aa r X i v : . [ m a t h . L O ] A p r Varieties of algebras without the amalgamationproperty
Tarek Sayed AhmedDepartment of Mathematics, Faculty of Science,Cairo University, Giza, Egypt.April 3, 2019
Abstract .
Let α be an ordinal and κ be a cardinal, both infinite, such that κ ≤ | α | . For τ ∈ α α , let sup ( τ ) = { i ∈ α : τ ( i ) = i } . Let G κ = { τ ∈ α α : | sup ( τ ) | <κ } . We consider variants of polyadic equality algebras by taking cylindrifications onΓ ⊆ α , | Γ | < κ and substitutions restricted to G κ . Such algebras are also enrichedwith generalized diagonal elements. We show that for any variety V containing theclass of representable algebas and satisfying a finite schema of equations, V fails tohave the amalgamation property. In particular, many varieties of Halmos’ quasi-polyadic equality algebras and Lucas’ extended cylindric algebras (including that ofthe representable algebras) fail to have the amalgamation property. The most generic examples of algebraisations of first order logic are Tarski’scylindric algebras and Halmos’ polyadic algebras. Both algebras are wellknown and widely used. Polyadic algebras were introduced by Halmos [12]to provide an algebraic reflection of the study of first order logic withoutequality. Later the algebras were enriched by diagonal elements to permitthe discussion of equality. That the notion is indeed an adequate reflectionof first order logic was demonstrated by Halmos’ representation theorem forlocally finite polyadic algebras (with and without equality). Tarski proved ananalogous result for locally finite cylindric algebras. Daigneault and Monkproved a strong extension of Halmos’ theorem, namely, every polyadic algebraof infinite dimension (without equality) is representable [9]. However, not ev-ery cylindric algebra is representable. In fact, the class of infinite dimensionalrepresentable algebras is not axiomatizable by any finite schema, a classicalresult of Monk. This is a point (among others) where the two theories devi-ate. Monk’s result was considerably strengthened by Andr´eka by showing thatthere is an inevitable degree of complexity in any axiomatization of the class of Mathematics Subject Classification.
Primary 03G15.
Key words : algebraic logic, polyadic algebras, amalgamation KL ). KL is aproper extension of first order logic without equality, obtained when the boundon the number of variables in formulas is relaxed; and accordingly allowing thefollowing as extra operations on formulas: Quantification on infinitely manyvariables and simultaneous substitution of (infinitely many) variables for vari-ables. Adding equality to KL , proved problematic as illustrated algebraicallyby Johnson [10]. In op.cit, Johnson showed that the class of representablepolyadic algebras with equality is not closed under ultraproducts, hence thisclass is not elementary, i.e. cannot be axiomatized by any set of first ordersentences. However one can still hope for a nice axiomatization of the varietygenerated by the class of polyadic equality algebras. A subtle recent (nega-tive) result in this direction is N´emeti - S´agi’s [18]: In sharp contrast to KL ,the validities of KL with equality cannot be recaptured by any set of schemasanalogous to Halmos’ schemas, let alone a finite one. In particular, the varietygenerated by the class of representable polyadic algebras with equality cannotbe axiomatized by a finite schema of equations. The latter answers a questionoriginally raised by Craig [7].It is interesting (and indeed natural) to ask for algebraic versions of modeltheoretic results, other than completeness. Examples include interpolation the-orems and omitting types theorems. Unlike the cylindric case, omitting typesfor polyadic algebras prove problematic. This is the case because polyadic al-gebras of infinite dimension have uncountably many operations, and omittingtypes arguments- Baire Category arguments at heart - are very much tied tocountability. On the other hand, Daigneault succeeded in stating and provingversions of Beth’s and Craig’s theorems. This was done by proving the alge-braic analogue of Robinson’s joint consistency theorem: Locally finite polyadicalgebras (with and without equality) have the amalgamation property. LaterJohnson removed the condition of local finiteness, proving that polyadic alge-bras without equality have the strong amalgamation property [11] . With thisstronger result, Robinson’s, Beth’s and Craig’s theorems hold for KL .Yet another