Many-Sorted First-Order Model Theory Lecture 4 12 th June, 2020 1 / - - PowerPoint PPT Presentation

Many-Sorted First-Order Model Theory

Lecture 4 12th June, 2020

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Ordinals

Ordinals are special sets constructed as follows: ◮ 0 = ∅ ◮ 1 = 0 ∪ {0} = {0} ◮ 2 = 1 ∪ {1} = {0, 1} . . . ◮ n + 1 = n ∪ {n} = {0, 1, . . . , n} . . . ◮ ω = {0, 1, 2, . . . } ◮ ω + 1 = {0, 1, 2, . . . , ω} ◮ ω + 2 = {0, 1, 2, . . . , ω + 1, ω + 2} . . . ◮ ω+n+1 = {0, 1, 2, . . . , ω, ω+1, ω+2, . . . , ω+n} . . . ◮ ω′ = {0, 1, 2 . . . , ω, ω + 1, ω + 2 . . . } All ordinal numbers greater than 0 are produced in this way, ◮ either by taking the successor of the last produced ordinal, or ◮ if there is no such last ordinal, by taking the set of all the ordinals produced so far (as in the case of ω) which yields a new limit ordinal.

Fact 1

◮ For all sets A there exists an ordinal i and a bijective mapping f : i → A. ◮ Any subclass of ordinals has a least element. ◮ For any ordinal i, the pair (i, ∈) is a total ordering. ◮ One cannot take the set of all ordinals, since this set would be a new limit ordinal, which is impossible, as we already defined them all (the collection of all ordinals is a class).

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Transfinite induction

Convention

Whenever it is convenient, we denote the membership relation ∈ among

• rdinals by <.

Ordinals support the following principle of Transfinite/Ordinal Induction: ◮ Let P(i) be a property defined for all ordinals i. ◮ Suppose that whenever P(j) is true for all j < i then P(i) is also true. ◮ Then transfinite induction tells us that P is true for all ordinals. Usually the proof is broken down into three cases: 1. Zero case Prove that P(0) is true. 2. Successor case Prove that P(i) implies P(i + 1). 3. Limit case Prove that for any limit ordinal i, P(i) follows from P(j) for all j < i.

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Cardinals

◮ A cardinal number (or simply cardinal) say how many of something there are, for example,

• ne, two, three, four, five, six.

◮ Cardinal numbers are equivalence classes, or representatives of equivalence classes, of sets under the bijection relation. ◮ The representative of each equivalence class is the least ordinal in the equivalence class, which exists since any set of ordinals has a least element. ◮ card(A) = card(B) iff there exists a bijection f : A → B. ◮ card(A) ≤ card(B) iff there exists an injection f : A → B. ◮ card(A) ≥ card(B) iff there exists a surjection f : A → B.

Lemma 2

For all sets A there is no bijection f : A → P(A).

Proof.

Suppose there exists a bijective mapping f : A → P(A). Let A0 := {a ∈ A | a ∈ f (a)} ∈ P(A). Since f is bijective, there exists a0 ∈ A such that f (a0) = A0. ◮ If a0 ∈ f (a0) then a0 ∈ A0, that is a0 ∈ f (a0). ◮ If a0 ∈ f (a0) then a0 ∈ A0, that is a0 ∈ f (a0).

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More on cardinals

◮ If card(A) = α then α < card(P(A)) = 2α. ◮ For each cardinal α, the least cardinal greater than α is denoted by α+. ◮ The Generalized Continuum Hypothesis states that 2α = α+ for all infinite cardinals α. In particular ω+ = 2ω = c, the cardinal of real numbers.

Exercise 1

◮ Any language defined over a countable alphabet is countable. Hint: prove that there exists an injection f : ω∗ → ω. ◮ Prove that ωn is countable for all n ∈ ω.

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More on cardinals

◮ Countable union of countable sets is countable too. Let {An}n∈ω be an ω-sorted set such that card(An) = ω for all n ∈ ω. Let f : ω2 →

n∈ω An defined by f (n, m) = an,m for all n, m ∈ ω,

where An = {an,0, an,1 . . . } for all n ∈ ω. One can easily notice that f is

• surjective. Hence,

n∈ω An is countable.

