Hintikka sets and their applications

Jouko Väänänen1,2

1Department of Mathematics and Statistics, University of Helsinki

2Institute for Logic, Language and Computation, University of Amsterdam

Beijing, June 2016

Jouko Väänänen (Helsinki and Amsterdam) Hintikka sets Beijing, June 2016 1 / 35

We learn a model construction method.

A preliminary convention: Pushing negation inside

φ¬ = ¬φ if φ atomic(¬φ)¬ = φ(φ ∧ ψ)¬ = ¬φ ∨ ¬ψ(φ ∨ ψ)¬ = ¬φ ∧ ¬ψ(∀xφ)¬ = ∃x¬φ(∃xφ)¬ = ∀x¬φ

DefinitionL countable, C a countable set of new constant symbols andL′ = L ∪ C, C = C. A Hintikka set is any set H of L′-sentences, whichsatisfies:

1 t = t ∈ H for every constant L′-term t . (Add ¬c = c′ for strictHintikka set whenever s(c) 6= s(c′).)

2 If φ(t) ∈ H, φ(t) atomic, and t = t ′ ∈ H, then φ(t ′) ∈ H.3 If ¬φ ∈ H, then φ¬ ∈ H.4 If φ ∨ ψ ∈ H, then φ ∈ H or φ ∈ H.5 If φ ∧ ψ ∈ H, then φ ∈ H and φ ∈ H.6 If ∃xφ(x) ∈ H, then φ(c) ∈ H for some c ∈ C7 If ∀xφ(x) ∈ H, then φ(c) ∈ H for all c ∈ C8 For every constant L′-term t there is c ∈ C such that t = c ∈ H.9 There is no atomic sentence φ such that φ ∈ H and ¬φ ∈ H.

The Hintikka set H is a Hintikka set for a sentence φ if φ ∈ H.

1 t = t ∈ H for every constant L′-term t .2 If φ(t) ∈ H, φ(t) atomic, and t = t ′ ∈ H, thenφ(t ′) ∈ H.

3 If ¬φ ∈ H, then φ¬ ∈ H.4 If φ ∧ ψ ∈ H, then φ ∈ H and ψ ∈ H.5 If φ ∨ ψ ∈ H, then φ ∈ H or ψ ∈ H.6 If ∃xφ(x) ∈ H, then φ(c) ∈ H for some c ∈ C7 If ∀xφ(x) ∈ H, then φ(c) ∈ H for all c ∈ C8 For every constant L′-term t there is c ∈ C such

that t = c ∈ H.9 There is no atomic φ such that φ ∈ H and ¬φ ∈ H.

A basic lemma

LemmaIf there is a Hintikka set for φ, then φ has a model.

Note: Conversely, if φ has a modelM, then there is a Hintikka set forφ, built directly from formulas true inM.

Note: A Hintikka set need not be complete. There may be φ such thatneither φ ∈ H nor ¬φ ∈ H.

Proof.M = {[c] : c ∈ C}, where c ∼ c′ is defined as c = c′ ∈ H.cM = [c].Let fM([ci1 ], . . . , [cin ]) = [c] for c ∈ C such thatf (ci1 , . . . , cin ) = c ∈ H.For any constant term t there is a c ∈ C such that t = c ∈ H. It iseasy to see that tM = [c].We letM |= R(t1, . . . , tn) if and only if R(t1, . . . , tn) ∈ H.By induction on φ(x1, . . . , xn): if d1 . . . ,dn ∈ C then:

1 If φ(d1, . . . ,dn) ∈ H, thenM |= φ(d1, . . . ,dn).2 If ¬φ(d1, . . . ,dn) ∈ H, thenM 6|= φ(d1, . . . ,dn).

In particular,M |= φ for the φ we started with, since φ ∈ H.

Whence Hintikka sets?

How to find useful Hintikka sets?The tool is: consistency property.A consistency property is a set of (usually) finite sets S which areconsistent and the consistency property has information abouthow to extend S to a Hintikka set, which will then give a model forS.

DefinitionLet L be a countable signature,C a countable set of new constant symbolsand L′ = L ∪ C. A consistency property is any set ∆ of countable sets S ofL-formulas, which satisfies the conditions:

1 If S ∈ ∆, then S ∪ {t = t} ∈ ∆ for every constant L′-term t . For strict ConsistencyProperty demand S ∪ {¬t = t ′} ∈ ∆ if s(t) 6= s(t ′).

