Interpretability logics and generalized Veltman semantics

TL;DR

This paper establishes modal completeness, decidability, and finite model properties for interpretability logics using generalized Veltman semantics, introducing new proof techniques and constructions.

Contribution

It provides new completeness proofs, shorter proofs for classical interpretability logics, and develops a construction potentially useful for future completeness results.

Findings

Modal completeness of ILP_0 and ILR w.r.t. generalized Veltman semantics

Decidability and finite model property for ILP_0 and ILR

Development of a construction for future completeness proofs

Abstract

We obtain modal completeness of the interpretability logics ILP_0 and ILR w.r.t. generalized Veltman semantics. Our proofs are based on the notion of smart (full) labels. We also give shorter proofs of completeness w.r.t. generalized semantics for many classical interpretability logics. We obtain decidability and finite model property w.r.t. generalized semantics for ILP_0 and ILR. Finally, we develop a construction that might be useful for proofs of completeness of extensions of ILW w.r.t. generalized semantics in the future, and demonstrate its usage with ILW* = ILM_0W.

Equations

$$\mathfrak{F}⊩\mathsf{X}\text{mbox}ifandonlyif\mathfrak{F}\models (\mathsf{X})\text{ (resp. }\mathfrak{F}\models \text{(X)textsubscriptgen}\text{)}.$$

$$u_S^{\square }$$

$$u_S^{⊡}$$

$$W$$

$$wRu$$

$$u{S}_{w}V$$

$$w⊩p$$

$$\mathfrak{M},w⊩G\text{mbox}ifandonlyifG\in w,$$

$$u{S}_{w}V\Rightarrow (\exists V^{\prime }\subseteq V)(u{S}_w{V}^{\prime }\&R[V^{\prime }]\subseteq R[u]).$$

$$wRuRx{S}_{w}V\Rightarrow (\exists V^{\prime }\subseteq V)(u{S}_w{V}^{\prime }\&R[V^{\prime }]\subseteq R[u])).$$

$$wR{w}^{\prime }Ru{S}_{w}V\Rightarrow (\exists V^{\prime }\subseteq V)u{S}_{w^{\prime }}{V}^{\prime }$$

$$wRxRu{S}_{w}V\&(\forall v\in V)R[v]\cap Z\ne \varnothing \Rightarrow (\exists Z^{\prime }\subseteq Z)u{S}_x{Z}^{\prime }$$

$$wRxRu{S}_{w}V\Rightarrow (\forall C\in \mathcal{C}(x,u))(\exists U\subseteq V)(x{S}_{w}U\&R[U]\subseteq C),$$

$$(\forall S)\big{(}w\prec_{S}x\ \Rightarrow\ (\exists v\in V)(w\prec_{S\cup x_{\emptyset}^{\square}}v\ \&\ R[v]\subseteq C)\big{)}.$$

$$w{\prec }_{S}u\Rightarrow (\exists G\in \mathcal{D})w{\prec }_{S\cup \{\square \neg G\}}u\text{ and }G\in u,$$

$$\mathfrak{F}\models \text{Not-}\text{W}\text{ if and only if }\mathfrak{F}⊮\text{W}.$$

$$u{S}_{w}V\Rightarrow (\exists V^{\prime }\subseteq V)(u{S}_w{V}^{\prime }\&R[V^{\prime }]\cap S_w^{-1}[V]=\varnothing ).$$

$$W$$

$$wRu$$

$$u{S}_{w}V$$

$$(a)V\cap \dot{R}[u]\ne \varnothing ;$$

$$(b)(\forall S)(w{\prec }_{S}u\Rightarrow (\exists v\in V)(\exists G\in \mathcal{D}\cap \bigcup \dot{R}[u])w{\prec }_{S\cup \{\square \neg G\}}v);$$

$$w⊩p$$

$$\mathfrak{M},w⊩G\text{mbox}ifandonlyifG\in w,$$

$$(\forall S)(w{\prec }_S{v}_0\Rightarrow (\exists x\in U_{v_0})(\exists G\in \mathcal{D}\cap \bigcup \dot{R}[v_0])w{\prec }_{S\cup \{\square \neg G\}}x).$$

$$(\forall S)(w{\prec }_{S}u\Rightarrow (\exists v\in V)(\exists G\in \mathcal{D}\cap \bigcup \dot{R}[u])w{\prec }_{S\cup \{\square \neg G\}}v).$$

$$(\forall S^{\prime })(w{\prec }_{S^{\prime }}{v}_0\Rightarrow (\exists x\in U_{v_0})(\exists G^{\prime }\in \mathcal{D}\cap \bigcup \dot{R}[v_0])w{\prec }_{S^{\prime }\cup \{\square \neg G^{\prime }\}}x).$$

$$\mathfrak{M},w⊩G\text{mbox}ifandonlyifG\in w.$$

$$u{S}_{w}V\Rightarrow (\exists V^{\prime }\subseteq V)(u{S}_w{V}^{\prime }\&R[V^{\prime }]\cap S_w^{-1}[V]=\varnothing ).$$

$$u{S}_{w}V\&(\forall V^{\prime }\subseteq V)(u{S}_w{V}^{\prime }\Rightarrow R[V^{\prime }]\cap S_w^{-1}[V]\ne \varnothing ).$$

$$y{S}_w^{i}U⟺y{S}_{w}U,{\mathcal{D}}_i\subseteq \bigcup \dot{R}[y],U\subseteq {\left[\underset{G\in {\mathcal{D}}_i}{\bigvee }\square \neg G\right]}_w.$$

$$(\exists U^{\prime }\subseteq U)(\exists i<n)y{S}_w^i{U}^{\prime }.$$

$$U^{\prime }=\{z\in U:(\exists G\in \mathcal{D}\cap \bigcup \dot{R}[y])w{\prec }_{\{\square \neg G\}}z\};$$

$${\mathcal{D}}_i=\{G\in \mathcal{D}\cap \bigcup \dot{R}[y]:(\exists z\in U)w{\prec }_{\{\square \neg G\}}z\}.$$

$$Y=\{x\in U^{\prime }:(\exists y\in R[x])(\exists Z\subseteq U^{\prime })y{S}_w^{m+1}Z\}.$$

Full text

Keywords: interpretability logic, Veltman semantics, completeness, filtration, finite model property, decidability

1 Introduction

1.1 Interpretability logics

The language of interpretability logics is given by $A::=p\,|\,\bot\,|\,A\to A\,|\,A\rhd A,$ where $p$ ranges over a countable set of propositional variables. Other Boolean connectives are defined as abbreviations, as usual. Since $\square A$ can be defined (over extensions of IL) as an abbreviation too (exapnded to $\neg A\rhd\bot$), we do not include $\square$ or $\Diamond$ in the language. If $A$ is constructed in this way, we will say that $A$ is a modal formula.

Definition 1**.**

Interpretability logic IL is given by the following list of axiom schemata.

classical tautologies (in the new language); 2. K

$\square(A\rightarrow B)\rightarrow(\square A\rightarrow\square B)$; 3. L

$\square(\square A\rightarrow A)\rightarrow\square A$; 4. J1

$\square(A\rightarrow B)\rightarrow A\rhd B$; 5. J2

$(A\rhd B)\wedge(B\rhd C)\ \rightarrow\ A\rhd C$; 6. J3

$(A\rhd C)\wedge(B\rhd C)\ \rightarrow\ A\vee B\rhd C$; 7. J4

$A\rhd B\rightarrow(\Diamond A\rightarrow\Diamond B)$; 8. J5

$\Diamond A\rhd A$.

Rules of inference are modus ponens and necessitation (generalization).

We treat $\rhd$ as having higher priority than $\to$, but lower than other logical connectives. Other interpretability logics are obtained by extending IL with further schemata (“principles of interpretability”).

Let us briefly describe the motivation behind studying interpretability logics. In a sufficiently strong formal theory $T$ in the language $\mathcal{L}_{T}$, one can construct a binary interpretability predicate $\mathsf{Int}_{T}$. This predicate expresses that one finite extension of $T$ interprets another finite extension of $T$. Any mapping $A\mapsto A^{*}$ where $A$ is a modal formula and $A^{*}\in\mathcal{L}_{T}$, such that:

• •

it commutes with logical connectives;

• •

if $p$ is a propositional variable, $p^{*}$ is a sentence;

• •

$(A\rhd B)^{*}$ = $\mathsf{Int}_{T}(\lceil A^{*}\rceil,\lceil B^{*}\rceil)$, where $\lceil X\rceil$ is the numeral of the Gödel number of $X$;

is called an arithmetical interpretation. The interpretability logic of $T$, denoted by IL(T), is the set of all modal formulas $A$ such that $T\vdash A^{*}$ for all arithmetical interpretations. Berarducci [1] and Shavrukov [9] independently proved that $\text{{IL}{(T)}}=\text{{IL}{M}}$, if $T$ is an essentially reflexive theory. Visser [10] proved that $\text{{IL}{(T)}}=\text{{IL}{P}}$, if $T$ is finitely axiomatizable and contains supexp.

One of the major open problems in the field is to determine IL(All), the interpretability logic of all “reasonable arithmetical theories.” The ongoing search for IL(All) is the main motivation behind studying extensions of IL. Studying modal properties of lower bounds of IL(All) turns out to be useful for finding new principles within IL(All). For example, the principle R was discovered while trying to prove modal completeness of ILP0W [6].

Definitions and furher details can be found in e.g. [11].

1.2 Semantics

The most commonly used semantics for the interpretability logic IL and its extensions is the Veltman semantics (or ordinary Veltman semantics).

Definition 2** ([3], Definition 1.2).**

A Veltman frame $\mathfrak{F}$ is a structure $(W,R,\{S_{w}:w\in W\})$, where $W$ is a non-empty set, $R$ is a transitive and converse well-founded binary relation on $W$ and for all $w\in W$ we have:

• a)

$S_{w}\subseteq R[w]^{2}$, where $R[w]=\{x\in W:wRx\}$;

• b)

$S_{w}$ is reflexive on $R[w]$;

• c)

$S_{w}$ is transitive;

• d)

if $wRuRv$ then $uS_{w}v$.