point where the two theories deviate, Pigozzi [19] proves thatthe class of representable cylindric algebras fails to have the amalgamationproperty. This shows that certain infinitary algebraisable extensions of firstorder logic, the so-called typless logics (or finitary logics with infinitary rela-tions) fail to have the interpolation property. Further negative results concern-ing various amalgamation properties for cylindric-like algebras of relations canbe found in [15], [16], [17].Motivated by the quest for algebraisations that posses the positive prop-2rties of both polyadic algebras and cylindric algebras, in this paper we show,using basically Pigozzi’s techniques appropriately modified, that the interpo-lation property fails for many variants of KL with equality, contrasting theequality free case [2]. In such variants, formulas of infinite length are allowed,but quantification and substitutions are only allowed for < κ many variableswhere κ is a fixed beforehand infinite cardinal. Also (generalized) equalityis available. Such logics are (natural) extensions of the typeless logics corre-sponding to cylindric algebras.Our proof is algebraic adressing the amalgamation property for certainvariants of the class of polyadic equality algebras, that are also proper expan-sions of cylindric algebras. From our proof it can be easily destilled that manyvarieties of algebraic logics existing in the literature fail to have the amalga-mation property. Examples include Halmos’ quasi-polyadic equality algebrasand Lucas’ extended cylindric algebras. These results are new. Let α be an ordinal and κ be a cardinal, both infinite, such that κ ≤ | α | . For τ ∈ α α , let sup ( τ ) = { i ∈ α : τ ( i ) = i } . Let G κ = { τ ∈ α α : | sup ( τ ) | < κ } .Clearly G k is a semigroup under the operation of composition; in fact it is amonoid. We write Γ ⊆ κ α if Γ ⊆ α and | Γ | < κ . Let N = { E ⊆ α × α : E is an equivalence relation on α and |{ i < α : i/ E = { i }}| < κ } . Definition 1.1.
By a κ generalized polyadic equality algebra dimension α , ora P EA κ,α for short, we understand an algebra of the form A = h A, + , · , − , , , c (Γ) , s τ , d E i Γ ⊆ κ α,τ ∈ G κ , E ∈ N where c (Γ) (Γ ⊆ κ α ) and s τ ( τ ∈ G κ ) are unary operations on A , d E ∈ A ( E ∈ N ), such that postulates below hold for x, y ∈ A , τ, σ ∈ G, Γ , ∆ ⊆ κ α , E ∈ N , and all i, j ∈ α .1. h A, + , · , − , , i is a boolean algebra2. c (Γ) x ≤ c (Γ) x c (Γ) ( x · c (Γ) y ) = c (Γ) x · c (Γ) y c (Γ) c (∆) x = c (Γ ∪ ∆) x s τ is a boolean endomorphism7. s Id x = x s σ ◦ τ = s σ ◦ s τ
9. if σ ↾ ( α ∼ Γ) = τ ↾ ( α ∼ Γ), then s σ c (Γ) x = s τ c (Γ) x
10. If τ − Γ = ∆ and τ ↾ ∆ is one to one, then c (Γ) s τ x = s τ c (∆) x d I = 1 where I = Id ↾ α × α c (Γ) d E = d F where F = E ∩ ( α ∼ Γ) ∪ Id ↾ α × α s τ d E = d F where F = { ( τ ( i ) , τ ( j )) : ( i, j ) ∈ E } ∪ Id ↾ α × α x · d ij ≤ s [ i | j ] x In the above definition, and elsewhere throughout the paper, d ij denotes theelement d E where E is the equivalence relation relating i to j , and everythingelse only to itself. For a class K of algebras, S K stands for the class of allsubalgebras of algebras in K , P K is the class of products of algebras in K and H K is the class of all homomorphic images of algebras in K . The class ofrepresentable algebras is defined via set - theoretic operations on sets of α -arysequences. Let U be a set. For Γ ⊆ α , τ ∈ α α, i, j ∈ α and E ∈ N , we set c (Γ) X = { s ∈ α U : ∃ t ∈ X t ( j ) = s ( j ) ∀ j / ∈ Γ } . s τ X = { s ∈ α U : s ◦ τ ∈ X } . d ij = { s ∈ α U : s i = s j } . d E = { s ∈ α U : s i = s j ∀ ( i, j ) ∈ E } . Note that d E = T ( i,j ) ∈ E d ij . For a set X , let B ( X ) be the boolean set algebra( ℘ ( X ) , ∩ , ∪ , ∼ ) . The class of representable G k polyadic equality algebras, or RP EA κ,α is defined by SP {h B ( α U ) , c (Γ) , s τ , d E i : E ∈ N, Γ ⊆ κ α, τ ∈ G κ , U a set } . We make the following observations: • RP EA κ,α ⊆ P EA κ,α , and the inclusion is proper [4]. • If A ∈ P EA κ,α then A has a cylindric reduct and indeed this reductis a cylindric algebra of dimension α . In fact, A has a quasipolyadicequality reduct obtained by restricting the operations to finite quantifiers(cylindrifications) , finite substitutions and ordinary diagonal elements,i.e. the d ij ’s. • if G κ contains one infinitary substitution then RP EA κ,α is not closedunder ultraproducts [20], hence is not closed under H , lest it be a variety.4or what follows, we need: Definition 1.2.
