposition 1.7.

$(2 \Rightarrow 3)$: Suppose $L \nvdash \boxdot \bigwedge \{B \rhd B: B \in \textrm{PSub}(A)\} \rightarrow A$. Then, by Theorem 1, there exist a finite $\textbf{IL}^-$-model $M = \langle W, R, \{S_w\}_{w \in W}, \Vdash \rangle$ and $r \in W$ such that $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ is valid in the frame of M and $r \Vdash \boxdot \bigwedge \{B \rhd B: B \in \textrm{PSub}(A)\} \wedge \lnot A$. By the Generated Submodel Lemma, we may assume that r is the root of M.

We define an $\textbf{IL}^-$-model $M' = \langle W', R', \{S_w'\}_{w \in W'}, \Vdash ' \rangle$ as follows:

• $W': = W$;

• $x R' y: \iff x R y$;

• $y S_x' z: \iff y S_x z$ or (xRy and $z = y$);

• $x \Vdash ' p: \iff x \Vdash p$.

Claim

$\textbf{IL}$ is valid in the frame of $M'$.

Proof

By Proposition 1.8, it suffices to show that $\textbf{J1}$, $\textbf{J2}_+$, and $\textbf{J5}$ are valid in the frame of $M'$.

$\textbf{J1}$: Suppose xRy. Then, $y S_x' y$ by the definition of $S_x'$. Thus, $\textbf{J1}$ is valid.

$\textbf{J4}_+$: Suppose $y S'_x z$. Then, $y S_x z$ or (xRy and $y = z$). If $y S_x z$, then xRz because $\textbf{J4}_+$ is valid in the frame of M. If xRy and $y = z$, then xRz. Hence, in either case, we have xRz. Therefore, $\textbf{J4}_+$ is valid.

$\textbf{J2}_+$: Suppose $y S'_x z$ and $z S'_x u$. We distinguish the following four cases.

• (Case 1): $y S_x z$ and $z S_x u$. Since $\textbf{J2}_+$ is valid in the frame of M, $y S_x u$.

• (Case 2): $y S_x z$, xRz and $z = u$. Then, $y S_x u$.

• (Case 3): xRy, $y = z$ and $z S_x u$. Then, $y S_x u$.

• (Case 4): xRy, $y = z$, xRz and $z = u$. Then, xRy and $y = u$.

In either case, we have $y S_x' u$. Since $\textbf{J4}_+$ is valid, we obtain that $\textbf{J2}_+$ is valid in the frame of $M'$.

$\textbf{J5}$: Suppose $x R' y$ and $y R' z$. Then, xRy and yRz. Since $\textbf{J5}$ is valid in the frame of M, $y S_x z$. Then, $y S_x' z$. Therefore, $\textbf{J5}$ is valid. $\square$

Claim

For any $B \in \textrm{Sub}(A)$ and $x \in W$, $x \Vdash B$ if and only if $x \Vdash ' B$.

Proof

We prove the claim by induction on the construction of B. We only give a proof of the case that B is $C \rhd D$.

$(\Rightarrow )$: Suppose $x \Vdash C \rhd D$. Let $y \in W$ be such that xRy and $y \Vdash ' C$. By the induction hypothesis, $y \Vdash C$. Then, there exists $z \in W$ such that $y S_x z$ and $z \Vdash D$. Then, $y S_x' z$ and by the induction hypothesis, $z \Vdash ' D$. Therefore, $x \Vdash ' C \rhd D$.

$(\Leftarrow )$: Suppose $x \Vdash ' C \rhd D$. Let $y \in W$ be such that xRy and $y \Vdash C$. By the induction hypothesis, $y \Vdash ' C$. Hence, there exists $z \in W$ such that $y S'_x z$ and $z \Vdash ' D$. By the induction hypothesis, $z \Vdash D$. By the definition of $S_x'$, we have either $y S_x z$ or (xRy and $y = z$). If $y S_x z$, then $x \Vdash C \rhd D$. If xRy and $y = z$, then xRy and $y \Vdash D$. Here either $x = r$ or rRw. Since $D \in \textrm{PSub}(A)$, we obtain $x \Vdash D \rhd D$ because $r \Vdash \boxdot \bigwedge \{B \rhd B: B \in \textrm{Sub}(A)\}$. Thus, for some $z' \in W$, $y S_x z'$ and $z' \Vdash D$. We conclude $x \Vdash C\ \rhd D$. $\square$

Since $r \nVdash A$, we obtain $r \nVdash ' A$ by the claim. Thus, A is not valid in some finite $\textbf{IL}^-$-frame in which all axioms of $\textbf{IL}$ are valid. $\square$

Proof of Theorem 3

Suppose $\textbf{IL}\vdash A \rightarrow B$. Then, by Proposition 10,

$$\begin{aligned} \vdash \boxdot \bigwedge \{C \rhd C: C \in \textrm{PSub}(A \rightarrow B)\} \rightarrow (A \rightarrow B). \end{aligned}$$

Since $\textrm{PSub}(A \rightarrow B) = \textrm{Sub}(A) \cup \textrm{Sub}(B)$, we have

$$\begin{aligned} \vdash \boxdot \bigwedge \{C \rhd C: C \in \textrm{Sub}(A)\} \wedge A \rightarrow \left( \boxdot \bigwedge \{C \rhd C: C \in \textrm{Sub}(B)\} \rightarrow B \right) . \end{aligned}$$

By Theorem 6, there exists a formula D such that $v(D) \subseteq v(A) \cap v(B)$,

$$\begin{aligned} \vdash \boxdot \bigwedge \{C \rhd C: C \in \textrm{Sub}(A)\} \wedge A \rightarrow D \end{aligned}$$

and

$$\begin{aligned} \vdash D \rightarrow \left( \boxdot \bigwedge \{C \rhd C: C \in \textrm{Sub}(B)\} \rightarrow B \right) . \end{aligned}$$

Then, by Proposition 1.7, we obtain $\textbf{IL}\vdash A \rightarrow D$ and $\textbf{IL}\vdash D \rightarrow B$. $\square$

5 The fixed point property

In this section, we investigate FPP and $\ell$FPP. First, we study FPP for the logic $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$. Then, we prove that $\textbf{IL}^-(\textbf{J4}, \textbf{J5})$ has $\ell$FPP.

