Interpolation and Beth definability in implicative fragments of IPC

(1)

Interpolation and Beth definability in implicative

fragments of IPC

UvA

Supervised by Nick Bezhanishvili and Tommaso Moraschini

.

Paul Maurice Dekker

June 28, 2020

(2)

Abstract

It is a classic result by Maksimova (1977) that there are exactly eight axiomatic extensions of the intuitionistic propositional calculus IPC that have the interpolation property. Also by Maksimova (2000) there are exactly sixteen axiomatic extensions of IPC with the projective Beth property. In this thesis we study interpolation and Beth definability in axiomatic extensions of the (∧, →)- and (∧, →, ⊥)-fragments of intuitionistic propositional logic, denoted IPC∧,→and IPC∧,→,⊥. We prove

that there are exactly four axiomatic extensions of IPC∧,→and exactly nine axiomatic extensions of

IPC_∧,→,⊥ that have the interpolation property. It is furthermore shown that there are exactly eight axiomatic extensions of IPC∧,→ and exactly thirty axiomatic extensions of IPC∧,→,⊥that have the

projective Beth property. The algebraic models of axiomatic extensions of IPC∧,→ and IPC∧,→,⊥

are varieties of implicative semilattices – in the second case with 0 –, whose finite members can be studied through our main tool that is K¨ohler duality. We obtain the results by characterizing all varieties that have the amalgamation property, which corresponds to the interpolation property, or that have strong epimorphism surjectivity, which corresponds to the projective Beth property. Relevant insights in the finite and infinite Beth definability properties are included.

(3)

Intermediate logics are logics situated between the intuitionistic propositional calculus IPC and the classical propositional calculus CPC. There is an isomorphism from the lattice of extensions of IPC to the lattice of all varieties of Heyting algebras. Under this correspondence, logical properties can be translated into algebraic terms where one can use powerful methods from universal algebra in order to characterize those properties. A classic example of this is Maksimova’s characterization of intermediate logics with the projective Beth property. In [Mak79] she found all eight intermediate logics with the interpolation property, and in [Mak03] she found all sixteen intermediate logics with the projective Beth property. It was proved in [BMR17] that continuum many, but not all intermediate logics have the infinite Beth property. Finally, it has been known since [Kre60] that all intermediate logics have the finite Beth property.

In general, interpolation and Beth definability are very important logical properties. For exam-ple, if we prove that there is a unique mathematical object X satisfying certain properties, we might like to find an ‘explicit’ formulation or definition of X , so that we can investigate some concrete properties of X . If we are working in a logic satisfying the relevant variant of Beth definability, we know that an explicit definition indeed can be found. For details, see [GM05], Chapter 1. The

(4)

finite Beth property is a weak version of the infinite one, while the projective Beth property is a strengthening of the infinite one.

Some logical properties are very sensitive to changes in the signature. In particular, if an intermediate logic above IPC has a property, it may lose that property when one removes the formulas that are not built up from a specified range of connectives; and vice versa. Let IPC→ be

the fragment of IPC that employs only the connective →; define IPC∧,→ and IPC∧,→,⊥ similarly.

Blok and Hoogland [BH06, Corollary 4.6] discovered that if one studies axiomatic extensions of IPC→ instead of the whole IPC, then no consistent logic has the infinite Beth property, instead of

continuum many.

In this thesis, the aim is to investigate interpolation and the three variants of the Beth property in extensions of IPC∧,→and IPC∧,→,⊥. Algebraic semantics for these logics is given by implicative

semilattices (ISL’s, with or without a zero element). The interpolation property in the logical point of view is equivalent to the amalgamation property for ISL’s. Similarly the various Beth properties correspond to algebraic properties that concern the surjectivity of epimorphisms. The translation of the infinite Beth property states that epimorphisms in a category (in our case a category of ISL’s) coincide with surjective maps.

We prove that there are exactly four varieties of ISL’s and nine varieties of ISL’s with zero with the amalgamation property (Theorems 44 and 49). As a result we obtain that there are exactly four logics above IPC∧,→ and nine logics above IPC∧,→,⊥ with the interpolation property. We

prove in addition that there are exactly eight varieties of ISL’s and thirty varieties of ISL’s with zero with the strong ES property (Theorems 85 and 91). It will follow that there are exactly eight logics above IPC∧,→ and thirty logics above IPC∧,→,⊥ with the projective Beth property. We

include two examples of logics that have interpolation or the projective Beth property but lack that property when restricting to the {∧, →}-fragment. Always applying the algebraic techniques, we show that there are continuum many axiomatic extensions of IPC_{∧,→(,⊥)}that have the infinite Beth property (Corollary 77) and that not all of them have this property (Corollary 73). We also give an alternative semantic proof of a known result proved proof theoretically by Kreisel [Kre60]: that all logics above IPC_{∧,→(,⊥)}have the finite Beth property (Corollary 71).

Our main tool is Köhler’s duality from [Köh81] between finite implicative semilattices and finite posets. Unlike other well-known dualities such as Priestley duality ([Pri70]) and Esakia duality (see [Esa74] and the recent English book [Esa19]), Köhler duality applies in the finite case only. However, in contrast to the variety of all Heyting algebras, the variety of all ISL’s is locally finite, which is in general desirable because every subvariety is consequently characterized by its finite members. Even better, we will explain that we only need to consider finite algebras in our study of amalgamation and weak and strong ES. However, as the standard ES property (not weak or strong) depends essentially on infinite algebras, the ES property remains the most difficult property to investigate.

The thesis is organized as follows. In Section 2 we recall the required facts from universal algebra, and present the basic theory of ISL’s. The link between logic and ISL’s will be made precise in Section 2.5. In Section 3 we prove our first main results by characterizing all varieties of ISL’s and all varieties of ISL’s with zero that have amalgamation. In Section 4 we define and give some properties of the concept of depth in implicative semilattices. This is needed for a good understanding of Theorem 74 in Section 5. In this section, we prove that all varieties of ISL’s (with or without zero) have the weak ES property; we prove that continuum many varieties, but not all, have the ES property; and we characterize all varieties of ISL’s and all varieties of ISL’s with zero

(5)

that have the strong ES property. At the end of the thesis is a table that summarizes the main results (Section 6).

2 Preliminaries

2.1 Universal algebra

As a basic reference for universal algebra we give [BS81].

Let P, H and S be the standard closure operations of taking products, homomorphic images and subalgebras.

Theorem 1 (Birkhoff ’s HSP theorem). A class K of algebras of a certain type is equationally defin-able iff

K = PK = HK = SK.

Classes of algebras as in Theorem 1 are called varieties. This terminology is familiar from e.g. the modal logic book [BRV01].

Theorem 2 (Tarski’s theorem). For any class K of algebras, HSPK is closed under H, S and P. Let VK denote the variety generated by K. Then Theorem 2 implies that HSPK = VK.

Furthermore, we will occasionally use the operator Pu of ultraproducts. It is well-known, as a

corollary to Ło´s’ theorem, that all elementary classes and hence every variety is closed under Pu.

The symbol Kf inwill denote the class of finite members of K.

A subdirect product is a subalgebraA of a product

∏

i∈I

Ai

such that the projection map πi :A → Ai is onto for each i ∈ I. An algebra A is subdirectly

irreducibleif whenever one expressesA as a subdirect product of various Ai, then there is exactly

one index i with the property |Ai| > 1. It turns out thatA is subdirectly irreducible iff A has a

least nontrivial congruence. (The diagonal congruence is the only congruence labeled trivial.) Theorem 3 (Birkhoff ’s subdirect representation theorem). Every algebraA is a subdirect product of subdirectly irreducible algebras that are homomorphic images ofA .

An algebra A is congruence-distributive if the lattice of congruences on A is a distributive lattice. It is congruence-permutable if Θ ◦ Ψ = Ψ ◦ Θ for all congruences Θ and Ψ onA , where

Θ ◦ Ψ =(a, c) ∈ A2: ∃b ∈ A : (a, b) ∈ Θ & (b, c) ∈ Ψ .

A variety is said to be congruence-distributive (or -permutable) if all its members have that prop-erty, and the variety is arithmetical if it has both properties.

Furthermore, an algebraA is finitely subdirectly irreducible (f.s.i.) if the intersection of every pair of nontrivial congruences ofA is nontrivial. Every subdirectly irreducible algebra is finitely subdirectly irreducible, since subdirectly irreducible algebras are those with a least nontrivial con-gruence. Every finite f.s.i. algebra is also subdirectly irreducible.

(6)

Lemma 4 (J´onsson’s Lemma). Let V be a congruence-distributive variety andAi∈ V, i ∈ I.

Sup-pose thatB ⊆ ∏_iAi, and θ is a congruence ofB such that B/θ is a nontrivial subdirectly

irre-ducible algebra. Then there is an ultrafilter U over I such that θU|B ⊆ θ ,

where θU is the congruence determined by U .

Proof. Essentially [BS81, Corollary IV.6.9].

Varieties are axiomatized by identities. A quasivariety is a class of similar algebras that is axiomatized by quasi-identities (t1≈ s1& . . . & tn≈ sn) =⇒ t ≈ s.