Let K ⊆ V be classes of algebras. K is said to have theamalgamation property, or AP for short, with respect to V , if for all A , A and A ∈ K , and all monomorphisms i and i of A into A , A , respectively,there exists A ∈ V , a monomorphism m from A into A and a monomorphism m from A into A such that m ◦ i = m ◦ i .We will show that for any variety K , RP EA κ,α ⊆ K ⊆ P EA κ,α , K failsto have the amalgamation property with respect to P EA κ,α . For motivationsof studying such algebras, and similar reducts of polyadic equality algebras,initiated by Craig [7], see [1], [2], [3], [20], [21]. Amalgamation in varieties canbe pinned down to congruences on free algebras. Congruences correspond toideals. This prompts:
Definition 1.3.
Let A ∈ P EA κ,α . A subset I of A in an ideal if the followingconditions are satisfied:(i) 0 ∈ I, (ii) If x, y ∈ I , then x + y ∈ I, (iii) If x ∈ I and y ≤ x then y ∈ I, (iv) For all Γ ⊆ κ α and τ ∈ G κ if x ∈ I then c (Γ) x and s τ x ∈ I .It can be checked that ideals function properly, that is ideals correspondto congruences the usual way. For X ⊆ A , the ideal generated by X , Ig A X is the smallest ideal containing X , i.e the intersection of all ideals containing X . We let Sg A X and sometimes A ( X ) denote the subalgebra of A generatedby X . Lemma 1.4.
Let A ∈ P EA κ,α and X ⊆ A . Then Ig A X = { y ∈ A : y ≤ c (Γ) ( x + . . . x k − ) } : for some k ∈ ω, x ∈ k X and Γ ⊆ κ α } . Proof.
Let H denote the set of elements on the right hand side. It is easyto check H ⊆ Ig A X . Conversely, assume that y ∈ H, Γ ⊆ κ α. It is clear that c (Γ) y ∈ H . H is closed under substitutions, since for any τ ∈ G κ , any x ∈ A there exists Γ ⊆ κ α such that s τ x ≤ c (Γ) x . Indeed sup ( τ ) is such a Γ. Now let z, y ∈ H . Assume that z ≤ c (Γ) ( x + . . . x k − ) and y ≤ c (∆) ( y + . . . y l − ) , then z + y ≤ c (Γ ∪ ∆) ( x + . . . x k − + y . . . + y l − ) . The Lemma is proved.Fixing α and κ throughout, in what follows we denote ( R ) P EA κ,α simplyby ( R ) P EA.