5.1 FPP for $\textbf{IL}^{-}(\textbf{J2}_{+}, \textbf{J5})$

From Theorem 6 and Lemma 2, we immediately obtain the following corollary.

Corollary 2

(FPP for $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5}))$ $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ has FPP.

Moreover, we give a syntactical proof of FPP for $\textbf{IL}^{-}(\textbf{J2}_{+}, \textbf{J5})$ by modifying de Jongh and Visser’s proof of FPP for $\textbf{IL}$. Since the Substitution Principle (Proposition 3) holds for extensions of $\textbf{IL}^-(\textbf{J4}_+)$, as usual, it suffices to prove that every formula of the form $A(p) \rhd B(p)$ has a fixed point in $\textbf{IL}^{-}(\textbf{J2}_{+}, \textbf{J5})$. As a consequence, we show that every formula A(p) which is modalized in p has the same fixed point in $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ as given by de Jongh and Visser. That is,

Theorem 7

For any formulas A(p) and B(p), $A(\top ) \rhd B(\Box \lnot A(\top ))$ is a fixed point of $A(p) \rhd B(p)$ in $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$.

Theorem 7 follows from the following five lemmas.

Lemma 9

Let L be any extension of $\textbf{IL}^{-}$. For any formulas A and B, if $L \vdash \Box \lnot A \rightarrow (A \leftrightarrow B)$, then $L \vdash (A \wedge \Box \lnot A) \leftrightarrow (B \wedge \Box \lnot B)$.

Proof

Suppose $L \vdash \Box \lnot A \rightarrow (A \leftrightarrow B)$. Then, $L \vdash \Box \lnot A \rightarrow (\Box \lnot A \leftrightarrow \Box \lnot B)$ and hence $L \vdash \Box \lnot A \rightarrow \Box \lnot B$. By combining this with our supposition, we obtain

$$\begin{aligned} L \vdash (A \wedge \Box \lnot A) \rightarrow (B \wedge \Box \lnot B). \end{aligned}$$

On the other hand, $L \vdash \lnot B \rightarrow (\Box \lnot A \rightarrow \lnot A)$. Hence, by the axiom scheme $\textbf{L3}$, $L \vdash \Box \lnot B \rightarrow \Box \lnot A$. Therefore, by our supposition,

$$\begin{aligned} L \vdash (B \wedge \Box \lnot B) \rightarrow (A \wedge \Box \lnot A). \end{aligned}$$

$\square$

Lemma 10

For any formulas A and C,

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}_{+}) \vdash (A(\top ) \wedge \Box \lnot A(\top )) \leftrightarrow (A(A(\top ) \rhd C) \wedge \Box \lnot A(A(\top ) \rhd C)). \end{aligned}$$

Proof

By Proposition 1.1, $\textbf{IL}^- \vdash \Box \lnot A(\top ) \rightarrow A(\top ) \rhd C$. Therefore, we obtain $\textbf{IL}^- \vdash \Box \lnot A(\top ) \rightarrow (\top \leftrightarrow (A(\top ) \rhd C) )$. Then, $\textbf{IL}^- \vdash \Box \lnot A(\top ) \rightarrow \boxdot (\top \leftrightarrow (A(\top ) \rhd C))$. Therefore, by Proposition 3.1, we obtain

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}_{+}) \vdash \Box \lnot A(\top ) \rightarrow (A(\top ) \leftrightarrow A(A(\top ) \rhd C)). \end{aligned}$$

The lemma directly follows from this and Lemma 9. $\square$

Lemma 11

For any formulas A, C, and D,

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}, \textbf{J4}_{+}, \textbf{J5}) \vdash (A(\top ) \rhd D) \leftrightarrow (A(A(\top ) \rhd C) \rhd D). \end{aligned}$$

Proof

By Lemma 10 and $\textbf{R2}$, we obtain

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}_{+}) \vdash ((A(\top ) \wedge \Box \lnot A(\top )) \rhd D) \leftrightarrow ((A(A(\top ) \rhd C) \wedge \Box \lnot A(A(\top ) \rhd C)) \rhd D). \end{aligned}$$

Therefore, by Lemma 1.1, we obtain

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}, \textbf{J4}_{+}, \textbf{J5}) \vdash (A(\top ) \rhd D) \leftrightarrow (A(A(\top ) \rhd C) \rhd D). \end{aligned}$$

$\square$

Lemma 12

For any formulas B and C, $\textbf{IL}^-(\textbf{J4}_{+})$ proves

$$\begin{aligned} (B(\Box \lnot C) \wedge \Box \lnot B(\Box \lnot C)) \leftrightarrow (B(C \rhd B(\Box \lnot C)) \wedge \Box \lnot B(C \rhd B(\Box \lnot C))). \end{aligned}$$

Proof

Since $\textbf{IL}^- \vdash \Box \lnot B(\Box \lnot C) \rightarrow \Box (\bot \leftrightarrow B(\Box \lnot C))$,

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}_{+}) \vdash \Box \lnot B(\Box \lnot C) \rightarrow (C \rhd \bot \leftrightarrow C \rhd B(\Box \lnot C)). \end{aligned}$$

Then, by $\textbf{J6}$, $\textbf{IL}^-(\textbf{J4}_{+}) \vdash \Box \lnot B(\Box \lnot C) \rightarrow (\Box \lnot C \leftrightarrow C \rhd B(\Box \lnot C))$ and hence $\textbf{IL}^-(\textbf{J4}_{+}) \vdash \Box \lnot B(\Box \lnot C) \rightarrow \boxdot (\Box \lnot C \leftrightarrow C \rhd B(\Box \lnot C))$. Therefore, by Proposition 3.1, we obtain

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}_{+}) \vdash \Box \lnot B(\Box \lnot C) \rightarrow (B(\Box \lnot C) \leftrightarrow B(C \rhd B(\Box \lnot C))). \end{aligned}$$

The lemma is a consequence of this with Lemma 9. $\square$

Lemma 13

For any formulas B, C, and D,

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}_{+}, \textbf{J5}) \vdash (D \rhd B(\Box \lnot C)) \leftrightarrow (D \rhd B(C \rhd B(\Box \lnot C))). \end{aligned}$$