. . . . . . . . . . . . . . . A2 a2,0 a2,1 a2,2 . . . A1 a1,0 a1,1 a1,2 . . . A0 a0,0 a0,1 a0,2 . . . ◮ Since ω∗ =

n∈ω ωn, we have ωn ⊆ ω∗ for all n ∈ ω;

it follows that card(ωn) = ω for all n ∈ ω. ◮ For any set A of infinite cardinality, let’s say α, we have card(A∗) = α. card(A∗) = card(

n∈ω An) = n∈ω card(A)n = ω ∗ αn = α.

◮ Let α and β be two infinite cardinals, and let {Ai}i∈β be a family of sets such that (a) β < α and (b) card(Ai) < α for all i ∈ β. Then card(

i∈β Ai) = i∈β card(Ai) < β ∗ α = α.

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Signature extensions

Lemma 3

Let Σ be a signature of power α, and C an S-sorted set of new constants such that card(Cs) = α for all sorts s ∈ S. ιC : Σ ֒ → Σ[C] For all enumerations {ρi ∈ Sen(Σ[C]) | i < α} there exists a chain of signature morphisms Σ[C0]

χ0,1

֒ → Σ[C1]

χ1,2

֒ → . . . Σ[Ci]

χi,i+1

֒ → . . . Σ[Cα] such that ◮ Ci ∈ Pα(C) for all i < α, and Cα = C, ◮ ρi = χi,α(γi) for some γi ∈ Sen(Σ[Ci]), and ◮ if γi is of the form ∃Xi · γ′

i then there

exists an injective mapping ϑi : Xi → Ci+1 \ Ci. Σ[Ci, Xi]

ϑi

• Xi

ϑi

• Σ[Ci]

χi,i+1

• ιXi
• Σ[Ci+1]

Ci+1 \ Ci

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Maximally consistent sets

Lemma 4

Let Σ be a signature of power α, and C an S-sorted set of new constants such that card(Cs) = α for all sorts s ∈ S. Any consistent set Γ of Σ-sentences can be ‘extended’ 1 to a maximally consistent set Γα of Σ[C]-sentences such that ◮ if ∨E ∈ Γα then e ∈ Γα for some e ∈ E, and ◮ if ∃X · ρ ∈ Γα then ϑ(ρ) ∈ Γα for some injective mapping ϑ: X → C.

1Here ‘extended’ means that ιC(Γ) ⊆ Γα, where ιC : Σ ֒

→ Σ[C].

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Proof of Lemma 5.

Given an enumeration {ρi ∈ Sen(Σ[C]) | i < α} of the Σ[C]-sentences, consider the chain of signature morphisms Σ[C0]

χ0,1

֒ → Σ[C1]

χ1,2

֒ → . . . Σ[Ci]

χi,i+1

֒ → . . . Σ[Cα] from Lemma 4 such that for all i < α, we have ρi = χi,α(γi) for some Σ[Ci]-sentence γi. We construct a chain of presentations morphisms γ0 γi γi+1 (Σ[C0], Γ0)

χ0,1 . . .

(Σ[Ci], Γi)

χi,i+1 (Σ[Ci+1], Γi+1) χi+1,i+2

. . .

(Σ[Cα], Γα) with the following properties:

• 1. (Σ[Ci], Γi) is consistent for each i ≤ α,
• 2. χi,j(Γi) ⊆ Γj for all ordinals i and j such that i < j ≤ α,
• 3. χi,i+1(γi) ∈ Γi+1 or ¬χi,i+1(γi) ∈ Γi+1 for all ordinals i < α, and
• 4. if χi,i+1(γi) ∈ Γi+1 and γi is of the form ∃Xi · γ′

i then ϑi(γ′ i ) ∈ Γi+1.

We proceed by induction on ordinals.

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Proof of Lemma 5.

