Many-Sorted First-Order Model Theory Lecture 10 9 th July, 2020 1 / - - PowerPoint PPT Presentation

Many-Sorted First-Order Model Theory

Lecture 10 9th July, 2020

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Ehrenfeucht-Fra¨ ıss´ e games: back-and-forth equivalence

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A theorem of Cantor’s

Theorem 1

Let A and B be dense linear orders without endpoints. If A and B are both countable, then A ∼ = B.

Proof.

◮ We construct an isomorphism f : A → B step by step. Fix some enumerations (ai : i < ω) and (bi : i < ω) of elements of A and B. ◮ Step 0. Put f (a0) = b0. Then f preserves and reflects <. ◮ Step 1. Consider b1. If b0 < b1, pick any aj with aj < a0; if b1 < b0, pick any ak with a0 < aj. Extend f putting f (aj) = (b1). ◮ Now, assume inductively that f constructed so far preserves and reflects <. ◮ Step i + 1, for even i. Consider bi. If bi = f (aj) for some j, we are done. Otherwise, let the elements we have chosen so far from B be bi0, . . . , bik , and wlog we have bi0 < · · · < bir < bi < bir+1 < · · · < bik . ◮ Then we have f −1(bir ) < f −1(bir+1). Pick aℓ with f −1(bir ) < aℓ < f −1(bir+1), and extend f putting f (aℓ) = bi. Still f preserves and reflects <. ◮ Step i + 1, for odd i. Proceed analogously with A in place of B. ◮ The map f constructed this way (formally f =

i<ω fi where fi is the partial map

constructed at stage i) is an isomorphism.

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A theorem of Cantor’s

Theorem 1

Let A and B be dense linear orders without endpoints. If A and B are both countable, then A ∼ = B.

Proof.

◮ We construct an isomorphism f : A → B step by step. Fix some enumerations (ai : i < ω) and (bi : i < ω) of elements of A and B. ◮ Step 0. Put f (a0) = b0. Then f preserves and reflects <. ◮ Step 1. Consider b1. If b0 < b1, pick any aj with aj < a0; if b1 < b0, pick any ak with a0 < aj. Extend f putting f (aj) = (b1). ◮ Now, assume inductively that f constructed so far preserves and reflects <. ◮ Step i + 1, for even i. Consider bi. If bi = f (aj) for some j, we are done. Otherwise, let the elements we have chosen so far from B be bi0, . . . , bik , and wlog we have bi0 < · · · < bir < bi < bir+1 < · · · < bik . ◮ Then we have f −1(bir ) < f −1(bir+1). Pick aℓ with f −1(bir ) < aℓ < f −1(bir+1), and extend f putting f (aℓ) = bi. Still f preserves and reflects <. ◮ Step i + 1, for odd i. Proceed analogously with A in place of B. ◮ The map f constructed this way (formally f =

i<ω fi where fi is the partial map

constructed at stage i) is an isomorphism.

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Intuitions from the proof

◮ The construction in the proof can be viewed as a game between two players, say, Abelard and Helo¨ ıse. ◮ Abelard chooses an element from A or B, and Helo¨ ıse must respond by choosing an element from the other structure while maintaining a partial isomorphism. ◮ Abelard can choose any element whatever, so he resembles ∀ (some writers call him ∀belard). ◮ Helo¨ ıse must choose carefully, to preserve isomorphism, so she resembles ∃ (some writers call her ∃lo¨ ıse; initial H is silent anyway). ◮ Helo¨ ıse wins if she can maintain isomorphism throughout the game (in this case forever as the game is infinite). ◮ “Maintaining partial isomorphism” can be cashed out as: sequences chosen so far satisfy precisely the same atomic sentences – in the signature expanded by (names of) the elements of these sequences.

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Ehrenfeucht-Fra¨ ıss´ e games

Comparing models by games

A method conceptually due to Fra¨ ıss´ e (1950), and formulated in game theoretic terms by Ehrenfeucht (1961). Two player game of perfect information ∀ ∃ Spoiler Duplicator Abelard Helo¨ ıse ∀belard ∃loise

Abelard and Helo¨ ıse as depicted in the 14th centure manuscript Roman de la Rose.