Proof

By Lemma 12 and $\textbf{R1}$, $\textbf{IL}^-(\textbf{J4}_{+})$ proves

$$\begin{aligned} D \rhd (B(\Box \lnot C) \wedge \Box \lnot B(\Box \lnot C)) \leftrightarrow D \rhd (B(C \rhd B(\Box \lnot C)) \wedge \Box \lnot B(C \rhd B(\Box \lnot C))). \end{aligned}$$

Therefore, by Lemma 1.2,

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}_{+}, \textbf{J5}) \vdash (D \rhd B(\Box \lnot C)) \leftrightarrow (D \rhd B(C \rhd B(\Box \lnot C))). \end{aligned}$$

$\square$

Proof of Theorem 7

Let $F \equiv A(\top ) \rhd B(\Box \lnot A(\top ))$. By Lemma 11 for $C \equiv D \equiv B(\Box \lnot A(\top ))$, we obtain

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}, \textbf{J4}_+, \textbf{J5}) \vdash F \leftrightarrow (A(F) \rhd B(\Box \lnot A(\top ))). \end{aligned}$$

Furthermore, by Lemma 13 for $C \equiv A(\top )$ and $D \equiv F$,

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}_+, \textbf{J5}) \vdash (A(F) \rhd B(\Box \lnot A(\top ))) \leftrightarrow (A(F) \rhd B(F)). \end{aligned}$$

We conclude

$$\begin{aligned} \textbf{IL}^-(\textbf{J2}_+, \textbf{J5}) \vdash F \leftrightarrow A(F) \rhd B(F). \end{aligned}$$

$\square$

5.2 $\ell$FPP for $\textbf{IL}^{-}(\textbf{J4}, \textbf{J5})$

From Lemma 11, we immediately obtain the following corollary.

Corollary 3

For any formulas A(p) and B, if $p \notin v(B)$, then $A(\top ) \rhd B$ is a fixed point of $A(p) \rhd B$ in $\textbf{IL}^-(\textbf{J2}, \textbf{J4}_+, \textbf{J5})$.

Therefore, $\textbf{IL}^-(\textbf{J2}, \textbf{J4}_+, \textbf{J5})$ has $\ell$FPP. Moreover, we prove the following theorem.

Theorem 8

$(\ell$FPP for $\textbf{IL}^-(\textbf{J4}, \textbf{J5}))$ For any formulas A(p) and B, if the formula $A(p) \rhd B$ is left-modalized in p, then $A(\Box \lnot A(\top )) \rhd B$ is a fixed point of $A(p) \rhd B$ in $\textbf{IL}^-(\textbf{J4}, \textbf{J5})$. Therefore, $\textbf{IL}^-(\textbf{J4}, \textbf{J5})$ has $\ell$FPP.

Before proving Theorem 8, we prepare two lemmas.

Lemma 14

For any formula A(p) such that $\Box A(p)$ is left-modalized in p,

$$\begin{aligned} \textbf{IL}^- \vdash \Box A(\top ) \leftrightarrow \Box A(\Box A(\top )). \end{aligned}$$

Proof

This is proved in a usual way by using Proposition 6. $\square$

Lemma 15

Let A(p) and B be any formulas such that $p \notin v(E)$ for any subformula $D \rhd E$ of A(p). Then,

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}, \textbf{J5}) \vdash (A(\Box \lnot A(p)) \rhd B) \leftrightarrow (A(A(p) \rhd B) \rhd B). \end{aligned}$$

Proof

By Proposition 1.1, $\textbf{IL}^- \vdash \Box \lnot A(p) \rightarrow A(p) \rhd B$. On the other hand, since $\textbf{IL}^-(\textbf{J4}) \vdash A(p) \rhd B \rightarrow (\Diamond A(p) \rightarrow \Diamond B)$, we have $\textbf{IL}^-(\textbf{J4}) \vdash \Box \lnot B \rightarrow (A(p) \rhd B \rightarrow \Box \lnot A(p))$. Hence, $\textbf{IL}^-(\textbf{J4}) \vdash \Box \lnot B \rightarrow (\Box \lnot A(p) \leftrightarrow A(p) \rhd B)$. Then,

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}) \vdash \Box \lnot B \rightarrow \boxdot (\Box \lnot A(p) \leftrightarrow A(p) \rhd B). \end{aligned}$$

By Proposition 6.1, we obtain

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}) \vdash \Box \lnot B \rightarrow (A(\Box \lnot A(p)) \leftrightarrow A(A(p) \rhd B)). \end{aligned}$$

Thus,

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}) \vdash (A(\Box \lnot A(p)) \vee \Diamond B) \leftrightarrow (A(A(p) \rhd B) \vee \Diamond B). \end{aligned}$$

By $\textbf{R2}$, we obtain

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}) \vdash ((A(\Box \lnot A(p)) \vee \Diamond B) \rhd B) \leftrightarrow ((A(A(p) \rhd B) \vee \Diamond B) \rhd B). \end{aligned}$$

Therefore, we conclude

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}, \textbf{J5}) \vdash (A(\Box \lnot A(p)) \rhd B) \leftrightarrow (A(A(p) \rhd B) \rhd B). \end{aligned}$$

$\square$

Proof of Theorem 8

Let $F: \equiv \Box \lnot A(\top )$. Since $\Box \lnot A(p)$ is left-modalized in p, $\textbf{IL}^- \vdash F \leftrightarrow \Box \lnot A(F)$ by Lemma 14. Since $\textbf{IL}^-(\textbf{J4}) \vdash \Box (F \leftrightarrow \Box \lnot A(F))$, by Proposition 6.2, we have

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}) \vdash (A(F) \rhd B) \leftrightarrow (A(\Box \lnot A(F)) \rhd B). \end{aligned}$$

By Lemma 15,

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}, \textbf{J5}) \vdash (A(\Box \lnot A(F)) \rhd B) \leftrightarrow (A(A(F) \rhd B) \rhd B). \end{aligned}$$

Therefore,

$$\begin{aligned} \textbf{IL}^-(\textbf{J4}, \textbf{J5}) \vdash (A(F) \rhd B) \leftrightarrow (A(A(F) \rhd B) \rhd B). \end{aligned}$$

$\square$

6 Failure of $\ell$FPP, FPP, and CIP

In this section, we provide counter models of $\ell$FPP for $\textbf{CL}$ and $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$, and also provide a counter model of FPP for $\textbf{IL}^-(\textbf{J1}, \textbf{J4}_+, \textbf{J5})$. We also show that CIP is not the case for our sublogics except $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$ and $\textbf{IL}$. Let $\omega$ be the set $\{0, 1, 2, \ldots \}$ of all natural numbers.