◮ i = 0 Γ0 := ιC0(Γ), where ιC0 : Σ ֒ → Σ[C0] is an inclusion. Since Γ is consistent, Γ0 is consistent as well. ◮ i ⇒ i + 1 There are two subcases. 1. Γi ∪ {γi} ⊢ ⊥ There are two subcases. 1.1 γi = ∃Xi · γ′

i

We have ιXi (Γi ∪ {∃Xi · γ′

i }) ∪ {γ′ i } ⊢ ⊥;

indeed if ιXi (Γi ∪ {∃Xi · γ′

i }) ∪ {γ′ i } ⊢ ⊥ then

by (QuantI ), Γi ∪ {∃Xi · γ′

i } ⊢ ⊥ which is a

contradiction. Since ϑi is injective, we have ϑi(ιXi (Γi ∪ {∃Xi · γ′

i }) ∪ {γ′ i }) ⊢ ⊥.

Γi+1 := χi,i+1(Γi) ∪ {χi,i+1(γi), ϑi(γ′

i )} is consistent.

Σ[Ci, Xi]

ϑi

• Σ[Ci]

χi,i+1

• ιXi
• Σ[Ci+1]

1.2 γi is not existentially quantified Γi+1 := χi,i+1(Γi) ∪ {χi,i+1(γi)} which is consistent, as Γi ∪ {γi} ⊢ ⊥ and χi,i+1 is injective. 2. Γi ∪ {γi} ⊢ ⊥ Γi+1 := χi,i+1(Γi) ∪ {¬χi,i+1(γi)}. Since Γi ⊢ ¬γi and Γi is consistent, Γi ∪ {¬γi} is consistent too. Since χi,i+1 is injective, Γi+1 = χi,i+1(Γi) ∪ {¬χi,i+1(γi)} is consistent. ◮ i ≤ α is a limit ordinal Γi :=

j<i χj,i(Γj). Suppose that Γi is not consistent. By

compactness, χj,i(Γj) is not consistent for some j < i. Since χi,j is injective, by (Cons), Γj is not consistent, which is a contradiction. Hence, Γi is consistent.

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Proof of Lemma 5.

We show that if ∨E ∈ Γα then e ∈ Γα for some e ∈ E. Suppose that e ∈ Γα for all e ∈ E, then: 1 ¬e ∈ Γα for all e ∈ E since Γα is maximally consistent 2 Γα ⊢ ¬e for all e ∈ E by (Monotonicity) 3 Γα ∪ {e} ⊢ ⊥ for all e ∈ E by (NegE ) 4 Γα ⊢ ∨E by (Monotonicity), since ∨E ∈ Γα 5 Γα ⊢ ⊥ from (3) and (4), by (DisjE ) 6 contradiction since Γα is consistent We show that if ∃X · ρ′ ∈ Γα then ϑ(ρ′) ∈ Γα for some injective mapping ϑ: X → C. Σ[Ci, Xi]

χ′

i,α

• ϑi
• Σ[C, X]

ϑ

• Σ[Ci]
• ιXi
• χi,i+1 Σ[Ci+1] χi+1,α Σ[C]
• ιX
• 1Σ[C]

Σ[C]

1 ∃X · ρ′ = χi,α(γi) for some i < α by Lemma 4 2 γi is of the form ∃Xi · γ′

i

by the definition of Sen(χi,α) 3 ϑi(γ′

i ) ∈ Γα

by the first part of the proof 4 let ϑ: X → C defined by ϑ(x, s, Σ[C]) = ϑ′

i(x, s, Σ[Ci])

for all (x, s, Σ[C]) ∈ X 5 ϑ(ρ′) = ϑ(χ′

i,α(γ′ i )) = χi+1,α(ϑi(γ′ i )) ∈ χi+1,α(Γi+1) ⊆ Γα

by the commutativity of the above diagrams and the first part of the proof

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Consistency vs. satisfiability

Theorem 5

Let Γ be a set of consistent set of sentences over a signature Σ of power α. Then for all signature extensions ιC : Σ → Σ[C], where C is a set of new constants such that card(Cs) = α for all s ∈ S, there exists a reachable model over Σ[C] which satisfies ιC (Γ).

Proof.