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Game EFγ(A, B)

Definition 2 (Game EFγ(A, B) for an ordinal γ)

◮ Fix similar structures A and B. The game EFγ(A, B) is played as follows. ◮ At each round ∀belard selects some a from A or some b from B. ◮ ∃loise responds by selecting an element from the other structure. ◮ These choices constitute a play. At the end of the play, sequences a = (ai : i < γ) and b = (bi : i < γ) have been chosen. ◮ ∃loise wins the play if aA ∼ = bB under the map ai → bi. ◮ ∃loise wins the game, if she can win every play, regardless of the moves ∀belard makes. If this is the case, we say that ∃loise has a winning strategy.

Definition 3

If ∃ has a winning strategy for EFω(A, B), we say that A and B are back-and-forth equivalent, and write A ∼ω B.

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Game EFγ(A, B)

Definition 2 (Game EFγ(A, B) for an ordinal γ)

◮ Fix similar structures A and B. The game EFγ(A, B) is played as follows. ◮ At each round ∀belard selects some a from A or some b from B. ◮ ∃loise responds by selecting an element from the other structure. ◮ These choices constitute a play. At the end of the play, sequences a = (ai : i < γ) and b = (bi : i < γ) have been chosen. ◮ ∃loise wins the play if aA ∼ = bB under the map ai → bi. ◮ ∃loise wins the game, if she can win every play, regardless of the moves ∀belard makes. If this is the case, we say that ∃loise has a winning strategy.

Definition 3

If ∃ has a winning strategy for EFω(A, B), we say that A and B are back-and-forth equivalent, and write A ∼ω B.

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Back-and-forth equivalence

Lemma 4

Let (a, b) be a play of EFγ(A, B). The following are equivalent:

• 1. ∃ wins the play (a, b).
• 2. A |

= ϕ(a) ⇔ B | = ϕ(b), for every atomic formula ϕ(x).

Proof.

By Diagram Lemma (Lecture 5, Lemma 12).

Theorem 5 (Isomorphism by back-and-forth game)

For countable structures A and B, we have A ∼ = B if and only if A ∼ω B.

Proof.

◮ (⇒) An isomorphism ι : A → B gives ∃ an obvious winning strategy. ◮ (⇐) If A ∼ω B, list the elements of A and B, and let ∀ choose alternately the first “fresh” element of A and B. By assumption ∃ has a winning strategy, so she uses it to pick her responses to win the resulting play (a, b). Thus, aA ∼ = bB. ◮ Since A and B are countable, all elements are listed in (a, b). So, A ∼ = B.

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Back-and-forth equivalence

Lemma 4

Let (a, b) be a play of EFγ(A, B). The following are equivalent:

• 1. ∃ wins the play (a, b).
• 2. A |

= ϕ(a) ⇔ B | = ϕ(b), for every atomic formula ϕ(x).

Proof.

By Diagram Lemma (Lecture 5, Lemma 12).

Theorem 5 (Isomorphism by back-and-forth game)

For countable structures A and B, we have A ∼ = B if and only if A ∼ω B.

Proof.

◮ (⇒) An isomorphism ι : A → B gives ∃ an obvious winning strategy. ◮ (⇐) If A ∼ω B, list the elements of A and B, and let ∀ choose alternately the first “fresh” element of A and B. By assumption ∃ has a winning strategy, so she uses it to pick her responses to win the resulting play (a, b). Thus, aA ∼ = bB. ◮ Since A and B are countable, all elements are listed in (a, b). So, A ∼ = B.

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Back-and-forth equivalence

Lemma 4

Let (a, b) be a play of EFγ(A, B). The following are equivalent:

• 1. ∃ wins the play (a, b).
• 2. A |

= ϕ(a) ⇔ B | = ϕ(b), for every atomic formula ϕ(x).

Proof.

By Diagram Lemma (Lecture 5, Lemma 12).

Theorem 5 (Isomorphism by back-and-forth game)

For countable structures A and B, we have A ∼ = B if and only if A ∼ω B.

Proof.

◮ (⇒) An isomorphism ι : A → B gives ∃ an obvious winning strategy. ◮ (⇐) If A ∼ω B, list the elements of A and B, and let ∀ choose alternately the first “fresh” element of A and B. By assumption ∃ has a winning strategy, so she uses it to pick her responses to win the resulting play (a, b). Thus, aA ∼ = bB. ◮ Since A and B are countable, all elements are listed in (a, b). So, A ∼ = B.

