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About the unification type of \(\mathbf {K}+\square \square \bot \)

Abstract

The unification problem in a propositional logic is to determine, given a formula φ, whether there exists a substitution σ such that σ(φ) is in that logic. In that case, σ is a unifier of φ. When a unifiable formula has minimal complete sets of unifiers, it is either infinitary, finitary, or unitary, depending on the cardinality of its minimal complete sets of unifiers. Otherwise, it is nullary. In this paper, we prove that in modal logic \(\mathbf {K}+\square \square \bot \), unifiable formulas are either finitary, or unitary.

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Notes

  1. 1.

    Since S is finite, the proof of the existence of \(\sim \) does not require the use of the axiom of choice.

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Acknowledgements

We are indebted to Silvio Ghilardi for his suggestion to consider the question of the unification type of L2. We also make a point of thanking for their feedback the referees of all preliminary versions of this paper: their useful suggestions have been essential for improving its correctness and its readability.

Funding

The preparation of this paper has been supported by the Project DN02/15/19.12.2016 of Bulgarian Science Fund — ‘Space, time and modality: relational, algebraic and topological models’ — and the Programme Professeurs invités 2018 of Université Paul Sabatier.

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Appendix

Appendix

Proof of Proposition 1:

Suppose S is nonempty. Let \(\sim \) be an equivalence relation on S.

Suppose \({\parallel }S/{\sim }{\parallel }{\leq }{\parallel }T{\parallel }{\leq }{\parallel }S{\parallel }\). Let h be a function from \(S/{\sim }\) to S such that for all αS, h([α])∈[α], i.e. h is a function selecting an element in each equivalence class modulo \(\sim \).Footnote 1 Obviously, h is injective. Let S0={h([α]) : αS}. Since h is injective, \({\parallel }S/{\sim }{\parallel }{=}{\parallel }S_{0}{\parallel }\). Since \({\parallel }S/{\sim }{\parallel }{\leq }{\parallel }T{\parallel }\), ∥S0∥≤∥T∥. Let T0 be a subset of T such that ∥T0∥=∥S0∥. Let f0 be a one-to-one correspondence between S0 and T0. Let T1=TT0. Notice that T0 and T1 make a partition of T. Since ∥T∥≤∥S∥ and ∥T0∥=∥S0∥, ∥T1∥≤∥SS0∥. Let S1 be a subset of SS0 such that ∥S1∥=∥T1∥. Let f1 be a one-to-one correspondence between S1 and T1. Let S2=(SS0)∖S1. Let f2 be the function from S2 to T such that for all αS2, f2(α)=f0(h([α])). Let f be the function from S to T defined by f|S0=f0, f|S1=f1 and f|S2=f2. □

Claim

f is surjective.

Proof

Let βT. We consider the following cases. □

  • Case βT0.

  • Since f0 is one-to-one, let αS0 be such that f0(α)=β. Thus, αS. Moreover, f(α)=f0(α). Since f0(α)=β, f(α)=β.

  • Case βT1.

  • Since f1 is one-to-one, let αS1 be such that f1(α)=β. Hence, αS. Moreover, f(α)=f1(α). Since f1(α)=β, f(α)=β.

Claim

For all α,βS, if f(α)=f(β) then \(\alpha {\sim }\beta \).

Proof

Let α,βS be such that f(α)=f(β). We consider the following cases.

Case αS0 and βS0. Consequently, f(α)=f0(α) and f(β)=f0(β). Since f(α)=f(β), f0(α)=f0(β). Since f0 is one-to-one, α=β. Thus, \(\alpha {\sim }\beta \).

Case αS0 and βS1. Consequently, f(α)=f0(α) and f(β)=f1(β). Since f(α)=f(β), f0(α)=f1(β). Since f0(α)∈T0 and f1(β)∈T1, T0 and T1 do not make a partition of T: a contradiction.

Case αS0 and βS2. Hence, f(α)=f0(α) and f(β)=f2(β). Since f(α)=f(β), f0(α)=f2(β). Thus, f0(α)=f0(h([β])). Since f0 is one-to-one, α=h([β]). Since h([β])∈[β], α∈[β]. Consequently, \(\alpha {\sim }\beta \).

Case αS1 and βS1. Hence, f(α)=f1(α) and f(β)=f1(β). Since f(α)=f(β), f1(α)=f1(β). Since f1 is one-to-one, α=β. Thus, \(\alpha {\sim }\beta \).

CaseαS1 and βS2. Hence, f(α)=f1(α) and f(β)=f2(β). Since f(α)=f(β), f1(α)=f2(β). Thus, f1(α)=f0(h([β])). Since f1(α)∈T1 and f0(h([β]))∈T0, T0 and T1 do not make a partition of T: a contradiction.

CaseαS2 andβS2. Hence, f(α)=f2(α) and f(β)=f2(β). Since f(α)=f(β), f2(α) =f2(β). Consequently, f0(h([α]))=f0(h([β])). Since f0 is one-to-one, h([α])=h([β]). Since h([α])∈[α] and h([β])∈[β], [α] ∩ [β]≠. Thus, \(\alpha {\sim }\beta \).

Suppose f is a surjective function from S to T such that for all α,βS, if f(α)=f(β) then \(\alpha {\sim }\beta \). For the sake of the contradiction, suppose either \({\parallel }S/{\sim }{\parallel }{>}{\parallel }T{\parallel }\), or ∥T∥>∥S∥. Since f is surjective, ∥T∥≤∥S∥. Since either \({\parallel }S/{\sim }{\parallel }{>}{\parallel }T{\parallel }\), or ∥T∥>∥S∥, \({\parallel }S/{\sim }{\parallel }{>}{\parallel }T{\parallel }\). Let \(p{\in }\mathbb {N}\) and β1,…,βpS be such that p>∥T∥ and for all \(q,r{\in }\mathbb {N}\), if 1≤q,rp and qr then \(\beta ^{q}{\not \sim }\beta ^{r}\). Hence, for all \(q,r{\in }\mathbb {N}\), if 1≤q,rp and qr then f(βq)≠f(βr). Thus, p≤∥T∥: a contradiction.