A Veltman model is a quadruple $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$, where the first three components form a generalized Veltman frame. The forcing relation $\Vdash$ is extended as usual in Boolean cases, and $w\Vdash A\rhd B$ holds if and only if for all $u$ such that $wRu$ and $u\Vdash A$ there exists $v$ such that $uS_{w}v$ and $v\Vdash B$.

In what follows we will mainly use the following semantics, which we will refer to as the generalized Veltman semantics. De Jongh defined this specific generalization of Veltman semantics. The main purpose of its introduction, and until recently the only usage, was to show independence of certain extensions of IL, by Verbrugge (private correspondence), Vuković [13] and Goris and Joosten [6].

Definition 3**.**

A generalized Veltman frame $\mathfrak{F}$ is a structure $(W,R,\{S_{w}:w\in W\})$, where $W$ is a non-empty set, $R$ is a transitive and converse well-founded binary relation on $W$ and for all $w\in W$ we have:

• a)

$S_{w}\subseteq R[w]\times\left(\mathcal{P}({R[w]})\setminus\{\emptyset\}\right)$;

• b)

$S_{w}$ is quasi-reflexive: $wRu$ implies $uS_{w}\{u\}$;

• c)

$S_{w}$ is quasi-transitive: if $uS_{w}V$ and $vS_{w}Z_{v}$ for all $v\in V$, then $uS_{w}(\bigcup_{v\in V}Z_{v})$;

• d)

if $wRuRv$, then $uS_{w}\{v\}$;

• e)

monotonicity: if $uS_{w}V$ and $V\subseteq Z\subseteq R[w]$, then $uS_{w}Z$.

A generalized Veltman model is a quadruple $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$, where the first three components form a Veltman frame. Now $w\Vdash A\rhd B$ holds if and only if for all $u$ such that $wRu$ and $u\Vdash A$ there exists $V$ such that $uS_{w}V$ and $V\Vdash B$. We write $V\Vdash B$ if $v\Vdash B$ for all $v\in V$.

1.3 Principles, completeness and decidability

Let us review some relevant results and approaches. When we need to refer to an extension of IL (an arbitrary extension if not stated otherwise), we will write ILX.

Let (X) (resp. (X)gen) denote a formula of first-order or higher-order logic such that for all ordinary (resp. generalized) Veltman frames $\mathfrak{F}$ the following holds:

[TABLE]

Formulas $(\mathsf{X})$ and (X)gen are called characteristic properties (or frame conditions) of the given logic ILX. The class of all ordinary (resp. generalized) Veltman frames $\mathfrak{F}$ such that $\mathfrak{F}\models(\mathsf{X})\text{\ (resp.\ }\mathfrak{F}\models\text{({{X}})\textsubscript{gen}}\text{)}$ is called the chararacteristic class of (resp. generalized) frames for ILX. If $\mathfrak{F}\models\text{({{X}})\textsubscript{gen}}$ we also say that the frame $\mathfrak{F}$ possesses the property (X)gen. We say that an ordinary (resp. generalized) Veltman model $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ is an ILX-model (resp. ILgenX-model), or that model $\mathfrak{M}$ possesses the property (X) (resp. (X)gen), if the frame $(W,R,\{S_{w}:w\in W\})$ possesses the property (X) (resp. (X)gen). A logic ILX will be said to be complete with respect to ordinary (resp. generalized) semantics if for all modal formulas $A$ we have that validity of $A$ over all ILX-frames (resp. all ILgenX-frames) implies $\text{{IL}{X}}\vdash A$.

We say that ILX has finite model property (FMP) w.r.t. ordinary (resp. generalized) semantics if for each formula $A$ satisfiable in some ILX-model (resp. ILgenX-model), $A$ is also satisfiable in some finite ILX-model (resp. ILgenX-model).

If we include results from the current paper, we have the following table. Here, o stands for the ordinary Veltman semantics, and g for the generalized Veltman semantics (as defined earlier).333If we adhere to our definitions, ILP0 has FMP w.r.t. ordinary semantics, but this is not a useful result since it is incomplete w.r.t. ordinary semantics.

[TABLE]

De Jongh and Veltman proved the completeness of the logics IL, ILM and ILP w.r.t. their characteristic classes of ordinary (and finite) Veltman frames in [3]. As is usual for extension of the provability logic GL, all completeness proofs suffer from compactness-related issues. One way to go about this is to define a (large enough) adequate set of formulas and let worlds be maximal consistent subsets of such sets (used e.g. in [3]). With interpretability logics and ordinary semantics, worlds have not been identified (just) with sets of formulas, because of issues described in e.g. [3]. In [4], de Jongh and Veltman proved completeness of the logic ILW w.r.t. its characteristic class of ordinary (and finite) Veltman frames.

Goris and Joosten introduced a more robust approach to proving completeness of interpretability logics, the construction method [5], [6]. In this type of proofs, one builds models step by step, and the final model is retrieved as a union. While closer to intuitions and more informative than the standard proofs, these proofs are hard to produce and verify due to their size. (They might have been shorter if tools from [2] have been used from the start.) For the purpose for which they were invented (completeness of ILM0 and ILW* w.r.t. ordinary semantics) they are still the only known tools.

The completeness of interpretability logics w.r.t. generalized semantics is usually an easy consequence of the completeness w.r.t. ordinary semantics. In [8] and [7], filtration technique was used to prove the finite model property of IL and its extensions ILM, ILM0 and ILW∗ w.r.t. generalized Veltman semantics. Generalized semantics was used because issues occurred when merging worlds of ordinary Veltman models. Those explorations yielded some decidability results.

The aim of this paper is to show completeness (w.r.t. generalized semantics) and decidability of some interpretability logics. We introduce a very direct type of proof for proving completeness; similar to [3] in their general approach. We use smart labels from [2] for this purpose. An example that illustrates benefits of using generalized semantics will be given in the section dedicated to ILM0.

The main new results of this paper are completeness and finite model property (and thus decidability) of ILR and ILP0. The principle R is important because it forms the basis of the, at the moment, strongest candidate for IL(All). Results concerning the principle ILP0 are interesting in a different way; they answer an old question: is there an unravelling technique that transforms generalized ILX-models to ordinary ILX-models, that preserves satisfaction of relevant characteristic properties? The answer is no: we find ILP0 to be complete w.r.t. generalized semantics, but it is known to be incomplete w.r.t. ordinary semantics.

Other results include reproving some known facts with (much) shorter proofs. Of particular interest is the logic ILW, which was known to be complete and decidable, but for which we nevertheless reprove completeness w.r.t. gerenalized semantics using our approach. We will explain our motivation for doing so in the section dedicated to ILW.

2 Completeness w.r.t. generalized semantics

In what follows, “formula” will always mean “modal formula”. If the ambient logic in some context is ILX, a maximal consistent set w.r.t. ILX will be called an ILX-MCS. Let us now introduce smart labels from [2].

Definition 4** ([2], slightly modified Definition 3.1).**

Let $w$, $u$ and $x$ be some ILX-MCS’s, and let $S$ and $T$ be arbitrary sets of formulas. We define $w\prec_{S}u$ if for any finite $S^{\prime}\subseteq S$ and any formula $A$ we have that $A\rhd\bigvee_{G\in S^{\prime}}\neg G\in w$ implies $\neg A,\square\neg A\in u.$

Note that the small differences between our Definition 4 and Definition 3.1 [2] do not affect results of [2] that we use.

Definition 5** ([2], page 4).**

Let $u$ be an ILX-MCS, and $S$ an arbitrary set of formulas. Put:

[TABLE]

Thus, $w\prec_{S}u$ if and only if $w_{S}^{\boxdot}\subseteq u$. If $S=\emptyset$ then $u_{\emptyset}^{\square}=\{\square\neg A:A\rhd\bot\in u\}$. Since $u$ is maximal consistent, usages of this set usually amount to the same as of $\{\square A:\square A\in u\}.$

We will usually write $w\prec u$ instead of $w\prec_{\emptyset}u$.

Lemma 6** ([2], Lemma 3.2).**

Let $w$, $u$ and $v$ be some ILX-MCS’s, and let $S$ and $T$ be some sets of formulas. Then we have:

• a)

if $S\subseteq T$ and $w\prec_{T}u$, then $w\prec_{S}u$;

• b)

if $w\prec_{S}u\prec v$, then $w\prec_{S}v$;

• c)

if $w\prec_{S}u$, then $S\subseteq u$.

We will tacitly use properties of the preceding lemma in most of our proofs.

The following two lemmas can be used to construct (or in our case, find) a MCS with the required properties.

Lemma 7** ([2], Lemma 3.4).**

Let $w$ be an ILX-MCS, and let $\neg(B\rhd C)\in w.$ Then there is an ILX-MCS $u$ such that $w\prec_{\{\neg C\}}u$ and $B,\square\neg B\in u.$

Lemma 8** ([2], Lemma 3.5).**

Let $w$ and $u$ be some ILX-MCS’s such that $B\rhd C\in w,$ $w\prec_{S}u$ and $B\in u.$ Then there is an ILX-MCS $v$ such that $w\prec_{S}v$ and $C,\square\neg C\in v.$

Let $B$ be a formula, and $w$ a world in a generalized Veltman model. Put $[B]_{w}=\{u:wRu$ and $u\Vdash B\}$.

In the remainder of the current paper, we will assume that $\mathcal{D}$ is always a finite set of formulas, closed under taking subformulas and single negations, and $\top\in\mathcal{D}$. The following definition is central to most of the results of this paper.

Definition 9**.**

Let X be a subset of $\{$M, M0, P, P0, R$\}$. We say that $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ is the ILX-structure for a set of formulas $\mathcal{D}$ if:

[TABLE]

Lemma 10**.**

If $\text{{IL}{X}}\nvdash\neg A$ then there is an ILX-MCS $w$ such that $A\wedge\square\neg A\in w.$

Proof.

We are to show that $\{A\wedge\square\neg A\}$ is an ILX-consistent set. Suppose $A,\square\neg A\vdash\bot.$ It follows that $\vdash\square\neg A\rightarrow\neg A.$ Applying generalization (necessitation) gives $\vdash\square(\square\neg A\rightarrow\neg A).$ The Löb axiom implies $\vdash\square\neg A.$ Now, $\vdash\square\neg A$ and $A,\square\neg A\vdash\bot$ imply $A\vdash\bot,$ i.e. $\vdash\neg A,$ a contradiction. ∎

Lemma 11**.**

Let X be a subset of $\{$M, M0, P, P0, R$\}$. The ILX-structure $\mathfrak{M}$ for a set formula $\mathcal{D}$ is a generalized Veltman model. Furthermore, the following holds:

[TABLE]

for all $G\in\mathcal{D}$ and $w\in W.$

Proof.

