Many-Sorted First-Order Model Theory Lecture 9 2 nd July, 2020 1 / - - PowerPoint PPT Presentation

Many-Sorted First-Order Model Theory

Lecture 9 2nd July, 2020

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Quantifier elimination example Dense linear orders w/o endpoints

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Eliminating one ∃

Lemma 1

Let T be the theory of dense linear orders without endpoints. Every formula ∃x · ϕ1 ∧ · · · ∧ ϕn, where each ϕi is atomic or negated atomic, is equivalent over T to a (positive) quantifier free formula.

Proof sketch.

◮ Wlog, assume that each ϕi contains an occurrence of x. ◮ Next, observe that T | = x = y ↔ x < y ∨ y < x, and T | = x < y ↔ x = y ∨ y < x. ◮ Using this and distributivity of ∃ over disjunction we can assume that each ϕi is atomic. ◮ If some ϕi is x = y, then we can replace all occurrences of x in ϕ1 ∧ · · · ∧ ϕn by y, and delete y = y, unless it is a single conjunct. Note that this can leave some conjuncts of the form z = y. They can harmlessly be brought out of the scope of ∃. ◮ Now we can assume all ϕi are of the form < x or x < . ◮ Group them into

j(vj < x) and k(x < vj), to get ∃x · j(vj < x) ∧ k(x < vk).

◮ We have T | = ∃x ·

j(vj < x) ∧ k(x < vk) ↔ j,k(vj < vk). To deal with special cases

we assume (as usual) that an empty conjunction is equivalent to ⊤ (or to x = x). ◮ We have shown more than we claimed: ∃x · ϕ1 ∧ · · · ∧ ϕn is equivalent over T to a positive quantifier free formula ψ, such that we have either T | = ψ or T | = ¬ψ, and it is decidable which. (Because ψ is quantifier free, we have T | = ψ iff T | = ∀x · ψ.)

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Eliminating one ∃

Lemma 1

Let T be the theory of dense linear orders without endpoints. Every formula ∃x · ϕ1 ∧ · · · ∧ ϕn, where each ϕi is atomic or negated atomic, is equivalent over T to a (positive) quantifier free formula.

Proof sketch.

◮ Wlog, assume that each ϕi contains an occurrence of x. ◮ Next, observe that T | = x = y ↔ x < y ∨ y < x, and T | = x < y ↔ x = y ∨ y < x. ◮ Using this and distributivity of ∃ over disjunction we can assume that each ϕi is atomic. ◮ If some ϕi is x = y, then we can replace all occurrences of x in ϕ1 ∧ · · · ∧ ϕn by y, and delete y = y, unless it is a single conjunct. Note that this can leave some conjuncts of the form z = y. They can harmlessly be brought out of the scope of ∃. ◮ Now we can assume all ϕi are of the form < x or x < . ◮ Group them into

j(vj < x) and k(x < vj), to get ∃x · j(vj < x) ∧ k(x < vk).

◮ We have T | = ∃x ·

j(vj < x) ∧ k(x < vk) ↔ j,k(vj < vk). To deal with special cases

we assume (as usual) that an empty conjunction is equivalent to ⊤ (or to x = x). ◮ We have shown more than we claimed: ∃x · ϕ1 ∧ · · · ∧ ϕn is equivalent over T to a positive quantifier free formula ψ, such that we have either T | = ψ or T | = ¬ψ, and it is decidable which. (Because ψ is quantifier free, we have T | = ψ iff T | = ∀x · ψ.)

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Eliminating all quantifiers

Theorem 2

Let T be the theory of dense linear orders without endpoints. Then for every formula ϕ there exist a quantifier free formula ϕ∗ such that ◮ T | = ϕ ↔ ϕ∗, ◮ ϕ∗ is effectively obtainable from ϕ.

Proof.