2 If φ(t) ∈ S ∈ ∆, φ(t) atomic, and t = t ′ ∈ S, then S ∪ {φ(t ′)} ∈ ∆.

3 If ¬φ ∈ S ∈ ∆, then S ∪ {φ¬} ∈ ∆.

4 If φ ∨ ψ ∈ S ∈ ∆, then S ∪ {φ} ∈ ∆ or S ∪ {ψ} ∈ ∆

5 If φ ∧ ψ ∈ S ∈ ∆, then S ∪ {φ} ∈ ∆ and S ∪ {ψ} ∈ ∆

6 If ∃xφ(x) ∈ S ∈ ∆, then S ∪ {φ(c)} ∈ ∆ for some c ∈ C.

7 If ∀xφ(x) ∈ S ∈ ∆, then S ∪ {φ(c)} ∈ ∆ for all c ∈ C.

8 For every constant L′-term t there is c ∈ C such that S ∪ {t = c} ∈ ∆.

9 There is no atomic formula φ such that φ ∈ S and ¬φ ∈ S.

The consistency property ∆ is a consistency property for a set T of L-sentences if for all S ∈ ∆ and all φ ∈ T we have S ∪ {φ} ∈ ∆.

Existence of Hintikka sets

Let T be a countable set of L-sentences. Suppose ∆ is a consistencyproperty for T . Then for any S ∈ ∆ there is a Hintikka set H for T suchthat S ⊆ H .

Proof:

Let Trm the set of all constant L′-terms.Let

T = {φn : n ∈ N}C = {cn : n ∈ N}

Trm = {tn : n ∈ N}.

Let {ψn : n ∈ N} be a list of all L′-formulas.We define H as the union of an increasing sequence S0,S1, ...,where S0 = S.Sn+1 = Sn, but:

1 If n = 3i , then Sn+1 is Sn ∪ {φi} ∈ ∆.

2 If n = 2 · 3i , then Sn+1 is Sn ∪ {ti = ti} ∈ ∆.

3 If n = 4 · 3i · 5j · 7k , ψi = (t = t ′) ∈ Sn, and ψj = φ(t) ∈ Sn with φ(t)atomic, then Sn+1 is Sn ∪ {φ(t ′)} ∈ ∆.

4 If n = 8 · 3i · 5k and ψi = ¬ψ ∈ Sn, then Sn+1 is Sn ∪ {ψ¬}.5 If n = 16 · 3i · 5k and ψi = ψ ∨ φ ∈ Sn, then Sn+1 is Sn ∪ {ψ} or Sn ∪ {φ},

whichever is in ∆.

6 If n = 32 · 3i · 5j · 7k , j ∈ {0,1} and ψi = φ0 ∧ φ1 ∈ Sn, then Sn+1 isSn ∪ {φj}(∈ ∆).

7 If n = 64 · 3i · 7k and ψi = ∃xφ ∈ Sn, then Sn+1 is Sn ∪ {φ(c)} for suchc ∈ C that Sn+1 ∈ ∆.

8 If n = 128 · 3i · 5j · 7k and ψi = ∀xφ ∈ Sn, then Sn+1 is Sn ∪ {φ(cj )}(∈ ∆).

9 If n = 256 · 3i , then Sn+1 is Sn ∪ {ti = c} for such c ∈ C that Sn+1 ∈ ∆.

Clearly⋃

n Sn is a Hintikka set for T .

Consistency property from “consistency"

LemmaThe set ∆ of finite sets S of sentences such that S 0 ⊥ is aconsistency property.

Proof:1 Clearly, if S ∈ ∆, then S ∪ {t = t} ∈ ∆ for every constant L′-term t .2 Clearly, if φ(t) ∈ S ∈ ∆, φ(t) atomic, and t = t ′ ∈ S, then

S ∪ {φ(t ′)} ∈ ∆.3 Suppose ¬φ ∈ S ∈ ∆, but S ∪ {φ¬} ` ⊥. Then S ` ⊥,

contradiction.

Suppose φ ∨ ψ ∈ S ∈ ∆ but S ∪ {φ} ` ⊥ and S ∪ {ψ} ` ⊥. ThenS ` ⊥, contradiction.Suppose φ ∧ ψ ∈ S ∈ ∆ but S ∪ {φ} ` ⊥ or S ∪ {ψ} ` ⊥. ThenS ` ⊥, contradiction.Suppose ∃xφ(x) ∈ S ∈ ∆ but S ∪ {φ(c)} ` ⊥ for all c ∈ C. ThenS ` ⊥, because we can choose c so that it does not occur in S.Contradiction.Suppose ∀xφ(x) ∈ S ∈ ∆ but S ∪ {φ(c)} ` ⊥ for some c ∈ C.Then S ` ⊥, contradiction.Let us consider a constant L′-term t . There is c ∈ C such thatS ∪ {t = c} ∈ ∆.There is no atomic formula φ such that φ ∈ S and ¬φ ∈ S,because {φ,¬φ} ` ⊥.

The Completeness Theorem

TheoremTFAE for all φ:

1 |= φ i.e. φ is true in all models.2 ` φ i.e. φ has a proof.