Theorem 5. (i) A class is a quasivariety iff it is closed under I, S, P and Pu.

(ii) The least quasivariety containing a class K is ISPPuK.

Proof. [BS81, Theorem V.2.25].

Lemma 6. Let K be a class of similar algebras. IfA is a subdirectly irreducible member of the quasivariety generated by K, thenA ∈ ISPuK.

Proof. See [CD90, Lemma 1.5].

An algebraA in a quasivariety K is relatively subdirectly irreducible provided that it satisfies the following condition: if we expressA as a subdirect product of members Aiof K, then exactly

one of thoseAiis nontrivial.

A variety is locally finite if all its members that are generated by a finite set, are finite.

Theorem 7. Let V be a variety of finite type. V is locally finite iff for every n ∈ N>0, the size of

subdirectly irreducible members of V with n generators is bounded. Proof. Bezhanishvili [Bez01, Theorem 3.7.4].

2.2 Implicative semilattices

The following definition was adopted from [K¨oh81]. An implicative semilattice (ISL) is an algebra A = (A,∧,→) such that:

• the reduct (A, ∧) is a meet-semilattice. This means that

a≤ b ⇔ a∧ b = a (a, b ∈ A)

defines a partial order on A such that for any a, b ∈ A the highest element max{c ∈ A : c ≤ a & c ≤ b}

(7)

• → is relative pseudo-complementation, i.e. the equivalence

c≤ a → b ⇔ c∧ a ≤ b

holds for all a, b, c ∈ A.

So in an ISL with domain A, the operator → is completely determined by the operator ∧. The maximal element 1 is equal to a → a for any a ∈ A. We have a ≤ b iff a → b = 1, so ≤ is also completely determined by →. As ≤ in turn implicitly defines ∧, it follows that the partial order, the meet operation and the relative pseudo-complementation each contain all information about an ISL.

Example 8. A finite chain of length n is a meet-semilattice, and one can show that this can be given the structure of an implicative semilattice, which we will denote byBn.

LetA be a finite ISL. Note that a join operation can be defined via a∨ b =^{c ∈ A : a ≤ c & b ≤ c}. This acts dually to the meet operation.

A homomorphism between implicative semilattices preserves ∧ and → by definition, so it also preserves 1 but it need not preserve ∨.

A meet-irreducible element a 6= 1 of an ISL has by definition the property that a = c ∧ d only if a = c or a = d. If A is finite and b is not the least element of A and it has the property that b= c ∨ d only if b = c or b = d, then b is called join-irreducible.

If an ISL has a largest element smaller than 1, this element is called opremum. An opremum is necessarily meet-irreducible. An ISL has an opremum iff it is subdirectly irreducible.1

The following result is essentially due to Diego [Die66], who proved it for so-called Hilbert algebras. For implicative semilattices, it was first stated by McKay [McK68].

Theorem 9. ISL is locally finite.

Proof. We want to apply Theorem 7. Proceed by induction on n. The case n = 1 is trivial. Let n> 1.

LetA be an algebra with opremum ε that is generated by a1, . . . , an.

The key observation is that ε ∈ {a1, . . . , an}. This is because A \ {ε} is the domain of a

sub-algebra. To prove the last claim, notice that if a, b < ε and a → b = ε, then a = a ∧ ε ≤ b so a→ b = 1.

Birkhoff’s subdirect representation theorem 3 and the inductive hypothesis imply that there is a bound M(n − 1) on the cardinality of all ISL’s with n − 1 generators. So |A \ {ε}| ≤ M(n − 1) and |A| ≤ M(n − 1) + 1.

This theorem tells in particular that every (quasi)variety K of ISL’s is fully determined by Kf in.

Lemma 10. Any finite homomorphic image of an ISLA is a subalgebra of A .

Proof. K¨ohler ([K¨oh81, Lemma 5.1]) proved that a homomorphic image of a finite ISL C is a subalgebra ofC .

(8)

Now, suppose that f :A B, where B is finite. For every b ∈ B, pick an inverse image a(b) ∈ Aunder f . Then {a(b) : b ∈ B} generates a finite subalgebraC of A , and B is a homomorphic image ofC . By K¨ohler’s result, B is a subalgebra of C and thus of A .

Corollary 11. A subclass of ISL is a variety iff it is a quasivariety.

Proof. Every variety is a quasivariety. Besides that, for any class K ⊆ ISL we have (HSPK)f in⊆ (ISSPK)f in= (ISPK)f in

by Lemma 10. It follows from Theorems 5ii and 9 that the quasivariety generated by K is VK. Proposition 12. ISL is arithmetical.

Proof. This was shown in [K¨oh81] around equation (1.12).

2.3 K¨ohler duality

K¨ohler [K¨oh81] constituted a link between finite implicative semilattices and finite posets. For this to be a categorical equivalence, one uses certain partial maps between finite posets. We next present this duality, with a small difference from the original paper. The deviation boils down to reversing all the finite posets, so that their maps are closer to Esakia morphisms2. This works nicely because a poset of upsets is allways the reverse of the poset of downsets.

Our category P has finite posets as its objects, and its morphisms are as follows. If P, Q ∈ P and α : P → Q is a partial mapping that preserves the strict order and satisfies

α [ p) = α(p) ∀p ∈ dom α, (1)

α is called a P-morphism. dom α is the domain of α and Im α is the full image α (dom α ).

Let P be a poset. Write [E) = {p ∈ P : ∃q ∈ E : q ≤ p} for E ⊆ P. An upset of P is a subset E such that [E) = E. The set U (P) of all upsets of P forms an implicative semilatticeU (P) when ordered by inclusion; the operations are given by

E∧ F = E ∩ F, E→ F = (E \ F]c. D(P) is defined to be the set of downsets of P.

If α : P → Q is a P-morphism we defineU (α) : U (Q) → U (P) by U (α)(E) = (α−1_Ec_]c_.

Proposition 13.U is a dual categorical equivalence between P and ISLf in.

Proof. LetR be the functor on P that reverses all orders. That is to say, R leaves the P-morphisms untouched but for every (P, ≤) ∈ P does R(P,≤) = (P,≥). So R replaces condition (1) for P-morphisms by its dual condition. If O is the functor that Köhler used (there also named O), we have O ◦ R = U . As Köhler proved that O is a coequivalence of categories ([Köh81, Theorem 3.3]), the claim follows.

(9)

A filter (resp. an ideal) of an ISLA is a non-empty upset (resp. downset) of A that is closed under ∧ (resp. ∨). Ideals are only defined ifA is finite. A filter is meet-irreducible if it is proper (as a subset) and it is not the intersection of two other filters.

Suppose that A is finite. The poset of filters of A , ordered by inclusion, is isomorphic to (A, ≥), and the poset of ideals of A , ordered by inclusion, is isomorphic to (A,≤). A filter F is prime if it is not the whole set A, and a ∨ b ∈ F implies a ∈ F or b ∈ F. The dual requirement holds for prime ideals. Write PrfA for the poset of prime filters and Pri(A ) for the poset of prime ideals. It is known that the complement of a prime filter is a prime ideal and vice versa. Thus, PrfA is dually isomorphic to Pri(A ). Also, a prime filter is the same as a meet-irreducible filter. Furthermore, it is not difficult to prove that a filter (ideal) in a finite ISL is prime iff its minimal (maximal) element is join-irreducible (meet-irreducible). So M A is the order dual of PrfA . While K¨ohler dualizes A as M A , we simply use PrfA instead.

It turns out that p 7→ [[p)) and a 7→ {F ∈ PrfA : a ∈ F} are respectively the isomorphisms

P ∼= PrfU (P) and A ∼=U (PrfA ). (2)

Lemma 14. Let P, Q ∈ P and α : P → Q a P-morphism. Then (i) U (α) is 1-1 iff Imα = Q;

(ii) U (α) is onto iff α is a full imbedding, i.e. domα = P and α is 1-1. Proof. This is Lemma 3.4 in [K¨oh81].

Lemma 15. LetA ,B be finite implicative semilattices. Then

(i) B is isomorphic to a subalgebra of A iff PrfB is an image of PrfA under a P-morphism; (ii) B is a homomorphic image of A iff PrfB is isomorphic to an upset of PrfA .

Proof. This is Lemma 3.5 in [K¨oh81].

Example 16. If A ,B,C are ISL’s and A is a subalgebra of B, a homomorphism A → C can usually not be extended to a homomorphism B → C . For example, consider the three posets in figure 1. By Lemma 15, this translates to three ISL’sA ,B,C with an embedding B → A and an isomorphismB → C . The latter homomorphism cannot be extended because the P-morphism ‘?’ cannot be found.

Let V ⊆ ISL be a variety.

Proposition 17. (i) Prf Vf in(= {PrfA : A ∈ Vf in}) is closed under subposets and finite disjoint

unions.

(ii) Let Q ⊆ P be any class. Then the following are equivalent: (a) For some variety W of ISL’s it holds thatU (Q) = Wf in.

(b) Q is closed under P-morphic images and finite disjoint unions.

(c) There exists a class R of posets such that Q is the class of P-morphic images of finite disjoint unions of posets in R.

(10)

Figure 1: this diagram will never commute Proof. First, we have

Prf(A × B) ∼= PrfA t PrfB because

J (A × B) ∼=JA t J B.