◮ We can assume ϕ is in prenex form, so ϕ = Q1 . . . Qnψ where ψ is quantifier free. ◮ Consider the innermost quantifier Qn; if Qn is ∀, replace it by ¬∃¬. Bring ¬ψ into disjunctive form, and distribute ∃ over disjunctions. ◮ Thus, we can assume Qnψ is (¬)(∃x · ρ1 ∨ · · · ∨ ∃x · ρm), where each ρi is a conjunction of atomic and/or negated atomic formulas. ◮ By Lemma 1 each ∃x · ρi is equivalent over T to a quantifier free σi. Replace them. ◮ Repeat the procedure n times, to remove all quantifiers and obtain ϕ∗. The procedure is clearly effective.

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Eliminating all quantifiers

Theorem 2

Let T be the theory of dense linear orders without endpoints. Then for every formula ϕ there exist a quantifier free formula ϕ∗ such that ◮ T | = ϕ ↔ ϕ∗, ◮ ϕ∗ is effectively obtainable from ϕ.

Proof.

◮ We can assume ϕ is in prenex form, so ϕ = Q1 . . . Qnψ where ψ is quantifier free. ◮ Consider the innermost quantifier Qn; if Qn is ∀, replace it by ¬∃¬. Bring ¬ψ into disjunctive form, and distribute ∃ over disjunctions. ◮ Thus, we can assume Qnψ is (¬)(∃x · ρ1 ∨ · · · ∨ ∃x · ρm), where each ρi is a conjunction of atomic and/or negated atomic formulas. ◮ By Lemma 1 each ∃x · ρi is equivalent over T to a quantifier free σi. Replace them. ◮ Repeat the procedure n times, to remove all quantifiers and obtain ϕ∗. The procedure is clearly effective.

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Decidabillity

Corollary 3

The theory T of dense linear orders without endpoints is decidable.

Proof.

◮ Let ϕ be a formula in the signature of T. By Lemma 1 and Theorem 2, ϕ is equivalent

• ver T to a Boolean combination of formulas ψj (1 ≤ j ≤ n) such that for each j we have

T | = ψj or T | = ¬ψj and it is decidable which. ◮ Hence the problem T | = ϕ reduces to propositional tautology checking.

Lemma 4

Every formula ϕ is equivalent over T to a disjunction of conjunctions of atomic formulas. Thus, ϕ(x) is equivalent over T to the statement x1 ≤ x2 ≤ · · · ≤ xn, for some suitable renumbering of x, where each ≤ is either < or =.

Proof.

By analysing the proofs of Lemma 1 and Theorem 2.

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Decidabillity

Corollary 3

The theory T of dense linear orders without endpoints is decidable.

Proof.

◮ Let ϕ be a formula in the signature of T. By Lemma 1 and Theorem 2, ϕ is equivalent

• ver T to a Boolean combination of formulas ψj (1 ≤ j ≤ n) such that for each j we have

T | = ψj or T | = ¬ψj and it is decidable which. ◮ Hence the problem T | = ϕ reduces to propositional tautology checking.

Lemma 4

Every formula ϕ is equivalent over T to a disjunction of conjunctions of atomic formulas. Thus, ϕ(x) is equivalent over T to the statement x1 ≤ x2 ≤ · · · ≤ xn, for some suitable renumbering of x, where each ≤ is either < or =.

Proof.

By analysing the proofs of Lemma 1 and Theorem 2.

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Decidabillity

Corollary 3

The theory T of dense linear orders without endpoints is decidable.

Proof.

◮ Let ϕ be a formula in the signature of T. By Lemma 1 and Theorem 2, ϕ is equivalent

• ver T to a Boolean combination of formulas ψj (1 ≤ j ≤ n) such that for each j we have

T | = ψj or T | = ¬ψj and it is decidable which. ◮ Hence the problem T | = ϕ reduces to propositional tautology checking.

Lemma 4

Every formula ϕ is equivalent over T to a disjunction of conjunctions of atomic formulas. Thus, ϕ(x) is equivalent over T to the statement x1 ≤ x2 ≤ · · · ≤ xn, for some suitable renumbering of x, where each ≤ is either < or =.

