aa r X i v : . [ m a t h . L O ] O c t ON THE BLOK-ESAKIA THEOREM FOR UNIVERSALCLASSES
MICHAŁ M. STRONKOWSKI
Abstract.
The Blok-Esakia theorem states that there is an isomor-phism from the lattice of intermediate logics onto the lattice of nor-mal extensions of Grzegorczyk modal logic. The extension for multi-conclusion consequence relations was obtained by Emil Jeřábek as anapplication of canonical rules. We show that Jeřábek’s result followsalready from Blok’s algebraic proof.We also prove that the properties of strong structural complete-ness and strong universal completeness are preserved and reflects bythe aforementioned isomorphism. These properties coincide with struc-tural completeness and universal completeness respectively for single-conclusion consequence relations and, in particular, for logics. Introduction
Let us consider the modal formula grz = ( ( p → p ) → p ) → p. Let GRZ be the normal extension of S4 modal logic axiomatized by grz [27].Let INT be the intuitionistic logic.The celebrated Blok-Esakia theorem states that there is an isomorphismfrom the lattice of extensions of INT onto the lattice of normal extensionsof GRZ [7, 19]. Algebraically, it means that there is an isomorphism fromthe lattice of varieties of Heyting algebras onto the lattice of varieties ofmodal Grzegorczyk algebras. (The reader may find a more detailed historyof classical interpretations of intuitionistic logic in papers [11, 37, 43].)Nowadays we have various proofs of this result. The one of Blok is purelyalgebraic. Esakia used Esakia spaces, i.e., certain topological relational struc-tures. There is also akharyashchev’s proof based on his canonical formulas[10, 44].There are many extensions of the Blok-Esakia theorem, see a recent survey[43] and the references therein. Let us recall one of them. In [31] Jeřábekextended the Blok-Esakia theorem to multi-conclusion consequence relations
Mathematics Subject Classification.
Key words and phrases.
Blok-Esakia theorem, multi-conclusion consequence relations,universal classes, Heyting algebras, Grzegorczyk algebras, structural completeness, deduc-tion theorem.The work was supported by the Polish National Science Centre grant no. DEC-2011/01/D/ST1/06136. (called there rule systems). It means that he allows not only axiomaticextensions, but also extensions obtained by adding new inference rules (withpossibly many conclusions). Algebraically Jeřábek’s result says that there isan isomorphism from the lattice of universal classes of Heyting algebras ontothe lattice of universal classes of modal Grzegorczyk algebras. Jeřábek’sproof is based on canonical rules, a tool which extents Zakharyashchev’scanonical formulas. The starting point of this project was an observationthat actually Jeřábek’s result follows directly from Blok’s algebraic proof.There is a quite developed algebraic theory for single-conclusion conse-quence relations, see e.g. [22, 21, 17]. However multi-conclusion conse-quence relations, despite being known for decades [40], seems to be ne-glected. They appear naturally in proof theory in the context of sequentcalculi [38]. But there they are used mainly as a handy tool for describ-ing logics and are not object of studies per se. The situation has beenchanged with Jeřábek’s paper [31] and his observation that multi-conclusioninference rules may be used for the canonical axiomatization of intermedi-ate and modal logics. This topic was recently undertaken in many papers[2, 4, 3, 5, 6, 13, 14, 15, 16, 24, 30, 29, 32].The paper contains a full presentation of an algebraic proof of Jeřábek’sresult. We simplify one point of Blok’s reasoning. Namely, we provide ashort proof that free Boolean extensions of Heyting algebras are Grzegorczykalgebras (this is mainly done in Section 3). Along the line, we also obtaina new proof of Blok’s characterization of Grzegorczyk algebras. Still, themain technical ingredient is the Blok lemma which connects free Booleanextensions with Grzegorczyk algebras (its full proof is given in the appendix).We failed to find any essential simplification of its tricky proof.We also undertake the problem of preservation and reflection of someproperties by the Blok-Esakia isomorphisms. For logics the topic was thor-oughly investigated in the past and summarized in the survey [11]. We verifypreservation and reflection of strong structural completeness and strong uni-versal completeness. For single-conclusion consequence relations and logicsthey are equivalent to structural completeness and universal completenessrespectively. For logics this fact was proved by Rybakov [39]. (We used itin [18] in order to construct a normal extension of S4 modal logic which isalmost structurally complete but does not have projective unification.)2.
Multi-conclusion consequence relations
Let us fix a language L (i.e., an infinite set of variables and a set of symbolsof operations with ascribed arrives). Let Form be the algebra of all formulas(terms) in L . An inference rule (in L ) is an ordered pair, written as Γ / ∆ ,of finite subsets of F orm . A set of inference rules, written as a relation ⊢ , is a multi-conclusion consequence relation , mcr in short (in [6, 31] it iscalled a rule system ) if for every finite subsets Γ , Γ ′ , ∆ , ∆ ′ of F orm , for every
N THE BLOK-ESAKIA THEOREM FOR UNIVERSAL CLASSES 3 ϕ ∈ F orm and for every substitution (i.e., an automorphism of
Form ) σ the following conditions are satisfied • { ϕ } ⊢ { ϕ } ; • if Γ ⊢ ∆ , then Γ ∪ Γ ′ ⊢ ∆ ∪ ∆ ′ ; • if Γ ⊢ ∆ ∪ { ϕ } and Γ ∪ { ϕ } ⊢ ∆ , then Γ ⊢ ∆ ; • if Γ ⊢ ∆ , then σ (Γ) ⊢ σ (∆) .In this paper we are interested only in intermediate and modal mcrs. Let K denotes modal logic K and IN T denotes the intuitionistic logic, bothinterpreted as set of formulas. Then an intermediate mcr is an mcr ⊢ in thelanguage of IN T such that ∅ ⊢ { ϕ } for every ϕ ∈ IN T and { p, p → , q } ⊢ q .And a modal mcr is an mcr ⊢ in the language of K such that ∅ ⊢ { ϕ } forevery ϕ ∈ K , { p, p → , q } ⊢ q and { p } ⊢ { p } . A (general, intermediateor modal) mcr is axiomatized by a set R of inference rules if it is a least(general, intermediate or modal receptively) mcr containing R .Intermediate and modal mcrs have algebraic semantics. Let us recall thata Heyting algebra is a bounded lattice endowed with the binary operation → such that a ∧ b c iff a b → c for every triple