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Back-and-forth equivalence

Lemma 4

Let (a, b) be a play of EFγ(A, B). The following are equivalent:

• 1. ∃ wins the play (a, b).
• 2. A |

= ϕ(a) ⇔ B | = ϕ(b), for every atomic formula ϕ(x).

Proof.

By Diagram Lemma (Lecture 5, Lemma 12).

Theorem 5 (Isomorphism by back-and-forth game)

For countable structures A and B, we have A ∼ = B if and only if A ∼ω B.

Proof.

◮ (⇒) An isomorphism ι : A → B gives ∃ an obvious winning strategy. ◮ (⇐) If A ∼ω B, list the elements of A and B, and let ∀ choose alternately the first “fresh” element of A and B. By assumption ∃ has a winning strategy, so she uses it to pick her responses to win the resulting play (a, b). Thus, aA ∼ = bB. ◮ Since A and B are countable, all elements are listed in (a, b). So, A ∼ = B.

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Properties of back-and-forth equivalence

Lemma 6

Back-and-forth equivalence is an equivalence relation on similar structures.

Proof.

◮ Reflexivity and symmetry are obvious. ◮ Assume A ∼ω B and B ∼ω C. We want a winning strategy for ∃ in EFω(A, C). ◮ Assume ∀ chooses a ∈ A. Then ∃ uses her strategy for EFω(A, B) to pick (in private) a response b ∈ B. ◮ Then she treats this response as ∀s move in EFω(B, C), and uses her strategy to pick (in public) c ∈ C. ◮ If ∀ chooses c ∈ C instead, she performs the same trick in reverse. ◮ Since ∃ uses her winning strategies, she cannot lose.

Exercise 1

Prove that (R, <) ∼ω (Q, <), as dense linear orders without endpoints. This shows that countability assumption in Theorem 5 is necessary.

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Properties of back-and-forth equivalence

Lemma 6

Back-and-forth equivalence is an equivalence relation on similar structures.

Proof.

◮ Reflexivity and symmetry are obvious. ◮ Assume A ∼ω B and B ∼ω C. We want a winning strategy for ∃ in EFω(A, C). ◮ Assume ∀ chooses a ∈ A. Then ∃ uses her strategy for EFω(A, B) to pick (in private) a response b ∈ B. ◮ Then she treats this response as ∀s move in EFω(B, C), and uses her strategy to pick (in public) c ∈ C. ◮ If ∀ chooses c ∈ C instead, she performs the same trick in reverse. ◮ Since ∃ uses her winning strategies, she cannot lose.

Exercise 1

Prove that (R, <) ∼ω (Q, <), as dense linear orders without endpoints. This shows that countability assumption in Theorem 5 is necessary.

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Properties of back-and-forth equivalence

Lemma 6

Back-and-forth equivalence is an equivalence relation on similar structures.

Proof.

◮ Reflexivity and symmetry are obvious. ◮ Assume A ∼ω B and B ∼ω C. We want a winning strategy for ∃ in EFω(A, C). ◮ Assume ∀ chooses a ∈ A. Then ∃ uses her strategy for EFω(A, B) to pick (in private) a response b ∈ B. ◮ Then she treats this response as ∀s move in EFω(B, C), and uses her strategy to pick (in public) c ∈ C. ◮ If ∀ chooses c ∈ C instead, she performs the same trick in reverse. ◮ Since ∃ uses her winning strategies, she cannot lose.

Exercise 1

Prove that (R, <) ∼ω (Q, <), as dense linear orders without endpoints. This shows that countability assumption in Theorem 5 is necessary.

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Properties of back-and-forth equivalence

Lemma 7

Let I be an infinite index set and U a nonprincipal ultrafilter on I. Then (Z, <) ∼ω (Z, <)I/U.

Proof.

◮ As we saw in Lecture 8, (Z, <)I /U consists of uncountably many copies of (Z, <). ◮ Let (Z, <)u and (Z, <)ℓ be two such copies, with (Z, <)ℓ < (Z, <)u. ◮ Now ∀ can perform a squeeze, picking alternately elements ever higher in (Z, <)ℓ and ever lower in (Z, <)u. ◮ Then ∃ has to respond by picking elements from (Z, <), and however far apart she starts, after a finite number of steps she will have no good choice to make.

