Many-Sorted First-Order Model Theory Lecture 9 2 nd July, 2020 1 / 37
Quantifier elimination example Dense linear orders w/o endpoints 2 / 37
Eliminating one ∃ Lemma 1 Let T be the theory of dense linear orders without endpoints. Every formula ∃ x · ϕ 1 ∧ · · · ∧ ϕ n , where each ϕ i is atomic or negated atomic, is equivalent over T to a (positive) quantifier free formula. Proof sketch. ◮ Wlog, assume that each ϕ i 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 ϕ i is atomic. ◮ If some ϕ i is x = y , then we can replace all occurrences of x in ϕ 1 ∧ · · · ∧ ϕ n 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 ϕ i are of the form < x or x < . ◮ Group them into � j ( v j < x ) and � k ( x < v j ), to get ∃ x · � j ( v j < x ) ∧ � k ( x < v k ). ◮ We have T | = ∃ x · � j ( v j < x ) ∧ � k ( x < v k ) ↔ � j , k ( v j < v k ). 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 · ϕ 1 ∧ · · · ∧ ϕ n 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 .) 3 / 37
Eliminating one ∃ Lemma 1 Let T be the theory of dense linear orders without endpoints. Every formula ∃ x · ϕ 1 ∧ · · · ∧ ϕ n , where each ϕ i is atomic or negated atomic, is equivalent over T to a (positive) quantifier free formula. Proof sketch. ◮ Wlog, assume that each ϕ i 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 ϕ i is atomic. ◮ If some ϕ i is x = y , then we can replace all occurrences of x in ϕ 1 ∧ · · · ∧ ϕ n 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 ϕ i are of the form < x or x < . ◮ Group them into � j ( v j < x ) and � k ( x < v j ), to get ∃ x · � j ( v j < x ) ∧ � k ( x < v k ). ◮ We have T | = ∃ x · � j ( v j < x ) ∧ � k ( x < v k ) ↔ � j , k ( v j < v k ). 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 · ϕ 1 ∧ · · · ∧ ϕ n 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 .) 4 / 37
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 = ϕ ↔ ϕ ∗ , ◮ T | ◮ ϕ ∗ is effectively obtainable from ϕ . Proof. ◮ We can assume ϕ is in prenex form, so ϕ = Q 1 . . . Q n ψ where ψ is quantifier free. ◮ Consider the innermost quantifier Q n ; if Q n is ∀ , replace it by ¬∃¬ . Bring ¬ ψ into disjunctive form, and distribute ∃ over disjunctions. ◮ Thus, we can assume Q n ψ is ( ¬ )( ∃ x · ρ 1 ∨ · · · ∨ ∃ x · ρ m ), where each ρ i is a conjunction of atomic and/or negated atomic formulas. ◮ By Lemma 1 each ∃ x · ρ i is equivalent over T to a quantifier free σ i . Replace them. ◮ Repeat the procedure n times, to remove all quantifiers and obtain ϕ ∗ . The procedure is clearly effective. 5 / 37
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 = ϕ ↔ ϕ ∗ , ◮ T | ◮ ϕ ∗ is effectively obtainable from ϕ . Proof. ◮ We can assume ϕ is in prenex form, so ϕ = Q 1 . . . Q n ψ where ψ is quantifier free. ◮ Consider the innermost quantifier Q n ; if Q n is ∀ , replace it by ¬∃¬ . Bring ¬ ψ into disjunctive form, and distribute ∃ over disjunctions. ◮ Thus, we can assume Q n ψ is ( ¬ )( ∃ x · ρ 1 ∨ · · · ∨ ∃ x · ρ m ), where each ρ i is a conjunction of atomic and/or negated atomic formulas. ◮ By Lemma 1 each ∃ x · ρ i is equivalent over T to a quantifier free σ i . Replace them. ◮ Repeat the procedure n times, to remove all quantifiers and obtain ϕ ∗ . The procedure is clearly effective. 6 / 37
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 over T to a Boolean combination of formulas ψ j (1 ≤ j ≤ n ) such that for each j we have T | = ψ j or T | = ¬ ψ j 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 x 1 ≤ x 2 ≤ · · · ≤ x n , for some suitable renumbering of x, where each ≤ is either < or = . Proof. By analysing the proofs of Lemma 1 and Theorem 2. 7 / 37
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 over T to a Boolean combination of formulas ψ j (1 ≤ j ≤ n ) such that for each j we have T | = ψ j or T | = ¬ ψ j 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 x 1 ≤ x 2 ≤ · · · ≤ x n , for some suitable renumbering of x, where each ≤ is either < or = . Proof. By analysing the proofs of Lemma 1 and Theorem 2. 8 / 37
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 over T to a Boolean combination of formulas ψ j (1 ≤ j ≤ n ) such that for each j we have T | = ψ j or T | = ¬ ψ j 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 x 1 ≤ x 2 ≤ · · · ≤ x n , for some suitable renumbering of x, where each ≤ is either < or = . Proof. By analysing the proofs of Lemma 1 and Theorem 2. 9 / 37
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 over T to a Boolean combination of formulas ψ j (1 ≤ j ≤ n ) such that for each j we have T | = ψ j or T | = ¬ ψ j 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 x 1 ≤ x 2 ≤ · · · ≤ x n , for some suitable renumbering of x, where each ≤ is either < or = . Proof. By analysing the proofs of Lemma 1 and Theorem 2. 10 / 37
Completeness Theorem 5 e Let A and B be models of T. If A ֒ → B , then A → 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 a 1 ≤ · · · ≤ a i ≤ b ≤ a i +1 ≤ . . . a n , with a renumbered suitably. ◮ If b is already in A there is nothing to do, so assume a i < b < a i +1 and b / ∈ A . By density of A , there is a c ∈ A with a i < c < a i +1 . Therefore, a 1 ≤ · · · ≤ a i ≤ c ≤ a i +1 ≤ . . . a n 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 e e card ( A ) ≤ card ( B ). By Theorem 5, A ֒ → B or B ֒ → A holds. In either case A ≡ B . 11 / 37
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