Proof.

By analysing the proofs of Lemma 1 and Theorem 2.

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Decidabillity

Corollary 3

The theory T of dense linear orders without endpoints is decidable.

Proof.

◮ Let ϕ be a formula in the signature of T. By Lemma 1 and Theorem 2, ϕ is equivalent

• ver T to a Boolean combination of formulas ψj (1 ≤ j ≤ n) such that for each j we have

T | = ψj or T | = ¬ψj and it is decidable which. ◮ Hence the problem T | = ϕ reduces to propositional tautology checking.

Lemma 4

Every formula ϕ is equivalent over T to a disjunction of conjunctions of atomic formulas. Thus, ϕ(x) is equivalent over T to the statement x1 ≤ x2 ≤ · · · ≤ xn, for some suitable renumbering of x, where each ≤ is either < or =.

Proof.

By analysing the proofs of Lemma 1 and Theorem 2.

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Completeness

Theorem 5

Let A and B be models of T. If A ֒ → B, then A

e

֒ → B.

Proof sketch.

◮ We use Tarski-Vaught criterion. Let ϕ(x, y) be a formula and let a be a tuple from A. Wlog, A ≤ B. Assume B | = ∃x · ϕ(x, a). Thus, B | = ϕ(b, a), for some b from B. ◮ By Lemma 4, this is equivalent to a1 ≤ · · · ≤ ai ≤ b ≤ ai+1 ≤ . . . an, with a renumbered suitably. ◮ If b is already in A there is nothing to do, so assume ai < b < ai+1 and b / ∈ A. By density

• f A, there is a c ∈ A with ai < c < ai+1. Therefore, a1 ≤ · · · ≤ ai ≤ c ≤ ai+1 ≤ . . . an

holds in A. ◮ By Lemma 4, we obtain A | = ϕ(c, a). Hence, A | = ∃x · ϕ(x, a) as required.

Corollary 6

All models of T are elementarily equivalent. Hence, T is complete.

Proof.

Let A and B be dense linear orders without endpoints. We have A ֒ → B iff card(A) ≤ card(B). By Theorem 5, A

e

֒ → B or B

e

֒ → A holds. In either case A ≡ B.

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Completeness

Theorem 5

Let A and B be models of T. If A ֒ → B, then A

e

֒ → B.

Proof sketch.

◮ We use Tarski-Vaught criterion. Let ϕ(x, y) be a formula and let a be a tuple from A. Wlog, A ≤ B. Assume B | = ∃x · ϕ(x, a). Thus, B | = ϕ(b, a), for some b from B. ◮ By Lemma 4, this is equivalent to a1 ≤ · · · ≤ ai ≤ b ≤ ai+1 ≤ . . . an, with a renumbered suitably. ◮ If b is already in A there is nothing to do, so assume ai < b < ai+1 and b / ∈ A. By density

• f A, there is a c ∈ A with ai < c < ai+1. Therefore, a1 ≤ · · · ≤ ai ≤ c ≤ ai+1 ≤ . . . an

holds in A. ◮ By Lemma 4, we obtain A | = ϕ(c, a). Hence, A | = ∃x · ϕ(x, a) as required.

Corollary 6

All models of T are elementarily equivalent. Hence, T is complete.

Proof.

Let A and B be dense linear orders without endpoints. We have A ֒ → B iff card(A) ≤ card(B). By Theorem 5, A

e

֒ → B or B

e

֒ → A holds. In either case A ≡ B.

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Completeness

Theorem 5

Let A and B be models of T. If A ֒ → B, then A

e

֒ → B.

Proof sketch.