Plan: show that ∼ω is strictly between ≡ and ∼ =

◮ Exercise 1 shows that ∼ω is strictly weaker than ∼ =. ◮ Lemma 7 will show that ∼ω is strictly stronger than ≡, once we prove (soon) that ∼ω implies ≡.

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Properties of back-and-forth equivalence

Lemma 7

Let I be an infinite index set and U a nonprincipal ultrafilter on I. Then (Z, <) ∼ω (Z, <)I/U.

Proof.

◮ As we saw in Lecture 8, (Z, <)I /U consists of uncountably many copies of (Z, <). ◮ Let (Z, <)u and (Z, <)ℓ be two such copies, with (Z, <)ℓ < (Z, <)u. ◮ Now ∀ can perform a squeeze, picking alternately elements ever higher in (Z, <)ℓ and ever lower in (Z, <)u. ◮ Then ∃ has to respond by picking elements from (Z, <), and however far apart she starts, after a finite number of steps she will have no good choice to make.

Plan: show that ∼ω is strictly between ≡ and ∼ =

◮ Exercise 1 shows that ∼ω is strictly weaker than ∼ =. ◮ Lemma 7 will show that ∼ω is strictly stronger than ≡, once we prove (soon) that ∼ω implies ≡.

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Properties of back-and-forth equivalence

Lemma 7

Let I be an infinite index set and U a nonprincipal ultrafilter on I. Then (Z, <) ∼ω (Z, <)I/U.

Proof.

◮ As we saw in Lecture 8, (Z, <)I /U consists of uncountably many copies of (Z, <). ◮ Let (Z, <)u and (Z, <)ℓ be two such copies, with (Z, <)ℓ < (Z, <)u. ◮ Now ∀ can perform a squeeze, picking alternately elements ever higher in (Z, <)ℓ and ever lower in (Z, <)u. ◮ Then ∃ has to respond by picking elements from (Z, <), and however far apart she starts, after a finite number of steps she will have no good choice to make.

Plan: show that ∼ω is strictly between ≡ and ∼ =

◮ Exercise 1 shows that ∼ω is strictly weaker than ∼ =. ◮ Lemma 7 will show that ∼ω is strictly stronger than ≡, once we prove (soon) that ∼ω implies ≡.

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Ehrenfeucht-Fra¨ ıss´ e games: elementary equivalence

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Unnested formulas

Definition 8

Let Σ be a signature, and x, x, y be variables. An unnested atomic formula is of one of the following forms: ◮ x = y ◮ c = y for some constant c of Σ ◮ f (x) = y for some function symbol f of Σ ◮ R(x) for some relation symbol R of Σ A formula ϕ is unnested if all atomic subformulas of ϕ are unnested.

Theorem 9

Every formula ϕ is logically equivalent to an unnested formula.

Proof by example.

◮ Let ϕ be R(f (c, x), g(y)). ◮ The formula c = v0 ∧ f (v0, x) = v1 ∧ g(y) = v3 ∧ R(v1, v3) is clearly equivalent to ϕ. ◮ A formal proof proceeds by induction on complexity of ϕ.

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Unnested formulas

Definition 8

Let Σ be a signature, and x, x, y be variables. An unnested atomic formula is of one of the following forms: ◮ x = y ◮ c = y for some constant c of Σ ◮ f (x) = y for some function symbol f of Σ ◮ R(x) for some relation symbol R of Σ A formula ϕ is unnested if all atomic subformulas of ϕ are unnested.

Theorem 9

Every formula ϕ is logically equivalent to an unnested formula.

Proof by example.

◮ Let ϕ be R(f (c, x), g(y)). ◮ The formula c = v0 ∧ f (v0, x) = v1 ∧ g(y) = v3 ∧ R(v1, v3) is clearly equivalent to ϕ. ◮ A formal proof proceeds by induction on complexity of ϕ.

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Unnested formulas

Definition 8

Let Σ be a signature, and x, x, y be variables. An unnested atomic formula is of one of the following forms: ◮ x = y ◮ c = y for some constant c of Σ ◮ f (x) = y for some function symbol f of Σ ◮ R(x) for some relation symbol R of Σ A formula ϕ is unnested if all atomic subformulas of ϕ are unnested.