6.1 A counter model of $\ell$FPP for $\textbf{CL}$

In this subsection, we prove that $\textbf{IL}^-$, $\textbf{IL}^-(\textbf{J1})$, $\textbf{IL}^-(\textbf{J4}_+)$, $\textbf{IL}^-(\textbf{J1}, \textbf{J4}_+)$, $\textbf{IL}^-(\textbf{J2}_+)$, and $\textbf{CL}$ have neither $\ell$FPP nor CIP.

Theorem 9

The formula $p \rhd q$ which is left-modalized in p has no fixed points in $\textbf{CL}$. That is, for any formula A which satisfies $v(A) \subseteq \{q\}$,

$$\begin{aligned} \textbf{CL}\nvdash A \leftrightarrow A \rhd q. \end{aligned}$$

Proof

We define an $\textbf{IL}^-$-frame ${\mathcal {F}} = \langle W, R, \{S_{w}\}_{w \in W}\rangle$ as follows:

• $W:= \{x_{i}, y_{i}: i \in \omega \}$;

• $R:= \{\langle x_{i}, x_{j}\rangle , \langle x_{i}, y_{j}\rangle , \langle y_{i}, x_{j}\rangle , \langle y_{i}, y_{j}\rangle \in W^2: i > j \}$;

• For each $i \in \omega$ and $w_i \in \{x_i, y_i\}$, $S_{w_i}:=\{\langle a, a\rangle : w_iRa\} \cup \{\langle a, b\rangle :$ there exists an even number $k < i-1$ such that $((a = x_{k}$ or $a=y_{k})$ and $b = x_{k+1})\}$.

For example, $S_{x_3}$, $S_{y_3}$, $S_{x_4}$, and $S_{y_4}$ are shown in the following figure (Fig. 3). In the figure, R relations and S relations are drawn by solid lines and broken lines, respectively. Since R is transitive, we draw only solid lines connecting the immediately preceding and succeeding elements.

Fig. 3

A counter model of $\ell$FPP for $\textbf{CL}$

It is easy to show that $\textbf{J1}$ and $\textbf{J2}_+$ are valid in ${\mathcal {F}}$. Thus, $\textbf{CL}$ is valid in ${\mathcal {F}}$ by Proposition 1.9. Let $\Vdash$ be a satisfaction relation on ${\mathcal {F}}$ such that for any $i \in \omega$, we have $x_i \Vdash q$ and $y_i \nVdash q$. For each $w \in W$, we say that $i \in \omega$ is an index of w if either $w = x_i$ or $w = y_i$.

Claim 1

For any formula A with $v(A) \subseteq \{q\}$, there exists an $n_A \in \omega$ satisfying the following two conditions:

• 1. Either $\forall m \ge n_A\, (x_{m} \Vdash A)$ or $\forall m \ge n_A\, (x_{m} \nVdash A)$;

• 2. Either $\forall m \ge n_A\, (y_{m} \Vdash A)$ or $\forall m \ge n_A\, (y_{m} \nVdash A)$.

Proof

We prove the claim by induction on the construction of A.

$A \equiv \bot$: Then, $n_A = 0$ and $\forall m \ge n_A\, (x_m \nVdash A$ and $y_m \nVdash A)$.

$A \equiv q$: Then, $n_A = 0$ and $\forall m \ge n_A\, (x_m \Vdash q$ and $y_m \nVdash q)$.

$A \equiv B \rightarrow C$: By the induction hypothesis, there exist $n_B, n_C \in W$ satisfying the statement of the claim for B and C, respectively. Let $n_A = \max \{n_B, n_C\}$. We distinguish the following three cases.

• $\forall m \ge n_A\, (x_m \nVdash B)$: Then, $\forall m \ge n_A\, (x_m \Vdash B \rightarrow C)$.

• $\forall m \ge n_A\, (x_m \Vdash C)$: Then, $\forall m \ge n_A\, (x_m \Vdash B \rightarrow C)$.

• $\forall m \ge n_A\, (x_m \Vdash B)$ and $\forall m \ge n_A\, (x_m \nVdash C)$: Then, $\forall m \ge n_A\, (x_m \nVdash B \rightarrow C)$.

In a similar way, it is proved that either $\forall m \ge n_A\, (y_{m} \Vdash B \rightarrow C)$ or $\forall m \ge n_A\, (y_{m} \nVdash B \rightarrow C)$.

$A \equiv \Box B$: We distinguish the following two cases.

• There exists an $n \in W$ such that either $x_n \nVdash B$ or $y_n \nVdash B$: Then, $n_A = n + 1$ and $\forall m \ge n_A\, (x_m \nVdash \Box B$ and $y_m \nVdash \Box B)$.

• For all $n \in W$, $x_n \Vdash B$ and $y_n \Vdash B$: Then, $n_A = 0$ and $\forall m \ge n_A\, (x_m \Vdash \Box B$ and $y_m \Vdash \Box B)$.

$A \equiv B \rhd C$: We distinguish the following five cases.

• (Case 1): There exists an even number k such that $x_{k} \Vdash B$, $x_{k} \nVdash C$, and $x_{k+1} \nVdash C$. Let $n_A = k + 2$ and $m \ge n_A$. Then, $x_{m} R x_{k}$ and $x_{k} \Vdash B$. For any $v \in W$ with $x_{k}S_{x_{m}} v$, either $v = x_{k}$ or $v = x_{k+1}$ by the definition of $S_{x_{m}}$. Thus, $v \nVdash C$. Therefore, we obtain $x_{m} \nVdash B \rhd C$. Since $y_{m} R x_{k}$, we also obtain $y_{m} \nVdash B \rhd C$ in a similar way.

• (Case 2): There exists an even number k such that $y_{k} \Vdash B$, $y_{k} \nVdash C$, and $x_{k+1} \nVdash C$. It is proved that $n_A = k+2$ witnesses the claim as in Case 1.