By Lemma 5, Γ can be ‘extended’ to a maximally consistent set Γα of Σ[C]-sentences (ιC (Γ) ⊆ Γα, where ιC : Σ ֒ → Σ[C]) such that ◮ if ∨E ∈ Γα then e ∈ Γα for some e ∈ E, and ◮ if ∃X · ρ′ ∈ Γα then ϑ(ρ′) ∈ Γα for some injective mapping ϑ: X → C. Let B ⊆ Γα be the set of all equations and relations from Γα. We show that AB | = ρ iff Γα ⊢ ρ for all Σ[C]-sentences ρ. We proceed by induction on the structure of ρ. ◮ for atomic sentences By atomic completeness, AB | = ρ iff B | = ρ iff B ⊢ ρ. ◮ If B ⊢ ρ then, by (Monotonicity) and (Transitivity), Γα ⊢ ρ. ◮ If Γα ⊢ ρ then since Γα is maximally consistent, ρ ∈ Γα; since ρ is atomic, ρ ∈ B; by (Monotonicity), B ⊢ ρ.

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Proof of Theorem 6.

◮ ¬ρ if AB | = ¬ρ iff AB | = ρ iff Γα ⊢ ρ; since Γα is maximally consistent Γα ⊢ ρ iff Γα ⊢ ¬ρ. ◮ ∨E ◮ if AB | = ∨E then AB | = e for some e ∈ E; by the induction hypothesis, Γα ⊢ e; by (DisjI ), Γα ⊢ ∨E. ◮ if Γα ⊢ ∨E then since Γα is maximally consistent, ∨E ∈ Γα; by Lemma 5, e ∈ Γα for some e ∈ E; by (Monotonicity), Γα ⊢ e; by the induction hypothesis, AB | = e; hence, AB | = ∨E. ◮ ∃X · ρ′ ◮ Assume that AB | = ∃X · ρ′. 1 A | = ρ′ for some ιX -expansion A of AB since AB | = ∃X · ρ′ 2 AB ↾θ= A for some substitution θ: X → TΣ[C] since AB is reachable 3 AB | = θ(ρ′) by the satisfaction condition for substitutions since A | = ρ′ 4 Γα ⊢ θ(ρ′) by the induction hypothesis 5 Γα ⊢ ∃X · ρ′ by (Sb) ◮ Assume thar Γα ⊢ ∃X · ρ′ 1 Γ ⊢ ϑ(ρ′) for some injective mapping ϑ: X → C by Lemma 5 2 AB | = ϑ(ρ′) by the induction hypothesis 3 AB ↾ϑ| = ρ′ by the satisfaction condition 4 AB | = ∃X · ρ′ since (AB ↾ϑ)↾ιX = AB It follows that AB | = Γα. In particular, AB | = ιC (Γ).

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Completeness

Corollary 6

The first-order entailment relation ⊢ is complete.

Proof.

By (Unions), it suffices to prove that Γ | =Σ γ implies Γ ⊢Σ γ, which is equivalent to Γ ⊢Σ γ implies Γ | =Σ γ. 1 assume that Γ ⊢Σ γ 2 Γ ∪ {¬γ} ⊢Σ ⊥ if Γ ∪ {¬γ} ⊢Σ ⊥ then Γ ⊢Σ ¬¬γ and we get Γ ⊢Σ γ, which is a contradiction 3 let α = card(Sen(Σ) and ιC : Σ ֒ → Σ[C] such that card(Cs) = α for all sorts s ∈ S 4 A | = ιC (Γ ∪ {¬γ}) for some reachable by Theorem 6 model A over the signature Σ[C] 5 A↾Σ| = Γ ∪ {¬γ} by the satisfaction condition 6 Γ | =Σ γ since A↾Σ| = Γ and A↾Σ| = γ

Fact 7

By soundness and completeness, we have that ⊢ = | =.

Exercise 2

Show that a set of sentences is consistent iff it is satisfiable.

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Compactness

The following result is a consequence of soundness and completeness.

Corollary 8

• 1. The satisfaction relation |

= is compact.

• 2. A set of sentences Γ is satisfiable iff every finite subset of Γ is satisfiable.

Proof.