◮ We use Tarski-Vaught criterion. Let ϕ(x, y) be a formula and let a be a tuple from A. Wlog, A ≤ B. Assume B | = ∃x · ϕ(x, a). Thus, B | = ϕ(b, a), for some b from B. ◮ By Lemma 4, this is equivalent to a1 ≤ · · · ≤ ai ≤ b ≤ ai+1 ≤ . . . an, with a renumbered suitably. ◮ If b is already in A there is nothing to do, so assume ai < b < ai+1 and b / ∈ A. By density

• f A, there is a c ∈ A with ai < c < ai+1. Therefore, a1 ≤ · · · ≤ ai ≤ c ≤ ai+1 ≤ . . . an

holds in A. ◮ By Lemma 4, we obtain A | = ϕ(c, a). Hence, A | = ∃x · ϕ(x, a) as required.

Corollary 6

All models of T are elementarily equivalent. Hence, T is complete.

Proof.

Let A and B be dense linear orders without endpoints. We have A ֒ → B iff card(A) ≤ card(B). By Theorem 5, A

e

֒ → B or B

e

֒ → A holds. In either case A ≡ B.

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Completeness

Theorem 5

Let A and B be models of T. If A ֒ → B, then A

e

֒ → B.

Proof sketch.

◮ We use Tarski-Vaught criterion. Let ϕ(x, y) be a formula and let a be a tuple from A. Wlog, A ≤ B. Assume B | = ∃x · ϕ(x, a). Thus, B | = ϕ(b, a), for some b from B. ◮ By Lemma 4, this is equivalent to a1 ≤ · · · ≤ ai ≤ b ≤ ai+1 ≤ . . . an, with a renumbered suitably. ◮ If b is already in A there is nothing to do, so assume ai < b < ai+1 and b / ∈ A. By density

• f A, there is a c ∈ A with ai < c < ai+1. Therefore, a1 ≤ · · · ≤ ai ≤ c ≤ ai+1 ≤ . . . an

holds in A. ◮ By Lemma 4, we obtain A | = ϕ(c, a). Hence, A | = ∃x · ϕ(x, a) as required.

Corollary 6

All models of T are elementarily equivalent. Hence, T is complete.

Proof.

Let A and B be dense linear orders without endpoints. We have A ֒ → B iff card(A) ≤ card(B). By Theorem 5, A

e

֒ → B or B

e

֒ → A holds. In either case A ≡ B.

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Quantifier elimination in general

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Quantifier elimination

Definition 7

Let Σ be a signature an K a class of Σ-structures. A set Φ of formulas is called an elimination set for K, if for every formula ϕ(x) of Σ there is a formula ϕ∗(x) such that ◮ ϕ∗(x) is a Boolean combination of formulas in Φ, ◮ For every A ∈ K and every tuple a from A we have A | = ϕ(a) iff A | = ϕ∗(a). If there is an elimination set Φ for K, such that all formulas in Φ are quantifier free, then we say that K admits quantifier elimination. If K = Mod(T) for some theory T, we also say that T admits quantifier elimination.

Lemma 8

If every formula ∃x · α1 ∧ · · · ∧ αn, where each αi is atomic or negated atomic, is equivalent over T to a quantifier free formula, then T admits quantifier elimination.

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Quantifier elimination

Definition 7

Let Σ be a signature an K a class of Σ-structures. A set Φ of formulas is called an elimination set for K, if for every formula ϕ(x) of Σ there is a formula ϕ∗(x) such that ◮ ϕ∗(x) is a Boolean combination of formulas in Φ, ◮ For every A ∈ K and every tuple a from A we have A | = ϕ(a) iff A | = ϕ∗(a). If there is an elimination set Φ for K, such that all formulas in Φ are quantifier free, then we say that K admits quantifier elimination. If K = Mod(T) for some theory T, we also say that T admits quantifier elimination.

Lemma 8

If every formula ∃x · α1 ∧ · · · ∧ αn, where each αi is atomic or negated atomic, is equivalent over T to a quantifier free formula, then T admits quantifier elimination.