• (Case 3): There exists an odd number k such that $x_{k} \Vdash B$ and $x_{k} \nVdash C$. Let $n_A = k + 1$ and $m \ge n_A$. Then, $x_{m} R x_{k}$ and $x_{k} \Vdash B$. For any $v \in W$ with $x_{k}S_{x_{m}} v$, we have $v = x_{k}$ by the definition of $S_{x_{m}}$. Thus, $v \nVdash C$. Therefore, we obtain $x_{m} \nVdash B \rhd C$. Since $y_{m} R x_{k}$, $y_{m} \nVdash B \rhd C$ is also proved.

• (Case 4): There exists an odd number k such that $y_{k} \Vdash B$ and $y_{k} \nVdash C$. It is proved that $n_A = k+1$ witnesses the claim as in Case 3.

• (Case 5): Otherwise, all of the following conditions are satisfied.

  • (I) For any even number k, if $x_{k} \Vdash B$, then either $x_{k} \Vdash C$ or $x_{k+1} \Vdash C$.

  • (II) For any even number k, if $y_{k} \Vdash B$, then either $y_{k} \Vdash C$ or $x_{k+1} \Vdash C$.

  • (III) For any odd number k, if $x_{k} \Vdash B$, then $x_{k} \Vdash C$.

  • (IV) For any odd number k, if $y_{k} \Vdash B$, then $y_{k} \Vdash C$.

  By the induction hypothesis, there exists an $n_{B} \in \omega$ which is a witness of the statement of the claim for B. Now, we prove that there exists a natural number $n_A \ge 1$ such that for each $i \ge n_A - 1$, we have $x_i \Vdash \lnot B \vee C$ and $y_i \Vdash \lnot B \vee C$. We distinguish the following four cases.

  • $\forall m \ge n_{B}\, (x_{m} \Vdash B$ and $y_{m} \Vdash B)$: Then, by (III) and (IV), there are infinitely many odd numbers k such that $x_{k} \Vdash C$ and $y_{k} \Vdash C$. Thus, by the induction hypothesis, there exists an $n_{C} \in \omega$ such that $\forall m \ge n_{C}\, (x_{m} \Vdash C$ and $y_{m} \Vdash C)$. Then, we define $n_A:= n_C + 1$.

  • $\forall m \ge n_{B}\, (x_{m} \Vdash B$ and $y_{m} \nVdash B)$: Then, by (III), there are infinitely many odd numbers k such that $x_{k} \Vdash C$. Thus, by the induction hypothesis, there exists an $n_{C} \in \omega$ such that $\forall m \ge n_{C}\, (x_{m} \Vdash C)$. Then, we define $n_A:= \max \{n_{B}, n_{C}\} + 1$.

  • $\forall m \ge n_{B}\, (x_{m} \nVdash B$ and $y_{m} \Vdash B)$: Then, by (IV), there are infinitely many odd numbers k such that $y_{k} \Vdash C$. Thus, by the induction hypothesis, there exists an $n_{C} \in \omega$ such that $\forall m \ge n_{C}\, (y_{m} \Vdash C)$. Then, we define $n_A: = \max \{n_{B}, n_{C}\} + 1$.

  • $\forall m \ge n_{B}\, (x_{m} \nVdash B$ and $y_{m} \nVdash B)$: We define $n_A: = n_{B} + 1$.

  We prove that $n_A$ witnesses the claim. Let $m \ge n_A$ and $z \in W$ be such that $x_{m} R z$ and $z \Vdash B$. We show that there exists a $v \in W$ such that $z S_{x_{m}} v$ and $v \Vdash C$. Let i be an index of z. If i is odd, then $z S_{x_{m}} z$ and $z \Vdash C$ by (III) and (IV). Assume that i is even. We distinguish the following two cases.

  • $n_A - 1 \le i < m$: We obtain $z \Vdash \lnot B \vee C$ by the choice of $n_A$. Since $z \Vdash B$, we have $z \Vdash C$. By the definition of $S_{x_{m}}$, we find $z S_{x_{m}} z$.

  • $i < n_A-1$: Then, $i < m-1$. Therefore, $z S_{x_{m}} z$ and $z S_{x_{m}} x_{i+1}$. Furthermore, by (I) and (II), we obtain $z \Vdash C$ or $x_{i+1} \Vdash C$.

  In any case, there exists $v \in W$ such that $z S_{x_{m}} v$ and $v \Vdash C$. Therefore, we obtain $x_{m} \Vdash B \rhd C$. Similarly, we have $y_m \Vdash B \rhd C$. $\square$

We suppose, towards a contradiction, that there exists a formula A such that $v(A) \subseteq \{q\}$ and $\textbf{CL}\vdash A \leftrightarrow A \rhd q$. Since $\textbf{CL}$ is valid in ${\mathcal {F}}$, $A \leftrightarrow A \rhd q$ is valid in ${\mathcal {F}}$. Moreover, the following claim holds.

Claim 2

For any $w \in W$ whose index is n, n is even if and only if $w \Vdash A$.

Proof

We prove the claim by induction on n. Let $w \in W$ be any element whose index is n.

For $n =0$, since there is no $w' \in W$ such that $w R w'$, we obtain $w \Vdash A \rhd q$ and hence, $w \Vdash A$. Suppose $n > 0$ and that the claim holds for any natural number less than n.

$(\Leftarrow )$: Assume that n is an odd number. Then, $w R y_{n-1}$. Since $n-1$ is even, $y_{n-1} \Vdash A$ by the induction hypothesis. Let v be any element in W satisfying $y_{n-1} S_{w} v$. By the definitions of $S_{w}$ and $\Vdash$, we obtain $v = y_{n-1}$ and $v \nVdash q$. Therefore, $w \nVdash A \rhd q$ and hence $w \nVdash A$.

$(\Rightarrow )$: Assume that n is an even number. Let v be any element in W with wRv and $v \Vdash A$. Let m be the index of v. Since $m < n$ and $v \Vdash A$, m is even by the induction hypothesis. Since n is also even, $m < n-1$ and hence $v S_{w} x_{m+1}$. Furthermore, $x_{m+1} \Vdash q$ by the definition of $\Vdash$. Therefore, we obtain $w \Vdash A \rhd q$ and hence, $w \Vdash A$. $\square$

This contradicts Claim 1. Therefore, for any formula A with $v(A) \subseteq \{q\}$, we obtain $\textbf{CL}\nvdash A \leftrightarrow A \rhd q$. $\square$

Corollary 4

Let L be any logic that is closed under substituting a formula for a propositional variable and satisfies $\textbf{IL}^{-} \subseteq L \subseteq \textbf{CL}$. Then, L has neither $\ell$FPP nor CIP.