Let us prove this last isomorphism. φ sends (a, b) to the unique element of these two that is non-zero. This is well-defined because (a, b) = (a, 0) ∨ (0, b) so at most one is join-irreducible, and if a= c ∨ d then (a, b) = (c, b) ∨ (d, b) and so c = a or d = a.

From the definition, it is clear that φ is a poset homomorphism.

φ is onto, because (a, 0) is join-irreducible inA ×B if a is join-irreducible in A , since (a,0) = (b, c) ∨ (d, e) implies c = e = 0 and a = b ∨ d.

Now, by Lemma 15i we have proved (i).

By Tarski’s Theorem 2 and Lemma 10, we have for any class K of ISL’s that (VK)f in= SP(Kf in).

It is easy to see that this is equal to the class of subalgebras of finite products of members of Kf in.

Therefore, our proof of (i) also yields (iic) ⇒ (iia). Finally, we already have (iia) ⇒ (iib) ⇒ (iic).

A finite ISL is subdirectly irreducible iff the poset of its prime filters has a least element (a root).

Lemma 18. Let K ⊆ ISL. IfA ∈ VK is subdirectly irreducible and finite, then A ∈ SK.

Proof. _{By Proposition 17ii, Prf(VK)}_{f in} is the class of P-morphic images of disjoint unions of members of Prf Kf in. So there is such a disjoint union Q and a P-morphism α : Q Prf A .

Choose q ∈ Q such that α(q) is the root of PrfA . Then PrfA is a P-morphic image of [q) ⊆ Q, so it is also the P-morphic image of a member of Prf Kf in. By Lemma 15i,A ∈ SKf in.

Definition 19. If P is a poset and p ∈ P, we denote by d(p) the depth of p in P, being the maximal cardinality of a chain in (p]. So minimal elements have depth 1. We define the nth level of P as

(11)

Figure 2: subposet problem

Note that every level is an antichain. The co-depth dc(p) is the depth with respect to the dual order instead of the normal order. Also

Lc(P, n) = {p ∈ P : dc(p) = n}

is the nth co-level of P. The depth of the entire poset P is sup

C⊆P a chain

|C|.

2.4 Adding zero

The behaviour of some nontrivial properties changes when we include 0 in the signature. The following terminology is from [Nem65].

Definition 20. A bounded ISL is an implicative semilattice with a smallest element 0. A homo-morphism between bounded ISL’s is required to preserve 0. The class of all bounded ISL’s is BISL.

Definition 21. If P, Q are finite posets, a P-0morphism α from P to Q is a P-morphism of which the dual is a homomorphism between bounded ISL’s, meaning that the domain of α contains all maximal elements of P.

The inverse images of maximal elements under a P-0morphism are precisely all maximal ele-ments.

Remark 22. A subposet is not necessarily a 0morphic image, however the domain of a P-0morphism always is. Another nuance is given by the following example, which will later be good to know when we study co-amalgamability.

Example 23. If α, β : P → Q are P-0morphisms and R ⊆ dom α ∪ dom β is a subset such that α (R) = β (R) = Q, then R is not necessarily a P-0morphic image of P.

See figure 2; let P be the displayed poset. If α, β are the P-0morphisms onto the two-element chain defined by 1 /∈ dom β and 2 /∈ dom α, then R = {1, 2, 3, 5} satisfies R ⊆ P = dom α ∪ dom β and also α(R) = β (R) = Q, but R is not a P-0morphic image of {1, . . . , 5}.

2.5 Implicative semilattices and logic

Let IPC be the intuitionistic propositional calculus over a proper class of proposition letters. Let For∧,→be the set of formulas defined inductively by

p,

φ , ψ φ ∧ ψ,

φ , ψ φ → ψ.

(12)

Let For∧,→,⊥ have the same definition with the added rule

⊥.

Define IPC∧,→= IPC ∩ For∧,→and IPC∧,→,⊥= IPC ∩ For∧,→,⊥. An axiomatic extension of IPC∧,→(,⊥)

is a set E ⊆ For_{∧,→(,⊥)}such that the superintuitionistic logic L axiomatized by E satisfies

L∩ For_{∧,→(,⊥)}= E.

We view axiomatic extensions of IPC_{∧,→(,⊥)} in this way so that we donot need a special proof system for the restricted signature.

Axiomatic extensions of IPC_{∧,→(,⊥)}are directly linked to the varieties of (bounded) ISL’s that we have been developing:

Theorem 24. The map

E7→ {A ∈ ISL (BISL) : ∀φ ∈ E : A |= φ ≈ 1}

is a lattice isomorphism from the set of all axiomatic extensions of IPC_{∧,→(,⊥)} to the class of all varieties of (bounded) ISL’s.

For a detailed explanation, see [Cur63].

3 Amalgamable varieties

In this section we study various necessary and sufficient conditions for a variety to be amalgamable. From section 3.1 on we will focus on ISL’s.

Definition 25 (amalgamation). Let K be a class of algebras. IfA0,A1,A2are members of K such

thatA0is a subalgebra of bothA1andA2, then (A0;A1,A2) is an amalgam in K. This amalgam

is amalgamated in K if there is anA3∈ K and embeddings φ1:A1→A3and φ :A2→A3such

that φ1|A0= φ2|A0. The class K is said to have the amalgamation property (AP) if every amalgam

in K is amalgamated in K.

A standard way of proving or refuting amalgamability uses the theory of universal algebra. Lemma 26. Let K be a variety. An amalgam (A0;A1,A2) can be amalgamated in K iff the

follow-ing holds:

(*) Given i ∈ {1, 2}, a, b ∈ Aiwith a 6= b, there exist an algebraA₃0∈ K and homomorphisms

φ0_j:Aj→A30for j = 1, 2 such that φ10|A0= φ20|A0and φi0(a) 6= φi0(b).

Proof. ⇒ is clear: if φ1, φ2are embeddings then φi0(a) 6= φi0(b).

⇐: For every pair of distinct elements a, b ∈ A_i, letA₃a,b, φ₁a,b and φ₂a,b be given from (*). We letA3be the direct product of all algebras A₃a,b. The embedding of Ai inA3 is provided by the

combination of all φ_ia,b.

LetA be a subalgebra of an algebra B. The algebra B is called an essential extension of A if any congruence onB whose restriction to A is trivial3is itself trivial.

(13)

Lemma 27. An essential extension of a finitely subdirectly irreducible algebra is finitely subdirectly irreducible.

Proof. Let A be a finitely subdirectly irreducible algebra and let B be an essential extension of A . Given nontrivial congruences Θ,Θ0_on_{B we want to show that Θ ∩ Θ}0_{is also nontrivial. Now}

Θ ∩ A2and Θ0∩ A2are nontrivial by assumption. Therefore (Θ ∩ A2) ∩ (Θ0∩ A2) is nontrivial, since A is f.s.i. But this is the same as (Θ ∩ Θ0_{) ∩ A}2_{. So Θ ∩ Θ}0_{must be nontrivial. This completes the}

proof.

A class K of algebras has the so-called congruence extension property if, wheneverA ,B ∈ K andA ⊆ B and Θ is a congruence on A , there is a congruence Ψ on B such that Ψ ∩ A2= Θ. Lemma 28. (i) LetA be a subalgebra of an algebra B, and let Θ be a congruence on B. Then

there is a congruence Φ onB such that Θ ⊆ Φ, Φ ∩ A2= Θ ∩ A2, and the extensionB/Φ of A /(Θ ∩ A2_{) is essential.}

(ii) Let K be a class of algebras having the congruence extension property. LetA ,B ∈ K, let A be a subalgebra ofB, and let Θ be a congruence on A. Then there is a congruence Φ on B such that Φ ∩ A2= Θ and such that the extensionB/Φ of A /Θ is essential.

Proof. One can consult [GL71]: Lemma 3(b) and its Corollary.

Theorem 29. Let K be a variety having the congruence extension property, and let every subalgebra of each subdirectly irreducible algebra in K be f.s.i. Then K has the amalgamation property iff every amalgam in K consisting of three f.s.i. algebras can be amalgamated in K.

Proof. If K has AP then these last mentioned amalgams can certainly be amalgamated in K. To prove the other direction, let (A0;A1,A2) be an amalgam in K and let a, b ∈ A1with a 6= b. Since

A1is a subdirect product of subdirectly irreducible algebras in K by Birkhoff’s Theorem 3, there is

a subdirectly irreducible algebraS1∈ K and a homomorphism φ₁00ofA1ontoS1such that φ₁00(a) 6=

φ₁00(b). LetS0denote the image ofA0under φ100. The algebraS0is f.s.i. by the assumption of the

theorem. S0is isomorphic to a quotient ofA0. Applying Lemma 28ii withA = A0,B = A2and

A /Θ = S0 we see that there is an essential extensionS2 ofS0and a homomorphism φ₂00 from

A2ontoS2such that φ100|A0= φ200|A0. By Lemma 27,S2is also f.s.i. The amalgam (S0;S1,S2)

is amalgamated in K by an algebraS3=A₃0via homomorphisms ψ1and ψ2. Let φ_i0= ψi◦ φ_i00. By

symmetry, condition (*) in Lemma 26 holds, implying that (A0;A1,A2) is amalgamated in K.