Theorem 9

Every formula ϕ is logically equivalent to an unnested formula.

Proof by example.

◮ Let ϕ be R(f (c, x), g(y)). ◮ The formula c = v0 ∧ f (v0, x) = v1 ∧ g(y) = v3 ∧ R(v1, v3) is clearly equivalent to ϕ. ◮ A formal proof proceeds by induction on complexity of ϕ.

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Game EFn[A, B]

Definition 10 (Game EFn[A, B] for a finite n)

◮ Fix similar structures A and B. The game EFn[A, B] is played exactly like the game EFn(A, B). ◮ ∃ wins the (a1, . . . , an, b1, . . . , bn) play if A | = ϕ(a) iff B | = ϕ(b) for every unnested atomic formula ϕ. ◮ ∃ wins the game, if she can win every play.

EFn(A, B) vs EFn[A, B]

The corresponding winning condition for EFn(A, B) is: ◮ ∃ wins the (a1, . . . , an, b1, . . . , bn) play if A | = ϕ(a) iff B | = ϕ(b) for every atomic formula ϕ.

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Game EFn[A, B]

Definition 10 (Game EFn[A, B] for a finite n)

◮ Fix similar structures A and B. The game EFn[A, B] is played exactly like the game EFn(A, B). ◮ ∃ wins the (a1, . . . , an, b1, . . . , bn) play if A | = ϕ(a) iff B | = ϕ(b) for every unnested atomic formula ϕ. ◮ ∃ wins the game, if she can win every play.

EFn(A, B) vs EFn[A, B]

The corresponding winning condition for EFn(A, B) is: ◮ ∃ wins the (a1, . . . , an, b1, . . . , bn) play if A | = ϕ(a) iff B | = ϕ(b) for every atomic formula ϕ.

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Comparing EFn(A, B) and EFn[A, B]

Definition 11

For a structure A and a set U ⊆ |A|, we define the restriction A|U of A to U, as a partial substructure of A with universe U and such that: ◮ all constants belong to U; ◮ relations and functions are restricted to U as relations. That is, if we have f (a) / ∈ U for some function f and a tuple a from U, then f (a) is undefined in A|U. ◮ Restrictions to finite sets are particularly useful. In universal algebra they are known as finite partial subalgebras. ◮ If there are no function symbols, restrictions are the same as substructures.

What EFn(A, B) and EFn[A, B] speak about

◮ The winning condition in EFn(A, B) is equivalent to aA ∼ = bB ◮ The winning condition in EFn[A, B] is equivalent to A|a ∼ = B|b.

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Comparing EFn(A, B) and EFn[A, B]

Definition 11

For a structure A and a set U ⊆ |A|, we define the restriction A|U of A to U, as a partial substructure of A with universe U and such that: ◮ all constants belong to U; ◮ relations and functions are restricted to U as relations. That is, if we have f (a) / ∈ U for some function f and a tuple a from U, then f (a) is undefined in A|U. ◮ Restrictions to finite sets are particularly useful. In universal algebra they are known as finite partial subalgebras. ◮ If there are no function symbols, restrictions are the same as substructures.

What EFn(A, B) and EFn[A, B] speak about

◮ The winning condition in EFn(A, B) is equivalent to aA ∼ = bB ◮ The winning condition in EFn[A, B] is equivalent to A|a ∼ = B|b.

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Comparing EFn(A, B) and EFn[A, B]

Definition 11

For a structure A and a set U ⊆ |A|, we define the restriction A|U of A to U, as a partial substructure of A with universe U and such that: ◮ all constants belong to U; ◮ relations and functions are restricted to U as relations. That is, if we have f (a) / ∈ U for some function f and a tuple a from U, then f (a) is undefined in A|U. ◮ Restrictions to finite sets are particularly useful. In universal algebra they are known as finite partial subalgebras. ◮ If there are no function symbols, restrictions are the same as substructures.

What EFn(A, B) and EFn[A, B] speak about

◮ The winning condition in EFn(A, B) is equivalent to aA ∼ = bB ◮ The winning condition in EFn[A, B] is equivalent to A|a ∼ = B|b.