a, b, c of its elements. Itappears that the class of all Heyting algebras form a variety which constitutesa semantics for IN T . A modal algebra is a Boolean algebra endowed withadditional unary operation such that for all its elements a, b we have ( a ∧ b ) = a ∧ b and . The variety of modal algebras gives asemantics for K .We say that a class U of algebras is universal iff it is axiomatizable by firstorder sentences of the form ( ∀ ¯ x )[ s ≈ s ′ ⊓ · · · ⊓ s m ≈ s ′ m ⇒ t ≈ t ′ ⊔ · · · ⊔ t n ≈ t ′ n ] , where n and m are natural numbers not both equal to zero and s i , s ′ i , t j , t ′ j are arbitrary terms (We use the symbol ⊓ for first order conjunction, ⊔ forfirst order disjunction and ⇒ for first order implication. The symbols ∧ , ∨ and → will denote operations in algebras.) We call such formulas disjunctiveuniversal sentences . When m = 0 we talk about disjunctive universal positivesentences , when n = 1 about quasi-identities , when m = 0 and n = 1 about identities . Recall also that a class of algebras is a universal positive class ifit is axiomatizable by disjunctive universal positive sentences, a quasivariety if it is axiomatizable by quasi-identities, and a variety if it is axiomatizableby identities.Let r be an inference rule { ϕ , . . . , ϕ m } / { ψ , . . . , ψ n } . By the translation of r we mean the disjunctive universal sentence T ( r ) given by ( ∀ ¯ x )[ ϕ ≈ ⊓ · · · ⊓ ϕ m ≈ ⇒ ψ ≈ ⊔ · · · ⊔ ψ n ≈ . An inference rule Γ / ∆ is a single-conclusion inference rule if | ∆ | = 1 , amulti-theorem if Γ = ∅ , and a theorem if Γ = ∅ and | ∆ | = 1 . The translationof a single-conclusion inference rule is a quasi-identity, of a multi-theorem isa disjunctive universal positive sentence, and of theorem is an identity. MICHAŁ M. STRONKOWSKI
The following completeness theorem follows from [31, theorem 2.2], seealso [6, Theorem 2.5 in Appendix] for the modal case.
Theorem 2.1.
Let ⊢ be an intermediate or modal mcr axiomatized by aset of inference rules R . Let U ⊢ be the universal class of Heyting or modalalgebras respectively axiomatized by T ( R ) . Then for every inference rule r we have r ∈ ⊢ if and only if U ⊢ | = T ( r ) . Note that the choice of the axiomatizing set R of inference rules in Theo-rem 2.1 is arbitrary. Indeed, whichever axiomatizing set for ⊢ we choose, weobtain the same class U ⊢ Moreover, the assignment ⊢ 7→ U ⊢ is injective. Inaddition, every disjunctive universal sentence in the language of Heyting ormodal algebras is equivalent to some T ( r ) in the class of all Heyting or modalalgebras. Thus the assignment ⊢ 7→ U ⊢ is also surjective. Hence the abovefacts allows us to switch completely in further considerations from mcrs touniversal classes We say that ⊢ and U ⊢ correspond to each other.3. Grzegorczyk algebras
A modal algebra M is called an interior algebra if for every a ∈ M itsatisfies a = a a. An element a of M is open M if a = a . Recall that for a modal algebrathere is one to one correspondence between its congruences and its openfilters, i.e., Boolean filters closed under operation. It is given by θ /θ and F θ F , where ( a, b ) ∈ θ F iff a ↔ b ∈ F . (Here θ is a congruence and F is an open filter.) In particular, for an interior algebra, an element b belongsto the open filter generated by a iff a b . It follows that an interior algebrais subdirectly irreducible iff it has a largest non-top open element. And aninterior algebra is simple iff it has exactly two open elements 0 and 1.An interior algebra M is a Grzegorczyk algebra if it also satisfies(Grz) ( ( a → a ) → a ) a for every a ∈ M [20].Recall that the variety of interior/Grzegorczyk algebras characterizes themodal logic S4/GRZ.The condition (Grz) is rather difficult for “intuitive understanding”. Nev-ertheless, some semantical characterizations were found. It is known thatmodal frame validates grz iff it is Noetherian partially ordered set [10, The-orem 3.38] (however see [33] for set theoretical subtleties). Zakharyaschevshowed that a general transitive frame validates grz iff it is not subreducibleto a modal frame which consists of one irreflexive point, and to a modalframe which consists of two points with the total relation (two-element clus-ter) [10, Proposition 9.3], see also [42] for an algebraic proof. Moreover, theclass of Grzegorczyk algebras is an intersection of two splitting subvarietiesof interior algebras. More precisely, Blok proved in [7] that an interior alge-bra is a Grzegorczyk algebra iff it does not have a subalgebra admitting a N THE BLOK-ESAKIA THEOREM FOR UNIVERSAL CLASSES 5 homomorphisms onto S or S , (Corollary 3.4). These algebras are depictedin Figure 1 (open elements are depicted by ). ◦ ◦ ❄❄❄❄❄❄❄ ⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧ ❄❄❄❄❄❄❄ ◦ ◦ ◦◦ ◦ ❄❄❄❄❄❄❄ ⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧❄❄❄❄❄❄❄❄ ⑧⑧⑧⑧⑧⑧⑧⑧❄❄❄❄❄❄❄⑧⑧⑧⑧⑧⑧⑧ ❄❄❄❄❄❄❄ Figure 1.
Interior algebras S and S , .It appears that the variety of Grzegorczyk algebras is generated, even asa universal class, by the class of all interior algebras which are generated byopen elements. Such algebras are isomorphic to free Boolean extensions ofHeyting algebras, see Section 4. In this section we provide a new proof of thefact that every interior algebra which is generated by open elements is a Grze-gorczyk algebra (Corollary 3.3). The fact that the universal class generatedby the free Boolean extensions of Heyting algebras is not a proper subclassof Grzegorczyk algebras follows from the Blok lemma. (Alternatively, onemay use filtration and the fact that all finite Grzegorczyk algebras are gen-erated by open elements. But we do not present this approach here). Asa byproduct, we obtain new algebraic proofs of the above mentioned Blok’scharacterization of Grzegorczyk algebras.Let us start with recalling a crucial notion of stable homomorphisms. Let M and N be modal algebras. We say that a mapping f : N → M is a stablehomomorphism from N into M if it is a Boolean homomorphism and f ( a ) f ( a ) holds for every a ∈ M . The reader may consult e.g. [2, 23] for the im-portance of stable homomorphisms in modal logic (in [23] they are called continuous morphism ). Note that a stable homomorphism does not need tobe a homomorphism. For instance S does not admit a homomorphism ontoa two-element interior algebra, but it admits such stable homomorphisms. Lemma 3.1.