Proof

By Theorem 9, every sublogic of $\textbf{CL}$ does not have $\ell$FPP. By Lemma 3, every logic L such that $\textbf{IL}^- \subseteq L \subseteq \textbf{CL}$ does not have CIP. $\square$

6.2 A counter model of $\ell$FPP for $\textbf{IL}^{-}(\textbf{J1}, \textbf{J5})$

In this subsection, we prove that $\textbf{IL}^-(\textbf{J5})$ and $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$ have neither $\ell$FPP nor CIP.

Theorem 10

The formula $p \rhd q$ which is left-modalized in p has no fixed point in $\textbf{IL}^{-}(\textbf{J1}, \textbf{J5})$. That is, for any formula A which satisfies $v(A) \subseteq \{q\}$,

$$\begin{aligned} \textbf{IL}^{-}(\textbf{J1}, \textbf{J5}) \nvdash A \leftrightarrow A \rhd q. \end{aligned}$$

Proof

We define an $\textbf{IL}^-$-frame ${\mathcal {F}} = \langle W, R, \{S_{w}\}_{w \in W}\rangle$ as follows:

• $W:= \omega \cup \{v\}$;

• $R:= \{\langle x, y\rangle \in W^2: x, y \in \omega$ and $x > y \}$;

• $S_{v}:= \emptyset$ and for each $n \in \omega$, $S_{n}:=\{\langle x, y\rangle \in W^2: n R x$ and ($y = x$ or xRy or (x is even, $x < n -1$ and $y = v)) \}$.

For instance, the relations $S_3$ and $S_4$ are shown in the following figure (Fig. 4). In the case of xRy for $x, y < n$, $x S_n y$ holds, and the corresponding broken lines are omitted in the figure.

Fig. 4

A counter model of $\ell$FPP for $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$

Then, $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$ is valid in ${\mathcal {F}}$. Let $\Vdash$ be a satisfaction relation on ${\mathcal {F}}$ such that $v \Vdash q$ and for each $n \in \omega$, $n \nVdash q$.

Claim 1

For any formula A with $v(A) \subseteq \{q\}$, there exists an $n_A \in \omega$ such that

$$\begin{aligned} \forall m \ge n_A\, (m \Vdash A) \ \text {or} \ \forall m \ge n_A\, (m \nVdash A). \end{aligned}$$

Proof

We prove the claim by induction on the construction of A. We only prove the case of $A \equiv B \rhd C$. We distinguish the following three cases.

• Case 1: There exists an even number k such that $k \Vdash B$, for all $j \le k$, $j \nVdash C$ and $v \nVdash C$: Let $n_A = k + 1$ and $m \ge n_A$. Then, mRk and $k \Vdash B$. For any $w \in W$ with $k S_{m} w$, since either $w \le k$ or $w = v$, we obtain $w \nVdash C$. Therefore, $m \nVdash B \rhd C$.

• Case 2: There exists an odd number k such that $k \Vdash B$ and for all $j \le k$, $j \nVdash C$: Let $n_A = k + 1$ and $m \ge n_A$. Then, mRk and $k \Vdash B$. For any $w \in W$ with $k S_{m} w$, we have $w \nVdash C$ because $w \le k$. Therefore, $m \nVdash B \rhd C$.

• Case 3: Otherwise: Then, the following conditions (I) and (II) are fulfilled.

  • (I) For any even number k, if $k \Vdash B$, then there exists a $j \le k$ such that $j \Vdash C$ or $v \Vdash C$.

  • (II) For any odd number k, if $k \Vdash B$, then there exists a $j \le k$ such that $j \Vdash C$.

  By the induction hypothesis, there exists an $n_B \in \omega$ such that $\forall m \ge n_B\, (m \Vdash B)$ or $\forall m \ge n_B\, (m \nVdash B)$. We may assume that $n_B$ is an odd number. We distinguish the following two cases.

  • $\forall m \ge n_B\, (m \Vdash B)$: Let $n_A = n_B + 1$ and $m \ge n_A$. Let k be any element in W satisfying mRk and $k \Vdash B$. Since $n_B$ is odd and $n_B \Vdash B$, there exists a $j_{0} \le n_B$ such that $j_{0} \Vdash C$ by (II). We distinguish the following three cases.

    • k is odd: By (II), there exists a $j_1 \le k$ such that $j_1 \Vdash C$. Then, $k S_{m} j_1$ and $j_1 \Vdash C$.

    • k is even and $k \ge n_B$: Since $k \ge j_{0}$, we have $k S_{m} j_{0}$ and $j_{0} \Vdash C$.

    • k is even and $k < n_B$: By (I), there exists a $j_1 \le k$ such that $j_1 \Vdash C$ or $v \Vdash C$. Since $k < n_B \le m -1$, we obtain $k < m-1$. Hence, $k S_{m} j_1$ and $k S_{m} v$.

    In any case, there exists a $w \in W$ such that $k S_{m} w$ and $w \Vdash C$. Therefore, $m \Vdash B \rhd C$.

  • $\forall m \ge n_B\, (m \nVdash B)$: Let $n_A = n_B + 1$ and $m \ge n_A$. Let k be any element in W satisfying mRk and $k \Vdash B$. Then, $k < n_B$ because $k \Vdash B$. We distinguish the following two cases.

    • k is odd: Since there exists a $j \le k$ such that $j \Vdash C$ by (II). Then, $k S_{m} j$.

    • k is even: By (I), there exists a $j \le k$ such that $j \Vdash C$ or $v \Vdash C$. Since $k < n_B \le m -1$, we obtain $k < m-1$ and hence $k S_{m} j$ and $k S_{m} v$.

  In any case, there exists a $w \in W$ such that $k S_{m} w$ and $w \Vdash C$. Therefore, $m \Vdash B \rhd C$.$\square$

We suppose, towards a contradiction, that there exists a formula A such that $v(A) \subseteq \{q\}$ and $\textbf{IL}^{-}(\textbf{J1}, \textbf{J5}) \vdash A \leftrightarrow A \rhd q$. Since $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$ is valid in ${\mathcal {F}}$, we have that $A \leftrightarrow A \rhd q$ is also valid in ${\mathcal {F}}$. Then, the following claim holds.