The above theorem provides a strategy for checking amalgamability, and its advantage is that not all algebras have to be considered. Unfortunately, this strategy cannot be applied to the case of implicative semilattices. This is because a subalgebra of a subdirectly irreducible ISL might not be f.s.i., as shown by the following example. The ISLB₂2⊕B1is subdirectly irreducible4, but the

subalgebraB₂2is not f.s.i. because it is finite and not subdirectly irreducible. The following proposition outlines a different strategy to handle amalgamation.

Proposition 30. Let V be a locally finite variety of algebras of type τ. Then V is amalgamable iff Vf in, the class of finite algebras in V, is amalgamable.

(14)

Proof. ⇒. Suppose that (A₀;A1,A2) is an amalgam in Vf in. There isA3∈ V and φi:Ai,→A3

such that

φ1|A0= φ2|A0.

We can replace A3 by the subalgebra generated by Im φ1∪ Im φ2, which is finite because V is

locally finite.

⇐. Suppose that (A0;A1,A2) is an amalgam in V. We define a first-order predicate language.

Let ci(a) be a constant for i = 1, 2 and a ∈Ai. We also use an n-ary function symbol γ for every

n-ary operator γ ∈ τ. Let Σ be the following set of formulas Σ = {c1(a) ≈ c2(a) : a ∈ A0}

∪ {ci(a) 6≈ ci(b) : i ∈ {1, 2} & a, b ∈ Ai& a 6= b}

∪ {γ(ci(a1), . . . , ci(an)) ≈ ci(γAi(a1, . . . , an)) : i ∈ {1, 2} & γ ∈ τ & aj∈ Ai}

∪ Th(V)

where Th(V) is the first-order theory of V, which contains in particular all equations that are valid in V. We are going to apply the compactness theorem of the first-order logic. Let ˜σ ⊆ Σ be a finite subset. Then only a finite set B1⊆ A1and B2⊆ A2are involved in ˜σ . LetA_i0be the subalgebra of

Aigenerated by Bi. These are finite because V is locally finite. Now B0:= A0∩ (A01∪ A02) is finite

and we chooseA₀0as the subalgebra generated by B0.

Because Vf inis amalgamable, we obtain aA₃0∈ Vf inand φ_i0:A_i0,→A₃0such that

φ₁0|A0₀= φ₂0|A0₀. (3)

We claim thatA₃0 is a model for ˜σ , with ci(a) being interpreted as a for i = 1, 2 and a ∈ Bi, and

function symbols being interpreted in the obvious way.

The first set of formulas in the definition of Σ, intersected with ˜σ , is satisfied because of (3). The second set of formulas in ˜σ is satisfied because φ_i0 are 1-1. The third set is satisfied because φ_i0is a homomorphism. The last set is satisfied becauseA₃0∈ V.

By the compactness theorem of first-order logic, there is a model A3 of Σ. Let φi map a to

the interpretation of ci(a) in A3. The structure A3 can be viewed as a τ-algebra. The first set

of formulas guarantees that φ1|A0= φ2|A0. The second set of formulas guarantees that φi is an

injection. The third set of formulas guarantees that φi is a morphism. The fourth set guarantees

thatA3∈ V.

3.1 Amalgamation with implicative semilattices

We turn to the more special case of implicative semilattices, regarding the amalgamation property. First recall Theorem 24 linking varieties of ISL’s to logics.

Definition 31. A logic ` is said to have the interpolation property if whenever φ ` ψ then there is a formula χ such that Var(χ) ⊆ Var(φ ) ∩ Var(ψ), φ ` χ and χ ` ψ.

Proposition 32. An axiomatic extension of IPC_{∧,→(,⊥)}has the interpolation property iff the corre-sponding variety has the amalgamation property.

(15)

We can apply Proposition 30 and study finite algebras only.

The main advantage of studying finite implicative semilattices is the applicability of K¨ohler’s duality. Proposition 13 and Lemma 14i imply that an amalgamable class of finite ISL’s has a dual translation that satisfies a similar property where all arrows are reversed. We call this co-amalgamability. The maps that are involved are then surjective P-morphisms.

Furthermore, we have the following lemma from which we will derive that L is amalgamable (defined below).

Lemma 33. A variety V of implicative semilattices is amalgamable if every amalgam (A0;A1,A2)

of finite ISL’s such thatA0is subdirectly irreducible, can be amalgamated in V.

Proof. By virtue of Proposition 30 because ISL is locally finite, let (A0;A1,A2) be a finite

amal-gam in V. Let A0= n

∏

j=1 A( j) 0

be a subdirect representation into subdirectly irreducible factors (Theorem 3). Because V is a variety of ISL’s, every A₀( j) is a subalgebra of A0 (Lemma 10). Now we apply the assumption

to every amalgam (A₀( j);A1,A2) to findA ( j) 3 and φ ( j) 1 , φ ( j) 2 such that φ ( j) 1 |A ( j) 0 = φ ( j) 2 |A ( j) 0 . Then simply let A3= n

∏

j=1 A( j) 3 and φi= (φ (1) i , . . . , φ (n) i ).

Proposition 34. ISL has the amalgamation property.

Proof. This will be the one of the few proofs in this treatise that use Heyting algebras, a.k.a. pseudo-boolean algebras.

By [Mak00], the class of Heyting algebras has the amalgamation property. For a finite ISLA , note that the poset of downsets D(A ) is a finite set closed under ∩ and ∪ and thence a Heyting algebra. Relative pseudo-complementation works as follows: for L, M ∈ D(A ), we have a ∈ L → Miff (a] ∩ L ⊆ M iff a ∧ b ∈ M for each maximal element b ∈ Lc(L, 1), so:

a∈ L → M ⇔ ∀b ∈ Lc(L, 1)∃c ∈ Lc(M, 1) : a ≤ b → c (4)

as a ∧ b ≤ c is equivalent to a ≤ b → c. An element c that witnesses this condition for particular a, b will be denoted c(a, b). Furthermore,

σ :A → D(A ) a7→ (a]

is an embedding of finite ISL’s – let us show that σ preserves →: a∈ (b] → (c] ⇔ a ≤ b → c ⇔ a ∈ (b → c] by (4).

(16)

Figure 3: transferring the amalgamation property

Given a finite amalgam (A0;A1,A2) in ISL, we will make the diagram in figure 3 commute,

all with ISL embeddings. So, we already have σi:Ai→D(Ai).

Now let

ψi:D(A0) →D(Ai)

L7→ L

Ai.

The map ψiis an embedding of Heyting algebras: the really nontrivial criterion is the preservation

of →. For L, M ∈ D(A0), by (4), x∈ ψi(L → M) ⇔ x ∈ L → M Ai ⇔ ∃a ∈ L → M : x ≤ a ⇔ ∃a ∈ L → M : x ≤ ^ b∈Lc_(L,1) b→ c(a, b).

The last equivalence uses the fact that L → M is a downset. It is equivalent to the fact that we can find such values c(x, b) for x, because the exact value of a is irrelevant. But L and ψi(L) have the

same maximal elements (analogously for M), thus this is to say that x ∈ ψi(L) → ψi(M). Clearly,

σiafter the inclusionA0→Ai, is ψi◦ σ0.

By the amalgamation property for Heyting algebras, D(A0);D(A1),D(A2) viewed as an

amalgam with the maps ψ1, ψ2, can be amalgamated by some Heyting algebraA3. Then we take

the (∧, →)-reduct ofA3and figure 3 commutes.

We move on to consider subvarieties of ISL that are ‘linear’. Let L ⊆ ISL be the variety generated by finite chains. It was proved by K¨ohler (in [K¨oh81] after Lemma 5.2) thatA ∈ L iff B2

2⊕B1∈ ISA . These algebras are also characterized by the identity/

(((x₁→ x₂) → x₃) ∧ ((x₂→ x₁) → x₃)) → x₃_{≈ 1.} (5) Definition 35. A forest is a poset of which all principal downsets5 are totally ordered. A finite forest is a finite disjoint union of finite trees. A co-forest is a forest (T, ≤) such that (T, ≥) is a forest.

Proposition 36. LetA ∈ ISLf in. ThenA ∈ L iff PrfA is a finite co-forest.

(17)

Proof. The duals of finite chains are finite chains. So by Proposition 17ii, A ∈ L iff PrfA is a P-morphic image of a finite disjoint union of finite chains. Such posets are seen to be precisely the finite co-forests: one the one hand a P-morphic image of a finite co-forest is again a co-forest, and on the other hand every finite co-forest is the P-morphic image of the disjoint union of all it’s maximal chains.

Recall from Example 8 that Bn is the n-element chain viewed as an ISL, and let Cn be the

n-element chain viewed as a poset.

Proposition 37. L and V(B2) are amalgamable.