The following about ideals will be frequently used.5 If A ⊆ B are P EA ’s and I is an ideal of A , then Ig B ( I ) = { b ∈ B : ∃ a ∈ I ( b ≤ a ) } . • If I and J are ideals of a P EA then the ideal generated by I ∪ J is { x : x ≤ i + j for i ∈ I, j ∈ J } . For a class K and a set X , Fr X K denotes the K algebra freely generatedby X , or the K free algebra on | X | generators. As a wide spread custom,we identify X with | X | . We understand the notion of free algebras in thesence of [13] Definition 0.4.19. In particular, free K algebras may not be in K . However, they are always in HSP ( K ), the variety generated by K . Wewrite R ∈ Co A if R is a congruence relation A . For X ⊆ A , then by ( A /R ) ( X ) we undertand the subalgebra of A /R generated by { x/R : x ∈ X } . Since ouralgebras have cylindric reducts, in what follows we use freely results of HenkinMonk and Tarski’s treatise [13] on the arithmetic of cylindric algebras. Wenow formulate and prove our main result:
Theorem 1.5.
Let K be a variety such that RP EA ⊆ K ⊆ P EA . Then K does not have AP with respect to P EA.
Proof.
The proof is an adaptation of Pigozzi’s techniques for showing failure ofthe amalgamation property for cylindric algebras [19]. Seeking a contradictionassume that K has AP with respect to P EA . Let A = Fr P EA.
Let r, s and t be defined as follows: r = c ( x · c y ) · c ( x · − c y ) ,s = c c ( c z · s c z · − d ) + c ( x · − c z ) ,t = c c ( c w · s c w · − d ) + c ( x · − c w ) , where x, y, z, and w are the first four free generators of A . Then r ≤ s · t . Thisinequality is proved by Pigozzi, whose proof we include. Indeed put a = x · c y · − c ( x · − c z ) ,b = x · − c y · − c ( x · − c z ) . Then we have c a · c b ≤ c ( x · c y ) · c ( x · − c y ) by [13]1.2.7= c x · c y · c x · − c y by [13] 1.2.11and so c a · c b = 0 . (1)6rom the inclusion x · − c z ≤ c ( x · − c z ) we get x · − c ( x · − c z ) ≤ c z. Thus a, b ≤ c z and hence, by [13] 1.2.9, c a, c b ≤ c z. (2)We now compute: c a · c b ≤ c c a · c c b by [13] 1.2.7= c c a · c s c b by [13] 1.5.8 (i), [13] 1.5.9 (i)= c ( c c a · s c b )= c c ( c a · s c b )= c c [ c a · s c b · ( − d + d )= c c [( c a · s c b · − d ) + ( c a · s c b. d )]= c c [( c a · s c b · − d ) + ( c a · c b · d )] by [13] 1.5.5= c c ( c a · s c b · − d ) by (1) ≤ c c ( c z · s c z · − d ) by (2), [13] 1.2.7We have proved that c [ x · c y · − c ( x · − c z )] · c [ x · − c y · − c ( x · − c z )] ≤ c c ( c z · s c z · − d ) . In view of [13] 1.2.11 this gives c ( x · c y ) · c ( x · − c y ) · − c ( x · − c z ) ≤ c c ( c z · s c z · − d ) . The conclusion now follows. Let X = { x, y } and X = { x, z, w } . Then A ( X ∩ X ) = Sg A { x } . (3)We have r ∈ A ( X ) and s, t ∈ A ( X ) . (4)Let R be an ideal of A such that A /R ∼ = Fr K α . (5)Since r ≤ s · t we have r ∈ Ig A { s · t } ∩ A ( X ) . (6)7et M = Ig A ( X [ { s · t } ∪ ( R ∩ A ( X ) )]; (7) N = Ig A ( X [( M ∩ A ( X ∩ X ) ) ∪ ( R ∩ A ( X ) )] . (8)Then we have R ∩ A ( X ) ⊆ M and R ∩ A ( X ) ⊆ N. (9)From the first of these inclusions we get M ∩ A ( X ∩ X ) ⊇ ( R ∩ A ( X ) ) ∩ A ( X ∩ X ) = ( R ∩ A ( X ) ) ∩ A ( X ∩ X ) . By (8) we have N ∩ A ( X ∩ X ) = M ∩ A ( X ∩ X ) . For R an ideal of A and X ⊆ A , by ( A /R ) ( X ) we understand the subalgebraof A /R generated by { x/R : x ∈ X } . Define θ : A ( X ∩ X ) → A ( X ) /N by a a/N. Then kerθ = N ∩ A ( X ∩ X ) and Imθ = ( A ( X ) /N ) ( X ∩ X ) . It follows that¯ θ : A ( X ∩ X ) /N ∩ A ( X ∩ X ) → ( A ( X ) /N ) ( X ∩ X ) defined by a/N ∩ A X ∩ X ) a/N is a well defined isomorphism. Similarly¯ ψ : A ( X ∩ X ) /M ∩ A ( X ∩ X ) → ( A ( X ) /M ) ( X ∩ X ) defined by a/M ∩ A X ∩ X ) a/M is also a well defined isomorphism. But N ∩ A ( X ∩ X ) = M ∩ A ( X ∩ X ) , Hence φ : ( A ( X ) /N ) ( X ∩ X ) → ( A ( X ) /M ) ( X ∩ X ) defined by a/N a/M
8s a well defined isomorphism. Now ( A ( X ) /N ) ( X ∩ X ) embeds into A ( X ) /N via the inclusion map; it also embeds in A ( X ) /M via i ◦ φ where i is alsothe inclusion map. For brevity let A = ( A ( X ) /N ) ( X ∩ X ) , A = A ( X ) /N and A = A ( X ) /M and j = i ◦ φ . Then A embeds in A and A via i and j respectively. Now observe that A , A and A are in K . So by assumption,there exists an amalgam, i.e there exists B ∈ P EA and monomorphisms f and g from A and A respectively to B such that f ◦ i = g ◦ j . Let¯ f : A ( X ) → B be defined by a f ( a/N )and ¯ g : A ( X ) → B be defined by a g ( a/M ) . Let B ′ be the algebra generated by Imf ∪ Img . Then ¯ f ∪ ¯ g ↾ X ∪ X → B ′ is a function since ¯ f and ¯ g coincide on X ∩ X . By freeness of A , there exists h : A → B ′ such that h ↾ X ∪ X = ¯ f ∪ ¯ g . Let P = kerh . Then it is not hard tocheck that P ∩ A ( X ) = N, (10)and P ∩ A ( X ) = M. (11)In view of (4) , (7) , (11) we have s · t ∈ P and hence by (6) r ∈ P . Conse-quently from (4) and (11) we get r ∈ N . From (8) there exist elements u ∈ M ∩ A ( X ∩ X ) (12)and b ∈ R such that r ≤ u + b. (13)Since u ∈ M by (7) there is a Γ ⊆ κ α and c ∈ R such that u ≤ c (Γ) ( s · t ) + c. Let { x ′ , y ′ , z ′ , w ′ } be the first four generators of D = Fr K . Let h be thehomomorphism from A to D be such that h ( i ) = i ′ for i ∈ { x, y, w, z } . Noticethat kerh = R . Then h ( b ) = h ( c ) = 0. It follows that h ( r ) ≤ h ( u ) ≤ c (Γ) ( h ( s ) .h ( t )) . r ′ = h ( r ), u ′ = h ( u ), s ′ = h ( s ) and t ′ = h ( t ) . Let B = ( ℘ ( α α ) , ∪ , ∩ , ∼ , ∅ , α α, c (Γ) , s τ , d E ) Γ ⊆ κ α,τ ∈ G κ , E ∈ N that is B is the full set algebra in the space α α . Let E be the set of allequivalence relations on α , and for each R ∈ E set X R = { ϕ : ϕ ∈ α α and for all ξ, η < α, ϕ ξ = ϕ η iff ξRη } . More succintly X R = { ϕ ∈ α α : kerϕ = R } . Let C = { [ R ∈ L X R : L ⊆ E } .C is clearly closed under the formation of arbitrary unions, and since ∼ [ R ∈ L X L = [ R ∈ E ∼ L X R for every L ⊆ E , we see that C is closed under the formation of complementswith respect to α α . Thus C is a Boolean subuniverse (indeed, a