Let M be an interior algebra and S be a simple interior algebra.Then the mapping f : M → S is a stable homomorphism if and only if it isa Boolean homomorphism and for every a ∈ M we have f ( a ) ∈ { , } .Proof. Assume that f is a stable homomorphism. By the definition, f is aBoolean homomorphism. Moreover, if f ( a ) < then f ( a ) = f ( a ) f ( a ) = 0 . For the opposite implication assume that f is a Booleanhomomorphism and f ( a ) = { , } . If f ( a ) = 0 then, clearly, f ( a ) f ( a ) . If f ( a ) = 1 then, since a a , f ( a ) = 1 , and so f ( a ) = 1 = f ( a ) . (cid:3) MICHAŁ M. STRONKOWSKI
Proposition 3.2.
Let M be an interior algebra and a ∈ M . If (Grz) for a in M , then there exists a stable homomorphism h from M onto S such that h ( a ) is a (co)atom in S .Proof. Let us consider the term t ( x ) = ( x → x ) → x. Let G be a maximal open filter of M with respect to the following conditions a G and t ( a ) ∈ G. Since the inequality (Grz) does not hold, the Zorn lemma guarantees theexistence of G . Let M ′ = M /G , g : M → M ′ ; d d/G , and a ′ = g ( a ) = a/G . Then (Grz) does not hold in M ′ for a ′ either. Furthermore, M ′ issubdirectly irreducible and a least nontrivial open filter of M is the openfilter generated by a ′ . It follows that all open non-top elements in M ′ are inthe interval [0 , a ′ ] .Let S be the interior algebra constructed in the following way. Its Booleanreduct is a quotient of Boolean reduct of M ′ divided by the Boolean filter F generated by ¬ a ′ . We equip S with only two open elements: the top andthe bottom ones. This means that b = 0 iff b < and in S . Let f : M ′ → S ; b b/F . Clearly, f is a Boolean surjective homomorphism.Moreover, f − (0) = [0 , a ′ ] . Hence Lemma 3.1 yields that f is a stablehomomorphism from M ′ onto S .Let a ′ b, c and b = c . Then a ′ b ↔ c < and hence ¬ a ′ b ↔ c .Thus the pair ( b, c ) is not in the Boolean congruence associated with F . Thisshows that restriction of f to the interval [ a ′ , is injective. We prove thatthis interval has more than two elements.For every b ∈ M ′ we have t ( b ) = b. But in M ′ we have t ( a ′ ) = 1 > a ′ . Thus a ′ is not open and hence a ′ < a ′ < . This and the fact that the restriction of f onto [ a ′ , is injective yieldsthat S has at least four elements. In particular, there exists a Booleanhomomorphism k from the Boolean reduct of S onto the Boolean reduct of S . Clearly, k is also a stable homomorphism. Also, we may choose k suchthat kf g ( a ) is a (co)atom in S .Finally, a composition of stable surjective homomorphisms is a stable sur-jective homomorphism. Thus the composite mapping kf g is a stable homomorphism from M onto S . (cid:3) The converse of Proposition 3.2 does not hold. Indeed, let M be an interioralgebra where the carrier M is the power set of the set N of natural numbers, N THE BLOK-ESAKIA THEOREM FOR UNIVERSAL CLASSES 7 the Boolean operations are the set theoretic operations, and a = ( if a = 1 { , . . . , k − } if a = 1 and k = min ¬ a . Note that M is the dual algebra for the modal frame ( N , > ) in the Jónsson-Tarski duality. Then M is a Grzegorczyk algebra admitting a stable homo-morphism onto S , see [42, Example 3.5] for details. However, the converse ofProposition 3.2 holds for finite interior algebras. Indeed, it follows from Za-kharyaschev’s characterization of Grzegorczyk frames, see also [42, Theorem2.4]. Corollary 3.3.
Let M be an interior algebra generated by its open element.Then M is a Grzegorczyk algebra.Proof. Let f be a stable homomorphism from M into S . By Lemma 3.1, forevery a ∈ M we have f ( a ) ∈ { , } . Since f preserves Boolean operations, { , } is closed under Boolean operations and in fact the algebra M is alsoBoolean generated by its open elements, f maps M onto { , } . Thus f isnot surjective. (cid:3) Let us also show that the proof of Proposition 3.2 may be slightly modifiedin order to obtain a new proof of Blok’s characterization for Grzegorczykalgebras.
Corollary 3.4 ([7, Example III.3.9]) . An interior algebra M is a Grzegor-czyk algebra if and only if it does not have a subalgebra with a homomorphicimage isomorphic to S or to S , .Proof. Assume that M is an interior algebra and a ∈ M are such that (Grz)does not hold for a . Let M ′ and a ′ be as in the proof of Proposition 3.2. Let N be a subalgebra of M ′ generated by a ′ . We claim that N is isomorphic to S or to S , .If a ′ = 0 , then N is simple and hence, since it is generated by a ′ , isisomorphic to S . So let us assume that < a ′ < a ′ < . Let P be the carrier of the eight-element Boolean algebra generated by thischain. This means that P = { a ′ , a ′ , ¬ a ′ , ¬ a ′ , ¬ a ∨ a ′ , a ∧ ¬ a ′ , , } . In order to show that N is isomorphic to S , it is enough to verify that ¬ a ′ = 0 and ( ¬ a ′ ∨ a ′ ) = a ′ . This, in particular, would show that P is closed under the operation and, since P is closed under the Booleanoperations, P = N .Here we will use the fact that a ′ is the largest not equal to 1 open elementin N . So ¬ a ′ = ¬ a ′ ∧ a ′ = ¬ a ′ ∧ a ′ = ( ¬ a ′ ∧ a ′ ) = MICHAŁ M. STRONKOWSKI and ( ¬ a ′ ∨ a ′ ) = ( ¬ a ′ ∨ a ′ ) ∧ a ′ = (( ¬ a ′ ∨ a ′ ) ∧ a ′ ) = a ′ = a ′ . For the converse, note that (Grz) fails for any coatom in S and in S , .Thus it fails for some element in every algebra having subalgebra with ahomomorphic image isomorphic to S or to S , . (cid:3) The Blok-Esakia theorem
The connection of Heyting algebra with interior algebra is given by thefollowing McKinsey-Tarski theorem [36, Section 1] (see also [7, Chapter 1],[8, Theorem 2.2] and [35, Section 3]). Recall that open elements of an interioralgebra M form the Heyting algebra O ( M ) with the order structure inheritedfrom M . Theorem 4.1.