Proof.If φ has a proof, then clearly |= φ. If φ does not have a proof, then¬φ 0 ⊥. So {¬φ} ∈ ∆ for the ∆ in the previous lemma. Hence ¬φ hasa model and 6|= φ.

Consistency property from “finitely consistent"

LemmaThe set ∆ of sets S of sentences such that only finitely manyconstants in C occur in S and every finite subset of S has a model, is aconsistency property.

The Compactness Theorem

TheoremIf T is a countable set of sentences such that every finite subset of Thas a model then T itself has a model.

Proof.By assumption S ∈ ∆ for the ∆ in the previous lemma. Hence S has amodel.

Interpolation

Craig Interpolation Theorem

Theorem

We assume that L1 and L2 are vocabularies. Suppose |= φ→ ψ,where φ is an L1-sentence and ψ is an L2-sentence. Then there is anL1 ∩ L2-sentence θ (“interpolant") such that

1 |= φ→ θ

2 |= θ → ψ

PropositionSuppose φ depends only on R in the sense that

M |= φ ⇐⇒ N |= φ

wheneverM and N have the same domain and the sameinterpretation of R. Then |= φ↔ ψ where no non-logical symbolsexcept R occurs in ψ.

Proof.Let S1, ...,Sn be the non-logical symbols in φ or ψ in addition to R. Letφ′ be the result of replacing each Si in φ by a new symbol S′i . Now

|= φ→ φ′.

By the Interpolation Theorem there is θ containing only R such that|= φ→ θ and |= θ → φ′. Hence |= φ↔ θ.

Theorem (Beth Definability Theorem)Suppose the predicate S depends only on R in the sense that

φ(R,S) ∧ φ(R,S′) |= ∀x(S(x)↔ S′(x)).

Then there is θ(R, x) where S does not occur such thatφ(R,S) |= ∀x(S(x)↔ θ(R, x)).

Proof.By assumption

|= (φ(R,S) ∧ S(c))→ (φ(R,S′)→ S′(c)).

Let θ(R, c) be such that

|= (φ(R,S) ∧ S(c))→ θ(R, c) and |= θ(R, c)→ (φ(R,S′)→ S′(c)).

Then φ(R,S) |= ∀x(S(x)↔ θ(R, x)).

ExampleInterpolation fails in finite models. Suppose φ is as follows:

∀x∃yS(x , y)∧∀x∀y∀y ′((S(x , y) ∧ S(x , y ′))→ y = y ′)∀x∀y(S(x , y)→ (¬x = y ∧ S(y , x)))

Suppose ψ is

∃z[S′(z, z) ∧ ∀x∃yS′(x , y)∧∀x∀y∀y ′((S′(x , y) ∧ S′(x , y ′))→ y = y ′)∀x∀y((S′(x , y) ∧ ¬x = z)→ (¬x = y ∧ S′(y , x)))]

Then |= φ→ ¬ψ but if θ is an interpolant, then θ is an identity-sentencewhich is true in exactly the finite models with even cardinality, which isimpossible.

Now: Proof of the Interpolation Theorem.

Craig Interpolation Theorem

Theorem

We assume that L1 and L2 are vocabularies. Suppose |= φ→ ψ,where φ is an L1-sentence and ψ is an L2-sentence. Then there is anL1 ∩ L2-sentence θ (“interpolant") such that

1 |= φ→ θ

2 |= θ → ψ

Proof of Interpolation

Let us assume that the claim of the theorem is false and derive acontradiction. Since |= φ→ ψ, the set {φ,¬ψ} has no models.(We may assume that φ is consistent and that ψ is not valid.) Weconstruct a consistency property for {φ,¬ψ}.Let L = L1 ∩ L2. Suppose C = {cn : n ∈ N} is a set of newconstant symbols.Given a set S of sentences, let S1 consists of all L1 ∪C-sentencesin S and let S2 consists of all L2 ∪ C-sentences in S.

Separation

DefinitionLet us say that θ separates S1 and S2 if

1 S1 |= θ,2 S2 |= ¬θ,3 θ is an L-sentence.

Our consistency property

DefinitionLet ∆ consist of all finite sets S of L-sentences such that S = S1 ∪ S2and:(?) There is no L ∪ C-sentence that separates S1 and S2.

∆ is a consistency property

Case 0. {φ,¬ψ} ∈ ∆. True by assumption.

Case 1. Suppose S ∈ ∆ and consider c = c, where, for example,c ∈ L1 ∪C. We let S′1 = S1 ∪ {c = c} and S′2 = S2 ∪ {c = c}. Supposeθ(c0, . . . , cm−1) separates S′1 and S′2. Then clearly also θ(c0, . . . , cm−1)separates S1 and S2, a contradiction.Case 2. Suppose φ(c) ∈ S ∈ ∆, φ(c) atomic, and c = c′ ∈ S. ClearlyS ∪ {φ(c′)} ∈ ∆.Case 3. Negation: trivial.