Let us verify that the ILX-structure $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ for $\mathcal{D}$ is a generalized Veltman model. Since $\text{{IL}{X}}\nvdash\bot$ and $\top\in\mathcal{D}$, Lemma 10 implies $W\neq\emptyset$ .

Transitivity of $R$ is immediate. To see converse well-foundedness, assume there are more than $|\mathcal{D}|$ worlds in an $R$-chain. Then there are $x$ and $y$ with $xRy$ and for some $G\in\mathcal{D}$, $G,\square\neg G\in x,y$. However, $\square\neg G\in x$ and $G\in y$ obviously contradict the assumption that $xRy$ ($x\prec y$).

Next, let us prove properties of $S_{w}$ for $w\in W$. Clearly $S_{w}\subseteq R[w]\times\mathcal{P}(R[w])$. If $xS_{w}V$, then $w\prec_{\emptyset}x$ implies there is at least one element $v$ in $V$ (with $w\prec_{\emptyset}v$). Quasi-reflexivity and monotonicity are obvious. Next, assume $wRxRu$ and $w\prec_{S}x$. Lemma 6 and $w\prec_{S}x\prec u$ imply $w\prec_{S}u.$ Thus $xS_{w}\{u\}$. It remains to prove quasi-transitivity. Assume $xS_{w}V$ and $vS_{w}U_{v}$ for all $v\in V$. Put $U=\bigcup_{v}U_{v}$. We claim that $xS_{w}U$. We have $U\subseteq R[w]$. Assume $w\prec_{S}x$. This and $xS_{w}V$ imply there is $v\in V$ such that $w\prec_{S}v$. This and $vS_{w}U_{v}$ imply there is $u\in U_{v}$ (thus also $u\in U$) such that $w\prec_{S}u$.

Let us prove the truth lemma with respect to formulas contained in $\mathcal{D}$. The claim is proved by induction on the complexity of $G\in\mathcal{D}.$ We will only consider the case $G=B\rhd C.$

Assume $B\rhd C\in w,$  $wRu$ and $u\Vdash B.$ Induction hypothesis implies $B\in u.$ We claim that $uS_{w}[C]_{w}$. Clearly $[C]_{w}\subseteq R[w]$. Assume $w\prec_{S}u$. Lemma 8 implies there is an ILX-MCS $v$ with $w\prec_{S}v$ and $C,\square\neg C\in v$ (thus also $wRv$ and $v\in W$). Induction hypothesis implies ${\mathfrak{M}},v\Vdash C$.

To prove the converse, assume $B\rhd C\notin w$. Lemma 7 implies there is $u$ with $w\prec_{\{\neg C\}}u$ and $B,\square\neg B\in u$ (thus $u\in W$). It is immediate that $wRu$ and the induction hypothesis implies that $u\Vdash B$. Assume $uS_{w}V.$ We are to show that $V\nVdash C$. Since $w\prec_{\{\neg C\}}u$ and $uS_{w}V$, there is $v\in V$ such that $w\prec_{\{\neg C\}}v$. Lemma 6 implies $\neg C\in v.$ The induction hypothesis implies $v\nVdash C$; thus $V\nVdash C$. ∎

Theorem 12**.**

Let $X\subseteq\{$M, M0, P, P0, R$\}$. Assume that for every set $\mathcal{D}$ the ILX-structure for $\mathcal{D}$ possesses the property (X)gen. Then ILX is complete w.r.t. ILgenX-models.

Proof.

Let $A$ be a formula such that $\nvdash\neg A$. Lemma 10 implies there is an ILX-MCS $w$ such that $A\wedge\square\neg A\in w.$ Let $\mathcal{D}$ have the usual properties, and contain $A$. Let ${\mathfrak{M}}=(W,R,\{S_{w}:w\in W\},\Vdash)$ be the ILX-structure for $\mathcal{D}$. Since $A\wedge\square\neg A\in w$ and $A\in\mathcal{D}$, we have $w\in W$. Lemma 11 implies $\mathfrak{M},w\nVdash\neg A.$ ∎

Corollary 13**.**

The logic IL is complete w.r.t. generalized Veltman models.

Note that the method presented in [14] now implies completeness of IL w.r.t. ordinary Veltman models. Unfortunately, this method does not preserve characteristic properties in general.

In the following sections we prove (or reprove) the completeness of the following logics w.r.t. generalized semantics: ILM, ILM0, ILP, ILP0, ILR, ILW and ILW*.

2.1 The logic ILM

Completeness of the logic ILM w.r.t. generalized semantics is an easy consqeuence of the completeness of ILM w.r.t. ordinary semantics, first proved by de Jongh and Veltman [3]. Another proof of the same result was given by Goris and Joosten, using the construction method in [6].

Verbrugge determined the characteristic property (M)gen in 1992. in an unpublished paper:

[TABLE]

Lemma 14** ([2], Lemma 3.7).**

Let $w$ and $u$ be some ILM-MCS’s, and let $S$ be a set of formulas. If $w\prec_{S}u$ then $w\prec_{S\cup u_{\emptyset}^{\square}}u$.

Theorem 15**.**

The logic ILM is complete w.r.t. ILgenM-models.

Proof.

Given Theorem 12, it suffices to show that for any set $\mathcal{D}$, the ILM-structure for $\mathcal{D}$ possesses the property (M)gen. Let $(W,R,\{S_{w}:w\in W\},\Vdash)$ be the ILM-structure for $\mathcal{D}.$

Let $uS_{w}V$ and take $V^{\prime}=\{v\in V:w\prec_{u_{\emptyset}^{\square}}v\}$. We claim $uS_{w}V^{\prime}$ and $R[V^{\prime}]\subseteq R[u].$ Suppose $w\prec_{S}u$. Lemma 14 implies $w\prec_{S\cup u_{\emptyset}^{\square}}u.$ Since $uS_{w}V$, we have that there is $v\in V$ with $w\prec_{S\cup u_{\emptyset}^{\square}}v.$ So, $v\in V^{\prime}$. Thus, $uS_{w}V^{\prime}$.

Now let $v\in V^{\prime}$ and $z\in W$ be such that $vRz$. Since $v\in V^{\prime}$, $w\prec_{u_{\emptyset}^{\square}}v$. Then for all $\square B\in u$ we have $\square B\in v.$ Since $vRz$, we have $B,\square B\in z$. So, $u\prec z$ i.e.  $uRz$. ∎

2.2 The logic ILM0

Modal completeness of ILM0 w.r.t. Veltman semantics was proved in [5] by E. Goris and J. J. Joosten. Certain difficulties encountered in this proof were our main motivation for using generalized Veltman semantics. We will sketch these difficulties and show in what way does generalized semantics overcome them.

Characteristic property (M0)gen (see [7]):

[TABLE]

Lemma 16** ([2], Lemma 3.9).**

Let $w$, $u$ and $x$ be ILM0-MCS’s, and $S$ an arbitrary set of formulas. If $w\prec_{S}u\prec x$ then $w\prec_{S\cup u_{\emptyset}^{\square}}x$.

To motivate our proving of completeness (of ILM0, but also in general) w.r.t. generalized semantics, let us sketch a situation for which there are clear benefits in working with generalized semantics. We do this only now because ILM0 is sufficiently complex to display (some of) these benefits. Suppose we are building models step by step (as in the construction method [5]), and worlds $w$, $u_{1}$, $u_{2}$ and $x$ occur in the configuration displayed in Figure 1. Furthermore, suppose we need to produce an $S_{w}$-successor $v$ of $x$.

With ordinary semantics, we need to ensure that for our $S_{w}$-successor $v$, for each $\square B_{1}\in u_{1}$ and $\square B_{2}\in u_{2}$, we have $\square B_{1},\square B_{2}\in v$. It is not obvious that such construction is possible. In case of ILM0, it was successfully solved in [5] by preserving the invariant that sets of boxed formulas in $u_{i}$ are linearly ordered. This way, finite (quasi-)models can always be extended by only looking at the last $u_{i}$. With generalized semantics, we need to produce a whole set of worlds $V$, but the requirements on each particular world are less demanding. For each $u_{i}$, there has to be a corresponding $V_{i}\subseteq V$ with $\square B_{i}$ contained (true) in every world of $V_{i}$. Lemma 16 gives a recipe for producing such worlds.

Theorem 17**.**

The logic ILM0 is complete w.r.t. ILgenM0-models.

Proof.

Given Theorem 12, it suffices to show that for any set $\mathcal{D}$, the ILM0-structure for $\mathcal{D}$ possesses the property (M0)gen. Let $(W,R,\{S_{w}:w\in W\},\Vdash)$ be the ILM0-structure for $\mathcal{D}.$

Assume $wRuRxS_{w}V$ and take $V^{\prime}=\{v\in V:w\prec_{u_{\emptyset}^{\square}}v\}$. We claim that $uS_{w}V^{\prime}$ and $R[V^{\prime}]\subseteq R[u]$. Obviously $V^{\prime}\subseteq V\subseteq R[w].$ Assume $w\prec_{S}u$. Lemma 16 and $w\prec_{S}u\prec x$ imply $w\prec_{S\cup u_{\emptyset}^{\square}}x$. Now $xS_{w}V$ and the definition of $S_{w}$ imply there is $v\in V$ such that $w\prec_{S\cup u_{\emptyset}^{\square}}v$. Lemma 6 implies $w\prec_{u^{\square}_{\emptyset}}v.$ So, $v\in V^{\prime}.$

It remains to verify that $R[V^{\prime}]\subseteq R[u]$. Let $v\in V^{\prime}$ and $z\in W$ be worlds such that $vRz$. Since $w\prec_{u_{\emptyset}^{\square}}v$, for all $\square B\in u$ we have $\square B\in v$, and since $vRz$, it follows that $\square B,B\in z$. Thus, $u\prec z$ i.e.  $uRz.$ ∎

2.3 The logic ILP

As in the case of the logic ILM, the completeness of ILP w.r.t. generalized semantics is an easy consqeuence of the completeness of ILP w.r.t. ordinary semantics, first proved by de Jongh and Veltman [3].

