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property

We say that a classK of one-step Heyting algebras (respectively one-step intuitionisitic Kripke frames) has theextension propertyif every member ofKhas a one-step extension belonging toK, cf. Section 2.5of Chapter 2.

In this section we prove that the extension property for finite conservative one-step Heyting algebras validating a hypersequent calculus HC obtains precisely when HC enjoys the bounded proof property. To this end the following sufficient criterion for the extension property will be useful.

Lemma 4.5. Let HC be a hypersequent calculus and let ConAlg_<ω(HC) be the class of finite conservative one-step Heyting algebras validating HC. If every H ∈ ConAlg_<ω(HC)

embeds into some standard Heyting algebra validating HC then the class ConAlg_<ω(HC)

has the extension property.

Proof. LetHbe a finite (conservative) one-step Heyting algebra and suppose that there exists an embedding (g0, g1) :H →Ainto some Heyting algebra Avalidating HC. Let- ting A be the bounded lattice reduct of A, we see that H00 _{= (D}

1, A, g1) is a one-step algebra validatingHC and extending H.

To obtain a finite conservative one-step Heyting algebra validating HC and extending H let D2 be the bounded distributive sublattice of A generated by the set {g1(a) → g1(b) : a, b∈ D1}. As the variety of bounded distributive lattices is locally finite D2 is finite. Moreover, we have Im(g1) ⊆ D2. Therefore, H0 = (D1, D2, g1) will be a finite conservative one-step algebra validating HC and extending H.

i) HC has the bounded proof property;

ii) The class ConAlg<ω(HC) of finite conservative one-step algebras validatingHC has

the extension property;

iii) The class ConF rm_<ω (HC) of finite conservative one-step frames validating HC has the extension property.

Proof. Item ii) and item iii) are easily seen to be equivalent by elementary considerations on the duality between finite conservative one-step algebras and finite conservative one- step frames.

To see that ii) implies i) let S ∪ {S} be a set of hypersequents of degree at most 1. We prove that if the class of finite conservative one-step algebras validating HC has the extension property then S 6`1

HC S implies that S 6`HC S. By Proposition 4.1this suffices to establish the bounded proof property for HC. We proceed by constructing a standard Heyting algebra Atogether with a valuation v such that (A, v) validatesHC and S but not the hypersequentS. From Lemma3.17it then follows that S 6`HCS. We use the modified Lindenbaum-Tarski construction from Proposition 4.2. Let P be the set of propositional variables occurring inS ∪ {S}. By Proposition4.2there exists a finite conservative one-step Heyting algebra H₀ := LT_HC(S, S) = (D0, D1, i0) val- idating HC and a valuation v0 such that (H0, v0) validates S but not S. Therefore, by the assumption thatConAlg_<ω(Ax) has the extension property, there exists a one-step extension H₁ = (D1, D2, i1) of H0 also validating HC. Moreover, letting v1 = (v01, v11) be the valuation on H1 given by v10(p) =g0(v00(p)) and v11(q) =g1(v01(q)), Lemma 3.20 ensures that (H₁, v1) S and (H₁, v1) 6 S. In this way we may recursively define a chain

D0 D1 . . . Dn Dn+1 . . . i0 i1 in−1 in in+1

(†)

in the categorybDist<ω with the property that for all n∈ω and alla, b∈Dn

in+1(in(a)→n+1 in(b)) =in+1(in(a))→n+2 in+1(in(b)). (‡)

LettingAbe the chain colimit of diagram (†) in the categorybDist<ωwe see by a similar argument to the one used in the proof of Theorem2.12thatAis in fact a Heyting algebra with Heyting implication defined by

[a]→[b] = [in,k+1(a)→k+2 im,k+1], k= max{m, n}

As all the step algebras Hn = (Dn, Dn+1, in) validate HC we must also have that A validates HC. Finally if we let v0 be the valuation on A be given by v₀0(p) = [v0

0(p)] and v₁0(q) = [v0

1(q)] then because the maps ik: Dk → Dk+1 are all injective lattice homomorphisms we must have that (A, v) validatesS but notS.