Case 4. Consider φ ∨ ψ, such that, for example, φ ∨ ψ ∈ S1. We claimthat either the sets S1 ∪ {φ} and S2 satisfy (?), or the sets S1 ∪ {φ} andS2 satisfy (?). Otherwise there some θ that separates S1 ∪ {φ} and S2,and some θ′ that separates S1 ∪ {ψ} and S2. Let θ∗ = θ ∨ θ′. Then θ∗

separates S1 and S2 contrary to assumption.

Case 5. Consider φ ∧ ψ where for example φ ∧ ψ ∈ S1. LetS′1 = S1 ∪ {φ} and S′2 = S2. If θ separates S′1 and S′2, then clearly θalso separates S1 and S2. Hence S ∪ {φ} ∈ ∆. Similarly for ψ.

Case 6. Consider S ∈ ∆ and ∃xφ(x) ∈ S1. Let c0 ∈ C be such that cdoes not occur in S. We claim that the sets S1 ∪ {φ(c0)} and S2 satisfy(?). Otherwise there is some θ(c0, . . . , cm−1) that separatesS1 ∪ {φ(c0)} and S2. Let1 θ′(c1, . . . , cm−1) = ∃xθ(x , c1, . . . , cm−1). Weshow that θ′(c1, . . . , cm−1) separates S1 and S2, a contradiction.Checking this:

S1 |= θ′(c1, . . . , cm−1):

S1 ∪ {φ(c0)} |= θ(c0, . . . , cm−1) by assumption

S1 |= φ(c0)→ θ(c0, . . . , cm−1)

S1 |= ∀x(φ(x)→ θ(x , c1, . . . , cm−1))

S1 |= ∃xφ(x)→ ∃xθ(x , c1, . . . , cm−1)

S1 |= ∃xθ(x , c1, . . . , cm−1) as S1 |= ∃xφ(x)

S1 |= θ′(c1, . . . , cm−1)

1If c0 does not occur in θ, then we take θ′ = θ.Jouko Väänänen (Helsinki and Amsterdam) Hintikka sets Beijing, June 2016 30 / 35

Case 6. (Contd.)S2 |= ¬θ′(c0, . . . , cm−1):

S2 |= ¬θ(c0, . . . , cm−1)

S2 |= ∀x¬θ(x , c1, . . . , cm−1)

S2 |= ¬∃xθ(x , c1, . . . , cm−1)

S2 |= ¬θ′(c1, . . . , cm−1)

Case 6. (Contd.) Consider S ∈ ∆ and ∃xφ(x) ∈ S2. Let c0 ∈ C besuch that c0 does not occur in S. We claim that the sets S1 andS2 ∪ {φ(c0)} satisfy (?). Otherwise there is some θ(c0, . . . , cm−1) thatseparates S1 and S2 ∪ {φ(c0)}. Let2

θ′(c1, . . . , cm−1) = ∀xθ(x , c1, . . . , cm−1). We show that θ′(c1, . . . , cm−1)separates S1 and S2, a contradiction. Checking this:

S1 |= θ′(c1, . . . , cm−1):

S1 |= θ(c0, . . . , cm−1)

S1 |= ∀xθ(x , c1, . . . , cm−1)

S1 |= θ′(c1, . . . , cm−1)

2If c0 does not occur in θ, we choose θ′ = θ.Jouko Väänänen (Helsinki and Amsterdam) Hintikka sets Beijing, June 2016 32 / 35

S2 |= ¬θ′(c1, . . . , cm−1):

S2 ∪ {φ(c0)} |= ¬θ(c0, . . . , cm−1)

S2 |= φ(c0)→ ¬θ(c0, . . . , cm−1)

S2 |= ∀x(φ(x)→ ¬θ(x , c1, . . . , cm−1))

S2 |= ∃xφ(x)→ ∃x¬θ(x , c1, . . . , cm−1)

S2 |= ∃x¬θ(x , c1, . . . , cm−1)

S2 |= ¬θ′(c1, . . . , cm−1)

Case 7. Consider S ∈ ∆, φ(c0), where c0 ∈ C and ∀xφ(x) ∈ S1.Exercise!

Case 7. (Contd.) Consider φ(c0), where c0 ∈ C and ∀xφ(x) ∈ S2.Exercise!

Case 8. We are given a term t . Let c be a constant not in S. ThenS ∪ {t = c} ∈ ∆.

Case 9. Suppose φ ∈ S and ¬φ ∈ S, where S ∈ ∆.

Case 1: φ ∈ S1 and ¬φ ∈ S1. Let θ be ¬∀x(x = x). Then θ separatesS1 and S2, contradiction.

Case 2: φ ∈ S2 and ¬φ ∈ S2. Let θ be ∀x(x = x). Then θ separates S1and S2, contradiction.

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