Verbrugge determined the characteristic property (P)gen in 1992. in an unpublished paper:

[TABLE]

Lemma 18** ([2] Lemma 3.8).**

Let $w$, $x$ and $u$ be some ILP-MCS’s, and let $S$ and $T$ be arbitrary sets of formulas. If $w\prec_{S}x\prec_{T}u$ then $w\prec_{S\cup x_{T}^{\boxdot}}u$.

Theorem 19**.**

The logic ILP is complete w.r.t. ILgenP-models.

Proof.

Given Theorem 12, it suffices to show that for any set $\mathcal{D}$, the ILP-structure for $\mathcal{D}$ possesses the property (P)gen. Let $(W,R,\{S_{w}:w\in W\},\Vdash)$ be the ILP-structure for $\mathcal{D}.$

Let $wRw^{\prime}RuS_{w}V$ and take $V^{\prime}=V\cap R[w^{\prime}]$. We claim $uS_{w^{\prime}}V^{\prime}.$ Let $T$ be arbitrary such that $w^{\prime}\prec_{T}u$. Lemma 18 and $w\prec_{\emptyset}w^{\prime}\prec_{T}u$ imply $w\prec_{{w^{\prime}}_{T}^{\boxdot}}u.$ Now, $uS_{w}V$ implies that there is a $v\in V$ with $w\prec_{{w^{\prime}}_{T}^{\boxdot}}v$. Let $A\rhd\neg\bigwedge T^{\prime}\in w^{\prime}$ for some finite $T^{\prime}\subseteq T$. Then $\neg A,\square\neg A\in{w^{\prime}}_{T}^{\boxdot}$. Lemma 6 and $w\prec_{{w^{\prime}}_{T}^{\boxdot}}v$ imply $\neg A,\square\neg A\in v$. Thus $w^{\prime}\prec_{T}v$. Finally, $V^{\prime}\subseteq R[w^{\prime}]$ holds by assumption, thus $uS_{w^{\prime}}V^{\prime}$. ∎

2.4 The logic ILP0

The interpretability principle $\textsf{P\textsubscript{0}}=A\rhd\Diamond B\to\square(A\rhd B)$ is introduced in J. J. Joosten’s master thesis in 1998.

In [6] it is shown that the interpretability logic ILP0 is incomplete w.r.t. Veltman models. Since we will show that ILP0 is complete w.r.t. generalized semantics, this is the first example of an interpretability logic complete w.r.t. the generalized semantics, but incomplete w.r.t. ordinary semantics.

Characteristic property (P0)gen was determined in [6] . A slightly reformulated version:

[TABLE]

The following technical lemma is almost obvious.

Lemma 20**.**

Let $x$ be an ILX-MCS, $A$ a formula, and $T$ a finite set of formulas. Let $B_{G}$ be an arbitrary formula, and $T_{G}$ an arbitrary finite set of formulas, for every $G\in T$. Furthermore, assume:

• a)

$A\rhd\bigvee_{G\in T}B_{G}\in x$;

• b)

$(\forall G\in T)\ B_{G}\rhd\bigvee_{H\in T_{G}}\neg H\in x$.

Then we have $A\rhd\bigvee_{H\in S^{\prime}}\neg H\in x$, where $S^{\prime}=\bigcup_{G\in T}T_{G}$.

Proof.

Let $G\in T$. Since $T_{G}\subseteq S^{\prime}$, clearly $\vdash\bigvee_{H\in T_{G}}\neg H\rhd\bigvee_{H\in S^{\prime}}\neg H$. The requirement b) and the axiom (J2) imply $B_{G}\rhd\bigvee_{H\in S^{\prime}}\neg H\in x$. Now $|T|-1$ applications of the axiom (J3) give $\bigvee_{G\in T}B_{G}\rhd\bigvee_{H\in S^{\prime}}\neg H\in x$. Finally, apply the requirement a) and the axiom (J2). ∎

Next we need a labelling lemma for ILP0. This is where we use the technical lemma above.

Lemma 21**.**

Let $w$, $x$ and $u$ be some ILP0-MCS’s, and let $S$ be a set of formulas. If $w\prec x\prec_{S}u$ then $w\prec_{x_{S}^{\square}}u$.

Proof.

Let $A$ be an arbitrary formula. Let $T\subseteq x_{S}^{\square}$ be a finite set such that $A\rhd\bigvee_{G\in T}\neg G\in w.$ We will prove that $\neg A,\square\neg A\in u.$ If $G\in T\ (\subseteq x_{S}^{\square})$, then $G=\square\neg B_{G}$, for some formula $B_{G}$. Thus $A\rhd\bigvee_{G\in T}\neg\square\neg B_{G}\in w$, and by easy inferences and maximal consistency: $A\rhd\bigvee_{G\in T}\Diamond B_{G}\in w$, and $A\rhd\Diamond\bigvee_{G\in T}B_{G}\in w$. Applying P0 gives $\square(A\rhd\bigvee_{G\in T}B_{G})\in w$. The assumption $w\prec x$ implies $A\rhd\bigvee_{G\in T}B_{G}\in x.$ For each $G\in T\ (\subseteq x_{S}^{\square})$ there is a finite subset $T_{G}$ of $S$ such that $B_{G}\rhd\bigvee_{H\in T_{G}}\neg H\in x.$ Let $S^{\prime}=\bigcup_{G\in T}T_{G}.$ Clearly $S^{\prime}$ is a finite subset of $S.$ Lemma 20 implies $A\rhd\bigvee_{H\in S^{\prime}}\neg H\in x.$ Finally, $S^{\prime}\subseteq S$ and the assumption $x\prec_{S}u$ imply $\neg A,\square\neg A\in u.$ ∎

The following simple observation is useful both for ILP0 and ILR.

Lemma 22**.**

Let $w$, $x$, $v$ and $z$ be some ILX-MCS’s, and let $S$ be a set of formulas. If $w\prec_{x_{S}^{\square}}v\prec z$ then $x\prec_{S}z$.

Proof.

Let $S^{\prime}$ be a finite subset of $S$ with $A\rhd\bigvee_{G\in S^{\prime}}\neg G\in x$. Then $\square\neg A\in x_{S}^{\square}$. Now $w\prec_{x_{S}^{\square}}v$ and Lemma 6 imply $\square\neg A\in v$. Since $v\prec z$, we have $\neg A,\square\neg A\in z$. ∎

Theorem 23**.**

The logic ILP0 is complete w.r.t. ILgenP0-frames.

Proof.

Given Theorem 12, it suffices to show that for any set $\mathcal{D}$, the ILP0-structure for $\mathcal{D}$ possesses the property (P0)gen. Let $(W,R,\{S_{w}:w\in W\},\Vdash)$ be the ILP0-structure for $\mathcal{D}.$

Assume $wRxRuS_{w}V$ and $R[v]\cap Z\neq\emptyset$ for each $v\in V$. We will prove that there is $Z^{\prime}\subseteq Z$ such that $uS_{x}Z^{\prime}$.

Let $S$ be a set of formulas such that $w\prec x\prec_{S}u$. Lemma 21 implies $w\prec_{x_{S}^{\square}}u.$ Since $uS_{w}V$, there is $v\in V$ such that $w\prec_{x_{S}^{\square}}v.$ Since $R[v]\cap Z\neq\emptyset$, choose a world $z_{S}\in R[v]\cap Z$. Now $w\prec_{x_{S}^{\square}}v\prec z_{S}$ and Lemma 22 imply $x\prec_{S}z_{S}$. Put $Z^{\prime}=\{z_{S}:S$ is a set of formulas such that $x\prec_{S}u\}$. Clearly $Z^{\prime}\subseteq Z.$ So, $Z^{\prime}\subseteq R[x]$, and since for each set $S$ such that $x\prec_{S}u$ we have $x\prec_{S}z_{S}$, it follow that $uS_{x}Z^{\prime}.$ ∎

In [14] a possibility was explored of transforming a generalized Veltman model to an ordinary Veltman model, such that these two models are bisimilar (in some aptly defined sense). A natural question is whether such transformation exists if we add the requirement that characteristic properties are preserved. The example of ILP0 shows that there are ILgenP0-models with no (bisimilar or otherwise) counterpart ILP0-models.

2.5 The logic ILR

Completeness of ILR w.r.t. ordinary Veltman semantics is an open problem (see [2]), but completeness w.r.t. the generalized semantics is not yet resolved either. In this section we will prove that ILR is complete w.r.t. the generalized semantics.

Characteristic property (R)gen was determined in [6]. A slightly reformulated version:

[TABLE]

where $\mathcal{C}(x,u)=\{C\subseteq R[x]:(\forall Z)(uS_{x}Z\Rightarrow Z\cap C\neq\emptyset)\}$ is the family of “choice sets”.

Lemma 24** ([2], Lemma 3.10).**

Let $w$, $x$ and $u$ be some ILR-MCS’s, and let $S$ and $T$ be arbitrary sets of formulas. If $w\prec_{S}x\prec_{T}u$ then $w\prec_{S\cup x_{T}^{\square}}u.$

Theorem 25**.**

The logic ILR is complete w.r.t. ILgenR-models.

Proof.

Given Theorem 12, it suffices to show that for any set $\mathcal{D}$, the ILR-structure for $\mathcal{D}$ possesses the property (R)gen. Let $(W,R,\{S_{w}:w\in W\},\Vdash)$ be the ILR-structure for $\mathcal{D}.$

Assume $wRxRuS_{w}V$ and $C\in\mathcal{C}(x,u)$. We are to show that $(\exists U\subseteq V)(xS_{w}U\ \&\ R[U]\subseteq C)$. We will first prove an auxiliary claim:

[TABLE]

So, let $S$ be arbitrary such that $w\prec_{S}x$, and suppose (for a contradiction) that for every $v\in V$ with $w\prec_{S\cup x_{\emptyset}^{\square}}v$, we have $R[v]\nsubseteq C$, that is, there is some $z_{v}\in R[v]\setminus C$. Let $Z=\{z_{v}:v\in V,w\prec_{S\cup x_{\emptyset}^{\square}}v\}.$ We claim that $uS_{x}Z$. Let $T$ be arbitrary such that $x\prec_{T}u$, and we should prove that there exists $z\in Z$ such that $x\prec_{T}z.$ From $w\prec_{S}x\prec_{T}u$ and Lemma 24 it follows that $w\prec_{S\cup x_{T}^{\square}}u$. Since $uS_{w}V$, there is $v\in V$ with $w\prec_{S\cup x_{T}^{\square}}v$. Now, $x_{\emptyset}^{\square}\subseteq x_{T}^{\square}$ and Lemma 6 imply $w\prec_{S\cup x_{\emptyset}^{\square}}v$, so there is a world $z_{v}\in Z$ as defined earlier. Furthermore, $w\prec_{x_{T}^{\square}}v\prec z_{v}$ and Lemma 22 imply $x\prec_{T}z_{v}$. To prove $uS_{x}Z$ it remains to verify that $Z\subseteq R[x]$. Let $z_{v}\in Z$ be arbitrary and apply Lemma 6 and Lemma 22 as before. Now, $uS_{x}Z$ and $C\in\mathcal{C}(x,u)$ imply $C\cap Z\neq\emptyset$, contradicting the definition of $Z$. This concludes the proof of the auxiliary claim.