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Equivalence relations from EFn[A, B] games

Definition 12

Let A, B be similar structures. We write A ≈n B if ∃ has a winning strategy for the game EFn[A, B].

Lemma 13

Let A, B be similar structures. Then

• 1. A ≈n B implies A ≈m B for n ≥ m,
• 2. A ∼ω B implies A ≈n B for every n < ω.

Proof.

We will prove (2). The proof of (1) is similar. ◮ Every play of EFn[A, B] can be regarded as an initial segment of a play of EFω(A, B). So, let (a1, . . . , an, b1, . . . , bn) be an initial segment of a play (a, b). ◮ Since ∃ has a winning strategy for EFω(A, B), we have that A | = ϕ(a) iff B | = ϕ(b), for every atomic ϕ. ◮ In particular, A | = ψ(a1, . . . , an) iff B | = ψ(b1, . . . , bn), for every unnested atomic ψ.

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Equivalence relations from EFn[A, B] games

Definition 12

Let A, B be similar structures. We write A ≈n B if ∃ has a winning strategy for the game EFn[A, B].

Lemma 13

Let A, B be similar structures. Then

• 1. A ≈n B implies A ≈m B for n ≥ m,
• 2. A ∼ω B implies A ≈n B for every n < ω.

Proof.

We will prove (2). The proof of (1) is similar. ◮ Every play of EFn[A, B] can be regarded as an initial segment of a play of EFω(A, B). So, let (a1, . . . , an, b1, . . . , bn) be an initial segment of a play (a, b). ◮ Since ∃ has a winning strategy for EFω(A, B), we have that A | = ϕ(a) iff B | = ϕ(b), for every atomic ϕ. ◮ In particular, A | = ψ(a1, . . . , an) iff B | = ψ(b1, . . . , bn), for every unnested atomic ψ.

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Equivalence relations from EFn[A, B] games

Definition 12

Let A, B be similar structures. We write A ≈n B if ∃ has a winning strategy for the game EFn[A, B].

Lemma 13

Let A, B be similar structures. Then

• 1. A ≈n B implies A ≈m B for n ≥ m,
• 2. A ∼ω B implies A ≈n B for every n < ω.

Proof.

We will prove (2). The proof of (1) is similar. ◮ Every play of EFn[A, B] can be regarded as an initial segment of a play of EFω(A, B). So, let (a1, . . . , an, b1, . . . , bn) be an initial segment of a play (a, b). ◮ Since ∃ has a winning strategy for EFω(A, B), we have that A | = ϕ(a) iff B | = ϕ(b), for every atomic ϕ. ◮ In particular, A | = ψ(a1, . . . , an) iff B | = ψ(b1, . . . , bn), for every unnested atomic ψ.

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Useful trick: parameters in EFn[A, B] games

◮ We write (A, a) ≈n (B, b) to mean that ∃ has a winning strategy in EFn[(A, a), (B, b)].

Lemma 14

Let A, B be similar structures and let k, n < ω. Let a, b be k-tuples from A, B. The following are equivalent:

• 1. (A, a) ≈n+1 (B, b).
• 2. For every c from A there exists d from B, and for every d from B

there exists c from A, such that (A, a, c) ≈n (B, b, d).

Proof.

◮ Assume (1). Any c ∈ |A| can be treated by ∃ as the first move in EFn+1[(A, a), (B, b)]. ◮ Using her winning strategy, ∃ responds by choosing d ∈ |B|. ◮ Now, switching the point of view, ∃ gets a winning strategy for EFn[(A, a, c), (B, b, d)]. ◮ Assume (2). Then ∃ has a response, say d, to every ∀s first move, say c, in EFn+1[(A, a), (B, b)]. ◮ Next ∃ follows her strategy for EFn[(A, a, c), (B, b, d)].

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Useful trick: parameters in EFn[A, B] games

◮ We write (A, a) ≈n (B, b) to mean that ∃ has a winning strategy in EFn[(A, a), (B, b)].

Lemma 14

Let A, B be similar structures and let k, n < ω. Let a, b be k-tuples from A, B. The following are equivalent:

• 1. (A, a) ≈n+1 (B, b).
• 2. For every c from A there exists d from B, and for every d from B

there exists c from A, such that (A, a, c) ≈n (B, b, d).

Proof.