Claim 2

For any $n \in \omega$, n is even if and only if $n \Vdash A$.

Proof

We prove the claim by induction on n.

For $n = 0$, since obviously $0 \Vdash A \rhd q$, we have $0 \Vdash A$. Suppose $n > 0$ and the claim holds for any natural number less than n.

$(\Leftarrow )$: Assume that n is odd. Then, $n R n-1$ and since $n-1$ is even, $n-1 \Vdash A$ by the induction hypothesis. Let w be the any element in W which satisfies $n-1 S_{n} w$. By the definition of $S_{n}$, we find $w \le n-1$ and hence $w \nVdash q$. Therefore, $n \nVdash A \rhd q$, and thus $n \nVdash A$.

$(\Rightarrow )$: Assume that n is even. Let m be the any element in W which satisfies nRm and $m \Vdash A$. By the induction hypothesis, m is even and hence $m < n-1$. Then, $m S_{n} v$ and $v \Vdash q$. Therefore, $n \Vdash A \rhd q$ and hence, $n \Vdash A$. $\square$

This contradicts Claim 1. Threfore, for any formula A with $v(A) \subseteq \{q\}$, we obtain $\textbf{IL}^{-}(\textbf{J1}, \textbf{J5}) \nvdash A \leftrightarrow A \rhd q$. $\square$

As in Corollary 4, we obtain the following corollary.

Corollary 5

Let L be any logic that is closed under substituting a formula for a propositional variable and satisfies $\textbf{IL}^{-} \subseteq L \subseteq \textbf{IL}^-(\textbf{J1}, \textbf{J5})$. Then, L has neither $\ell$FPP nor CIP.

6.3 A counter model of FPP for $\textbf{IL}^-(\textbf{J1}, \textbf{J4}_+, \textbf{J5})$

In Theorems 9 and 10, we proved that the logics $\textbf{CL}$ and $\textbf{IL}^-(\textbf{J1}, \textbf{J5})$ do not have $\ell$FPP. On the other hand, we proved in Theorem 8 that $\textbf{IL}^-(\textbf{J4}, \textbf{J5})$ has $\ell$FPP. Thus, we cannot provide a counter model of $\ell$FPP for extensions of $\textbf{IL}^-(\textbf{J4}, \textbf{J5})$. In this subsection, we prove that the logics $\textbf{IL}^-(\textbf{J4}_+, \textbf{J5})$ and $\textbf{IL}^-(\textbf{J1}, \textbf{J4}_+, \textbf{J5})$ have neither FPP nor CIP.

Theorem 11

The formula $\top \rhd \lnot p$ has no fixed point in $\textbf{IL}^{-}(\textbf{J1}, \textbf{J4}_+, \textbf{J5})$. That is, for any formula A with $v(A) = \emptyset$,

$$\begin{aligned} \textbf{IL}^{-}(\textbf{J1}, \textbf{J4}_{+}, \textbf{J5}) \nvdash A \leftrightarrow \top \rhd \lnot A. \end{aligned}$$

Proof

We define an $\textbf{IL}^-$-frame ${\mathcal {F}} = \langle W, R, \{S_{w}\}_{w \in W}\rangle$ as follows:

• $W: = \omega$;

• $x R y: \iff x > y$;

• For each $n \in W$, $S_{n}:=\{\langle x, y\rangle \in W^2: x, y < n$ and $(x \ge y$ or $(x =0$ and (y is even or $y = n-1))) \}$.

We draw the relations $S_3$ and $S_4$. As in the proof of Theorem 10, in the case of xRy for $x, y < n$, $x S_n y$ holds, and the corresponding broken lines are omitted in the figure (Fig. 5).

Fig. 5

A counter model of FPP for $\textbf{IL}^-(\textbf{J1}, \textbf{J4}_+, \textbf{J5})$

Then, $\textbf{IL}^{-}(\textbf{J1}, \textbf{J4}_{+}, \textbf{J5})$ is valid in ${\mathcal {F}}$. Let $\Vdash$ be an arbitrary satisfaction relation on ${\mathcal {F}}$.

Claim 1

For any formula A with $v(A) = \emptyset$, there exists an $n_A \in W$ such that

$$\begin{aligned} \forall m \ge n_A\, (m \Vdash A) \ \text {or} \ \forall m \ge n_A\, (m \nVdash A). \end{aligned}$$

Proof

This is proved by induction on the construction of A. We prove only the case of $A \equiv B \rhd C$. We distinguish the following three cases.

• Case 1: There exists an $n > 0$ such that $n \Vdash B$ and for all $k \le n$, $k \nVdash C$. Let $n_A = n + 1$ and $m \ge n_A$. Then, mRn and $n \Vdash B$. Also, for any $k \in W$, if $nS_{m} k$, then $k \le n$ because $n \ne 0$. Therefore, $k \nVdash C$. Thus, $m \nVdash B \rhd C$.

• Case 2: $0 \Vdash B$ and for all even numbers k, $k \nVdash C$. By the induction hypothesis, there exists an $n_C \in W$ such that $\forall m \ge n_C\, (m \Vdash C)$ or $\forall m \ge n_C\, (m \nVdash C)$. Since there are infinitely many even numbers $k \in W$ such that $k \nVdash C$, we obtain $\forall m \ge n_C\, (m \nVdash C)$. Let $n_A = n_C + 1$. Then, for any $m \ge n_A$, we have mR0 and $0 \Vdash B$. Let $k \in W$ be such that $0 S_{m} k$. Then, k is even or $k = m-1$ by the definition of $S_{m}$. By our supposition, if k is even, then $k \nVdash C$. If $k = m-1$, then $m-1 \nVdash C$ because $m-1 \ge n_C$. Therefore, in either case, $k \nVdash C$. Thus, $m \nVdash B \rhd C$.

• Case 3: Otherwise: Then, the following conditions (I) and (II) are fulfilled.

  • (I) For any $n > 0$, if $n \Vdash B$, then there exists a $k \in W$ such that $k \le n$ and $k \Vdash C$.

  • (II) If $0 \Vdash B$, then there exists an even number $k \in W$ such that $k \Vdash C$.

  We distinguish the following two cases.