Let $U=\{v\in V:w\prec_{x_{\emptyset}^{\square}}v\ \mbox{ and }\ R[v]\subseteq C\}$. Auxiliary claim implies $U\neq\emptyset$. If $w\prec_{S}x$, auxiliary claim implies there is $v\in U$ such that $w\prec_{S\cup x_{\emptyset}^{\square}}v$ and $R[v]\subseteq C$, so $v\in U$. Thus $xS_{w}U$. It is clear that $R[U]\subseteq C$. ∎

3 The logics ILW and ILW*

To prove that ILW is complete, one could try to find a sufficiently strong “labelling lemma” and utilise Definition 9. One candidate might be the following condition:

[TABLE]

where $\mathcal{D}$ is finite, closed under subformulas and such that each $w\in W$ contains $A_{w}$ and $\Box\neg A_{w}$ for some $A_{w}\in\mathcal{D}$.

Since we weren’t successful in finding a sufficiently strong labelling lemma for ILW, we will use a modified version of Definition 9 to work with ILW and its extensions. This way we won’t require a labelling lemma, but we lose generality in the following sense. To prove the completeness of ILXW, for some $X$, it no longer suffices to simply show that the structure defined in Definition 9 has the required characteristic property (when each world is an ILX-MCS). Instead, the characteristic property of ILX has to be shown to hold even on the modified structure. So, to improve compatibility with proofs based on Definition 9, we should prove the completeness of ILW with as similar definition to Definition 9 as possible. That is what we do in the remainder of this section. This approach turns out to be good enough for ILW* (ILM0W). We didn’t succeed in using it to prove the completeness of ILRW. However, to the best of our knowledge, ILRW might not be complete at all.

In [6] the (complement of the) characteristic class for ILW is given by the condition Not-W such that for any generalized Veltman frame $\mathfrak{F}$ we have that

[TABLE]

Another condition is (W)gen from [7]:

[TABLE]

We will use (this formulation of) (W)gen in what follows. In the proof of the completeness of logic ILW we will use the following two lemmas. In what follows, ILWX denotes an arbitrary extension of ILW.

Lemma 26** ([2], Lemma 3.12).**

Let $w$ be an ILWX-MCS, and $B$ and $C$ formulas such that $\neg(B\rhd C)\in w.$ Then there is an ILWX-MCS $u$ such that $w\prec_{\{\square\neg B,\neg C\}}u$ and $B\in u.$

Lemma 27** ([2], Lemma 3.13).**

Let $w$ and $u$ be some ILWX-MCS, $B$ and $C$ some formulas, and $S$ a set of formulas such that $B\rhd C\in w,$ $w\prec_{S}u$ and $B\in u.$ Then there is an ILWX-MCS $v$ such that $w\prec_{S\cup\{\square\neg B\}}v$ and $C,\square\neg C\in v.$

Given a binary relation $R$, let $\dot{R}[x]=R[x]\cup\{x\}.$

Definition 28**.**

Let X be W or W*. We say that $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ is the ILX-structure for a set of formulas $\mathcal{D}$ if:

[TABLE]

Lemma 29**.**

Let X be W or W. ILX-structure $\mathfrak{M}$ for $\mathcal{D}$ is a generalized Veltman model. Furthermore, the following holds:*

[TABLE]

for each $G\in\mathcal{D}$ and $w\in W.$

Proof.

Let us first verify that the ILX-structure $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ for $\mathcal{D}$ is a generalized Veltman model. All the properties, except for quasi-transitivity, have easy proofs (see the proof of Lemma 11).

Let us prove quasi-transitivity. Assume $uS_{w}V$, and $vS_{w}U_{v}$ for all $v\in V$. Put $U=\bigcup_{v\in V}U_{v}$. We claim that $uS_{w}U$. Clearly $U\subseteq R[w]$. To prove $uS_{w}U$ we will distinguish the cases (a) and (b) from the definition of the relation $S_{w}$ for $uS_{w}V.$

In the case (a), we have $v_{0}\in V$ for some $v_{0}\in\dot{R}[u]$. We will next distinguish two cases from the definition of $v_{0}S_{w}U_{v_{0}}$.

In the case (aa) we have $x\in U_{v_{0}}$ for some $x\in\dot{R}[v_{0}]$. Since $v_{0}\in\dot{R}[u]$, we then have $x\in\dot{R}[u]$. Since $x\in U_{v_{0}}\subseteq U$, then $U\cap\dot{R}[u]\neq\emptyset.$ So, we have $uS_{w}U$, as required.

In the case (ab) we have:

[TABLE]

To prove $uS_{w}U$ in this case, we will use the case (b) from the definiton of the relation $S_{w}$. Assume $w\prec_{S}u$. Then we have $w\prec_{S}u\prec v_{0}$ or $w\prec_{S}u=v_{0}$. Either way, possibly using Lemma Lemma 6, we have $w\prec_{S}v_{0}$, and so there are $x\in U_{v_{0}}$ and $G\in\mathcal{D}\cap\bigcup\dot{R}[v_{0}]$ with $w\prec_{S\cup\{\square\neg G\}}x$. Since $uRv_{0}$ or $u=v_{0}$, we have $\dot{R}[v_{0}]\subseteq\dot{R}[u].$ So, the claim follows.

In the case (b), we have:

[TABLE]

To prove $uS_{w}U$ we will use the case (b) from the definition of the relation $S_{w}$. Assume $w\prec_{S}u$. Then there are $v_{0}\in V$ and $G\in\mathcal{D}\cap\bigcup\dot{R}[u]$ such that $w\prec_{S\cup\{\square\neg G\}}v_{0}$. From $v_{0}\in V$ it follows that $v_{0}S_{w}U_{v_{0}}$. We will next distinguish the possible cases in the definition of $v_{0}S_{w}U_{v_{0}}$.

In the first case (ba) we have $U_{v_{0}}\cap\dot{R}[v_{0}]\neq\emptyset,$ i.e. there is $x\in U_{v_{0}}\cap\dot{R}[v_{0}]$. Then $w\prec_{S\cup\{\square\neg G\}}v_{0}=x$ or $w\prec_{S\cup\{\square\neg G\}}v_{0}\prec x$. In both cases (possibly using Lemma 6) we have $w\prec_{S\cup\{\square\neg G\}}x$.

In the case (bb):

[TABLE]

From $w\prec_{S\cup\{\square\neg G\}}v_{0}$ it follows that there are some $x\in U_{v_{0}}$ and $G^{\prime}\in\mathcal{D}\cap\bigcup\dot{R}[v_{0}]$ such that $w\prec_{S\cup\{\square\neg G,\square\neg G^{\prime}\}}x$. Lemma 6 implies $w\prec_{S\cup\{\square\neg G\}}x$, as required.

We claim that for each formula $G\in\mathcal{D}$ and each world $w\in W$ the following holds:

[TABLE]

The claim is proved by induction on the complexity of $G$. The only non-trivial case is when $G=B\rhd C.$

Assume $B\rhd C\in w,$  $wRu$ and $u\Vdash B$. Induction hypothesis implies $B\in u.$ We claim that $uS_{w}[C]_{w}$. Clearly $[C]_{w}\subseteq R[w]$. Assume $w\prec_{S}u$. Lemma 27 implies that there is an ILX-MCS $v$ with $w\prec_{S\cup\{\square\neg B\}}v$ and $C,\square\neg C\in v$ (thus $v\in W$). Since $C\in v$, the induction hypothesis implies $v\Vdash C.$ Since $w\prec v$, i.e. $wRv$, then $v\in[C]_{w}$. Now, $B\in\mathcal{D}$ and $B\in u$ imply $B\in\mathcal{D}\cap\bigcup\dot{R}[u]$. Thus $uS_{w}[C]_{w}$ holds, by the clause (b) from the definition.

To prove the converse, assume $B\rhd C\notin w$. Since $w$ is ILX-MCS, $\neg(B\rhd C)\in w.$ Lemma 26 implies there is $u$ with $w\prec_{\{\square\neg B,\neg C\}}u$ and $B\in u.$ Lemma 6 implies $\square\neg B\in u.$ So, $B\wedge\square\neg B\in u$; thus $u\in W.$ The induction hypothesis implies $u\Vdash B$. Let $V\subseteq R[w]$ be such that $uS_{w}V$. We will find a world $v\in V$ such that $w\prec_{\{\neg C\}}v$. We will distinguish the cases (a) and (b) from the definition of the relation $S_{w}$. Consider the case (a). Let $v$ be an arbitrary node in $V\cap\dot{R}[u]$. If $v=u$, clearly $w\prec_{\{\square\neg B,\neg C\}}v$. If $uRv$, then we have $w\prec_{\{\square\neg B,\neg C\}}u\prec v.$ Lemma 6 implies $w\prec_{\{\square\neg B,\neg C\}}v$. Consider the case (b). From $w\prec_{\{\square\neg B,\neg C\}}u$ and the definition of $S_{w}$ it follows that there is $v\in V$ and a formula $D\in\mathcal{D}$ such that $w\prec_{\{\square\neg B,\neg C,\square\neg D\}}v$. In both cases we have $w\prec_{\{\neg C\}}v$; thus $C\notin v$. Induction hypothesis implies $v\nVdash C$; whence $V\nVdash C$, as required. ∎

Theorem 30**.**

The logic ILW is complete w.r.t. ILgenW-models.

Proof.

In the light of Lemma 29, it suffices to show that the ILW-structure $\mathfrak{M}$ for $\mathcal{D}$ possesses the property (W)gen. Recall the characteristic property (W)gen:

[TABLE]

Suppose for a contradiction that there are $w$, $u$ and $V$ such that:

[TABLE]

Let $\mathcal{V}$ denote all such sets $V$ (for arbitrary but fixed $w$ and $u$).

Let $n=2^{|\mathcal{D}|}$. Fix any enumeration $\mathcal{D}_{0},\dots,\mathcal{D}_{n-1}$ of $\mathcal{P}(\mathcal{D})$ that satisfies $\mathcal{D}_{0}=\emptyset.$ Denote a new relation $S_{w}^{i}$ for all $0\leq i<n$, $y\in W$ and $U\subseteq W$ as follows:

[TABLE]

Let $y\in W$ and $U\subseteq W$ be arbitrary. Let us prove that $yS_{w}U$ implies the following:

[TABLE]

If $yS_{w}U$ holds by (a) from the definition of $S_{w}$, the set $U\cap\dot{R}[y]$ is non-empty. Pick arbitrary $z\in U\cap\dot{R}[y]$ and put $U^{\prime}=\{z\}$. We have either $wRyRz$ or $y=z$. If $wRyRz$, we have $yS_{w}\{z\}.$ Otherwise $y=z$. Now quasi-reflexivity implies $yS_{w}\{z\}.$ Since $y\in W$, there is a formula $G\in\mathcal{D}$ such that $G\wedge\square\neg G\in y$. Fix $i<n$ such that $\mathcal{D}_{i}=\{G\}.$ Clearly $\mathcal{D}_{i}\subseteq\bigcup\dot{R}[y].$ Since $z\in U$ and $yS_{w}U$, clearly $U^{\prime}\subseteq R[w]$. Since $yRz$, we also have $\square\neg G\in z$. Truth lemma implies $U^{\prime}\Vdash\square\neg G$; since if $zRt$, $G\notin t$, (truth lemma is applied here) $t\nVdash G$, so $z\Vdash\square\neg G$. Thus $U^{\prime}\subseteq[\square\neg G]_{w}$, and $yS_{w}^{i}U^{\prime}$.

If $yS_{w}U$ holds by (b) from the definition of $S_{w}$, take:

[TABLE]

[TABLE]

In other words, $U^{\prime}$ is the image of the mapping that is implicitly present in the definition of the relation $S_{w}$ (clause (b)): for each $S$, pick a world $v_{S}$ (to be included in $U^{\prime}$), and a formula $G_{S}$ (to be included in $\mathcal{D}_{i}$).

Let $m<n$ be maximal such that there are $U\in\mathcal{V}$ and $U^{\prime}\subseteq U$ with the following properties:

1. (i)

$(\forall x\in U)[(\exists y\in R[x])(\exists Z\subseteq U)(\exists i\leq m)\ yS_{w}^{i}Z\Rightarrow x\notin U^{\prime}]$; 2. (ii)

$(\forall x\in W)(xS_{w}U\Rightarrow xS_{w}U^{\prime})$.

Since $\mathcal{D}_{0}=\emptyset,$ we have $[\bigvee_{G\in\mathcal{D}_{0}}\square\neg G]_{w}=[\bot]_{w}=\emptyset.$ So there are no $Z\subseteq[\bigvee_{G\in\mathcal{D}_{0}}\square\neg G]_{w}$ such that $yS_{w}Z$ for some $y\in W$. So, if we take $m=0$ and $U^{\prime}=U$ for any $U\in\mathcal{V}$, (i) and (ii) are trivially satisfied.

Since $n$ is finite and conditions (i) and (ii) are satisfied for at least one value $m$, there must be a maximal $m<n$ with the required properties.

Let us first prove that $m<n-1$. Assume the opposite, that is, $m=n-1$. Then there are $U\in\mathcal{V}$ and $U^{\prime}\subseteq U$ such that the conditions (i) and (ii) are satisfied for $m=n-1.$ Since $U\in\mathcal{V}$, we have $uS_{w}U.$ The condition (ii) implies $uS_{w}U^{\prime}.$ Now $U\in\mathcal{V},$  $U^{\prime}\subseteq U$ and $uS_{w}U^{\prime}$ imply $R[U^{\prime}]\cap S_{w}^{-1}[U]\neq\emptyset$. Thus there are $x\in U^{\prime}$ and $y\in R[x]$ such that $yS_{w}U$. Now (ii) implies $yS_{w}U^{\prime}$. The earlier remark (2) implies that there is $Z\subseteq U^{\prime}$ and $i<n$ such that $yS_{w}^{i}Z.$ Since $m=n-1$, it follows that $i\leq m.$ The condition (i) implies $x\not\in U^{\prime},$ a contradiction. Thus $m<n-1$.

Let us now prove that $m$ is not maximal, by showing that $m+1$ satisfies (i) and (ii). Let $U\in\mathcal{V}$ and $U^{\prime}\subseteq U$ be some sets such that the conditions (i) and (ii) are satisfied for $m$. Denote:

[TABLE]

Let us prove that $m+1$ also satisfies (i) and (ii) with $U^{\prime}$ instead of $U$, and $U^{\prime}\setminus Y$ instead of $U^{\prime}$. We should first show that $U^{\prime}\in\mathcal{V}$. So, suppose that $uS_{w}T\subseteq U^{\prime}$. Now, $T\subseteq U^{\prime}\subseteq U$ and $U\in\mathcal{V}$ imply that there are some $v\in T$ and $z\in R[v]$ such that $zS_{w}U$. The property (ii) for $m$ (with sets $U$ and $U^{\prime}$) implies $zS_{w}U^{\prime}$. So, $R[T]\cap S_{w}^{-1}[U^{\prime}]\neq\emptyset$, as required.

Now let us verify the property (i) for the newly defined sets. Let $x\in U^{\prime},y\in R[x],Z\subseteq U^{\prime},i\leq m+1$ be arbitrary such that $yS_{w}^{i}Z$. If $i\leq m$, the property (i) for $m$ implies $x\notin U^{\prime}$, so in particular, $x\notin U^{\prime}\setminus Y$. If $i=m+1$, then $x\in Y$. Thus $x\notin U^{\prime}\setminus Y$ and the conditon (i) is satisfied.

It remains to prove (ii). Take arbitrary $x\in W$ such that $xS_{w}U^{\prime}$. For every $y\in Y$, the definition of $Y$ implies existence of some $z_{y}\in R[y]$ and $U_{y}\subseteq U^{\prime}$ such that $z_{y}S_{w}^{m+1}U_{y}$. From the definiton of the relation $S_{w}^{m+1}$ we have $\mathcal{D}_{m+1}\subseteq\bigcup\dot{R}[z_{y}].$ Now, $yRz_{y}$ and the truth lemma imply $y\Vdash\Diamond G,$ for each $G\in\mathcal{D}_{m+1}.$ From the definiton of the relation $S_{w}^{m+1}$ and $z_{y}S_{w}^{m+1}U_{y}$ we have $U_{y}\subseteq[\bigvee_{G\in\mathcal{D}_{m+1}}\square\neg G]_{w}.$ So, the following holds:

[TABLE]

for all $y\in Y$. Thus, $U_{y}\cap Y=\emptyset,$ for every $y\in Y.$ For every $y\in U^{\prime}\setminus Y$ put $U_{y}=\{y\}$. Again, $U_{y}\cap Y=\emptyset$. Note that $\bigcup_{y\in U^{\prime}}U_{y}=U^{\prime}\setminus Y$. Now $xS_{w}U^{\prime}$ and quasi-transitivity imply $xS_{w}U^{\prime}\setminus Y$.

The fact that (i) and (ii) hold for $m+1$ contradicts the maximality of $m$. ∎

Goris and Joosten proved in [5] the completeness of ILW* (ILWM0) w.r.t. ordinary Veltman semantics.

Theorem 31**.**

The logic ILW is complete w.r.t. ILgenW*-models.*

Proof.

With Lemma 29, it suffices to prove the ILW*-structure for $\mathcal{D}$ possesses the properties (W)gen and (M0)gen, for each appropriate $\mathcal{D}.$ So, let $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ be the ILW*-structure for $\mathcal{D}$. Theorem 30 shows that the model $\mathfrak{M}$ possesses the property (W)gen. It remains to show that it posesses the property (M0)gen.

Assume $wRuRxS_{w}V$. We claim that there is $V^{\prime}\subseteq V$ such that $uS_{w}V^{\prime}$ and $R[V^{\prime}]\subseteq R[u]$.

First, consider the case when $xS_{w}V$ holds by the clause (a) from the definition of $S_{w}$. So there is $v\in V$ such that $x=v$ or $xRv$. In both cases, $wRuRv$, and so $uS_{w}\{v\}.$ It is clear that $R[v]\subseteq R[x]\subseteq R[u]$. So it suffices to take $V^{\prime}=\{v\}$.

Otherwise, $xS_{w}V$ holds by the clause (b). Take $V^{\prime}=\{v\in V:w\prec_{u_{\emptyset}^{\square}}v\}$. Clearly, $V^{\prime}\subseteq V\subseteq R[w]$. Assume $w\prec_{S}u$. Now $w\prec_{S}u\prec x$ and Lemma 16 imply $w\prec_{S\cup u_{\emptyset}^{\square}}x$. The definition of $xS_{w}V$ (clause (b)) implies there is $G\in\mathcal{D}\cap\bigcup\dot{R}[x]$ (so $G\in\mathcal{D}\cap\bigcup\dot{R}[u]$) and $v\in V$ such that $w\prec_{S\cup u_{\emptyset}^{\square}\cup\{\square\neg G\}}v$, thus also $v\in V^{\prime}$. In particular, $w\prec_{S\cup\{\square\neg G\}}v$. Since $S$ was arbitrary, $uS_{w}V^{\prime}$. It remains to verify that $R[V^{\prime}]\subseteq R[u]$. Assume $V^{\prime}\ni vRz$. Since $w\prec_{u_{\emptyset}^{\square}}v$, for all $\square B\in u$ we have $\square B\in v$, and since $vRz$, it follows that $\square B,B\in z$. Thus, $u\prec z$ i.e.  $uRz.$ ∎

In [7] it is shown that ILW* possesses finite model property w.r.t. generalized Veltman models. To show decidability, (stronger) completeness w.r.t. ordinary Veltman models was used, but the Theorem 31 would suffice for this purpose.