Proof. Let LHbe the variety of Heyting algebras generated by all finite chains. Because ISL’s have

only less structure then Heyting algebras, the finite chains generate only less Heyting algebras then implicative semilattices. More precisely, the {∧, →}-reduct of any member of LH is in L. Also,

applying the very last remark of the proof of Proposition 36 to the Kripke frames that correspond to the finite members of LH, shows that all finite co-forests are among these Kripke frames. Thus

every finite member of L is also in LH, when viewed as a Heyting algebra. By [CZ97, Theorem

14.22], LH has AP. Now by virtue of Lemma 33 let a finite amalgam (A0;A1,A2) in L be given

such that A0 is subdirectly irreducible. Then A0 is a finite chain. For i = 0, 1, 2, letAi0 be the

algebraAi with a new element 0 attached below. It is easy to show that A00 is a Heyting

subal-gebra of A₁0 and A₂0. So the amalgam (A₀0;A₁0,A₂0) is amalgamated by some A3 ∈ LH. Then

A3 also amalgamates (A0;A1,A2), and the {∧, →}-reduct of A3 is in L. This proves that L is

amalgamable.

An analogous proof holds for V(B2).

Remark 38. If an amalgam in L consists of finite chains, there exists a resolution which is also a finite chain. This follows from the argument that was used in [CZ97] to prove that LH has AP (see

Theorem 14.17 there).

V(B2) is the class of Boolean algebras if naturally enriched with the other operators ∨ and ¬.

Proposition 39. V(B1), V(B2) and L are the only subvarieties of L that are amalgamable.

Proof. The proof follows the idea of the proof of [CZ97, Theorem 14.19].

Let V ⊆ L be an amalgamable variety. Let Q ⊆ P be the corresponding dual class. Let P ∈ P \ Prf V(B2)f in. Then there are points p < q in P. So there is a P-morphism from P onto C2and

hence C2∈ Q.

We will prove by induction on n that Cn∈ Q for all n. This implies V = L.

Let n ≥ 3, T0 = Cn−2 and T1 = T2 = Cn−1. Define surjective P-morphisms αi : Ti → T0 for

i= 1, 2 by rejecting the lowest element c of Cn−1 from dom α1 and rejecting the highest element

of Cn−1from dom α2. There is a forest T3with βi: T3 Tifor i = 1, 2 such that α1◦ β1= α2◦ β2.

There is a chain {c1< · · · < cn−1} ⊆ T3 that is mapped onto T2 by β2. But α2(β2(c1)) = c so

c= α1(β1(c1)) whence β1(c1) is the second element of T1. So there is c0< c1 because β1 is a

P-morphism (condition (1)). This makes a chain of length n in T3. Thus we found a P-morphism

from T3onto Cn.

(18)

Figure 4: T (4)

Lemma 40. Let P be a finite poset, let α : P → Q be a P-morphism to a finite forest Q and α(p) = q. Then there is a subposet of [p)P that is mapped isomorphically to [q)Qby α and that contains p.

Proof. By induction on n we obtain that for all r ∈ L([q)Q, n) there is p(r) ∈ [p(u)) with α(p(r)) =

r, where u is the immediate predecessor of r in Q. Now p(·) is an isomorphism, as r < r0implies p(r) < p(r0) by construction and p(r) < p(r0) implies r = α(p(r)) < α(p(r0)) = r0.

Let T (m) be the full binary tree of depth m; T (m) has a bottom element ⊥ and every element of depth < m has precisely two successors which are not successing any other element. See figure 4. For any N ∈ N>0, the full N-ary tree of depth m is defined analogously.

Proposition 41. Every finite forest is embedded in some finite complete binary tree. Proof. Nearly trivial, as shown in [Bie90, Example 1].

Proposition 42 (unraveling). Given an arbitrary finite poset P, there is a forest T such that P is a P-homomorphic image of T . The forest T can be chosen as the set of downsets of all maximal chains of P, ordered by inclusion.

Proof. This is in essention explained in [BRV01, Proposition 2.15].

Lemma 43. Let V ⊆ ISL be an amalgamable variety that is not a subvariety of L. Then V = ISL. Proof. Write Q = Prf Vf in.

We would like to prove the lemma in four steps: 1. T (2) ∈ Q,

2. T (m) ∈ Q for all m, 3. any finite forest is in Q, 4. any finite poset is in Q.

(19)

Figure 5: a co-amalgam

Figure 6: A(4, k) for k = 2

Quick proof of step 4. By the co-amalgam in figure 5, the poset of figure 11 is in Q. By Proposi-tion 47 Q is co-amalgamable with P-0morphisms. At the end of the proof of Theorem 49, we will show that if such a class Q contains the poset of figure 11, then Q = P.

Direct proof of step 2. Proceed with induction on m. Suppose that T (m − 1) ∈ Q and write n = 2m−2. We would like to add 2n elements to T (m − 1), namely the elements of the mth level. Let L(T (m − 1), m − 1) = {p0, . . . , pn−1}; for each i add two new elements qi and ri such that

(qi] = (pi] ∪ {qi} and (ri] = (pi] ∪ {ri}. We will show by induction on k that

A(m, k) := T (m − 1) ∪ {qi, ri: i < k} ∈ Q (6)

for all k ≤ n and this will complete the overall induction step because A(m, n) = T (m).

(6) is obvious for k = 0. Suppose that (6) holds for k − 1. Let Q0be the subposet of A(m, k − 1)

that contains precisely the elements that are not strictly smaller than x := ak−1. Also let Q1=

A(m, k − 1). The point x is a maximal element of Q1. Thus x is even an isolated element of Q0. Let

x₀= qk−1, x1= rk−1 and let Q2= Q0∪ {x0, x1}. The point x is the only element of Q2 to which

these two new points are related. Q2∈ Q, because it is a disjoint union of Q0\ {x} and T (2), see

figure 6. In order to apply co-amalgamability, we need P-morphisms αi: Qi→ Q0 for i = 1, 2.

Observe that Q0 = Q1∩ Q2. Then αi is the partial map whose domain is Q0 and which is the

(20)

Now we receive a Q3∈ Q and βi: Q3 Qifor i = 1, 2 such that α1◦ β1= α2◦ β2. Choose u ∈

Q3such that β1(u) =⊥. By Lemma 40, there is a subposet Q4of [u) that is mapped isomorphically

to Q1 by β1. Let y be the pre-image of x in Q4. We have α1(β1(y)) = x so α2(β2(y)) = x so

β2(y) = x. Thus there are y0, y1> y such that β2(y0) = x0and β2(y1) = x1. Let Q5= Q4∪ {y0, y1}.

Note that yj∈ Q/ 4because else would imply β1(yj) > x and x is maximal in Q1.

The claim is that Q5∼= A(m, k).

This isomorphism φ is given by (β1|Q4) ∪ {(y0, x0), (y1, x1)}. The image of φ is clearly

Q₁∪ {x0, x1} = A(m, k).

Why is φ an isomorphism? The points y0 and y1 are incomparible because x0 and x1 are

incom-parible. Furthermore, y is maximal in Q4. So what is left to prove is that for any z ∈ Q4 and

j∈ {0, 1},

z< yjiff β1(z) < xj.

If β1(z) < xj, then β1(z) ≤ x. Because β1is an isomorphism on Q4, this implies z ≤ y whence

z< y_j. If z < yjand not β1(z) < xj, then not β1(z) < x, so β1(z) ∈ Q0. This yields β1(z) = β2(z) <

β2(yj) = xjin Q2. So β2(z) ≤ x and β1(z) ≤ x and β1(z) < xj. This establishes (6).

Step 3 follows from Proposition 41, and we are ready to prove step 4. As Q contains all finite forests, Proposition 42 gives V = ISL.

Theorem 44. The only amalgamable subvarieties of ISL are V(B1) ⊆ V(B2) ⊆ L ⊆ ISL .

Proof. Proposition 39 states that the three mentioned linear varieties are the only amalgamable ones. In Proposition 34 and Lemma 43 we proved that ISL is the only other amalgamable variety.

We can apply this theorem to logic.

Corollary 45. There are exactly four axiomatic extensions of IPC∧,→ with the interpolation

prop-erty.

Proof. Theorem 44 and Proposition 32. Example 46. Let bd0= x0and

bd_n=

(xn→ bdn−1) → xn → xn

.

It is an easy exercise to show that an intuitionistic time frame validates bdniff it is of depth at most

n. By [CZ97, Theorem 14.22], the logic axiomatized by IPC +bd2 has the interpolation property.

However by Corollary 50, the logic in the {∧, →}-fragment axiomatized by IPC∧,→+bd2does not

have the interpolation property.

The strong amalgamation property (for a quasivariety) is equivalent to the conjunction of the amalgamation property and the weak ES property (which is defined in section 5 below), by [GR15, Theorem 9.1]. A theorem by G. Kreisel [Kre60] implies that all varieties of ISL’s have the weak ES property – see Proposition 68 and [GR15, Theorem 12.3]. It follows that the four amalgamable varieties of ISL’s have the strong amalgamation property.

(21)

3.2 Adding zero

Proposition 47. Let V ⊆ BISL be a variety such that

(A, ∧, →) : (A, ∧, →, 0) ∈ V is amalgamable. Then V is amalgamable.