• 1. Assume that Γ |

= T and let Tf ⊆ T finite. We show that Γf | = Tf for some Γf ⊆ Γ finite: 1 Γ | = Tf since Γ | = T and Tf ⊆ T 2 Γ ⊢ Tf by completeness 3 Γf ⊢ Tf for some Γf ⊆ Γ finite since ⊢ is compact 4 Γf | = Tf by soundness

• 2. Since the first-order entailment relation ⊢ is compact,

Γ ⊢ ⊥ iff Γf ⊢ ⊥ for some finite subset Γf ⊆ Γ. It follows that Γ ⊢ ⊥ iff Γf ⊢ ⊥ for all finite subsets Γf ⊆ Γ. Since soundness and completeness, we get Γ is satisfiable iff each finite subset Γf ⊆ Γ is satisfiable.

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Downwards L¨

• wenheim-Skolem (DLS) Theorem

The following result is a consequence of Theorem 6.

Corollary 9

Every satisfiable set of sentences Γ over a signature Σ of power α has a model A of power at most α, i.e. card(As) ≤ α for all s ∈ S.

Proof.

Since Γ is satisfiable, Γ is consistent. Let C be an S-sorted set of constants such that card(Cs) = α for all s ∈ S. 1 there exists a reachable Σ[C]-model B by Theorem 6 such that B | = ιC (Γ) 2 card(Bs) ≤ α for all s ∈ S (a) card(TΣ[C],s) ≤ α for all s ∈ S, and (b) TΣ[C] → B is surjective 3 A | = Γ, where A = B↾Σ by the satisfaction condition 4 card(As) ≤ α for all s ∈ S since card(Bs) ≤ α for all s ∈ S

Exercise 3

Every satisfiable set of sentences Γ over a signature Σ of power α has a model A of power α, i.e. card(As) = α for all s ∈ S. Hint: Let C be an S-sorted set of constants such that card(Cs) = α for all s ∈ S. Show that ιC (Γ) ∪ {¬(c = d) | s ∈ S and c, d ∈ Cs such that c = d} is satisfiable using compactness.

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Countable models

A model A over a signature Σ = (S, F, P) is called countable iff As is countable for all sorts s ∈ S. The following result is a consequence of Theorem 6.

Corollary 10

Every countable and satisfiable set of sentences has a countable model.

Proof.

Let Γ be a countable and satisfiable set of sentences over a signature Σ = (S, F, P). There exists a subsignature Σ′ ⊆ Σ which consists of a countable number of symbols such that Γ ⊆ ι(Sen(Σ′)), where ι: Σ′ ֒ → Σ is an inclusion. Let Γ′ = ι−1(Γ). Let C be a set of new constant symbols such that card(Cs) = ω for all sorts s ∈ S. By Theorem 6, there exists a reachable Σ′[C]-model A′ such that A′ | = ιC (Γ′). Notice that card(A′

s′) ≤ ω for all s′ ∈ S′,

which means that A′ is countable. It follows that A′ ↾Σ′ is countable too. By the satisfaction condition, A′ ↾Σ′| = Γ′. Let m be an element which doesn’t belong to the universe |A′|. We define the Σ-model A as follows: ◮ As = A′

s

for all s ∈ S′ A′

s = {m}

for all s ∈ S \ S′ ◮ For all σ : w → s ∈ F ′, the function σA : Aw → As is defined by σA(a) = σA′. For all σ : w → s ∈ F \ F ′, the function σA : Aw → As is defined randomly. ◮ For all π : w ∈ P′, πA = πA′. For all π : w ∈ P \ P′, πA = ∅. Since A′ ↾Σ′ is countable, A is countable too. Since ι: Sen(Σ′) → Sen(Σ) is injective, ι(ι−1(Γ)) = Γ. By the satisfaction condition, A | = ι(Γ′) = ι(ι−1(Γ)) = Γ .

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Exercise 4

Let Γ be a set of sentences over a signature Σ = (S, F, P) that has arbitrarily large finite models, i.e. for each n ∈ ω there exists a model An such that An | = Γ and card(An

s ) ≥ n for all s ∈ S. Show that Γ has an

infinite model, i.e. there exists a Σ-model A such that A | = Γ and card(As) ≥ ω for all s ∈ S.

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