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Dense linear orders again

Exercise 1

Let K be the class of dense linear orders. Define Φ to be

◮ ∃x · ∀y · x = y ∨ x < y ◮ ∃x · ∀y · x = y ∨ y < x ◮ ∀y · x = y ∨ x < y ◮ ∀y · x = y ∨ y < x ◮ x < y

Prove that Φ is an elimination set for K.

Exercise 2

Compare K with dense linear orders without endpoints. Let A, B ∈ K. ◮ Does A ֒ → B imply A

e

֒ → B? ◮ Are A and B elementarily equivalent?

Exercise 3

Repeat Exercise 2 for K′ – the class of dense linear orders with endpoints.

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Dense linear orders again

Exercise 1

Let K be the class of dense linear orders. Define Φ to be

◮ ∃x · ∀y · x = y ∨ x < y ◮ ∃x · ∀y · x = y ∨ y < x ◮ ∀y · x = y ∨ x < y ◮ ∀y · x = y ∨ y < x ◮ x < y

Prove that Φ is an elimination set for K.

Exercise 2

Compare K with dense linear orders without endpoints. Let A, B ∈ K. ◮ Does A ֒ → B imply A

e

֒ → B? ◮ Are A and B elementarily equivalent?

Exercise 3

Repeat Exercise 2 for K′ – the class of dense linear orders with endpoints.

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Dense linear orders again

Exercise 1

Let K be the class of dense linear orders. Define Φ to be

◮ ∃x · ∀y · x = y ∨ x < y ◮ ∃x · ∀y · x = y ∨ y < x ◮ ∀y · x = y ∨ x < y ◮ ∀y · x = y ∨ y < x ◮ x < y

Prove that Φ is an elimination set for K.

Exercise 2

Compare K with dense linear orders without endpoints. Let A, B ∈ K. ◮ Does A ֒ → B imply A

e

֒ → B? ◮ Are A and B elementarily equivalent?

Exercise 3

Repeat Exercise 2 for K′ – the class of dense linear orders with endpoints.

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Model completeness

Definition 9

A theory T is model complete if every embedding between models of T is

• elementary. That is, for every A, B ∈ Mod(T) we have that A ֒

→ B implies A

e

֒ → B.

Theorem 10

If a theory T admits quantifier elimination, then T is model complete.

Proof.

◮ Let A, B be models of T such that (wlog) A ≤ B. Let ϕ(x) be a formula, and let a be a tuple from A. ◮ By quantifier elimination, we have (⋆) T | = ϕ(x) ↔ ϕ∗(x, y), for some quantifier-free ϕ∗. ◮ Then, ϕ∗ is both Π1 and Σ1, so it is preserved by substructures and extensions (by

• Lo´

s-Tarski), so we have A | = ϕ∗(a, y) if and only if B | = ϕ∗(a, y). ◮ Combining this with (⋆), we get A | = ϕ(a) if and only if B | = ϕ(a), as required.

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Model completeness

Definition 9

A theory T is model complete if every embedding between models of T is

• elementary. That is, for every A, B ∈ Mod(T) we have that A ֒

→ B implies A

e

֒ → B.

Theorem 10

If a theory T admits quantifier elimination, then T is model complete.

Proof.

◮ Let A, B be models of T such that (wlog) A ≤ B. Let ϕ(x) be a formula, and let a be a tuple from A. ◮ By quantifier elimination, we have (⋆) T | = ϕ(x) ↔ ϕ∗(x, y), for some quantifier-free ϕ∗. ◮ Then, ϕ∗ is both Π1 and Σ1, so it is preserved by substructures and extensions (by

• Lo´

s-Tarski), so we have A | = ϕ∗(a, y) if and only if B | = ϕ∗(a, y). ◮ Combining this with (⋆), we get A | = ϕ(a) if and only if B | = ϕ(a), as required.

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Model completeness

Definition 9

A theory T is model complete if every embedding between models of T is

• elementary. That is, for every A, B ∈ Mod(T) we have that A ֒

→ B implies A

e

֒ → B.