For every Heyting algebra H there is an interior algebra B ( H ) such that (1) OB ( H ) = H ; (2) for every interior algebra M , if H O ( M ) , then B ( H ) is isomorphicto the subalgebra of M generated by H ; (3) for every interior algebra M and every homomorphism f : H → O ( M ) there is a unique homomorphism ¯ f : B ( H ) → M extending f . The algebra B ( H ) is called the free Boolean extension of H . We will treat O and B as a class operators. This is, for a class K of Heyting algebrasand a class M of interior algebras we put B ( K ) = { B ( H ) | H ∈ K} and O ( M ) = { O ( M ) | M ∈ M} . From Theorem 4.1, we immediately obtain OB ( K ) = K and BO ( M ) ⊆ S ( M ) . Let us list their basic properties with respect to other class operators: H homomorphic image, S subalgebra, P product, and P U ultraproduct classoperators. (We tacitly assume that all class operators are composed withthe isomorphic image class operator, i.e, S = IS and so on.) Lemma 4.2.
Let K be a class of Heyting algebras and M be a class ofinterior algebras. Then OH ( M ) = HO ( M ) , BH ( K ) = HB ( K ) , OS ( M ) = SO ( M ) , BS ( K ) = SB ( K ) , OP ( M ) = PO ( M ) , BP ( K ) ⊆ SPB ( K ) , OP U ( M ) = P U O ( M ) , BP U ( K ) ⊆ SP U B ( K ) . Proof.
The equalities from the first column follows directly from the defini-tions. Let us illustrate this by verifying the fourth one. Let M = Q i M i /U and H = Q i H i /U be ultraproducts of interior algebras and Heyting algebrasrespectively, where H i = O ( M i ) . Let α and β be congruences of Q i M i and N THE BLOK-ESAKIA THEOREM FOR UNIVERSAL CLASSES 9 Q i H i respectively induced by the ultrafilter U . Then β = α ∩ ( Q i H i ) . Itfollows that the mapping f : c/β c/α is an embedding of H into O ( M ) . Inorder to see that f is surjective, note that for c ∈ Q i M i we have the equiv-alences: c/α is open iff { i | c i = c i } ∈ U iff c α c ′ , where c ′ i = c i if c i = c i and c ′ i = 1 otherwise. Hence, if c/α is open, then c/α = c ′ /α = f ( c ′ /β ) .The equalities from the second column follows from Theorem 4.1. For thefirst containment, note that BP ( K ) = BPOB ( K ) = BOPB ( K ) ⊆ SPB ( K ) . Indeed, here the first equality and the containment follows from the proper-ties of B and O recalled before the lemma, and the second equality followsfrom the third equality in the first column. The proof of the second inclusionis analogical. (cid:3) Let H be the variety of all Heyting algebras, and G be the variety of allGrzegorczyk algebras. Let L V ( H ) and L V ( G ) denote the lattices of vari-eties of Heyting algebras and Grzegorczyk algebras respectively. Recall twooperators on these lattices: ρ : L V ( G ) → L V ( H ); W 7→ O ( W ) σ : L V ( H ) → L V ( G ); V 7→ VB ( V ) , where V ( K ) denotes a least variety containing K . (The analog of this op-erator for logics was introduced in [35].) Notice that ρ is well defined, i.e.,when W is a variety of interior algebras, then ρ ( W ) is indeed the variety ofHeyting algebras. Indeed, this follows from Lemma 4.2. A much less obvi-ous fact is that if V is a variety of Heyting algebras, then σ ( V ) is a varietyof Grzegorczyk algebras. This fact follows from the following Proposition(based on the previous section). Proposition 4.3 ([7, Corollary III.7.9]) . Let H be a Heyting algebra. Then B ( H ) is a Grzegorczyk algebra.Proof. By Theorem 4.1 the class of interior algebras isomorphic to algebrasof the form B ( H ) coincides with the class of interior algebras generated byits open elements. Thus the assertion follows from Corollary 3.3. (cid:3) Let us now formulate the Blok-Esakia theorem [7, Theorem 7.10], [20,Theorem 7.11] (see aslo [1, Section 2], [10, Theorem 9.66], [43, Section 3]) inalgebraic terms.
Blok-Esakia Theorem.
The mappings σ and ρ are mutually inverse iso-morphisms between the lattices L V ( H ) and L V ( G ) . The hardness of the algebraic proof of the Blok-Esakia theorem lies in inthe fact that not all Grzegorczyk algebras are representable as B ( H ) . Thefollowing lemma, doe to Blok, shows how to overcome this difficulty. (Theoriginal, rather technical formulation, may be found in [7, Lemma III.7.6],[43, Lemma 2]. Our presentation is simpler in use.) Blok Lemma.
Let M be a Grzegorczyk algebra. Then M embeds into someelementary extension of BO ( M ) . Our formulation follows from the original one. However, for the seek ofcompleteness, we provide its proof in Appendix. Note that it is essentiallythe same proof. It is just written in a bit different fashion. Here we providea simple proof for the finite case [7, Theorem II.2.11]. We failed to find asemantical transparent proof for the general case.
Blok Lemma: finite case.