Proof. Let αi: Qi Q0 be P-0morphisms for i = 1, 2. Apply the lemma’s assumption to find a

poset Q3 and surjective P-morphisms βi. We can let Q3 be a forest by Proposition 42. Assume

that Q3= dom β1∪ dom β2. We want to adjust β1and β2so that their domains contain all maximal

elements of Q3. Consider a maximal point q of Q3that is for example in dom β1but not in dom β2.

Let r be a maximal predecessor of q which is in dom β2. If β2(r) is not maximal, this allows us

to extend β2 by letting β2(q) be a successor of β2(r). If β2(r) is maximal, let β2(q) = β2(r) and

remove r from the domain of β2. Repeating this process as many times as necessary, we are assured

that the domains of β1 and β2 can be made to contain all maximal elements, without damageing

the original properties of β1and β2.

If Q is a class of posets, let

1. Q+1=Q ∪ {>} : Q ∈ Q, > > q ∀q ∈ Q ,

2. Q+2=Q ∪ {>1, >2} : Q ∈ Q, >iare independent and >i> q ∀q ∈ Q and

3. Q+<ω =Q ∪ {>1, . . . , >n} : Q ∈ Q, n ∈ N>0 similarly.

For a poset P let P− be the adaptation with the maximal points chopped off, so P− = P \ Lc_{(P, 1). Let V(B}1) 6= V ⊆ ISL be a variety, Q = Prf Vf in and let Qsi be the class of rooted

members of Q.

Lemma 48. Suppose that in every poset in Qsi, maximal elements are above all other elements.

Then Q is co-amalgamable (with P-0morphisms) iff the following conditions hold.

I If a poset Q ∈ Qsi has three or more maximal elements, then even the same posets with any

amount of maximal elements are also in Qsi. In other words, we can add a maximal point that

is incomparible to the existing maximal points, and the result will still be in Qsi.

II If a poset in Qsi has n maximal elements, then any poset in Qsi with only m ≤ n maximal

elements can be given an additional n − m maximal elements and the result will still be in Qsi.

III Q−= {P−: P ∈ Q} is co-amalgamable with P-morphisms, and Q−= Prf V−_{f in}for some variety V−⊆ ISL.

IV The diamond (the lower poset in figure 9) is not in Q.

Proof. ⇒. We start with I. The idea here is given by figure 7. First delete all but two of the maximal elements of Q and call the result P. Label the two remaining maximal points p1and p2. For i = 1, 2,

let αi: Q → P be the P-0morphism that is the identity on P and that sends all other points to pi.

In this way, (P; Q, Q) is a co-amalgam the resolution Q0, β1, β2 of which is in Q. There is a set

(22)

and |Ai| ≥ n − 2. Because dom α1= dom α2= Q, we have dom β1= dom β2∈ Q by Remark 22. If

Qhad n maximal elements, it is clear that dom β1has 2 + (n − 2) + (n − 2) = 2n − 2 > n maximal

elements. Let Q0 be the poset Q except that the first co-level has 2n − 2 elements (instead of n). Then it is clear that β1|(dom β1)− can be extended to a P-morphism dom β1 Q0. So Q0∈ Q.

We now give the proof of II. The thinking is roughly given by figure 8 where n = 3 and m = 1. Suppose that Q, P ∈ Q such that Q has n maximal elements and P has m maximal elements. It follows that the rooted poset T of depth 2 with n maximal elements is in Q. Create the co-amalgam (C2; T, P) where the P-morphisms α1 and α2 map maximal elements to maximal elements and

roots to roots. Other elements of P will not be in the domain of α2. Let R be a poset that completes

this co-amalgam (C2; T, P) with morphisms β1 and β2. Then R has at least n maximal elements.

Therefore β2|(dom β2)− can be extended to a P-morphism dom β2 P0, where P0 is the poset P

except that the first co-level has n elements instead of m. Thus P0∈ Q.

Claims I-II say that Qsi (and thus Q) is characterized by on the one side that each poset can

have either (i) at most one, (ii) at most two, or (iii) arbitrarily many

maximal elements, and on the other side of course by the ‘bodies’ of the posets, which consist of the non-maximal elements.

(23)

Let us prove III. Let a co-amalgam (Q0; Q1, Q2) with P-morphisms α1, α2 be given. Let

Q+_i be Qi with one maximal point attached on top. Then Qi∈ Q. α1 and α2 easily extend to

surjective P-morphisms α_i+ : Q+_i → Q+₀. So, there is Q3∈ Q and βi: Q3→ Q+_i for i = 1, 2 such

that α₁+◦ β1= α2+◦ β2. Then Q−3 co-amalgamates (Q0; Q1, Q2). The second claim in III is very

straightforward.

We finally consider IV. If the diamond is in Q, then figure 9 presents a co-amalgam in Q. Imagine that R ∈ Q co-amalgamates this diagram with certain P-morphisms. Some point r ∈ R is mapped to 1, and some successors s,t of r are mapped to 2 and 3. Then s and t are mapped to 5 and 6 by the other P-morphism. So there must be points x > s and y > t that are mapped to 7 and 8. These x and y are maximal points of R, but x 6> t and y 6> s. This contradicts assumption 1.

This finishes ⇒. Let us prove ⇐: assume I-IV, and we will prove that Q is co-amalgamable. In preparation, recall that a poset P is connected if

P,{p, q} ⊆ P : p ≤ q | q ≤ p

is a connected graph. In this case, if E and F are upsets of P with the property E ∪ F = P, then E∩ F 6= /0.

Now note that the initial assumption of the lemma and condition (i)/(ii)/(iii) (whichever applies to Qsi) carry over to all posets Q ∈ Q such that Q− is connected; this can be shown by induction

on the number of minimal elements as follows. Suppose that Q ∈ Q and Q− is connected. Let q∈ L(Q, 1), Q1= [q) and Q2=L(Q, 1) \ {q}. Then Q is a P-morphic image of Q1t Q2, and

by induction Q1and Q2satisfy the desired properties – i.e.: maximal elements are above all other

elements and they each have at most n ∈ {1, 2, ω} maximal elements. Pick an element of Q−₁ ∩ Q−₂ by virtue of the previous paragraph. Then we see that Q1, Q2 and Q have the same maximal

elements, because Q1 and Q2 are upsets. It follows that Q also has at most n maximal elements.

For the property 1, let a maximal element p ∈ Q and a non-maximal element x ∈ Q be given. Then there is an i such that p, x ∈ Qi. Therefore x < p.

Let a co-amalgam (Q0; Q1, Q2) in Q with P-0morphisms αi : Qi Q0 be given. The

co-amalgam (Q−₀; Q−₁, Q−₂) with surjections αi|Q_i− can be co-amalgamated in Q− by some Q03, β10, β20.

Because the diamond is not in Q and the poset of figure 11 is neither in Q, observe that members of Qsisatisfy the property that forks only occur in the two top co-levels. In other words, Q−contains

(24)

assume that Q0₃is such a union of finite chains. Now, if we are in case (i), let k(C) = 1 for every component C of Q0₃. In case (ii), let k(C) = 2 for all C. And in case (iii), let k(C) be a large number. More concretely, let

k(C) = maxn1,

_∑

i=1,2

number of successors of βi(maxC)

o

– if maxC /∈ dom βi, the term in the sum is intended to be zero. In each of the cases (i)-(iii), attach

k(C) pairwise incomparible top elements >_C1, . . . , >_Ck(C) to every component C and let

C+= C ∪ {>_C1, . . . , >_Ck(C)}. With this define

Q₃= {1, 2} ×G{C+ : C a component of Q0₃}

t {isolated points from Q1} t {isolated points from Q2};

Q₃ contains two isolated copies of each C+. We will prove that Q3 co-amalgamates the original

co-amalgam with P-0morphisms.

β₁0 and β₂0 can be extended to P-morphisms β_i00: Q3→ Qias follows: first let βi00( j, q) = βi0(q)

for q ∈ Q0₃ and i, j = 1, 2 whenever it is either the case that i = j or dom(α1◦ β10) intersects the

component of Q0₃ that contains q. Also, let β_i00 map the isolated points of Qi to themselves. To

complete the definition of β_i00, we will map the top points of Q3 to top points of Qi in such a way

that β_i00 is a P-morphism and for q ∈ Q3, the equation

α1 β100(q) = α2 β200(q)

(7) holds whenever it makes sense. Let C be a component of Q0₃. If Di:= β_i00 { j} × C = /0 we leave

all >_C1, . . . , >_Ck(C) out of the domain of β_i00. Suppose on the other hand that Di6= /0. Then D is a

chain in Q−_i . Say there are m elements above Di;

• if i = 1, we postulate that j, >1

C, . . . , j, >Cm be mapped onto these m elements by βi00;

• if i = 2, we postulate that j, >_Ck(C), . . . , j, >k(C)−m+1_C be mapped onto these m elements by β_i00;

• if we are in case (i) or (ii) and both D1and D2 are nonempty, we do this in such a way that

α1 β₁00( j, >_Cr) = α2 β₂00( j, >_Cr) for each r ∈ {1, 2}. Why is this possible? We have in this case that dom(α1◦ β₁0) = dom(α2◦ β₂0) intersects C; so dom αiintersects Diand the elements

above D1and D2are mapped by αionto the maximal elements above α1(D1) = α2(D2). If

there is only one such maximal element, the claim follows immediately. If there are two such maximal elements x1, x2, then we can let β_i00( j, >r_C) be the point above Dithat is mapped to

x_r by αi.