Theorem 10

If a theory T admits quantifier elimination, then T is model complete.

Proof.

◮ Let A, B be models of T such that (wlog) A ≤ B. Let ϕ(x) be a formula, and let a be a tuple from A. ◮ By quantifier elimination, we have (⋆) T | = ϕ(x) ↔ ϕ∗(x, y), for some quantifier-free ϕ∗. ◮ Then, ϕ∗ is both Π1 and Σ1, so it is preserved by substructures and extensions (by

• Lo´

s-Tarski), so we have A | = ϕ∗(a, y) if and only if B | = ϕ∗(a, y). ◮ Combining this with (⋆), we get A | = ϕ(a) if and only if B | = ϕ(a), as required.

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Some consequences of model completeness

Definition 11

A model A of a theory T is called algebraically prime if A embeds into every model of T.

Example 12

Let T be the theory of fields of characteristic p (for a prime p). Then GF(p) is a (unique) algebraically prime model of T.

Theorem 13

Let T be a model complete theory. If T has an algebraically prime model, then T is complete.

Proof.

◮ Let A be an algebraically prime model of T, and let B, C be arbitrary models of T. ◮ By model completeness of T we have A

e

֒ → B and A

e

֒ → C. ◮ It follows that B ≡ C (exercise). ◮ Thus, all models of T are elementarily equivalent, so T is complete.

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Some consequences of model completeness

Definition 11

A model A of a theory T is called algebraically prime if A embeds into every model of T.

Example 12

Let T be the theory of fields of characteristic p (for a prime p). Then GF(p) is a (unique) algebraically prime model of T.

Theorem 13

Let T be a model complete theory. If T has an algebraically prime model, then T is complete.

Proof.

◮ Let A be an algebraically prime model of T, and let B, C be arbitrary models of T. ◮ By model completeness of T we have A

e

֒ → B and A

e

֒ → C. ◮ It follows that B ≡ C (exercise). ◮ Thus, all models of T are elementarily equivalent, so T is complete.

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Some consequences of model completeness

Definition 11

A model A of a theory T is called algebraically prime if A embeds into every model of T.

Example 12

Let T be the theory of fields of characteristic p (for a prime p). Then GF(p) is a (unique) algebraically prime model of T.

Theorem 13

Let T be a model complete theory. If T has an algebraically prime model, then T is complete.

Proof.

◮ Let A be an algebraically prime model of T, and let B, C be arbitrary models of T. ◮ By model completeness of T we have A

e

֒ → B and A

e

֒ → C. ◮ It follows that B ≡ C (exercise). ◮ Thus, all models of T are elementarily equivalent, so T is complete.

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Some consequences of model completeness

Definition 11

A model A of a theory T is called algebraically prime if A embeds into every model of T.

Example 12

Let T be the theory of fields of characteristic p (for a prime p). Then GF(p) is a (unique) algebraically prime model of T.

Theorem 13

Let T be a model complete theory. If T has an algebraically prime model, then T is complete.

Proof.

◮ Let A be an algebraically prime model of T, and let B, C be arbitrary models of T. ◮ By model completeness of T we have A

e

֒ → B and A

e

֒ → C. ◮ It follows that B ≡ C (exercise). ◮ Thus, all models of T are elementarily equivalent, so T is complete.

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Some consequences of model completeness

Theorem 14

Let T be a model complete theory. Then T is a Π2 theory.

Proof.

◮ We will show that T is closed under unions of chains, and use Theorem 10 of Lecture 6 (Chang- Lo´ s-Suszko). ◮ Let (Ai : i < κ) be a chain of models of T, and let A =

i<κ Ai.

◮ By model completeneness, it is an elementary chain, so by Lemma 7 of Lecture 6, A is an elementary extension of each Ai. ◮ So, A | = T and hence T is closed under unions of chains.