Let M be a finite Grzegorczyk algebra. Then M is isomorphic to BO ( M ) .Proof. Let us first prove that if a is an element in M and a < , theneither covers a or there exists an element b in M such that a < b < .(This holds for an arbitrary Grzegorczyk algebras.) So assume that it is notthe case, i.e., that there exists an interval [ a, in M with more than elements and with exactly two open elements a and . We may assumethat a < a . Then (Grz) fails in M for a . Indeed, since a < a < , a < ¬ a ∨ a < . Hence ( a → a ) = a and ( ( a → a ) → a ) = a. This and the finiteness of | M | yields that either M is trivial or has an opencoatom. We use this fact to show that there is a maximal chain in M whoseall elements are open. This and the finiteness then imply that M is generatedby open elements.We proceed by induction on the cardinality of M . If | M | = 1 then M has only one element which is open. Thus the thesis holds. So assumethat M is a finite nontrivial Grzegorczyk algebra and the thesis holds forall smaller Grzegorczyk algebras. Let c be an open coatom in M and let F = { , c } . Then F is an open filter. By the induction assumption, thereis a maximal chain { a /F, . . . , a k /F } in M /F consisting of open elementsin M /F . Assume that each a i is chosen as a minimal element in a i /F , i.e., a i c . Then a i /F = { a i , a i ∨ ¬ c } and at least one of this two elementsis open. Actually, every a i is open, since if a i ∨ ¬ c is open then a i = (( a i ∨ ¬ c ) ∧ c ) = ( a i ∨ ¬ c ) ∧ c = ( a i ∨ ¬ c ) ∧ c = a i . Thus { , a , . . . , a k } is a maximal chain in M consisting of open elements. (cid:3) Let L U ( H ) and L U ( G ) be the lattices of all universal classes of Heytingalgebras and Grzegorczyk algebras respectively. Let us define two operatorson these lattices. For U ∈ L U ( H ) and Y ∈ L U ( G ) put ρ ′ : L U ( G ) → L U ( H ); Y 7→ { O ( M ) | M ∈ Y} σ ′ : L U ( H ) → L U ( G ); U 7→ U ( { B ( H ) | H ∈ U } ) , where U = SP U . Recall that if K is a class of algebras in the same signature,then SP U ( K ) is a least universal class containing K [9, Theorem V.2.20]. N THE BLOK-ESAKIA THEOREM FOR UNIVERSAL CLASSES 11
Since O commutes with P U and S (Lemma 4.2), ρ ( Y ) is an universal classfor Y ∈ L U ( G ) . Clearly, thus defined ρ ′ is an extension of previously definedoperator ρ for varieties. Note also that, by Proposition 4.3, σ ′ ( Y ) ∈ L U ( G ) for Y ∈ L U ( H ) . It follows from Lemma 4.5 that σ ′ is also an extension of σ . Lemma 4.4.
Let
U ∈ L U ( H ) . Then ρ ′ σ ′ ( U ) = U . In particular, M ∈ σ ′ ( U ) implies O ( M ) ∈ U .Proof. By the equalities from Lemma 4.2 and Theorem 4.1 point (1), wehave ρ ′ σ ′ ( U ) = OSP U B ( U ) = SP U OB ( U ) = SP U ( U ) = U . (cid:3) Lemma 4.5.
Let K be a universal class of Heyting algebras. Then K is aquasivariety/universal positive class/variety if and only if σ ′ ( K ) is quasivari-ety/universal positive class/variety respectively. In particular, σ ′ ( V ) = σ ( V ) for every variety V of Heyting algebras.Proof. Recall first that a class of algebras closed under isomorphic imagesis a quasivariety iff it is closed under S , P and P U [9, Theorem V.2.25], is auniversal positive class iff it is closed under S , H and P U [25, Corollary 2 ofTheorem 4 in Section 43], a variety if is closed under S , H and P [9, TheoremII.11.9].Thus the backward implication follows from Lemmas 4.2 and 4.4.For the forward implication it is enough to show that if K is a universalclass closed under P or H , then σ ′ ( K ) is also closed under P or H respectively.Assume first that K is closed under the P operator. Let M i ∈ σ ′ ( K ) and M = Q M i . By Lemma 4.4 and the assumption, O ( M ) = Q O ( M i ) ∈ K .Thus by the Blok lemma, M ∈ σ ′ ( K ) .Assume now that K is closed under the H operator. Let h : N → M , were N ∈ σ ′ ( K ) , be a surjective homomorphism. Let O ( h ) be the restriction of h to the carrier of O ( M ) . Then O ( h ) : O ( N ) → O ( M ) is also a surjectivehomomorphism. By Lemma 4.4, O ( N ) ∈ K , and by the assumption, O ( M ) ∈K . Hence the Blok lemma yields that M ∈ σ ′ ( K ) . (cid:3) From now on, we notationally identify σ ′ with σ and ρ ′ with ρ . Blok-Esakia Theorem extended version ([31, Theorem 5.5]) . The map-pings σ and ρ are mutually inverse isomorphisms between the lattices L U ( H ) and L U ( G ) .Proof. For
Y ∈ L U ( G ) the inclusion Y ⊆ σρ ( Y ) follows from the Blok lemma.The inclusion σρ ( Y ) ⊆ Y follows from Theorem 4.1 point (2) and the factthat Y is closed under taking subalgebras. For U ∈ L U ( H ) the equality U = ρσ ( U ) follows from Lemma 4.4. (cid:3) Let L Q ( H ) / L U + ( H ) and L Q ( G ) / L U + ( G ) be lattices of quasivarieties/universalpositive classes of Heyting algebras and Grzegorczyk algebras respectively. Blok-Esakia Theorem restricted versions ([31, Theorem 5.5]) . The ap-propriate restrictions of σ and ρ are mutually inverse isomorphisms betweenthe lattices L U ( H ) / L U + ( H ) / L Q ( H ) / L V ( H ) and the lattices L U ( G ) / L U + ( G ) / L Q ( G ) / L V ( G ) respectively.Proof. It follows from the Blok-Esakia theorem extended version and Lemma4.5. (cid:3) Strong structural completeness
Structural completeness and universal completeness are well establishedproperties for single-conclusion consequence relations and quasivarieties. How-ever its extension to multi-conclusion consequence relations and to universalclasses is recent [30]. Here we focus on a connected new properties of strongstructural completeness and strong universal completeness. For dealing alge-braically with the structural and universal completenesses for mcrs we needto introduce a new notion of free families. It is done in a separate paper [41].We prove that the Blok-Esakia isomorphism for universal preserves andreflects strong structural and universal completenesses. For (quasi)-varietiesstrong versions are equivalent to standard ones. For varieties preservationand reflection of structural completeness was proved in [39, Theorem 5.4.7].In our opinion, the proof presented here is simpler and more general.For an mcr ⊢ and an inference rule r let ⊢ r be a least mcr extending ⊢ and containing r . An inference rule r is weakly admissible for ⊢ if { ψ ∈ F orm | ∅ ⊢ { ψ }} = { ψ | ∅ ⊢ r { ψ }} , and admissible for ⊢ if { ∆ a finite subset of F orm | ∅ ⊢ ∆ } = { ∆ | ∅ ⊢ r ∆ } . In other words, r is (weakly) admissible if ⊢ and ⊢ r share the same (multi-)theorems.A mcr is structurally/universally complete if the sets of its admissible andderivable single-conclusion/multi-conclusion inference rules coincide. Anda mcr is strongly structurally/universally complete if the sets of its weaklyadmissible and derivable single/multi-conclusion inference rules coincide.Let us summarize these properties in the following table.weakly admissible admissiblesingle-conclusion strong structural completeness structural completenessmulti-conclusion strong universal completeness universal completenessAll these notions have algebraic counterparts. A universal disjunctivesentence v is weakly admissible for a universal class U if the sets of identitiessatisfied in U and in the class of all algebras from U satisfying v coincide.And v is admissible for U if the sets of positive universal disjunctive sentencessatisfied in U and in the class of all algebras from U satisfying v coincide. N THE BLOK-ESAKIA THEOREM FOR UNIVERSAL CLASSES 13
Analogically, a universal class U is ( strongly ) structurally complete if theset of (weakly) admissible for U quasi-identities and the set of quasi-identitieswhich hold in U coincide. And U is ( strongly ) universally complete if the setof (weakly) admissible for U disjunctive universal sentences and the set ofdisjunctive universal sentences which hold in U coincide. The following factfollows from Theorem 2.1. Fact 5.1.