This completes the definition of β_i00. β_i00 is a P-morphism: it preserves the strict order and if β_i00( j, q) = β0(q) < p then there are two cases: if p ∈ Q−_i , there is p < r ∈ Q0₃such that β_i00( j, r) = β0(r) = p. If p ∈ Lc(Qi, 1), there is some top element >Cr > ( j, q) that is mapped to p by βi00.

(25)

Figure 10: summary of amalgamation

Note that β_i00 indeed satisfies (7) whenever defined, because β_i0 satisfies it and it nicely works for maximal points q thanks to the bullet points: if we are in case (iii), then β₁00(q) and β₂00(q) cannot both be defined, and if we are in case (i) or (ii) then we already took care of the demand (7). Also, because Im β_i0= Q−_i , observe that Im β_i00= Qi.

Next, we would like to extend β_i00 to P-0morphisms βi: Q3→ Qi such that α1◦ β1= α2◦ β2.

This will work in two steps: we first define β_i00 ⊆ γi and then γi ⊆ βi. For this, we only need to

define βi on the maximal elements of Q3. Let q ∈ Lc(Q3, 1). If neither β₁00(q) nor β₂00(q) exists,

distinguish two cases. If there is i such that (q] ∩ dom β_i00 6= /0, let γi(q) be any maximal element

of Q1such that for all p < q, if p ∈ dom βi00 then βi00(p) < γi(p); this is possible because βi00 is an

extension of β_i0. If there is no such i, let γ1(q) be any maximal element of Q1. Then γi are still

P-morphisms and satisfy the same property (7) as β_i00. Furthermore dom γ1∪ dom γ2 contains all

maximal elements of Q3. We proceed to the definition of βi. If q ∈ Lc(Q3, 1) and γi(q) does not

exist but γ`(q) does, let { j} ×C = (q] \ {q}. If Di= /0, let βi(q) be any element of Qithat is mapped

to α` γ`(q) by αi. If Di6= /0, we have D`6= /0 by construction of γ1, γ2, so by construction of β₁00, β₂00

it follows that dom(α1◦ β10) intersects C. Therefore we can let βi(q) be an element above Dithat

is mapped to α` γ`(q) by αi. βiis a P-0morphism, and Q3co-amalgamates (Q0; Q1, Q2).

See figure 10. The dots signify any finite number of points. i, ii and vii are singleton sets of posets. iii is the set (a set of representatives) of all rooted posets of depth 2. iv is the set of all Cn.

v is the set of all posets that are a finite chain with two incomparible elements attached on top. vi is the set of all posets that are a finite chain with a certain finite number of pairwise incomparible elements attached on top.

Theorem 49. There are exactly nine amalgamable varieties of bounded ISL’s. These are V(B1),

BISL and the varieties generated by the duals of the ten sets of finite posets in figure 10. Proof. There are two possibilities.

1 Every P ∈ Qsihas the property that maximal elements are above all other elements.

(26)

Figure 11: a poset

For, if 1 fails, there is P ∈ Qsi and maximal p, p0 ∈ P and q ∈ P such that p > q but not p0> q.

There is a P-0morphism from P to {min P, p, p0, q} (min P is the least element of P) that sends every maximal element to p except p0which is sent to p0.

So, let us start with the first case.

Because there are four amalgamable varieties of ISL’s, claims I-III in Lemma 48 yield twelve candidate amalgamable classes of bounded ISL’s. Namely, if Q− is co-amalgamable with P-morphisms then Q is either (Q−)+1, (Q−)+2 or (Q−)+<ω. We have the four classes of posets from Theorem 44:

• { /0} gives us the class of discrete posets vii in figure 10. The classes { /0}+1 _{and { /0}}+2 _are

not closed under disjoint unions; • the class of discrete posets gives i-iii; • the class of co-trees gives iv-vi;

• the class of all posets gives three classes that are wrong because they contain the diamond. So together these are seven amalgamable varieties.

It is left to show that in case 2, if V is amalgamable, V = BISL. Because of figure 15 with omission of the center points of the second level, we may assume that the poset of figure 17 is in Q. Now we have that the posets of figure 13 are in Q: the poset on the right of figure 13 is the P-0morphic image of a resolution S of the co-amalgam in figure 12, because S contains an isomorphic copy of C3, and the middle point has two successors that are maximal points of S. Figure 13 shows

that there is a rooted poset with three maximal points. . . in the resolution R ∈ Q (with two certain P-0morphisms), there must be an isomorphic copy of {1, . . . , 4}, say {r1, . . . , r4}. Then r2 and

r₃ are mapped to 6 by the other P-0morphism, so in R there are r7> r2 and r8 > r2 which are

mapped to 4 so that r76> r3and r86> r3. So there is a rooted poset in Q with three maximal points.

If figure 14 we see another co-amalgam. The resolution contains a point ⊥ with three maximal successors. Furthermore the point a in the lower poset of figure 14 must also be reached by some point x in [⊥). It follows that x is incomparible to the maximal points of [⊥) that are mapped to the point b. We conclude the posets in figure 15 are in Q. A resolution of this co-amalgam contains a 3-fork y1, y2, y3> y where yi are maximal, and points y < x1< y1 and y < x3< y3. This x1 must

be incomparible to y2and y3; x3is incomparible to y1and y2. We see, again applying figure 15 but

now to {y, x1, x3, y1, y2, y3}, that figure 16 applies; in particular, the upper right poset of figure 16

is in Q. Here, points with the same label are mapped to the same label. Figure 16 guarantees the existence of the full ternary tree of depth 3 in Q. Because Q contains posets of every width, the same arguments can be applied to prove that all trees of depth 3 are in Q, as a subtree of a tree in Q is also in Q.

(27)

To construct a full N-ary tree of depth 4, let Q0 consist of its second and fourth level, let Q1

consist of its first, second and fourth level and let Q2consist of its second, third and fourth level.

A poset that co-amalgamates (Q0; Q1, Q2) contains the demanded tree. By this argument for every

depth, Q contains all trees so Q = P.

Corollary 50. There are exactly nine axiomatic extensions of IPC∧,→,⊥with the interpolation

prop-erty.

Proof. Theorem 49 and Proposition 32. The next lemma will be useful later.

Lemma 51. Let K be a co-amalgamable class of finite posets (with P-morphisms or P-0morphisms) containing C2and the disjoint union of two 2-chains, such that K is closed under P-morphic resp.

P-0morphic images. Then K is closed under disjoint unions.

Proof. See figure 17 – we have {2, 3, 4, 5} ∈ K (the point 1 is not relevant for this proof).

Let P1, P2 ∈ K. Define α1 : P1 → {2, 4} as follows: send maximal elements of P1 to 4, and

send minimal elements that are not maximal to 2. (If such minimal elements do not exist, use

(28)

(29)

α1: P1→ {4} instead.) We also have the projecting partial map α2: {2, 3, 4, 5} → {2, 4}. Since

K is co-amalgamable, we find a resolution R ∈ K of this co-amalgam with P-(0)morphisms β1, β2.

Note that

β₂−1{3, 5} ∩ dom β1= /0.

Therefore the poset P1t {3, 5} is a P-(0)morphic image of R so it is in K.

Let α₂0: P2→ {3, 5} be like α1. Combining this with the projection α10 : P1t {3, 5} → {3, 5} we

obtain a similar co-amalgam to the one in the previous paragraph. Conclude that P1t P2∈ K.

4 Varieties of ISL’s of bounded depth

In this section we study what it means for varieties of ISL’s to be of a certain depth, and we will count the amount of varieties that are of finite depth.

Definition 52. The depth of an ISL A is defined to be the depth of the poset of meet-irreducible filters ofA (see Definition 19).

Note that the trivial algebra is of depth 0.

Proposition 55 explains why this is a natural definition.

Lemma 53. An ultraproduct of subdirectly irreducible ISL’s is subdirectly irreducible. Proof. The class of subdirectly irreducible ISL’s is elementary, because

∃ε : ∀x : x < ε ↔ x 6≈ 1

describes the existence of an opremum. Every elementary class is closed under the ultraproduct operator.

Lemma 54. LetB be an ISL, A a subalgebra of B, F a filter of A and G a filter of B with the property G ∩ A ⊆ F.

(i) G can be extended to a filter H that is maximal w.r.t. the condition H ∩ A = F. (ii) If F is meet-irreducible inA , then H from (i) is meet-irreducible in B.

(iii) If j ≤ ω and {Fi: i < j} is a chain of meet-irreducible filters inA , then there exists a chain

of meet-irreducible filters {Gi: i < j} inB for which Gi∩ A = Fi.

Proof. (i) We first need to prove that the filter H0= hG ∪ Fi satisfies H0∩ A = F. The inclusion ⊇ is trivial. Suppose that a ∈ H0∩ A. Then there are b ∈ G, c ∈ F such that b ∧ c ≤ a. It follows that c → a ≥ b. However, c → a ∈ A and G ∩ A ⊆ F, so c → a ∈ F. Consequently, a≥ c ∧ (c → a) ∈ F. This proves ⊆.