Example 15

The converse of Theorem 13 does not hold. Consider the theory of dense linear orders and Exercise 2.

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Some consequences of model completeness

Theorem 14

Let T be a model complete theory. Then T is a Π2 theory.

Proof.

◮ We will show that T is closed under unions of chains, and use Theorem 10 of Lecture 6 (Chang- Lo´ s-Suszko). ◮ Let (Ai : i < κ) be a chain of models of T, and let A =

i<κ Ai.

◮ By model completeneness, it is an elementary chain, so by Lemma 7 of Lecture 6, A is an elementary extension of each Ai. ◮ So, A | = T and hence T is closed under unions of chains.

Example 15

The converse of Theorem 13 does not hold. Consider the theory of dense linear orders and Exercise 2.

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Some consequences of model completeness

Theorem 14

Let T be a model complete theory. Then T is a Π2 theory.

Proof.

◮ We will show that T is closed under unions of chains, and use Theorem 10 of Lecture 6 (Chang- Lo´ s-Suszko). ◮ Let (Ai : i < κ) be a chain of models of T, and let A =

i<κ Ai.

◮ By model completeneness, it is an elementary chain, so by Lemma 7 of Lecture 6, A is an elementary extension of each Ai. ◮ So, A | = T and hence T is closed under unions of chains.

Example 15

The converse of Theorem 13 does not hold. Consider the theory of dense linear orders and Exercise 2.

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Connections to decidability

Theorem 16

Let T be a complete, recursively axiomatisable theory. Then T is decidable.

Proof.

◮ Since T is recursively axiomatisable, we can recursively enumerate the consequences of T. ◮ Since T is complete, for any formula ϕ, either ϕ or ¬ϕ will appear in the enumeration.

Theorem 17

Let T be a theory with a reduction set Φ. Assume that the following hold: ◮ For every formula ϕ, there is an algorithm for constructing a Boolean combination of formulas from Φ, such that the resulting formula ϕ∗ is equivalent to ϕ over T. ◮ For every formula ϕ ∈ Φ it is decidable whether T | = ϕ. Then, T is decidable.

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Connections to decidability

Theorem 16

Let T be a complete, recursively axiomatisable theory. Then T is decidable.

Proof.

◮ Since T is recursively axiomatisable, we can recursively enumerate the consequences of T. ◮ Since T is complete, for any formula ϕ, either ϕ or ¬ϕ will appear in the enumeration.

Theorem 17

Let T be a theory with a reduction set Φ. Assume that the following hold: ◮ For every formula ϕ, there is an algorithm for constructing a Boolean combination of formulas from Φ, such that the resulting formula ϕ∗ is equivalent to ϕ over T. ◮ For every formula ϕ ∈ Φ it is decidable whether T | = ϕ. Then, T is decidable.

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Connections to decidability

Theorem 16

Let T be a complete, recursively axiomatisable theory. Then T is decidable.

Proof.

◮ Since T is recursively axiomatisable, we can recursively enumerate the consequences of T. ◮ Since T is complete, for any formula ϕ, either ϕ or ¬ϕ will appear in the enumeration.

Theorem 17

Let T be a theory with a reduction set Φ. Assume that the following hold: ◮ For every formula ϕ, there is an algorithm for constructing a Boolean combination of formulas from Φ, such that the resulting formula ϕ∗ is equivalent to ϕ over T. ◮ For every formula ϕ ∈ Φ it is decidable whether T | = ϕ. Then, T is decidable.

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Theories admitting quantifier elimination

The following classes/theories admit quantifier elimination: ◮ Dense linear orders without endpoints (Tarski). ◮ Presburger arithmetic (Presburger). Equivalent to Peano arithmetic without multiplication. It is complete and decidable. ◮ Algebraically closed fields (Tarski). Fields satisfying

◮ ∀x1 . . . xn · ∃y · y n + x1y n−1 + · · · + xn−1y + xn = 0, for every n ∈ N

◮ Ordered real closed fields (Tarski). Fields with ordering relation compatible with the operations, satisfying

◮ ∀x1 . . . xn : x2

1 + · · · + x2 n = −1, for every n ∈ N

◮ ∀x · ∃y · x = y 2 ∨ −x = y 2 ◮ ∀x1 . . . xn∃y : y n + x1y n−1 + · · · + xn−1y + xn = 0, for every odd n.