Let ⊢ be a modal or intermediate mcr, U be a universal classof modal or Heyting algebras respectively, and let r be an inference rulein the language of ⊢ . Assume that ⊢ and U correspond to each other (inthe sense of Theorem 2.1). Then r is (weakly) admissible for ⊢ if andonly if its translation T ( r ) is (weakly) admissible for U . Consequently, ⊢ is (strongly) structurally/universally complete if and only if U is (strongly)structurally/universally complete. We recall the notion of free algebras. Let V be a set. Let F be an algebrasuch that V ⊆ F and K be an arbitrary class of algebra in the same signatureas F . We say that F has the universal mapping property for K over V if forevery algebra A ∈ K and every mapping f : V → A there exists a uniquehomomorphism f ′ : F → A such that f ′ | V = f . Clearly, such algebra F isnot uniquely determined. Moreover, it does not need to be the case that F ∈ K . Therefore additional condition is considered. An algebra F is freefor K over V if it has the universal mapping property for K over V and F belongs to the variety generated by K . Then for every nonempty set V and every class K of algebras in the same language containing a nontrivialalgebra there exists a free algebra F for K over V . Moreover, an algebra F is uniquely determined up to isomorphism whose restriction to V is anidentity.For a class K of algebras in a fixed language, we write Q ( K ) and V ( K ) todenote a least quasivariety and a least variety respectively containing K . Incase when K consist of one algebra A we also write Q ( A ) and V ( A ) . Let ussummarize the needed properties of free algebras in the following fact. Fact 5.2.
Let F be free for K over a nonempty set V . Then (1) F ∈ Q ( K ) ([9, Theorem II.10.12]) ; (2) F is also free over V for V ( K ) ([9, Corollary II.11.10] , see also [34,Proposition 4.8.9] for a more direct argument ) ; (3) if V is infinite and ( ∀ ¯ x ) e (¯ x ) is an identity, then K | = ( ∀ ¯ x ) e (¯ x ) ifand only if F | = ( ∀ ¯ x ) e (¯ x ) if and only if F | = e (¯ v ) , where ¯ v is a tupleof mutually distinct elements from V of the same length as ¯ x . ([9,Theorem II.11.4]) . For a sentence ϕ let us denote by Mod( ϕ ) the class of all algebras, in thesame fixed language as ϕ , satisfying ϕ . Proposition 5.3.
Let U be a universal class, F be a free algebra for U overan infinite set, and v be a universal sentence. Then the following conditionsare equivalent (1) v is weakly admissible for U ; (2) V ( U ∩
Mod( v )) = V ( U ) ; (3) F ∈ Q ( U ∩
Mod( v )) .Proof. The definition of weak admissibility may be stated as { e is an identity | U ∩ Mod( v ) | = e } ⊆ { e is an identity | U | = e } . This shows that (1) is equivalent to (2). By Fakt 5.2 point (3), the condition(2) is equivalent to F ∈ V ( U ∩
Mod( v )) . Since F is free for U , it has the universal mapping property for U ∩
Mod( v ) .Thus the last condition is equivalent to F being free for U ∩
Mod( v ) . ThusFact 5.2 point (1) yields that it is equivalent to (3). (cid:3) Proposition 5.4.
Let U be a universal class and F be a free algebra for U over an infinite set. Then the following conditions are equivalent (1) U is strongly structurally complete; (2) for every quasi-identity q F ∈ Q ( U ∩
Mod( q )) yields U | = q ; (3) for every subquasivariety Q of Q ( U ) F ∈ Q ( U ∩ Q ) yields U ⊆ Q . Proof.
The equivalence (1) ⇔ (2) follows directly from Proposition 5.3 andthe definition of strong structural completeness.The implication (3) ⇒ (2) may be obtained by considering the quasivariety Q = Q ( U ) ∩ Mod( q ) .For the implication (2) ⇒ (3), let us assume that F ∈ Q ( U ∩ Q ) . Let Σ bethe set of all quasi-identities satisfied in Q . Then for every q ∈ Σ , we have F ∈ Q ( U ∩
Mod( q )) . Thus (2) yields U ⊆ T q ∈ Σ Mod( q ) = Q . (cid:3) Similarly, one may prove the following fact.
Proposition 5.5.
Let U be a universal class and F be a free algebra for U over an infinite set. Then the following conditions are equivalent (1) U is strongly universally complete; (2) for every disjunctive universal sentence v F ∈ Q ( U ∩
Mod( v )) yields U | = v ; (3) for every universal subclass W of U F ∈ Q ( W ) yields U = W . Lemma 5.6.