Conversely, to see that item i) implies item ii) assume that HC is a hypersequent calculus with the bounded proof property and let H = (D0, D1, i) be a finite conservative one- step Heyting algebra validating HC. As Hrefutes the diagram SH/SH it follows from

Proposition3.19thatSH6`1_HCSH. Therefore, since by assumption HC has the bounded

proof property,SH6`HCSH. Hence by Proposition3.17there exists a Heyting algebraA

which refutesSH/SH. Consequently by Proposition4.4there must exist an embedding

(g0, g1) : H → A and so by Lemma 4.5 it follows that the class ConAlg_<ω(HC) has the extension property.

In practice it can be somewhat cumbersome to work with one-step extensions of one-step algebras and frames. However, in what follows we show that under the assumption that the hypersequent calculus has the finite model property we obtain a version of Theorem 4.6which avoids the notion of one-step extensions.

Definition 4.7. We say that a hypersequent calculus HC has the (global) finite model property if for each set S ∪ {S} of hypersequents, S 6`HC S iff there exists a finite Heyting algebra A validating HC and a valuation v on A such that (A, v) validates all the hypersequents fromS but not the hypersequentS.

Proposition 4.8. A hypersequent calculus HChas the finite model property iff for each setS∪{S}of hypersequents,S 6`_HCSiff there exists a finite intuitionistic Kripke frame F validating HC and a valuation v on F such that (F, v) validates all the hypersequents from S but not the hypersequentS.

Proof. Immediate by the duality between finite Heyting algebras and finite intuitionistic Kripke frames.

Lemma 4.9. LetHCbe a hypersequent calculus. ThenHC has the finite model property iff if for each setS ∪ {S} of hypersequents of degree at most 1,S 6`_HCS iff there exists a finite Heyting algebra A together with a valuation v such (A, v) validates HC and all the hypersequents from S but not the hypersequentS.

Proof. Similar to the proof of Proposition4.1.

IfF= (P,≤) is an intuitionistic Kripke frame andS= (P1, P0, f) is a one-step frame we say thatS is therelative open image of Fif there exists a surjective one-step map from

F viewed as a standard one-step frame intoS. Evidently this is equivalent to requiring that there exist an f-open order preserving surjectiong:P →P1.

In the presence of the finite model property we may prove a strengthened version of Theorem4.6.

Theorem 4.10. Let HC be a hypersequent calculus. Then the following are equivalent.

i) HC has the bounded proof property and the finite model property;

ii) Every finite conservative one-step algebra validating HC embeds into some finite (standard) Heyting algebra validating HC;

iii) Every finite conservative one-step frame validating HC is the relative open image of some finite (standard) intuitionistic Kripke frame validating HC.

Proof. As before the equivalence between item ii) and iii) is an easy exercise in duality.

To see that item i) implies item ii) assume that HC is a hypersequent calculus with the bounded proof property and the finite model property and letHbe a finite conservative one-step Heyting algebra validating HC. Then as H refutes the diagramSH/SH with

the natural valuation we have by Proposition3.19 that SH6`1_HCSH. So as HC has the

bounded proof property we obtain that SH 6`HC SH. Consequently by the assumption

that HC has the finite model property there exists a finite Heyting algebra A which validates HC and refutes the diagram SH/SH. Therefore by Proposition 4.4 the one-

step algebra His embeddable into A.

To see that item ii) implies item i) we first note that if item ii) obtains then by Lemma 4.5and Theorem 4.6the calculus HC has the bounded proof property. To see that HC also has the finite model property we know by Lemma 4.9 that it suffices to consider setsS ∪ {S}of hypersequents of degree at most 1 such that S 6`HCS.

Given such a set we consider the one-step Lindenbaum-Tarski algebraH=LT_HC(S, S). By Lemma 4.2 this is a finite conservative one-step algebra validating HC, such that under the natural valuation H validates S but not S. Now by hypothesis there exists a finite Heyting algebra A validating HC together with an embedding (g0, g1) :H →A

and so lettingv0(p) =g0(v₀0(p)) andv1(q) =g1(v0₁(q)), where v0 is the natural valuation on the one-step Lindenbaum-Tarski algebra H, we obtain a valuation on A witnessing thatS 6`HCS.