4 Finite model property and decidability

For IL, ILM, ILP and ILW, original completeness proofs were proofs of completeness w.r.t. appropriate finite models [3], [4]. For these logics, FMP w.r.t. ordinary semantics and decidability are immediate (and completeness and FMP w.r.t. generalized semantics are easily shown to follow from these results). These completeness proofs use truncated maximal consistent sets, that is, sets that are maximal consistent with respect to the so-called adequate set. The principal requirement is that this set is finite. Already with ILM, defining adequacy is not trivial (see [3]).

For more complex logics, not much is known about FMP w.r.t. ordinary semantics. The filtration method can be used with generalized models to obtain finite models. This approach was successfully used to prove FMP of ILM0 and ILW* w.r.t. generalized semantics [8], [7]. A drawback of this approach is in that FMP w.r.t. ordinary semantics does not follow from FMP w.r.t. generalized semantics. Decidability can be obtained from FMP w.r.t. either semantics. At the moment it is not clear whether the choice of semantics would affect our ability to produce results regarding computational complexity of provability and consistency of ILX.

Let us overview basic notions and results of [8] and [7]. Let $A$ be a formula. If $A$ equals $\neg B$ for some $B$, then ${\sim}A$ is $B$, otherwise ${\sim}A$ is $\neg B$. We need to slightly extend the definition of adequate sets444Note that this is a different notion of adequacy than the one used for completeness proofs in [3], [4], and [5]. that was used in [8]. The modified version will satisfy all the old properties.

Definition 32**.**

Let $\mathcal{D}$ have the usual the properties: a finite set of formulas that is closed under taking subformulas and single negations, and $\top\in\mathcal{D}.$ We say that a set of formulas $\Gamma_{\mathcal{D}}$ is an adequate set w.r.t. $\mathcal{D}$ if it satisfies the following conditions:

$\Gamma_{\mathcal{D}}$ is closed under taking subformulas; 2. 2.

if $A\in\Gamma_{\mathcal{D}}$ then ${\sim}A\in\Gamma_{\mathcal{D}};$ 3. 3.

$\bot\rhd\bot\in\Gamma_{\mathcal{D}};$ 4. 4.

$A\rhd B\in\Gamma_{\mathcal{D}}$ if $A$ is an antecedent or succedent of some $\rhd$-formula in $\Gamma_{\mathcal{D}},$ and so is $B;$ 5. 5.

if $A\in\mathcal{D}$ then $\square\neg A\in\Gamma_{\mathcal{D}}.$

Since $\mathcal{D}$ is finite, $\Gamma_{\mathcal{D}}$ is finite too. Next we require the concept of bisimulations between generalized models.

Definition 33** ([12]).**

A bisimulation between generalized Veltman models $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ and $\mathfrak{M}^{\prime}=(W^{\prime},R^{\prime},\{S^{\prime}_{w^{\prime}}:w^{\prime}\in W^{\prime}\},\Vdash)$ is a non-empty relation $Z\subseteq W\times W^{\prime}$ such that:

• (at)

if $wZw^{\prime}$, then $w\Vdash p$ if and only if $w^{\prime}\Vdash p$, for all propositional variables $p;$

• (forth)

if $wZw^{\prime}$ and $wRu$, then there is $u^{\prime}\in W^{\prime}$ such that $w^{\prime}R^{\prime}u^{\prime}$, $uZu^{\prime}$ and for all $V^{\prime}\subseteq W^{\prime}$ such that $u^{\prime}S^{\prime}_{w^{\prime}}V^{\prime}$ there is $V\subseteq W$ such that $uS_{w}V$ and for all $v\in V$ there is $v^{\prime}\in V^{\prime}$ with $vZv^{\prime};$

• (back)

if $wZw^{\prime}$ and $w^{\prime}R^{\prime}u^{\prime}$, then there is $u\in W$ such that $wRu$, $uZu^{\prime}$ and for all $V\subseteq W$ such that $uS_{w}V$ there is $V^{\prime}\subseteq W^{\prime}$ such that $u^{\prime}S^{\prime}_{w^{\prime}}V^{\prime}$ and for all $v^{\prime}\in V^{\prime}$ there is $v\in V$ with $vZv^{\prime}.$

Given a generalized Veltman model $\mathfrak{M}$, the union of all bisimulations on $\mathfrak{M}$, denoted by $\sim_{\mathfrak{M}}$, is the largest bisimulation on $\mathfrak{M}$, and $\sim_{\mathfrak{M}}$ is an equivalence relation [12].

An $\sim_{\mathfrak{M}}$-equivalence class of $w\in W$ will be denoted by $[w]$. For any set of worlds $V$, put $\widetilde{V}=\{[w]:w\in V\}$.

A filtration of $\mathfrak{M}$ through $\Gamma_{\mathcal{D}},\ \sim_{\mathfrak{M}}$ is any generalized Veltman model $\widetilde{\mathfrak{M}}=(\widetilde{W},\widetilde{R},\{\widetilde{S}_{[w]}:w\in W\},\Vdash)$ such that for all $w\in W$ and $A\in\Gamma_{\mathcal{D}}$ we have $w\Vdash A$ if and only if $[w]\Vdash A$ (we denote both forcing relations as $\Vdash$, as there is no risk of confusion).

The following lemma combines key results of [8] (Lemma 2.3, Theorem 2.4., Theorem 3.2).

Lemma 34**.**

Let $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ be a generalized Veltman model, and $\sim_{\mathfrak{M}}$ the largest bisimulation on $\mathfrak{M}$. Define:

(1)* $[w]\widetilde{R}[u]$ if and only if for some $w^{\prime}\in[w]$ and $u^{\prime}\in[u]$, $w^{\prime}Ru^{\prime}$ and there is $\square A\in\Gamma_{\mathcal{D}}$ such that $w^{\prime}\nVdash\square A$ and $u^{\prime}\Vdash\square A;$*

(2)* $[u]\widetilde{S}_{[w]}\widetilde{V}$ if and only if $[w]\widetilde{R}[u]$, $\widetilde{V}\subseteq\widetilde{R}[[w]]$, and for all $w^{\prime}\in[w]$ and $u^{\prime}\in[u]$ such that $w^{\prime}Ru^{\prime}$ we have $u^{\prime}S_{w^{\prime}}V^{\prime}$ for some $V^{\prime}$ such that $\widetilde{V^{\prime}}\subseteq\widetilde{V};$*

(3)* for all propositional variables $p\in\Gamma_{\mathcal{D}}$ put $[w]\Vdash p$ if and only if $w\Vdash p,$ and interpret propositional variables $q\not\in\Gamma_{\mathcal{D}}$ arbitrarily (e.g. put $[w]\nVdash q$ for all $[w]\in\widetilde{W}).$*

Then $\widetilde{\mathfrak{M}}=(\widetilde{W},\widetilde{R},\{\widetilde{S}_{[w]}:w\in W\},\Vdash)$ is a filtration of $\mathfrak{M}$ through $\Gamma_{\mathcal{D}},\sim_{\mathfrak{M}}$. The model $\widetilde{\widetilde{\mathfrak{M}}}$ is finite.

Lemma 34 implies that IL has FMP w.r.t. generalized semantics. To prove that a specific extension has FMP, it remains to show that filtration preserves its characteristic property.

Since we are going to use ILX-structures as the starting models $\mathfrak{M}$, we can make use of their properties. In particular, we do not have to make sure that there is a formula $\square A$ such that $x\nVdash\square A$ and $y\Vdash\square A$ when we want to show that $xRy$ implies $[x]{\widetilde{R}}[y].$

Lemma 35**.**

Let $X$ be a subset of $\{$M, M0, P, P0, R, W$\}$ and $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ the ILX-structure for a set $\mathcal{D}.$ If $xRy$ then $[x]{\widetilde{R}}[y]$.

Proof.

We should find a formula $\square A\in\Gamma_{\mathcal{D}}$ such that $x\nVdash\square A$ and $y\Vdash\square A$. Since $y\in W$, there is a formula $B\in\mathcal{D}$ such that $B\wedge\square\neg B\in y$. Lemma 11 implies $y\Vdash B\wedge\square\neg B$ ($\square\neg B$ might not be in $\mathcal{D}$, but ${\sim}B$ is; since $R[y]\Vdash{\sim}B$, we must have $y\Vdash\square\neg B$). Since $xRy$, we have $x\nVdash\square\neg B$. Since $B\in\mathcal{D}$, we have $\square\neg B\in\Gamma_{\mathcal{D}}$. Thus we can take $A=\neg B$. ∎

Note that the proof above applies not only when transforming ILX structures $\mathfrak{M}$ to $\widetilde{\mathfrak{M}}$, but also when transforming $\widetilde{\mathfrak{M}}$ to $\widetilde{\widetilde{\mathfrak{M}}}$.

Lemma 36**.**

Let $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ be the ILP0-structure for $\mathcal{D}$ and $\sim_{\mathfrak{M}}$ be the largest bisimulation on $\mathfrak{M}.$ Then the filtration $\widetilde{\mathfrak{M}}$ as defined in Lemma 34 possesses the property (P0)gen.

Proof.