Let K be a class of Heyting algebras. Then σ V ( K ) = VB ( K ) and σ Q ( K ) = QB ( K ) . Hence, if U is a universal class of Heyting algebras, then σ V ( U ) = V σ ( U ) and σ Q ( U ) = Q σ ( U ) . N THE BLOK-ESAKIA THEOREM FOR UNIVERSAL CLASSES 15
Proof.
Recall that V ( K ) = HSP ( K ) and Q ( K ) = SPP U ( K ) [9, TheoremsII.9.5 and V.2.25]. Thus the inclusions σ V ( K ) ⊆ VB ( K ) and σ Q ( K ) ⊆ QB ( K ) follows from Lemma 4.2. By Lemma 4.5, σ V ( K ) is a variety and σ Q ( K ) is aquasivariety both containing B ( K ) . Thus the opposite inclusion follows. (cid:3) The following lemma, in the case when U is the class of all Heyting algebra,for was proved in [36, Theorem 3.16] Lemma 5.7.
Let U be a universal class of Heyting algebra, F be a freealgebra for U over V and F σ be a free algebra for σ ( U ) also over V . Then B ( F ) embeds into F σ .Proof. By Fact 5.2 point (2), F is free for V ( U ) over V and F σ is free for V σ ( U ) over V . Hence, by Lemma 5.6, F σ is free for σ V ( U ) over V . Thuswe may assume that U is a variety. Under this assumption, F ∈ U and F σ ∈ σ ( U ) .By Lemma 4.4, O ( F σ ) ∈ U . Thus the universal mapping property yieldsthat there exists a homomorphism f : F → O ( F σ ) such that f ( v ) = v forevery v ∈ V . By Proposition 4.1 Point (3), there exists a homomorphism f ′ : B ( F ) → F σ extending f .Further, B ( F ) ∈ σ ( U ) . Thus the universal mapping property yields theexistence of a homomorphism g : F σ → B ( F ) such that g ( v ) = v for every v ∈ V .We claim that gf ′ : B ( F ) → B ( F ) is an identity mapping. Indeed, sinceall elements from V are open in B ( F ) , for every v ∈ V we have gf ′ ( v ) = g ( v ) = g ( v ) = v = v . Thus, by the uniqueness of a homomorphicextension in the universal mapping property, gf ′ | F : F → F is the identitymapping on F . And similarly, by the uniqueness in Proposition 4.1 Point(3), gf ′ is an identity mapping. Hence f ′ is injective. (cid:3) Lemma 5.8.
Let F and F σ be as in Lemma 5.7, where V is infinite. Then F σ ∈ QB ( F ) .Proof. As in the proof of Lemma 5.7, we may assume that U and σ ( U ) arevarieties. Then, by Fact 5.2 point (3) and Lemma 5.6, σ ( U ) = σ V ( F ) = VB ( F ) . Thus Fact 5.2 point (1) yields that F σ ∈ QB ( F ) . (cid:3) Theorem 5.9.
Let U be a universal class of Heyting algebras. Then (1) U is strongly structurally complete if and only if σ ( U ) is stronglystructurally complete; (2) U is strongly universally complete if and only if σ ( U ) is strongly uni-versally complete.Proof. We prove (1). The proof of (2) is similar. Let F and F σ be as inLemma 5.7, where V is infinite. Assume that U is strongly structurally structurally complete. Let us verifythe condition (3) from Proposition 5.4 for σ ( U ) . Let Q ′ be a subquasivarietyof Q σ ( U ) and assume that F σ ∈ Q ( σ ( U ) ∩ Q ′ ) . By the Blok-Esakia theorem restricted version, there is a subquasivariety Q of Q ( U ) such that Q ′ = σ ( Q ) . By the Blok-Esakia theorem extended versionand Lemma 5.6, F σ ∈ Q ( σ ( U ) ∩ σ ( Q )) = Q σ ( U ∩ Q ) = σ Q ( U ∩ Q ) . Thus Lemma 5.7 yields that B ( F ) ∈ σ Q ( U ∩ Q ) . Now Lemma 4.4 and Theorem 4.1 point (1) gives that F ∈ Q ( U ∩ Q ) . Now we may apply the assumption and obtain the inclusion
U ⊆ Q . Finally,by the Blok-Esakia theorem extended version, σ ( U ) ⊆ Q ′ .For the opposite implication we proceed similarly. Assume that σ ( U ) isstrongly structurally complete and F ∈ Q ( U ∩ Q ) for some subquasivariety Q of Q ( U ) . By the Blok-Esakia theorem extended version and Lemma 5.6, B ( F ) ∈ Q ( σ ( U ) ∩ σ ( Q )) , and by Lemma 5.8 F σ ∈ Q ( σ ( U ) ∩ σ ( Q )) . Thus the assumption yields that σ ( U ) ⊆ σ ( Q ) , and the Blok-Esakia theoremextended version yields that U ⊆ Q . (cid:3) Appendix: Proof of the Blok Lemma
We say that a Boolean algebra A is a Boolean subalgebra of an interioralgebra M if A is a subalgebra of the Boolean reduct of M . Let M and N beinterior algebras, A be a Boolean subalgebra of M and B be a Boolean sub-algebra of N . We say that a mapping f : A → B is a -homomorphism from A into B if it is a Boolean homomorphisms and f ( a ) = f ( a ) whenever a ∈ A , and a ∈ A (note that then f ( a ) ∈ B ). Lemma 5.10.
Let M be a Grzegorczyk algebra, B , C be its finite Booleansubalgebras and g ∈ B . Assume that • B is generated (as a Boolean algebra) by C ∪ { g } , • B and C share the same open elements from M .Then the identity mapping on C may be extended to a -homomorphismfrom B into D , where D is a finite Boolean subalgebra of M generated by C and possibly some open elements. N THE BLOK-ESAKIA THEOREM FOR UNIVERSAL CLASSES 17
Proof.