Proof of Proposition 2:

From [11, Proposition 2.29] and from the fact that for all φFORn, there exists ψFORn such that n(ψ)< 2 and \(\varphi \leftrightarrow \psi {\in }\mathbf {L}_{2}\). □

Proof of Proposition 3:

From [11, Proposition 2.6], from [11, Lemma 4.21] and from the fact that for all maximal consistent sets w of n-formulas, there exists (α,S)∈MODn such that the submodel of the canonical model of L2 generated by w is isomorphic to (α,S). □

Proof of Proposition 4:

Let (α,S)∈MODn and βBITn. For the sake of the contradiction, suppose α=β not-iff \((\alpha ,S){\models _{n}}\bar {x}^{\beta }\). Hence, either α=β and \((\alpha ,S){\not \models _{n}}\bar {x}^{\beta }\), or αβ and \((\alpha ,S){\models _{n}}\bar {x}^{\beta }\). In the former case, for all i∈{1,…,n}, αi=βi. Thus, for all i∈{1,…,n}, \((\alpha ,S){\models _{n}}x_{i}^{\beta _{i}}\). Consequently, \((\alpha ,S){\models _{n}}\bar {x}^{\beta }\): a contradiction. In the latter case, let i∈{1,…,n} be such that αiβi. Hence, \((\alpha ,S){\not \models _{n}}x_{i}^{\beta _{i}}\). Thus, \((\alpha ,S){\not \models _{n}}\bar {x}^{\beta }\): a contradiction. □

Proof of Proposition 5:

From [20, Theorem 32], from the fact that for all (α,S)∈MODn, forn(α,S) characterizes n-models modulo bisimulation and from the fact that for all (α,S), (β,T)∈MODn, if (α,S) and (β,T) are bisimilar then (α,S)=(β,T). □

Proof of Proposition 6:

Let (α,S)∈MODk. Let β be the n-tuple of bits such that for all i∈{1,…,n}, if (α,S)⊮kσ(xi) then βi= 0 else βi= 1. Let T be the least set of n-tuples of bits such that for all γS, there exists δT such that for all i∈{1,…,n}, if (γ,)⊮kσ(xi) then δi= 0 else δi= 1. By induction on φFORn, the reader may easily verify that (α,S)⊧kσ(φ) iff (β,T)⊧nφ. Since by Proposition 5, (β,T)⊧nforn(β,T), (α,S)⊧kσ(forn(β,T)). □

Proof of Proposition 7:

Let (β,T), (γ,U)∈MODn. Suppose (α,S)⊧kσ(forn(β,T)) and (α,S)⊧kσ(forn(γ,U)). Hence, \((\alpha ,S){\models _{k}}\sigma (\bar {x}^{\beta })\) and \((\alpha ,S){\models _{k}}\sigma (\bar {x}^{\gamma })\). Thus, β=γ. Let \(\beta ^{\prime }{\in }T\) be arbitrary. Since (α,S)⊧kσ(forn(β,T)), let \(\alpha ^{\prime }{\in }S\) be such that \((\alpha ^{\prime },\emptyset ){\models _{k}}\sigma (\bar {x}^{\beta ^{\prime }})\). Since (α,S)⊧kσ(forn(γ,U)), let \(\gamma ^{\prime }{\in }U\) be such that \((\alpha ^{\prime },\emptyset ){\models _{k}}\sigma (\bar {x}^{\gamma ^{\prime }})\). Since \((\alpha ^{\prime },\emptyset ){\models _{k}}\sigma (\bar {x}^{\beta ^{\prime }})\), \(\beta ^{\prime }{=}\gamma ^{\prime }\). Consequently, \(\beta ^{\prime }{\in }U\). Since \(\beta ^{\prime }\) is arbitrary, \(T{\subseteq }U\). Reciprocally, the reader may easily verify that \(U{\subseteq }T\). Hence, T=U. Since β=γ, (β,T)=(γ,U). □

Proof of Proposition 8:

Let (k,σ)∈SUBn. For the sake of the contradiction, suppose g(k,σ) is not a (k,n)-morphism. Hence, let (α,S)∈MODk and (β,T)∈MODn be such that g(k,σ)(α,S)=(β,T) — and therefore (α,S)⊧kσ(forn(β,T)) — and either there exists γS such that for all δT, g(k,σ)(γ,)≠(δ,), or there exists δT such that for all γS, g(k,σ)(γ,)≠(δ,). In the former case, let \(\delta ^{\prime }{\in }T\) be such that \((\gamma ,\emptyset ){\models _{k}}\sigma (\bar {x}^{\delta ^{\prime }})\). Thus, \(g_{(k,\sigma )}(\gamma ,\emptyset ){\not =}(\delta ^{\prime },\emptyset )\). Since \((\gamma ,\emptyset ){\models _{k}}\sigma (\bar {x}^{\delta ^{\prime }})\), \(g_{(k,\sigma )}(\gamma ,\emptyset ){=}(\delta ^{\prime },\emptyset )\): a contradiction. In the latter case, let \(\gamma ^{\prime }{\in }S\) be such that \((\gamma ^{\prime },\emptyset ){\models _{k}}\sigma (\bar {x}^{\delta })\). Consequently, \(g_{(k,\sigma )}(\gamma ^{\prime },\emptyset ){\not =}(\delta ,\emptyset )\). Since \((\gamma ^{\prime },\emptyset ){\models _{k}}\sigma (\bar {x}^{\delta })\), \(g_{(k,\sigma )}(\gamma ^{\prime },\emptyset ){=}(\delta ,\emptyset )\): a contradiction. □

Proof of Proposition 9:

Let (k,σ)∈SUBn and (α,S), (β,T)∈MODk. Suppose g(k,σ)(α,S)=g(k,σ)(β,T). Hence, let (γ,U)∈MODn be such that g(k,σ)(α,S)=(γ,U) and g(k,σ)(β,T)=(γ,U). Thus, (α,S)⊧kσ(forn(γ,U)) and (β,T)⊧kσ(forn(γ,U)). Consequently, \((\alpha ,S){\models _{k}}\sigma (\bar {x}^{\gamma })\) and \((\beta ,T){\models _{k}}\sigma (\bar {x}^{\gamma })\). Hence, for all i∈{1,…,n}, (α,S)⊧kσ(xi) iff (β,T)⊧kσ(xi). □

Proof of Proposition 10:

For the sake of the contradiction, suppose f(β,T)=(γ,U) and the image by f of T ×{} is not equal to U ×{}. Hence, either the image by f of T ×{} is not included in U ×{}, or the image by f of T ×{} does not include to U ×{}. In the former case, let δT be such that f(δ,)∉U ×{}. Since f is a (k,n)-morphism and f(β,T)=(γ,U), let \(\epsilon ^{\prime }{\in }U\) be such that \(f(\delta ,\emptyset ){=}(\epsilon ^{\prime },\emptyset )\). Thus, f(δ,)∈U ×{}: a contradiction. In the latter case, let 𝜖U be such that (𝜖,)∉f(T ×{}). Since f is a (k,n)-morphism and f(β,T)=(γ,U), let \(\delta ^{\prime }{\in }T\) be such that \(f(\delta ^{\prime },\emptyset ){=}(\epsilon ,\emptyset )\). Consequently, (𝜖,)∈f(T ×{}): a contradiction. □

Proof of Proposition 11:

For the sake of the contradiction, suppose f(β,T)≠(γ,U), \(f(\beta ,T){\models _{n}}\bar {x}^{\gamma }\), for all δT, there exists 𝜖U such that f(δ,)=(𝜖,) and for all 𝜖U, there exists δT such that f(δ,)=(𝜖,). Let \((\gamma ^{\prime },U^{\prime }){\in }\mathbf {MOD}_{n}\) be such that \(f(\beta ,T){=}(\gamma ^{\prime },U^{\prime })\). Since \(f(\beta ,T){\models _{n}}\bar {x}^{\gamma }\), by Proposition 4, \(\gamma {=}\gamma ^{\prime }\). Since f(β,T)≠(γ,U) and \(f(\beta ,T){=}(\gamma ^{\prime },U^{\prime })\), \(U{\not =}U^{\prime }\). Hence, either \(U{\not \subseteq }U^{\prime }\), or \(U{\not \supseteq }U^{\prime }\). In the former case, let \(\epsilon ^{\prime }{\in }U\) be such that \(\epsilon ^{\prime }{\not \in }U^{\prime }\). Since for all 𝜖U, there exists δT such that f(δ,)=(𝜖,), let \(\delta ^{\prime }{\in }T\) be such that \(f(\delta ^{\prime },\emptyset ){=}(\epsilon ^{\prime },\emptyset )\). Since f is a (k,n)-morphism and \(f(\beta ,T){=}(\gamma ^{\prime },U^{\prime })\), \(\epsilon ^{\prime }{\in }U^{\prime }\): a contradiction. In the latter case, let \(\epsilon ^{\prime \prime }{\in }U^{\prime }\) be such that \(\epsilon ^{\prime \prime }{\not \in }U\). Since f is a (k,n)-morphism and \(f(\beta ,T){=}(\gamma ^{\prime },U^{\prime })\), let \(\delta ^{\prime \prime }{\in }T\) be such that \(f(\delta ^{\prime \prime },\emptyset ){=}(\epsilon ^{\prime \prime },\emptyset )\). Since for all δT, there exists 𝜖U such that f(δ,)=(𝜖,), \(\epsilon ^{\prime \prime }{\in }U\): a contradiction. □

Proof of Proposition 12:

Let (n,σ) and (n,τ) be the n-substitutions defined by:

  • \(\sigma (x_{1}){=}\square \bot \vee x_{1}\) and \(\tau (x_{1}){=}\lozenge \top \wedge x_{1}\),

  • for all i∈{2,…,n}, σ(xi)=xi and τ(xi)=xi.

Obviously, \(\sigma (\lozenge x_{1}\rightarrow \square x_{1}){\in }\mathbf {L}_{2}\) and \(\tau (\lozenge x_{1}\rightarrow \square x_{1}){\in }\mathbf {L}_{2}\). Hence, (n,σ) and (n,τ) are n-unifiers of \(\lozenge x_{1}\rightarrow \square x_{1}\). In order to prove that \(\lozenge x_{1}\rightarrow \square x_{1}\) is n-finitary, it suffices to prove that {(n,σ), (n,τ)} is a minimal n-complete set of n-unifiers of \(\lozenge x_{1}\rightarrow \square x_{1}\). □

n-completeness of {(n,σ), (n,τ)}:

Let (k,υ) be an arbitrary n-unifier of \(\lozenge x_{1}\rightarrow \square x_{1}\). Thus, \(\upsilon (\lozenge x_{1}\rightarrow \square x_{1}){\in }\mathbf {L}_{2}\). By using the semantics of L2, it follows that either \(\square \bot \rightarrow \upsilon (x_{1}){\in }\mathbf {L}_{2}\), or \(\upsilon (x_{1})\rightarrow \lozenge \top {\in }\mathbf {L}_{2}\). Indeed, for the sake of the contradiction, suppose neither \(\square \bot \rightarrow \upsilon (x_{1}){\in }\mathbf {L}_{2}\), nor \(\upsilon (x_{1})\rightarrow \lozenge \top {\in }\mathbf {L}_{2}\). Consequently, by Proposition 3, neither \(\square \bot \rightarrow \upsilon (x_{1})\) is k-valid, nor \(\upsilon (x_{1})\rightarrow \lozenge \top \) is k-valid. Hence, let \((\alpha ,\emptyset ),(\beta ,\emptyset ){\in }\mathbf {MOD}_{k}^{\mathtt {deg}}\) be such that (α,)⊮kυ(x1) and (β,)⊧kυ(x1). Let γBITk. Since (α,)⊮kυ(x1) and (β,)⊧kυ(x1), \((\gamma ,\{\alpha ,\beta \}){\not \models _{k}}\upsilon (\lozenge x_{1}\rightarrow \square x_{1})\). Thus, \(\upsilon (\lozenge x_{1}\rightarrow \square x_{1})\) is not k-valid. Consequently, by Proposition 3, \(\upsilon (\lozenge x_{1}\rightarrow \square x_{1}){\not \in }\mathbf {L}_{2}\): a contradiction. In the former case where \(\square \bot \rightarrow \upsilon (x_{1}){\in }\mathbf {L}_{2}\), it follows immediately that υ(σ(x1))≡kυ(x1). Hence, \((n,\sigma ){\preccurlyeq _{n}}(k,\upsilon )\). In the latter case where \(\upsilon (x_{1})\rightarrow \lozenge \top {\in }\mathbf {L}_{2}\), it follows immediately that \(\upsilon (\tau (x_{1})){\equiv _{\mathbf {L}_{2}}}\upsilon (x_{1})\). Thus, \((n,\tau ){\preccurlyeq _{n}}(k,\upsilon )\).

Minimality of {(n,σ), (n,τ)}:

For the sake of the contradiction, suppose {(n,σ), (n,τ)} is not minimal. Consequently, either \((n,\sigma ){\preccurlyeq _{n}}(n,\tau )\), or \((n,\tau ){\preccurlyeq _{n}}(n,\sigma )\). In the former case, there exists an n-substitution (n,υ) such that υ(σ(x1))≡nτ(x1). Hence, \(\square \bot \vee \upsilon (x_{1}){\equiv _{n}}\lozenge \top \wedge x_{1}\). In the latter case, there exists a substitution (n,υ) such that υ(τ(x1))≡nσ(x). Thus, \(\lozenge \top \wedge \upsilon (x_{1}){\equiv _{n}}\square \bot \vee x_{1}\). In both cases, \(\square \bot \rightarrow \lozenge \top {\in }\mathbf {L}_{2}\). Consequently, \(\lozenge \top {\in }\mathbf {L}_{2}\): a contradiction.

Proof of Proposition 13:

Suppose φ is n-π-reasonable. Let Σ be the set of all n-unifiers of φ. Notice that Σ is n-complete. Let \({\Sigma }^{\prime }\) be the set of n-substitutions obtained from Σ by keeping only the n-substitutions (k,σ) such that kπ. Since φ is n-π-reasonable and Σ is n-complete, \({\Sigma }^{\prime }\) is n-complete. Let \({\Sigma }^{\prime \prime }\) be the set of n-substitutions obtained from \({\Sigma }^{\prime }\) by keeping only one representative of each equivalence class modulo \(\simeq _{n}\). Since \({\Sigma }^{\prime }\) is n-complete, \({\Sigma }^{\prime \prime }\) is n-complete. Moreover, by Proposition 2, \({\Sigma }^{\prime \prime }\) is finite. Hence, either φ is n-finitary, or φ is n-unitary. □

Proof of Lemma 1:

(1) For the sake of the contradiction, suppose \({\parallel }\mathbf {MOD}_{k}^{\mathtt {deg}}/{\sim _{k}}{\parallel }{>}\) \({\parallel }\mathbf {MOD}_{n}^{\mathtt {deg}}{\parallel }\). Let \(p{\in }\mathbb {N}\) and \((\alpha ^{1},\emptyset ),\ldots ,(\alpha ^{p},\emptyset ){\in }\mathbf {MOD}_{k}^{\mathtt {deg}}\) be such that \(p{>}{\parallel }\mathbf {MOD}_{n}^{\mathtt {deg}}{\parallel }\) and for all \(q,r{\in }\mathbb {N}\), if 1≤q,rp and qr then \((\alpha ^{q},\emptyset ){\not \sim _{k}}(\alpha ^{r},\emptyset )\). Hence, for all \(q,r{\in }\mathbb {N}\), if 1≤q,rp and qr then g(k,σ)(αq,)≠g(k,σ)(αr,). Since g(k,σ) is a (k,n)-morphism, let \(\beta ^{1},\ldots ,\beta ^{p}{\in }\mathbf {BIT}_{n}\) be such that g(k,σ)(α1,)=(β1,), …, g(k,σ)(αp,)=(βp,). Since for all \(q,r{\in }\mathbb {N}\), if 1≤q,rp and qr then g(k,σ)(αq,)≠g(k,σ)(αr,), for all \(q,r{\in }\mathbb {N}\), if 1≤q,rp and qr then βqβr. Thus, p≤ 2n. Since \({\parallel }\mathbf {MOD}_{n}^{\mathtt {deg}}{\parallel }{=}2^{n}\), \(p{\leq }{\parallel }\mathbf {MOD}_{n}^{\mathtt {deg}}{\parallel }\): a contradiction. □

(2) Since \({\parallel }\mathbf {MOD}_{n}^{\mathtt {deg}}{\parallel }{=}2^{n}\), \({\parallel }\mathbf {MOD}_{k}^{\mathtt {deg}}{\parallel }{=}2^{k}\) and kn, \({\parallel }\mathbf {MOD}_{n}^{\mathtt {deg}}{\parallel }{\leq }{\parallel }\mathbf {MOD}_{k}^{\mathtt {deg}}{\parallel }\).