◮ Assume (1). Any c ∈ |A| can be treated by ∃ as the first move in EFn+1[(A, a), (B, b)]. ◮ Using her winning strategy, ∃ responds by choosing d ∈ |B|. ◮ Now, switching the point of view, ∃ gets a winning strategy for EFn[(A, a, c), (B, b, d)]. ◮ Assume (2). Then ∃ has a response, say d, to every ∀s first move, say c, in EFn+1[(A, a), (B, b)]. ◮ Next ∃ follows her strategy for EFn[(A, a, c), (B, b, d)].

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Useful trick: parameters in EFn[A, B] games

◮ We write (A, a) ≈n (B, b) to mean that ∃ has a winning strategy in EFn[(A, a), (B, b)].

Lemma 14

Let A, B be similar structures and let k, n < ω. Let a, b be k-tuples from A, B. The following are equivalent:

• 1. (A, a) ≈n+1 (B, b).
• 2. For every c from A there exists d from B, and for every d from B

there exists c from A, such that (A, a, c) ≈n (B, b, d).

Proof.

◮ Assume (1). Any c ∈ |A| can be treated by ∃ as the first move in EFn+1[(A, a), (B, b)]. ◮ Using her winning strategy, ∃ responds by choosing d ∈ |B|. ◮ Now, switching the point of view, ∃ gets a winning strategy for EFn[(A, a, c), (B, b, d)]. ◮ Assume (2). Then ∃ has a response, say d, to every ∀s first move, say c, in EFn+1[(A, a), (B, b)]. ◮ Next ∃ follows her strategy for EFn[(A, a, c), (B, b, d)].

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Characterising A ≈n B by formulas: preliminaries

Definition 15 (Quantifier rank)

◮ Atomic formulas ϕ have quantifier rank qr(ϕ) = 0, ◮ qr(ϕ1 ∧ ϕ2) = qr(ϕ1 ∨ ϕ2) := max{qr(ϕ1), qr(ϕ2)}, ◮ qr(¬ϕ) := qr(ϕ), ◮ qr(∀x · ϕ) = qr(∃x · ϕ) := qr(ϕ) + 1.

Definition 16

For structures A, B write A ≡n B if for every sentence ϕ of quantifier rank at most n we have A | = ϕ if and only if B | = ϕ.

Lemma 17

A ≡ B is the same as A ≡n B for all n < ω.

Proof.

Immediate from the definitions.

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Characterising A ≈n B by formulas: preliminaries

Definition 15 (Quantifier rank)

◮ Atomic formulas ϕ have quantifier rank qr(ϕ) = 0, ◮ qr(ϕ1 ∧ ϕ2) = qr(ϕ1 ∨ ϕ2) := max{qr(ϕ1), qr(ϕ2)}, ◮ qr(¬ϕ) := qr(ϕ), ◮ qr(∀x · ϕ) = qr(∃x · ϕ) := qr(ϕ) + 1.

Definition 16

For structures A, B write A ≡n B if for every sentence ϕ of quantifier rank at most n we have A | = ϕ if and only if B | = ϕ.

Lemma 17

A ≡ B is the same as A ≡n B for all n < ω.

Proof.

Immediate from the definitions.

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Characterising A ≈n B by formulas: preliminaries

Definition 15 (Quantifier rank)

◮ Atomic formulas ϕ have quantifier rank qr(ϕ) = 0, ◮ qr(ϕ1 ∧ ϕ2) = qr(ϕ1 ∨ ϕ2) := max{qr(ϕ1), qr(ϕ2)}, ◮ qr(¬ϕ) := qr(ϕ), ◮ qr(∀x · ϕ) = qr(∃x · ϕ) := qr(ϕ) + 1.

Definition 16

For structures A, B write A ≡n B if for every sentence ϕ of quantifier rank at most n we have A | = ϕ if and only if B | = ϕ.

Lemma 17

A ≡ B is the same as A ≡n B for all n < ω.

Proof.

Immediate from the definitions.