Since B is finite, a Boolean homomorphic extension of the identitymapping on C into M always exists. Indeed, by [26, Theorem 130], eachsuch extension is given by a possible value p for f ( g ) from the nonemptyinterval in M h _ { c | c ∈ C, c g } , ^ { c | c ∈ C, g c } i = [ p ∗ , p ∗ ] The point is to find an extension which also preserves the operation.For every c ∈ C define p c = ( g ∨ c ) ∧ ¬ c. Then p c ( g ∨ c ) ∧ ¬ c = g ∧ ¬ c g, and ( g ∨ c ) > ( p c ∨ c ) = (( ( g ∨ c ) ∧ ¬ c ) ∨ c )= ( ( g ∨ c ) ∨ c ) > ( g ∨ c ) = ( g ∨ c ) . Thus(P) ( p c ∨ c ) = ( g ∨ c ) . Next, for c ∈ C let u = ¬ g ∨ c and define p ′ c = ¬ (cid:0) ( u → u ) → u (cid:1) . Since M is an interior algebra, we have ( u → u ) → u > ( u → u ) → u = u. Hence p ′ c ¬ ( ¬ g ∨ c ) g. This, in particular, gives that ( ¬ p ′ c ∨ c ) > ( ¬ g ∨ c ) . On the other, side we have ( ¬ p ′ c ∨ c ) = (cid:16) ¬ ( u → u ) ∨ u ∨ c (cid:17) (cid:16) ¬ ( u → u ) ∨ u (cid:17) u = ( ¬ g ∨ c ) . Indeed, the first inequality follows from the inequalities u, c u and themonotonicity of , and the second inequality follows from (Grz). Thus(P’) ( ¬ p ′ c ∨ c ) = ( ¬ g ∨ c ) . Finally, define p = p ∗ ∨ _ c ∈ C p c ∨ _ c ∈ C p ′ c . Since p ∗ , p c , p ′ c g p ∗ for every c ∈ C , we have p ∗ p p ∗ . Thus theidentity mapping on C extends to a Boolean homomorphism f from B intothe Boolean reduct of M , where f ( g ) = p . Let us show that f also preservesthe operation. Let b ∈ B be such that b ∈ B . The fact that B and C share the same open elements yields that b ∈ C , and hence f ( b ) = b .Since B is generated as a Boolean algebra by C ∪ { g } , there are c , c ∈ C such that b = ( g ∨ c ) ∧ ( ¬ g ∨ c ) and f ( b ) = ( p ∨ c ) ∧ ( ¬ p ∨ c ) . Thus, sincein M the operations commutes with ∧ , b = ( g ∨ c ) ∧ ( ¬ g ∨ c ) . Further by (P) and (P’), since is monotone and p c , p ′ c p g , b = ( p ∨ c ) ∧ ( ¬ p ∨ c ) . Hence f ( b ) = b = f ( b ) . Finally, note that there is a finite set of open elements X such that p belongsto the Boolean subalgebra of M generated by C ∪ X . For D we may takethis algebra. (cid:3) Lemma 5.11.
Let M be a Grzegorczyk algebra. Let A be a finite Booleansubalgebra of M . Then there exists a -homomorphism f : A → BO ( M ) such that f ( a ) = a for every open element a from A .Proof. Assume that A is generated (as a Boolean algebra) by a set { g , . . . , g n } of non-open (in M ) elements and some open elements. Let us define a se-quence of Boolean subalgebras A i of M and a sequence of -homomorphisms f i from A i − into A i such that • A i is a finite Boolean subalgebra of M generated by { g , . . . , g n − i } and some open elements, • f i ( a ) = a for every open element a from A i − .In particular, A n is a Boolean subalgebra of BO ( M ) . Once these sequencesare defined, we may define f as the composition f = f n · · · f . Let A = A . We proceed recursively, so let us assume that A i − is alreadydefined. By Lemma 5.10, there is a -homomorphisms f i : A i − → A i ,where A i is a Boolean subalgebra of M generated by { g , . . . , g n − i } andsome open elements ( A i − is B , the Boolean subalgebra of A i − generatedby { g , . . . , g n − i } ∪ O i − , where O i − is the set of open elements from A i − ,is C , and A i is D in Lemma 5.10).At the end, note that f i ( a ) = a for every i and every open element a from A i − . This yields that f ( a ) = a for every open element a from A . (cid:3) Now we are in position to provide the last step of the proof of the BlokLemma.Let M be a Grzegorczyk algebra and N be an extension of BO ( M ) suchthat O ( M ) = O ( N ) (in particular, N might be M ). Let B be the carrierof O ( M ) . Let N B be the expansion of N obtained by considering everyelement a ∈ B a new constant c a . Then every homomorphism f : M B → N B is injective. It follows that we only need to verify the existence of any N THE BLOK-ESAKIA THEOREM FOR UNIVERSAL CLASSES 19 homomorphism from M B into some elementary extension of BO ( M ) B . Wedo this with the aid of basic model theory [12, 28].We show that the set Th el ( BO ( M ) B ) ∪ diag + ( M ) is satisfiable. Here Th el ( BO ( M ) B ) is the elementary theory of BO ( M ) B ,i.e., the set of firstorder sentences which are valid in BO ( M ) B . Note that every model of Th el ( BO ( M ) B ) is an elementary extension of BO ( M ) B . Further, diag + ( M ) is the positive diagram of M . With every element a ∈ M we associate asymbol of a constant c a . Then diag + ( M ) consists of all equations of theform c a ∧ c b ≈ c d , where a, b, d are such that a ∧ b = d in M , ¬ c a ≈ c ,where a, b are such that ¬ a = b in M , and c a ≈ c b , where a, b are such that a = b in M . Here some caution is needed: In diag + ( M ) all symbols of con-stants from Th el ( BO ( M ) B ) appear and they correspond to open elements in B . There remaining symbols of constants appearing in diag + ( M ) are thosecorresponding to elements in M − B .By compactness theorem, it is enough to show that every set of the form Th el ( BO ( M ) B ) ∪ Σ , where Σ is a finite subset of diag + ( M ) , is satisfiable.Let A be a Boolean subalgebra of M generated by the set of elements cor-responding to symbols of constants appearing in Σ . By Lemma 5.11, thereexists a -homomorphism f : A → BO ( M ) which fixes all elements from A ∩ B . Then Th el ( BO ( M ) B ) ∪ Σ holds in the expansion of BO ( M ) B in whichevery symbol of a constant occurring in Σ and corresponding to an element a in A − B is interpreted as f ( a ) . References [1] Guram Bezhanishvili. The universal modality, the center of a Heyting algebra, andthe Blok-Esakia theorem.
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