Proof of Lemma 2:

Let S,T be nonempty sets of k-tuples of bits. Suppose the images by f of S ×{} and T ×{} are equal. For the sake of the contradiction, suppose the images by g(k,σ) of S ×{} and T ×{} are not equal. Since g(k,σ) is a (k,n)-morphism, let \((\gamma ,\emptyset ){\in }\mathbf {MOD}_{n}^{\mathtt {deg}}\) be such that either (γ,) is in the image by g(k,σ) of S ×{} without being in the image by g(k,σ) of T ×{}, or (γ,) is in the image by g(k,σ) of T ×{} without being in the image by g(k,σ) of S ×{}. Without loss of generality, suppose (γ,) is in the image by g(k,σ) of S ×{} without being in the image by g(k,σ) of T ×{}. Hence, let αS be such that g(k,σ)(α,)=(γ,). Since the images by f of S ×{} and T ×{} are equal, let βT be such that f(α,)=f(β,). Thus, \((\alpha ,\emptyset ){\sim _{k}}(\beta ,\emptyset )\). Consequently, g(k,σ)(α,)=g(k,σ)(β,). Since g(k,σ)(α,)=(γ,), (γ,) is in the image by g(k,σ) of T ×{}: a contradiction. □

Proof of Lemma 3:

Let E be a nonempty set of n-tuples of bits. (1) For the sake of the contradiction, suppose \({\parallel }f^{\circ }(E)/{\sim _{k}}{\parallel }{>}{\parallel }f^{\bullet }(E){\parallel }\). Let \(p{\in }\mathbb {N}\) and \((\alpha ^{1},S_{1}),\ldots ,(\alpha ^{p},S_{p}){\in }\mathbf {MOD}_{k}\setminus \mathbf {MOD}_{k}^{\mathtt {deg}}\) be such that p>∥f(E)∥, the images by f of S1 ×{},…,Sp ×{} are equal to E ×{} and for all \(q,r{\in }\mathbb {N}\), if 1≤q,rp and qr then \((\alpha ^{q},S_{q}){\not \sim _{k}}(\alpha ^{r},S_{r})\). Hence, for all \(q,r{\in }\mathbb {N}\), if 1≤q,rp and qr then g(k,σ)(αq,Sq)≠g(k,σ)(αr,Sr). Since g(k,σ) is a (k,n)-morphism and the images by f of S1 ×{},…,Sp ×{} are equal to E ×{}, let \(\beta ^{1},\ldots ,\beta ^{p}{\in }\mathbf {BIT}_{n}\) be such that g(k,σ)(α1,S1)=(β1,E), …, g(k,σ)(αp,Sp)=(βp,E). Since for all \(q,r{\in }\mathbb {N}\), if 1≤q,rp and qr then g(k,σ)(αq,Sq)≠g(k,σ)(αr,Sr), for all \(q,r{\in }\mathbb {N}\), if 1≤q,rp and qr then βqβr. Thus, p≤ 2n. Since ∥f(E)∥= 2n, p≤∥f(E)∥: a contradiction. □

(2) Since kn, ∥f(E)∥≥ 2k and ∥f(E)∥= 2n, ∥f(E)∥≤∥f(E)∥.

Proof of Lemma 4:

Let βBITn and T be a set of n-tuples of bits. We consider the following cases. □

Case \((\beta ,T){\in }\mathbf {MOD}_{n}^{\mathtt {deg}}\). Since f is surjective, let αBITk be such that f(α,)=(β,). Hence, f(α,)=(β,T).

Case \((\beta ,T){\in }\mathbf {MOD}_{n}\setminus \mathbf {MOD}_{n}^{\mathtt {deg}}\). Thus, (β,T)∈f(T). Since fT is surjective, let (α,S)∈f(T) be such that fT(α,S)=(β,T). Consequently, \((\alpha ,S){\in }\mathbf {MOD}_{k}\setminus \mathbf {MOD}_{k}^{\mathtt {deg}}\) and the image by f of S ×{} is equal to T ×{}. Hence, f(α,S)=fT(α,S). Since fT(α,S)=(β,T), f(α,S)=(β,T).

Proof of Lemma 5:

For the sake of the contradiction, suppose f is not a (k,n)-morphism. Hence, let (α,S)∈MODk and (β,T)∈MODn be such that f(α,S)=(β,T) and either there exists \(\gamma ^{\prime }{\in }S\) such that for all \(\delta ^{\prime }{\in }T\), \(f(\gamma ^{\prime },\emptyset ){\not =}(\delta ^{\prime },\emptyset )\), or there exists \(\delta ^{\prime \prime }{\in }T\) such that for all \(\gamma ^{\prime \prime }{\in }S\), \(f(\gamma ^{\prime \prime },\emptyset ){\not =}(\delta ^{\prime \prime },\emptyset )\). In the former case, S. Thus, \((\alpha ,S){\in }\mathbf {MOD}_{k}\setminus \mathbf {MOD}_{k}^{\mathtt {deg}}\) and, E being the nonempty set of n-tuples of bits such that the image by f of S ×{} is equal to E ×{}, f(α,S)=fE(α,S). Since f(α,S)=(β,T), fE(α,S)=(β,T). Consequently, T=E. Since the image by f of S ×{} is equal to E ×{}, the image by f of S ×{} is equal to T ×{}. Hence, let δT be such that \(f(\gamma ^{\prime },\emptyset ){=}(\delta ,\emptyset )\). Since for all \(\delta ^{\prime }{\in }T\), \(f(\gamma ^{\prime },\emptyset ){\not =}(\delta ^{\prime },\emptyset )\), \(f(\gamma ^{\prime },\emptyset ){\not =}(\delta ,\emptyset )\): a contradiction. In the latter case, T. Since f(α,S)=(β,T), S. Thus, \((\alpha ,S){\in }\mathbf {MOD}_{k}\setminus \mathbf {MOD}_{k}^{\mathtt {deg}}\) and, E being the nonempty set of n-tuples of bits such that the image by f of S ×{} is equal to E ×{}, f(α,S)=fE(α,S). Since f(α,S)=(β,T), fE(α,S)=(β,T). Consequently, T=E. Since the image by f of S ×{} is equal to E ×{}, the image by f of S ×{} is equal to T ×{}. Hence, let γS be such that \(f(\gamma ,\emptyset ){=}(\delta ^{\prime \prime },\emptyset )\). Since for all \(\gamma ^{\prime \prime }{\in }S\), \(f(\gamma ^{\prime \prime },\emptyset ){\not =}(\delta ^{\prime \prime },\emptyset )\), \(f(\gamma ,\emptyset ){\not =}(\delta ^{\prime \prime },\emptyset )\): a contradiction. □

Proof of Lemma 6:

Let (α,S), (β,T)∈MODk. Suppose f(α,S)=f(β,T). We consider the following cases.