Quantifier elimination for industry

The theory of ordered real closed fields has a lot of industrial applications. The main focus (as far as I know) is on efficient algorithms for quantifier

• elimination. Ask Prof. Hirokazu Anai.

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Theories admitting quantifier elimination

The following classes/theories admit quantifier elimination: ◮ Dense linear orders without endpoints (Tarski). ◮ Presburger arithmetic (Presburger). Equivalent to Peano arithmetic without multiplication. It is complete and decidable. ◮ Algebraically closed fields (Tarski). Fields satisfying

◮ ∀x1 . . . xn · ∃y · y n + x1y n−1 + · · · + xn−1y + xn = 0, for every n ∈ N

◮ Ordered real closed fields (Tarski). Fields with ordering relation compatible with the operations, satisfying

◮ ∀x1 . . . xn : x2

1 + · · · + x2 n = −1, for every n ∈ N

◮ ∀x · ∃y · x = y 2 ∨ −x = y 2 ◮ ∀x1 . . . xn∃y : y n + x1y n−1 + · · · + xn−1y + xn = 0, for every odd n.

Quantifier elimination for industry

The theory of ordered real closed fields has a lot of industrial applications. The main focus (as far as I know) is on efficient algorithms for quantifier

• elimination. Ask Prof. Hirokazu Anai.

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A weak version of Hilbert’s Nullstellensatz

Theorem 18 (Hilbert)

Let F be an algebraically closed field, and E a finite system of equations and inequations over F in variables x. Suppose E has a solution in some field G extending F. Then, E has a solution in F.

Proof.

◮ The assumeption that G extends F, means (wlog) F ≤ G. ◮ Let G be the algebraic closure of G. Then, we have F ≤ G ≤ G. ◮ Now, E = {p1(x) = 0, . . . , pn(x) = 0, q1(x) = 0, . . . , qm(x) = 0} for some polynomials p1, . . . , pn, q1, . . . , qm with coefficients from F. ◮ Let ψ(x, c) be the conjunction of all formulas in E, with c being the coefficients. ◮ Then, E has a solution in G means G | = ∃x · ψ(x, c). This is an existential formula, so by

• Lo´

s-Tarski G | = ∃x · ψ(x, c). ◮ The theory of algebraically closed fields is model complete (see previous slide). ◮ Thus, the embedding F ≤ G is elementary, that is, F G. ◮ So, F | = ∃x · ψ(x, c) as required.

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A weak version of Hilbert’s Nullstellensatz

Theorem 18 (Hilbert)

Let F be an algebraically closed field, and E a finite system of equations and inequations over F in variables x. Suppose E has a solution in some field G extending F. Then, E has a solution in F.

Proof.

◮ The assumeption that G extends F, means (wlog) F ≤ G. ◮ Let G be the algebraic closure of G. Then, we have F ≤ G ≤ G. ◮ Now, E = {p1(x) = 0, . . . , pn(x) = 0, q1(x) = 0, . . . , qm(x) = 0} for some polynomials p1, . . . , pn, q1, . . . , qm with coefficients from F. ◮ Let ψ(x, c) be the conjunction of all formulas in E, with c being the coefficients. ◮ Then, E has a solution in G means G | = ∃x · ψ(x, c). This is an existential formula, so by

• Lo´

s-Tarski G | = ∃x · ψ(x, c). ◮ The theory of algebraically closed fields is model complete (see previous slide). ◮ Thus, the embedding F ≤ G is elementary, that is, F G. ◮ So, F | = ∃x · ψ(x, c) as required.

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