  • $0 \nVdash B$: Let $n_A = 0$ and $m \ge n_A$. For any $n \in W$ satisfying mRn and $n \Vdash B$, since $n \ne 0$, there exists a $k \le n$ such that $k \Vdash C$ by the condition (I). Since $n S_{m} k$, we obtain $m \Vdash B \rhd C$.

  • $0 \Vdash B$: By the condition (II), there exists an even number k such that $k \Vdash C$. Let $n_A = k + 1$ and $m \ge n_A$. Let $n \in W$ be such that mRn and $n \Vdash B$. If $n \ne 0$, then there exists a $k' \le n$ such that $k' \Vdash C$ and $n S_{m}k'$ by the condition (I). If $n =0$, then since k is even and $k < m$, we obtain $n S_{m} k$ and $k \Vdash C$. Therefore, $m \Vdash B \rhd C$. $\square$

We suppose, towards a contradiction, that there exists a formula A such that $v(A) = \emptyset$ and $\textbf{IL}^{-}(\textbf{J1}, \textbf{J4}_{+}, \textbf{J5}) \vdash A \leftrightarrow \top \rhd \lnot A$. Then, $A \leftrightarrow \top \rhd \lnot A$ is valid in ${\mathcal {F}}$ because so is $\textbf{IL}^{-}(\textbf{J1}, \textbf{J4}_{+}, \textbf{J5})$. Then, the following claim holds.

Claim 2

For any $n \in W$, n even if and only if $n \Vdash A$.

Proof

We prove the by induction on n. For $n = 0$, obviously $0 \Vdash A$. Suppose $n > 0$ and the claim holds for any natural number less than n.

$(\Leftarrow )$: Assume that n is odd. Then, nR0. For any $k \in W$ which satisfies $0 S_{n} k$, since n is odd, k is even and $k < n$. By the induction hypothesis, $k \Vdash A$. Thus, we obtain $n \nVdash \top \rhd \lnot A$ and hence, $n \nVdash A$.

$(\Rightarrow )$: Assume that n is even. Let $m \in W$ be such that nRm. We distinguish the following three cases.

• $m=0$: Then, $0 S_{n} n-1$. Since $n-1$ is odd, $n-1 \Vdash \lnot A$ by the induction hypothesis.

• m is even and $m \ne 0$: Then, $m S_{n} m-1$. Since $m-1$ is odd, $m-1 \Vdash \lnot A$ by the induction hypothesis.

• m is odd: Then, $m S_{n} m$. Since m is odd, $m \Vdash \lnot A$ by the induction hypothesis.

In any case, there exists a $w \in W$ such that $m S_{n} w$ and $w \Vdash \lnot A$. Therefore, we obtain $n \Vdash \top \rhd \lnot A$ and hence, $n \Vdash A$. $\square$

This contradictions Claim 1. Therefore, there is no formula A such that $v(A) = \emptyset$ and $\textbf{IL}^{-}(\textbf{J1}, \textbf{J4}_{+}, \textbf{J5}) \nvdash A \leftrightarrow \top \rhd \lnot A$. $\square$

Corollary 6

Every sublogic of $\textbf{IL}^{-}(\textbf{J1}, \textbf{J4}_{+}, \textbf{J5})$ does not have FPP. Furthermore, if L is closed under substituting a formula for a propositional variable and satisfies $\textbf{IL}^{-}(\textbf{J4}_{+}) \subseteq L \subseteq \textbf{IL}^{-}(\textbf{J1}, \textbf{J4}_{+}, \textbf{J5})$, then L does not have CIP.

Proof

By Theorem 11, every sublogic of $\textbf{IL}^{-}(\textbf{J1}, \textbf{J4}_{+}, \textbf{J5})$ does not have FPP. By Lemma 2, every logic L that is closed under substituting a formula for a propositional variable and satisfies $\textbf{IL}^{-}(\textbf{J4}_{+}) \subseteq L \subseteq \textbf{IL}^{-}(\textbf{J1}, \textbf{J4}_{+}, \textbf{J5})$ does not have CIP. $\square$

7 Concluding remarks

In this paper, we provided a complete description of twelve sublogics of $\textbf{IL}$ concerning UFP, FPP, and CIP. In particular, for these sublogics L, we proved that L has FPP if and only if L contains $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5})$. On the other hand, there are many other logics between $\textbf{IL}^-$ and $\textbf{IL}$. For instance, Kurahashi and Okawa [9] introduced eight sublogics such as $\textbf{IL}^-(\textbf{J2}, \textbf{J4}_+, \textbf{J5})$ that are not in Fig. 1, and proved that these eight logics are not complete with respect to usual Veltman semantics but complete with respect to generalized Veltman semantics. Then, it is natural to investigate a sharper threshold for FPP in a larger class of sublogics. Then, for example, we propose a question if $\textbf{J2}_+$ can be weakened by $\textbf{J2}$ in the statement of Corollary 2.

Problem 2

Does the logic $\textbf{IL}^-(\textbf{J2}, \textbf{J4}_+, \textbf{J5})$ have FPP?

In our proofs of Theorems 6, 7, and 8, the use of the axiom scheme $\textbf{J5}$ seems inevitable. In fact, $\textbf{CL}$ ($= \textbf{IL}^-(\textbf{J1}, \textbf{J2}_+)$) fails to have $\ell$FPP. From this observation, in an earlier version of the present paper, we had proposed the question whether $\textbf{J5}$ is necessary or not for $\ell$FPP and FPP. Later on, the third author of the present paper settled this question. For each $n \ge 1$, let $\textbf{J5}^n$ be the following axiom scheme:

$\textbf{J5}^n$:

$\Diamond ^n A \rhd A$.

Concerning $\textbf{J5}^n$, the following result is established.

Theorem 12

(Okawa [11]) Let $n \ge 1$.

• 1. $\textbf{IL}^-(\textbf{J5}^n) \vdash \textbf{J5}^{n+1}$ and $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5}^{n+1}) \nvdash \textbf{J5}^{n}$.

• 2. $\textbf{IL}^-(\textbf{J2}_+, \textbf{J5}^n)$ and $\textbf{IL}^-(\textbf{J4}, \textbf{J5}^n)$ have FPP and $\ell$FPP, respectively.

Furthermore, the authors have already developed several studies related to the present paper (cf. [6, 10]).