Case S=. Since f(α,S)=f(β,T), by Lemma 5, T=. Since S=, \((\alpha ,S),(\beta ,T){\in }\mathbf {MOD}_{k}^{\mathtt {deg}}\). Hence, f(α,S)=f(α,S) and f(β,T)=f(β,T). Since f(α,S)=f(β,T), f(α,S)=f(β,T). Thus, \((\alpha ,S){\sim _{k}}(\beta ,T)\). Consequently, g(k,σ)(α,S)=g(k,σ)(β,T).

Case S. Since f(α,S)=f(β,T), by Lemma 5, T. Since S, \((\alpha ,S),(\beta ,T){\in }\mathbf {MOD}_{k}\setminus \mathbf {MOD}_{k}^{\mathtt {deg}}\). Hence, f(α,S)=fE(α,S) and f(β,T)=fF(β,T), E being the nonempty set of n-tuples of bits such that the image by f of S ×{} is equal to E ×{} and F being the nonempty set of n-tuples of bits such that the image by f of T ×{} is equal to F ×{}. Since f(α,S)=f(β,T), fE(α,S)=fF(β,T). Thus, E=F. Since fE(α,S)=fF(β,T), \((\alpha ,S){\sim _{k}}(\beta ,T)\). Consequently, g(k,σ)(α,S)=g(k,σ)(β,T). □

Proof of Lemma 7:

By induction on ψ. Let (β,T)∈MODn. We consider the following cases.

Case ψ=xi for some i∈{1,…,n}.(i)⇒(ii) Suppose f(α,S)=(β,T) and (α,S)⊧kσ(xi) for some (α,S)∈MODk. Let \((\alpha ^{\prime },S^{\prime }){\in }\mathbf {MOD}_{k}\). Suppose \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\). Since f(α,S)=(β,T), \(f(\alpha ,S){=}f(\alpha ^{\prime },S^{\prime })\). Hence, \(g_{(k,\sigma )}(\alpha ,S){=}g_{(k,\sigma )}(\alpha ^{\prime },S^{\prime })\). Thus, by Proposition 9, (α,S)⊧kσ(xi) iff \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (x_{i})\). Since (α,S)⊧kσ(xi), \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (x_{i})\).

(ii)⇒(iii) Suppose for all (α,S)∈MODk, if f(α,S)=(β,T) then (α,S)⊧kσ(xi). Since f is surjective, let \((\alpha ^{\prime },S^{\prime }){\in }\mathbf {MOD}_{k}\) be such that \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\). Since for all (α,S)∈MODk, if f(α,S)=(β,T) then (α,S)⊧kσ(xi), \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (x_{i})\). Consequently, \(\mathbf {for}_{n}(f(\alpha ^{\prime },S^{\prime }))\) is one of the disjuncts of τ(xi). Since \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\), by Proposition 5, (β,T)⊧nτ(xi).

(iii)⇒(i) Suppose (β,T)⊧nτ(xi). Hence, let (α,S)∈MODk be such that (α,S)⊧kσ(xi) and (β,T)⊧nforn(f(α,S)). Thus, by Proposition 5, f(α,S)=(β,T).

Case ψ=⊥.(i)⇒(ii) Obviously, the condition “f(α,S)=(β,T) and (α,S)⊧kσ(⊥) for some (α,S)∈MODk cannot hold.

(ii)⇒(iii) Since f is surjective, the condition “for all (α,S)∈MODk, if f(α,S)=(β,T) then (α,S)⊧kσ(⊥)” cannot hold.

(iii)⇒(i) Obviously, the condition “(β,T)⊧nτ(⊥)” cannot hold.

Case ψχ.(i)⇒(ii) Suppose f(α,S)=(β,T) and (α,S)⊧kσχ) for some (α,S)∈MODk. Let \((\alpha ^{\prime },S^{\prime }){\in }\mathbf {MOD}_{k}\). Suppose \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\). For the sake of the contradiction, suppose \((\alpha ^{\prime },S^{\prime }){\not \models _{k}}\sigma (\neg \chi )\). Consequently, \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (\chi )\). Since f(α,S)=(β,T) and \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\), by induction hypothesis, (α,S)⊧kσ(χ). Hence, (α,S)⊮kσχ): a contradiction.

(ii)⇒(iii) Suppose for all (α,S)∈MODk, if f(α,S)=(β,T) then (α,S)⊧kσχ). For the sake of the contradiction, suppose (β,T)⊮nτχ). Thus, (β,T)⊧nτ(χ). Since f is surjective, let \((\alpha ^{\prime },S^{\prime }){\in }\mathbf {MOD}_{k}\) be such that \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\). Since for all (α,S)∈MODk, if f(α,S)=(β,T) then (α,S)⊧kσχ), \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (\neg \chi )\). Consequently, \((\alpha ^{\prime },S^{\prime }){\not \models _{k}}\sigma (\chi )\). Since \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\), by induction hypothesis, (β,T)⊮nτ(χ): a contradiction.

(iii)⇒(i) Suppose (β,T)⊧nτχ). Since f is surjective, let (α,S)∈MODk be such that f(α,S)=(β,T). For the sake of the contradiction, suppose (α,S)⊮kσχ). Hence, (α,S)⊧kσ(χ). Since f(α,S)=(β,T), by induction hypothesis, (β,T)⊧nτ(χ). Thus, (β,T)⊮nτχ): a contradiction.