Assume $[w]\widetilde{R}[x]\widetilde{R}[u]\widetilde{S}_{[w]}V$ and $\widetilde{R}[[v]]\cap Z\neq\emptyset$ for each $[v]\in V.$ We claim that there exists $Z^{\prime}\subseteq Z$ such that $[u]\widetilde{S}_{[x]}Z^{\prime}.$

Since $[w]{\widetilde{R}}[x]$, there are $w_{0}\in[w]$ and $x_{0}\in[x]$ such that $w_{0}Rx_{0}$. Let $x^{\prime}\in[x]$ and $u^{\prime}\in[u]$ be any worlds such that $x^{\prime}Ru^{\prime}$. The condition (back) implies that there is a world $u_{x^{\prime},u^{\prime}}$ such that $x_{0}Ru_{x^{\prime},u^{\prime}}$ and $u_{x^{\prime},u^{\prime}}\sim_{\mathfrak{M}}u^{\prime}$. Now, $[u]\widetilde{S}_{[w]}V$, $u_{x^{\prime},u^{\prime}}\in[u]$ and $w_{0}Ru_{x^{\prime},u^{\prime}}$ imply there is a set $V_{x^{\prime},u^{\prime}}$ such that $u_{x^{\prime},u^{\prime}}S_{w_{0}}V_{x^{\prime},u^{\prime}}$ and $\widetilde{V_{x^{\prime},u^{\prime}}}\subseteq V.$ Since $\widetilde{R}[[v]]\cap Z\neq\emptyset$ for each $[v]\in V$, we have $\widetilde{R}[[v]]\cap Z\neq\emptyset$ for each $v\in V_{x^{\prime},u^{\prime}}$. For each $v\in V_{x^{\prime},u^{\prime}}$, choose a world $z_{v}$ such that $[z_{v}]\in\widetilde{R}[[v]]\cap Z$. Now $[v]\widetilde{R}[z_{v}]$ implies that there are some $v^{\prime}\in[v]$ and $z^{\prime}_{v}\in[z_{v}]$ such that $v^{\prime}Rz_{v}^{\prime}$. Applying (back), we can find a world $z^{\prime\prime}_{v}$ such that $vRz^{\prime\prime}_{v}$ and $z^{\prime}_{v}\sim_{\mathfrak{M}}z^{\prime\prime}_{v}$. Put $Z_{x^{\prime},u^{\prime}}=\{z^{\prime\prime}_{v}:v\in V_{x^{\prime},u^{\prime}}\}$. Note that we have $R[v]\cap Z_{x^{\prime},u^{\prime}}\neq\emptyset$ for each $v\in V_{x^{\prime},u^{\prime}}.$

Applying (P0)gen gives $u_{x^{\prime},u^{\prime}}S_{x_{0}}Z^{\prime}_{x^{\prime},u^{\prime}}$ for some $Z^{\prime}_{x^{\prime},u^{\prime}}\subseteq Z_{x^{\prime},u^{\prime}}$. Clearly $\widetilde{Z^{\prime}_{x^{\prime},u^{\prime}}}\subseteq\widetilde{Z_{x^{\prime},u^{\prime}}}\subseteq Z$. Continuing our first application of (back), there is a set $Z^{\prime\prime}_{x^{\prime},u^{\prime}}$ such that $u^{\prime}S_{x^{\prime}}Z^{\prime\prime}_{x^{\prime},u^{\prime}}$, and for each $z^{\prime\prime}\in Z^{\prime\prime}_{x^{\prime},u^{\prime}}$ there is $z^{\prime}\in Z^{\prime}_{x^{\prime},u^{\prime}}$ such that $z^{\prime}\sim_{\mathfrak{M}}z^{\prime\prime}$. This implies $\widetilde{Z^{\prime\prime}_{x^{\prime},u^{\prime}}}\subseteq\widetilde{Z^{\prime}_{x^{\prime},u^{\prime}}}$. Let $T=\bigcup_{x^{\prime}\in[x],u^{\prime}\in[u],x^{\prime}Ru^{\prime}}Z^{\prime\prime}_{x^{\prime},u^{\prime}}$ and $Z^{\prime}=\widetilde{T}$. It is easy to see that $Z^{\prime}\subseteq Z$. Lemma 35 implies $Z^{\prime}\subseteq\widetilde{R}[[x]]$. We have $u^{\prime}S_{x^{\prime}}Z^{\prime\prime}_{x^{\prime},u^{\prime}}$ with $\widetilde{Z^{\prime\prime}_{x^{\prime},u^{\prime}}}\subseteq Z^{\prime}$ for all $x^{\prime}\in[x]$ and $u^{\prime}\in[u]$ with $x^{\prime}Ru^{\prime}$. Thus, $[u]\widetilde{S}_{[x]}Z^{\prime}.$ ∎

Corollary 37**.**

ILP0* is decidable.*

Let us prove the same for ILR.

Lemma 38**.**

Let $\mathfrak{M}=(W,R,\{S_{w}:w\in W\},\Vdash)$ be the ILR-structure for $\mathcal{D}$ and $\sim_{\mathfrak{M}}$ be the largest bisimulation on $\mathfrak{M}.$ Then the filtration $\widetilde{\mathfrak{M}}$ as defined in Lemma 34 possesses the property (R)gen.

Proof.

Assume $[w]\widetilde{R}[x]\widetilde{R}[u]\widetilde{S}_{[w]}{V}$, and let $C\in\mathcal{C}([x],[u])$ be an arbitrary choice set. We are to prove that there is a set $U$ such that $\widetilde{U}\subseteq V,$ $[x]\widetilde{S}_{[w]}\widetilde{U}$ and $\widetilde{R}[\widetilde{U}]\subseteq C$.

Put $C_{x^{\prime}}=\{c\in R[x^{\prime}]:[c]\in C\}$ for all $x^{\prime}\in[x]$.

Let us first prove that for some $x_{0}\in[x],u_{0}\in[u]$ with $x_{0}Ru_{0}$ we have $C_{x_{0}}\in\mathcal{C}(x_{0},u_{0})$. Suppose not. Then for all $x^{\prime}\in[x],u^{\prime}\in[u]$ with $x^{\prime}Ru^{\prime}$, there is a set $Z_{x^{\prime},u^{\prime}}$ such that $u^{\prime}S_{x^{\prime}}Z_{x^{\prime},u^{\prime}}$ with $Z_{x^{\prime},u^{\prime}}\cap C_{x^{\prime}}=\emptyset$. Put $Z=\bigcup_{x^{\prime}\in[x],u^{\prime}\in[u],x^{\prime}Ru^{\prime}}Z_{x^{\prime},u^{\prime}}$. Lemma 35 implies $\widetilde{Z}\subseteq\widetilde{R}[[x]]$. Thus $[u]\widetilde{S}_{[x]}\widetilde{Z}$. Since $C\in\mathcal{C}([x],[u])$, there is $z\in Z$ such that $[z]\in C\cap\widetilde{Z}$. Thus $z\in Z_{x^{\prime},u^{\prime}}$ for some $x^{\prime}\in[x],u^{\prime}\in[u]$ and $x^{\prime}Ru^{\prime}$. The definition of $C_{x^{\prime}}$ implies $z\in C_{x^{\prime}}$. Thus, $Z_{x^{\prime},u^{\prime}}\cap C_{x^{\prime}}\neq\emptyset$, a contradiction.

Now we claim that for all $y\in[x]$ there is $u_{y}\sim u_{0}$ with $yRu_{y}$ and $C_{y}\in\mathcal{C}(y,u_{y})$. Since $y\sim_{\mathfrak{M}}x_{0}$ and $x_{0}Ru_{0}$, the (back) condition implies that there is a world $u_{y}$ such that $u_{y}\sim_{\mathfrak{M}}u_{0}$ and $yRu_{y}$ (and other properties that we will return to later). We will show that $C_{y}\in\mathcal{C}(y,u_{y})$. Let $Z^{\prime}$ be such that $u_{y}S_{y}Z^{\prime}$, and we are to prove that $C_{y}\cap Z^{\prime}\neq\emptyset$. The earlier instance of (back) condition for $u_{y}$ further implies that there is a set $Z$ with $u_{0}S_{x_{0}}Z$, and for all $z\in Z$ there is $z^{\prime}\in Z^{\prime}$ with $z\sim_{\mathfrak{M}}z^{\prime}$. Let $z\in Z\cap C_{x_{0}}$ be an arbitrary element (which exists because, as we proved, $C_{x_{0}}$ is a choice set). So there is $z^{\prime}\in Z^{\prime}$ such that $z^{\prime}\sim_{\mathfrak{M}}z$. Since $[z]\in C$, equivalently $[z^{\prime}]\in C$, we have $z^{\prime}\in C_{y}$. In particular, $Z^{\prime}\cap C_{y}\neq\emptyset$. Thus $C_{y}\in\mathcal{C}(y,u_{y})$.

Let us prove that there is a set $U$ such that $\widetilde{U}\subseteq V,$ $[x]\widetilde{S}_{[w]}\widetilde{U}$ and $\widetilde{R}[\widetilde{U}]\subseteq C.$ Let $w^{\prime}\in[w]$ and $y\in[x]$ be such that $w^{\prime}Ry$. Since $[u]\widetilde{S}_{[w]}V$, there is a set $V_{w^{\prime},y}$ such that $u_{y}S_{w^{\prime}}V_{w^{\prime},y}$ and $\widetilde{V_{w^{\prime},y}}\subseteq V$. Applying (R)gen with $C_{y}$, there is $U_{w^{\prime},y}\subseteq V_{w^{\prime},y}$ such that $yS_{w^{\prime}}U_{w^{\prime},y}$ and $R[U_{w^{\prime},y}]\subseteq C_{y}$. Let $U=\bigcup_{w^{\prime}\in[w],y\in[x],w^{\prime}Ry}U_{w^{\prime},y}$. Clearly $\widetilde{U}\subseteq V.$ Lemma 35 implies $\widetilde{U}\subseteq\widetilde{R}[[w]].$ The definition of $\widetilde{S}_{[w]}$ implies $[x]\widetilde{S}_{[w]}\widetilde{U}$.

It remains to verify that $\widetilde{R}[\widetilde{U}]\subseteq C$. Let $t\in U$ and $z\in W$ be such that $[t]\widetilde{R}[z]$. Then we have $t\in U_{w^{\prime},y}$ for some $w^{\prime}\in[w]$ and $y\in[x]$. Since $[t]\widetilde{R}[z]$, there are $t^{\prime}\in[t]$ and $z^{\prime}\in[z]$ with $t^{\prime}Rz^{\prime}$. The condition (forth) implies that there is $z^{\prime\prime}$ such that $tRz^{\prime\prime}$ and $z^{\prime}\sim_{\mathfrak{M}}z^{\prime\prime}.$ Since $R[U_{w^{\prime},y}]\subseteq C_{y})$ and $z^{\prime\prime}\in R[U_{w^{\prime},y}]$, we have $z^{\prime\prime}\in C_{y}$. Definition of $C_{y}$ implies $[z^{\prime\prime}]\in C$, or equivalently, $[z]\in C$. ∎

Corollary 39**.**

ILR* is decidable.*