completeBoolean subuniverse) of B ; moreover, it is obvious that X R is an atom of ( C, ∪ , ∩ , ∼ , , α α ) for each R ∈ E. (14)For all E ∈ N we have d E = S { X R : E ⊆ R ∈ E } and hence d E ∈ C . Also, c (Γ) X R = [ { X S : S ∈ E, ( α ∼ Γ) ∩ S = ( α ∼ Γ) ∩ R } for any Γ ⊆ κ α and R ∈ E . Thus, because c (Γ) is completely additive, C isclosed under the operation c (Γ) for every Γ ⊆ κ α . It is easy to show that C isclosed under substitutions. For any τ ∈ G κ , s τ X R = [ { X S : S ∈ E, ∀ i, j < α ( iRj ⇐⇒ τ ( i ) Sτ ( j ) } . The set on the right may of course be empty. Since s τ is also completelyadditive, therefore, we have shown that C is a subuniverse of B . (15)We now show that there is a subset Y of α α such that X Id ∩ f ( r ′ ) = 0 for every f ∈ Hom ( D , B )such that f ( x ′ ) = X Id and f ( y ′ ) = Y, (16)10nd also that for every Γ ⊆ κ α , there are subsets Z, W of α α such that X Id ∼ c (Γ) g ( s ′ · t ′ ) = 0 for every g ∈ Hom ( D , B )such that g ( x ′ ) = X Id , g ( z ′ ) = Z and g ( w ′ ) = W. (17)Here Hom ( D , B ) stands for the set of all homomorphisms from D to B . Let σ ∈ α α be such that σ = 0, and σ κ = κ + 1 for every non-zero κ < ω and σj = j otherwise. Let τ = σ ↾ ( α ∼ { } ) ∪ { (0 , } . Then σ, τ ∈ X Id . Take Y = { σ } . Then σ ∈ X Id ∩ c Y and τ ∈ X Id ∼ c Y and hence σ ∈ c ( X Id ∩ c Y ) ∩ c ( X Id ∼ c Y ) . (18)Therefore, we have σ ∈ f ( r ′ ) for every f ∈ Hom ( D , B ) such that f ( x ′ ) = X Id and f ( y ′ ) = Y , and that (16) holds. We now want to show that for any givenΓ ⊆ κ α , there exist sets Z, W ⊆ α α such that (17) holds; it is clear that nogenerality is lost if we assume that 0 , ∈ Γ, so we make this assumption. Take Z = { ϕ : ϕ ∈ X Id , ϕ < ϕ } ∩ c (Γ) { Id } and W = { ϕ : ϕ ∈ X Id , ϕ > ϕ } ∩ c (Γ) { Id } . We show that Id ∈ X Id ∼ c (Γ) g ( s ′ · t ′ ) (19)for any g ∈ Hom ( D , B ) such that g ( x ′ ) = X Id , g ( z ′ ) = Z , and g ( w ′ ) = W ;to do this we simply compute the value of c (Γ) g ( s ′ · t ′ ). This part of the proofis taken verbatim from Pigozzi [19]. For the purpose of this computation wemake use of the following property of ordinals: if ∆ is any non-empty set ofordinals, then T ∆ is the smallest ordinal in ∆, and if, in addition, ∆ is finite,then S ∆ is the largest element ordinal in ∆. Also, in this computation weshall assume that ϕ always represents an arbitrary sequence in α α . Then,setting ∆ ϕ = Γ ∼ ϕ [Γ ∼ { , } ]for every ϕ , we successively compute: c Z = { ϕ : | ∆ ϕ | = 2 , ϕ = \ ∆ ϕ } ∩ c (Γ) { Id } , X Id ∼ c Z ) ∩ c (Γ) { Id } = { ϕ : | ∆ ϕ | = 2 , ϕ = [ ∆ ϕ, ϕ = \ ∆ ϕ } ∩ c (Γ) { Id } , and, finally, c ( X Id ∼ c Z ) ∩ c (Γ) { Id } = { ϕ : | ∆ ϕ | = 2 , ϕ = \ ∆ ϕ } ∩ c (Γ) { Id } . (20)Similarly, we obtain c ( X Id ∼ c W ) ∩ c (Γ) { Id } = { ϕ : | ∆ ϕ | = 2 , ϕ = [ ∆ ϕ } ∩ c (Γ) { Id } . The last two formulas together give c ( X Id ∼ c Z ) ∩ c ( X Id ∼ c W ) ∩ c (Γ) { Id } = 0 . (21)Continuing the computation we successively obtain: c Z ∩ d = { ϕ : | ∆ ϕ | = 2 , ϕ = ϕ = \ ∆ ϕ } ∩ c (Γ) { Id } , s c Z = { ϕ : | ∆ ϕ | = 2 , ϕ = \ ∆ ϕ } ∩ c (Γ) { Id } , c Z ∩ s c Z = { ϕ : | ∆ ϕ | = 2 , ϕ = ϕ = \ ∆ ϕ } ∩ c (Γ) { Id } ;hence we finally get c c ( c Z ∩ s c Z ∩ ∼ d ) = c c , (22)and similarly we get c c ( c W ∩ s c W ∩ ∼ d ) = 0 . (23)Now take g to be any homomorphism from D into B such that g ( x ′ ) = X Id , g ( z ′ ) = Z and g ( w ′ ) = W . Let a = g ( s ′ · t ′ ). Then from the above a ∩ c (Γ) { Id } = ∅ . Then applying c (Γ) to both sides of this equation we get c (Γ) a ∩ c (Γ) { Id } = ∅ . Thus (19) holds. Now there exists Γ ⊆ κ α and an interpolant u ′ ∈ D ( x ′ ) , thatis r ′ ≤ u ′ ≤ c (Γ) ( s ′ · t ′ ) . Y, Z, W ⊆ α α such that (16) and (17) hold. Take any k ∈ Hom ( D , B ) such that k ( x ′ ) = X Id , k ( y ′ ) = Y , k ( z ′ ) = Z , and k ( w ′ ) = W. This is possible by the freeness of D . Then using the fact that X Id ∩ k ( r ′ ) isnon-empty by (16) we get X Id ∩ k ( u ′ ) = k ( x ′ · u ′ ) ⊇ k ( x ′ · r ′ ) = 0 . And using the fact that X Id ∼ c (Γ) k ( s ′ · t ′ ) is non-empty by (17) we get X Id ∼ k ( u ′ ) = k ( x ′ · − u ′ ) ⊇ k ( x ′ · − c (Γ) ( s ′ · t ′ )) = 0 . However, in view of (14), it is impossible for X Id to intersect both k ( u ′ ) andits complement since k ( u ′ ) ∈ C and X Id is an atom; to see that k ( u ′ ) is indeedcontained in C recall that u ′ ∈ D ( x ′ ) , and then observe that because of (15)and the fact that X Id ∈ C we must have k [ D ( x ′ ) ] ⊆ C. This contradictionshows that K does not have the amalgamation property with respect to P EA .By this the proof is complete.Other algebraic logics to which our proof applies are Halmos’ quasi-polyadicequality algebras and Lucas’ κ extended cylindric algebras [14] p.267. In par-ticular, many varieties of those fail to have the amalgamation property. Werecall that Halmos quasi-polyadic algebras are of the form A = h A, + , · , − , , , c (Γ) , s τ , d ij i i,j ∈ α, Γ ⊆ ω α,τ ∈ G ω while Lucas, κ extended cylindric algebras are of the form A = h A, + , · , − , , , c (Γ) , d E i Γ ⊆ κ α, E ∈ N . Both classes of (abstract) algebras are defined by a finite schema analogousto Halmos’ schemas restricted to the appropriate similarity type, cf. Def 1.1.The representable algebras are defined as subdirect product of set algebras. Inthose two cases the class of representable algebras, as opposed to the class ofabstract algebras, is not finite schema axiomatizable. The methods of Andrekain [5] can be used to prove this (the proof though is not trivial). But in thosetwo cases the class of representable algebras forms a variety and using our proofit can be easily shown that any variety containing the representable algebrassuch that its cylindric reduct satisfies the cylindric axioms fails to have theamalgamation property. In particular, in both of these cases, both the varietyof abstract algebras as well as that of the representable algebras fail to havethe amalgamation property.
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