Case ψ=χ𝜃.(i)⇒(ii) Suppose f(α,S)=(β,T) and (α,S)⊧kσ(χ𝜃) for some (α,S)∈MODk. Let \((\alpha ^{\prime },S^{\prime }){\in }\mathbf {MOD}_{k}\). Suppose \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\). For the sake of the contradiction, suppose \((\alpha ^{\prime },S^{\prime }){\not \models _{k}}\sigma (\chi \vee \theta )\). Consequently, neither \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (\chi )\), nor \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (\theta )\). Since f(α,S)=(β,T) and \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\), by induction hypothesis, neither (α,S)⊧kσ(χ), nor (α,S)⊧kσ(𝜃). Hence, (α,S)⊮kσ(χ𝜃): a contradiction.

(ii)⇒(iii) Suppose for all (α,S)∈MODk, if f(α,S)=(β,T) then (α,S)⊧kσ(χ𝜃). For the sake of the contradiction, suppose (β,T)⊮nτ(χ𝜃). Thus, neither (β,T)⊧nτ(χ), nor (β,T)⊧nτ(𝜃). Since f is surjective, let \((\alpha ^{\prime },S^{\prime }){\in }\mathbf {MOD}_{k}\) be such that \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\). Since for all (α,S)∈MODk, if f(α,S)=(β,T) then (α,S)⊧kσ(χ𝜃), \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (\chi \vee \theta )\). Consequently, either \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (\chi )\), or \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (\theta )\). Since \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\), by induction hypothesis, either (β,T)⊧nτ(χ), or (β,T)⊧nτ(𝜃): a contradiction.

(iii)⇒(i) Suppose (β,T)⊧nτ(χ𝜃). Since f is surjective, let (α,S)∈MODk be such that f(α,S)=(β,T). For the sake of the contradiction, suppose (α,S)⊮kσ(χ𝜃). Hence, neither (α,S)⊧kσ(χ), nor (α,S)⊧kσ(𝜃). Since f(α,S)=(β,T), by induction hypothesis, neither (β,T)⊧nτ(χ), nor (β,T)⊧nτ(𝜃). Thus, (β,T)⊮nτ(χ𝜃): a contradiction.

Case \(\psi {=}\square \chi \).(i)⇒(ii) Suppose there exists (α,S)∈MODk such that f(α,S)=(β,T) and \((\alpha ,S){\models _{k}}\sigma (\square \chi )\). Let \((\alpha ^{\prime },S^{\prime }){\in }\mathbf {MOD}_{k}\). Suppose \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\). For the sake of the contradiction, suppose \((\alpha ^{\prime },S^{\prime }){\not \models _{k}}\sigma (\square \chi )\). Consequently, let \(\gamma ^{\prime }{\in }S^{\prime }\) be such that \((\gamma ^{\prime },\emptyset ){\not \models _{k}}\sigma (\chi )\). Since f is a (k,n)-morphism and \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\), let δT be such that \(f(\gamma ^{\prime },\emptyset ){=}(\delta ,\emptyset )\). Since f is a (k,n)-morphism and f(α,S)=(β,T), let γS be such that f(γ,)=(δ,). Since \((\gamma ^{\prime },\emptyset ){\not \models _{k}}\sigma (\chi )\) and \(f(\gamma ^{\prime },\emptyset ){=}(\delta ,\emptyset )\), by induction hypothesis, (γ,)⊮kσ(χ). Hence, \((\alpha ,S){\not \models _{k}}\sigma (\square \chi )\): a contradiction.

(ii)⇒(iii) Suppose for all (α,S)∈MODk, if f(α,S)=(β,T) then \((\alpha ,S){\models _{k}}\sigma (\square \chi )\). For the sake of the contradiction, suppose \((\beta ,T){\not \models _{n}}\tau (\square \chi )\). Thus, let δT be such that (δ,)⊮nτ(χ). Since f is surjective, let \((\alpha ^{\prime },S^{\prime }){\in }\mathbf {MOD}_{k}\) be such that \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\). Since for all (α,S)∈MODk, if f(α,S)=(β,T) then \((\alpha ,S){\models _{k}}\sigma (\square \chi )\), \((\alpha ^{\prime },S^{\prime }){\models _{k}}\sigma (\square \chi )\). Since f is a (k,n)-morphism and \(f(\alpha ^{\prime },S^{\prime }){=}(\beta ,T)\), let \(\gamma ^{\prime }{\in }S^{\prime }\) be such that \(f(\gamma ^{\prime },\emptyset ){=}(\delta ,\emptyset )\). Since (δ,)⊮nτ(χ), by induction hypothesis, \((\gamma ^{\prime },\emptyset ){\not \models _{k}}\sigma (\chi )\). Consequently, \((\alpha ^{\prime },S^{\prime }){\not \models _{k}}\sigma (\square \chi )\): a contradiction.

(iii)⇒(i) Suppose \((\beta ,T){\models _{n}}\tau (\square \chi )\). Since f is surjective, let (α,S)∈MODk be such that f(α,S)=(β,T). For the sake of the contradiction, suppose \((\alpha ,S){\not \models _{k}}\sigma (\square \chi )\). Hence, let γS be such that (γ,)⊮kσ(χ). Since f is a (k,n)-morphism and f(α,S)=(β,T), let δT be such that f(γ,)=(δ,). Since (γ,)⊮kσ(χ), by induction hypothesis, (δ,)⊮nτ(χ). Thus, \((\beta ,T){\not \models _{n}}\tau (\square \chi )\): a contradiction. □

Proof of Lemma 8:

Let (β,T)∈MODk and all i∈{1,…,n}. For the sake of the contradiction, suppose either (β,T)⊧kν(xi) and f(β,T)⊮nxi, or (β,T)⊮kν(xi) and f(β,T)⊧nxi. In the former case, by definition of ν, let (α,S)∈MODk be such that f(α,S)⊧nxi and (β,T)⊧kfork(α,S). Hence, by Proposition 5, (β,T)=(α,S). Since f(α,S)⊧nxi, f(β,T)⊧nxi: a contradiction. In the latter case, by definition of ν, fork(β,T) is one of the disjuncts of ν(xi). Since by Proposition 5, (β,T)⊧fork(β,T), (β,T)⊧kν(xi): a contradiction. □

Proof of Lemma 9:

Let (β,T)∈MODk and (γ,U)∈MODn. For the sake of the contradiction, suppose either f(β,T)=(γ,U) and (β,T)⊮kν(forn(γ,U)), or f(β,T)≠(γ,U) and (β,T)⊧kν(forn(γ,U)). In the former case, since by Proposition 4, \((\gamma ,U){\models _{n}}\bar {x}^{\gamma }\), \(f(\beta ,T){\models _{n}}\bar {x}^{\gamma }\). Hence, by Lemma 8, \((\beta ,T){\models _{k}}\nu (\bar {x}^{\gamma })\). Since f is a (k,n)-morphism and by Proposition 4, for all \(\gamma ^{\prime \prime \prime },\gamma ^{\prime \prime \prime \prime }{\in }U\), \((\gamma ^{\prime \prime \prime },\emptyset ){\models _{n}}\bar {x}^{\gamma ^{\prime \prime \prime \prime }}\) iff \(\gamma ^{\prime \prime \prime }{=}\gamma ^{\prime \prime \prime \prime }\), for all \(\beta ^{\prime }{\in }T\), there exists \(\gamma ^{\prime }{\in }U\) such that \(f(\beta ^{\prime },\emptyset ){\models _{n}}\bar {x}^{\gamma ^{\prime }}\) and for all \(\gamma ^{\prime \prime }{\in }U\), there exists \(\beta ^{\prime \prime }{\in }T\) such that \(f(\beta ^{\prime \prime },\emptyset ){\models _{n}}\bar {x}^{\gamma ^{\prime \prime }}\). Thus, by Lemma 8, for all \(\beta ^{\prime }{\in }T\), there exists \(\gamma ^{\prime }{\in }U\) such that \((\beta ^{\prime },\emptyset ){\models _{k}}\nu (\bar {x}^{\gamma ^{\prime }})\) and for all \(\gamma ^{\prime \prime }{\in }U\), there exists \(\beta ^{\prime \prime }{\in }T\) such that \((\beta ^{\prime \prime },\emptyset ){\models _{k}}\nu (\bar {x}^{\prime \prime })\). Since \((\beta ,T){\models _{k}}\nu (\bar {x}^{\gamma })\), (β,T)⊧kν(forn(γ,U)): a contradiction. In the latter case, \((\beta ,T){\models _{k}}\nu (\bar {x}^{\gamma })\). Moreover, for all \(\beta ^{\prime }{\in }T\), there exists \(\gamma ^{\prime }{\in }U\) such that \((\beta ^{\prime },\emptyset ){\models _{k}}\nu (\bar {x}^{\gamma ^{\prime }})\) and for all \(\gamma ^{\prime \prime }{\in }U\), there exists \(\beta ^{\prime \prime }{\in }T\) such that \((\beta ^{\prime \prime },\emptyset ){\models _{k}}\nu (\bar {x}^{\prime \prime })\). Consequently, by Lemma 8, for all \(\beta ^{\prime }{\in }T\), there exists \(\gamma ^{\prime }{\in }U\) such that \(f(\beta ^{\prime },\emptyset ){\models _{n}}\bar {x}^{\gamma ^{\prime }}\) and for all \(\gamma ^{\prime \prime }{\in }U\), there exists \(\beta ^{\prime \prime }{\in }T\) such that \(f(\beta ^{\prime \prime },\emptyset ){\models _{n}}\bar {x}^{\gamma ^{\prime \prime }}\). Since f is a (k,n)-morphism and by Proposition 4, for all \(\gamma ^{\prime \prime \prime },\gamma ^{\prime \prime \prime \prime }{\in }U\), \((\gamma ^{\prime \prime \prime },\emptyset ){\models _{n}}\bar {x}^{\gamma ^{\prime \prime \prime \prime }}\) iff \(\gamma ^{\prime \prime \prime }{=}\gamma ^{\prime \prime \prime \prime }\), for all \(\beta ^{\prime }{\in }T\), there exists \(\gamma ^{\prime }{\in }U\) such that \(f(\beta ^{\prime },\emptyset ){=}(\gamma ^{\prime },\emptyset )\) and for all \(\gamma ^{\prime \prime }{\in }U\), there exists \(\beta ^{\prime \prime }{\in }T\) such that \(f(\beta ^{\prime \prime },\emptyset ){=}(\gamma ^{\prime \prime },\emptyset )\). Since \((\beta ,T){\models _{k}}\nu (\bar {x}^{\gamma })\), by Proposition 11, f(β,T)=(γ,U): a contradiction. □

Proof of Lemma 10:

Let (β,T)∈MODk and i∈{1,…,n}. For the sake of the contradiction, suppose either (β,T)⊧kν(τ(xi)) and (β,T)⊮kσ(xi), or (β,T)⊮kν(τ(xi)) and (β,T)⊧kσ(xi). In the former case, by definition of τ, let (α,S)∈MODk be such that (α,S)⊧kσ(xi) and (β,T)⊧kν(forn(f(α,S))). Hence, by Lemma 9, f(β,T)=f(α,S). Thus, g(k,σ)(β,T)=g(k,σ)(α,S). Consequently, by Proposition 9, (β,T)⊧kσ(xi) iff (α,S)⊧kσ(xi). Since (α,S)⊧kσ(xi), (β,T)⊧kσ(xi): a contradiction. In the latter case, by definition of τ, forn(f(β,T)) is one of the disjuncts of τ(xi). Since by Lemma 9, (β,T)⊧kν(fornf(β,T)), (β,T)⊧kν(τ(xi)): a contradiction. □

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Balbiani, P., Gencer, Ç., Rostamigiv, M. et al. About the unification type of \(\mathbf {K}+\square \square \bot \). Ann Math Artif Intell (2021). https://doi.org/10.1007/s10472-021-09768-w

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Keywords

  • Propositional modal logics
  • Locally tabular modal logics
  • Unification problem
  • Unification types

Mathematics Subject Classification (2010)

  • 03B45
  • 03B70
  • 68T27