Note that the union of a chain of filters H00with the property H00∩ A = F, is again a filter with the same property. Applying Zorn’s lemma to all such H00 yields the result.

(ii) If H = H0∩ H00 where H0and H00 are other filters ofA , then by (i) we have H0∩ A 6= F and H00∩ A 6= F. So, F = H ∩ A = (H0∩ A) ∩ (H00∩ A) is not meet-irreducible.

(30)

(iii) Apply (ii) with F = F0, G = {1} and let G0= H. Recursively, suppose that Gi has been

defined. Then Gi∩ A = Fi⊆ Fi+1. So, apply (ii) again with F = Fi+1 and G = Gi, and now

let Gi+1= H.

Then the filters Giare clearly like desired.

As we already noted in the proof of Theorem 9, if an ISLA has opremum ε then A\{ε} is the domain of a subalgebra ofA .

Proposition 55. (i) LetA be a subdirectly irreducible ISL with opremum ε. Then A is of depth at most n + 1 iffA \ {ε} is of depth at most n.

(ii) LetA be an ISL. Then the following conditions are equivalent. (a) A is of depth at most n.

(b) Every subdirectly irreducible imageB of A is of depth at most n. (c) Bn+2∈ S(A )./

Moreover, ifA is finite, the above conditions are equivalent to the demand that PrfA is of depth at most n.

(iii) The ISL’s of depth at most n form a variety.

Proof. If A is finite, (iia) is clearly equivalent to the demand that PrfA is of depth at most n, because in this case prime filters coincide with the meet-irreducible ones.

We prove (i) and (iii) as well as the equivalence of (iia)-(iic) by induction on n. The base case is clear; now suppose that n > 0.

We start off by proving (i). Let F(A ) be the set of all filters on A , and let M(A ) be the subset of all meet-irreducible filters. Then {1} ∈ M(A ), and

φ : F A \ {ε} → F(A ) \ {1} : F 7→ F ∪ {ε}

is a bijection, from which we see that φ is still a bijection if we replace F by M in the definition of φ . Due to this correspondence, it is clear that there is a chain of length n in M A \ {ε} iff there is a chain of length n + 1 in M(A ).

We will now prove the contraposition of (iib) ⇐ (iic). Pick a subdirectly irreducible homomor-phic image D of A such that D is not of depth at most n, and let ε be the opremum of D. By (i), D \ {ε} is not of depth at most n − 1. Apply the inductive hypothesis to choose a subalgebra Bn+1∼=C ⊆ D \{ε}. Then by Lemma 10, E = C ∪{ε} ∼=Bn+2is isomorphic to a subalgebra of

the algebra f−1E and hence E is isomorphic to a subalgebra of A .

(iib) ⇒ (iia). LetA ⊆ B = ∏_i∈IAi be a subdirect representation into subdirectly irreducible

factors.

We first prove the implication for a subdirectly irreducible algebra, then for the direct product B and then for general A . Note that the case of a subdirectly irreducible algebra is trivial. We proceed to the proof of (iib) ⇒ (iia) for the productB.

(31)

For J ⊆ I let πJ:B → ∏j∈JAjbe the projection. For a meet-irreducible filter F ⊆ B let

W(F) = {J ⊆ I : πJ|F is onto}.

W(F) is a maximal ideal over I:

• W (F) does not contain I, because F 6= B. • W (F) is downwards closed.

• If J1, J2∈ W (F) and J = J1∪ J2, let πJ(a) ∈ πJ(A). There are by assumption a1, a2 ∈ F

such that πJi(a

i_{) = π}

Ji(a). Then πJ(a

1_{∧ a}2_{) ≤ π}

J(a). As there exists a3≥ a1∧ a2such that

πJ(a3) = πJ(a), it follows that J ∈ W (F).

• If J ⊆ I, suppose for a contradiction that J /∈ W (F) and Jc_{∈ W (F). Choose witnesses π}_/ J(a)

and πJc(b) for these facts. We may assume that

b_i= 1 for i ∈ J, a_i= 1 for i /∈ J. (8)

Then with the filters

G= hF ∪ {a}i, H= hF ∪ {b}i,

we claim that F = G ∩ H. (Note that G, H 6= F by construction.) Suppose that c ∈ G ∩ H. There are d, e ∈ F such that c ≥ d ∧ a and c ≥ e ∧ b. So, c ≥ d ∧ e ∧ a and c ≥ d ∧ e ∧ b. By (8) and distributivity, c ≥ d ∧ e so c ∈ F. This contradicts the fact that F is meet-irreducible. Now for two meet-irreducible filters F ⊆ G ⊆ B it holds that W (F) ⊆ W (G). Because these are maximal, it follows that W (F) = W (G). Now let C be a chain of meet-irreducible filters inB. Then W = W (F) for F ∈ C gives rise to an ultraproductB/U of the various Ai, where U = Wc.

For each F ∈ C, F/U is a filter in B/U, and F/U is in fact meet-irreducible: suppose that F/U = G0∩ H0. Let G be the union of all classes in G0, and H similarly. Then G and H are filters. If b ∈ G ∩ H, we have [b]U ∈ F/W , so there is a ∈ F for which J := [a 6= b] ∈ W . Then c ∈ B

defined by

c_i= (

bi if i ∈ J,

1 otherwise,

is in F and satisfies c ∩ a ≤ b, whence b ∈ F. So F/U is meet-irreducible. Furthermore, F/U = G/U with F, G ∈ C implies F/U = G/U ∩ G/U so F = G by the previous paragraph.

By Lemma 53,B/U is subdirectly irreducible. By assumption, every subdirectly irreducible image ofB/U is of depth at most n. So by the previous case, the existence of the chain {F/U : F∈ C} is not possible.

The general case follows from the previous case by Lemma 54iii. (iic) ⇐ (iia).

LetBn+2∼=E ⊆ A . E contains a chain of n+1 meet-irreducible filters. By Lemma 54iii, this

chain can be adjusted to a chain inA .

We next prove (iii). From the second step of our proof of (iib) ⇒ (iia) and from (iia) ⇔ (iic) it follows that the class of ISL’s of depth at most n is closed under P and S. Next, if f : B A andB is of depth at most n, then A is of depth at most n: suppose that A has a subalgebra C that is isomorphic toBn+2. By Lemma 10, C is isomorphic to a subalgebra of f−1C and hence to a

(32)

Example 56.Bn+2 ∈ H(A ) is not an equivalent condition in Proposition 55ii. For example, if/

T = T (2) is a 2-fork, then

B3∈ H U (T )/

by Lemma 15ii.

Proposition 57. LetA ∈ ISLf in. ThenA ∈ V(Bn) iff PrfA is a finite co-forest of depth ≤ n − 1.

Proof. ⇒. Propositions 36 and 55.

⇐. Unraveling PrfA (Proposition 42) yields a set of chains of length at most n − 1, so A ∈ V(Bn).

Now that we well understand what depth means, we would like to count the varieties of bounded depth. There are at most 2ℵ0 _{varieties of ISL’s in total, because there exist only countably} many finite posets.

Proposition 58. There are 2ℵ0 _{varieties of ISL’s of depth 3.}

Proof. For k ≥ 3, let Pkbe a poset with one point in the first level, k points in the second level and

kpoints in the third level. The ith point of the second level of P_kwill be below the jth point of the third level iff j − i 6≡ 0, 1 mod k. See figure 18.

We claim that A = {Pk: k ≥ 3} is a set of posets such that Pk is not a P-morphic image of P`if

k6= `. From Lemma 18 it will follow thatU E generates a different variety for every E ⊆ A. So, suppose that k < ` and α : P` Pkis a P-morphism. By rejecting some points from dom α

if necessary, we may assume that

L(P`, 2) ∩ dom α

= k.

We claim that the same holds for the third level. For, suppose α maps p and q to the jth point of L(P_k, 3). Choose r, s ∈ P_`such that α(r) is the j − 1th point of L(Pk, 3) (modulo k) and α(s) the jth

point. Then neither p nor q was above r or s, which implies p = q by construction of P`.

Next, let p ∈ L(P`, 2) ∩ dom α, say, the ith point. There are two non-successors of α(p), so the

ith point of L(P`, 3) and also the (i + 1)th point of L(P`, 3) are in dom α. But

L(P`, 3) ∩ dom α = L(P`, 2) ∩ dom α .

Apparently, L(P`, 3) ∩ dom α is closed under shifting one step to the right. It follows that this set

has size `, a contradiction.

Because every finite poset is the P-homomorphic image of a finite tree, one might expect that the same idea as in the proof of Proposition 58 can be executed with trees. This, however, is not the case, as we we will prove below depending on a nontrivial fact about trees.

Call a sequence p1, p2, . . . in a partial pre-order never increasing if pi≤ pjimplies j ≤ i. Partial

pre-orders whithout such sequences are called well-quasi-orders; see [DPR81] for details. The following was proved in [Nas08].

Theorem 59. The class of all trees is a well-quasi-order with respect to subtrees.

Lemma 60. The class of varieties of ISL’s of depth at most n that are generated by the duals of certain finite trees, is a well-quasi-order.

Interpolation and Beth definability in implicative fragments of IPC