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Characterising A ≈n B: game-normal formulas

Definition 18

Let Σ be a finite signature. For each n, k < ω we define a set Θn,k of unnested formulas by simultaneous induction on n, k: ◮ For k = 0 and arbitrary n < ω there are finitely many unnested atomic formulas: ϕ0, . . . , ϕm−1. Let Θn,0 be the set of all conjunctions ∗ϕ0 ∧ · · · ∧ ∗ϕm−1, where ∗ is either a blank or ¬. So Θn,0 contains 2m formulas. ◮ Assuming Θn+1,k has been defined, list the formulas in Θn+1,k as χ0(x, xn), . . . , χj−1(x, xn), where x = (x0, . . . , xn−1). ◮ Then define Θn,k+1 to be the set of all formulas

• i∈X

∃xn · χi(x, xn) ∧ ∀xn ·

• i∈X

χi(x, xn) where X ranges over all subsets of {0, . . . , j − 1}.

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Game-normal formulas

The intuition for formulas

i∈X ∃xn · χi(x, xn) ∧ ∀xn · i∈X χi(x, xn) is

• this. List all members of Θn+1,k:

χ0(x, xn), χ1(x, xn), . . . , χj−1(x, xn) Fix a tuple of parameters a from a model, say A, and try to fit elements of A in the “empty slot” to satisfy the χs. Take all χs that can be satisfied, and put their indices in a set E. Then, A | =

• i∈E

∃xn · χi(a, xn) Next consider all the χs with the property that every element of A satisfies at least one of them. Put their indices in a set U. We have A | = ∀xn ·

• i∈U

χi(a, xn) In general E and U are independent. The set we want is X = E ∩ U.

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Game-normal formulas

The intuition for formulas

i∈X ∃xn · χi(x, xn) ∧ ∀xn · i∈X χi(x, xn) is

• this. List all members of Θn+1,k:

χ0(a, xn), χ1(a, xn), . . . , χj−1(a, xn) Fix a tuple of parameters a from a model, say A, and try to fit elements of A in the “empty slot” to satisfy the χs. Take all χs that can be satisfied, and put their indices in a set E. Then, A | =

• i∈E

∃xn · χi(a, xn) Next consider all the χs with the property that every element of A satisfies at least one of them. Put their indices in a set U. We have A | = ∀xn ·

• i∈U

χi(a, xn) In general E and U are independent. The set we want is X = E ∩ U.

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Game-normal formulas

The intuition for formulas

i∈X ∃xn · χi(x, xn) ∧ ∀xn · i∈X χi(x, xn) is

• this. List all members of Θn+1,k:

χ0(a, xn), χ1(a, xn), . . . , χj−1(a, xn) Fix a tuple of parameters a from a model, say A, and try to fit elements of A in the “empty slot” to satisfy the χs. Take all χs that can be satisfied, and put their indices in a set E. Then, A | =

• i∈E

∃xn · χi(a, xn) Next consider all the χs with the property that every element of A satisfies at least one of them. Put their indices in a set U. We have A | = ∀xn ·

• i∈U

χi(a, xn) In general E and U are independent. The set we want is X = E ∩ U.

40 / 48

Game-normal formulas

The intuition for formulas

i∈X ∃xn · χi(x, xn) ∧ ∀xn · i∈X χi(x, xn) is

• this. List all members of Θn+1,k:

χ0(a, xn), χ1(a, xn), . . . , χj−1(a, xn) Fix a tuple of parameters a from a model, say A, and try to fit elements of A in the “empty slot” to satisfy the χs. Take all χs that can be satisfied, and put their indices in a set E. Then, A | =

• i∈E

∃xn · χi(a, xn) Next consider all the χs with the property that every element of A satisfies at least one of them. Put their indices in a set U. We have A | = ∀xn ·

• i∈U

χi(a, xn) In general E and U are independent. The set we want is X = E ∩ U.

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Fra¨ ıss´ e-Hintikka Theorem

Theorem 19

Let Σ be a finite signature. For every n, k < ω we can effectively find a finite set Θn,k of unnested formulas of quantifier rank at most k such that:

• 1. For every Σ-model A and each n-tuple a from A there is exactly one

formula θ ∈ Θn,k such that A | = θ(a).

• 2. For all n, k < ω, any Σ-models A and B, and any n-tuples a and b

from A and B, the following are equivalent:

◮ (A, a) ≈k (B, b), ◮ there is θ ∈ Θn,k such that A | = θ(a) and B | = θ(b).

• 3. For every k < ω and every unnested formula ϕ(x) with n free

variables and quantifier rank at most k, we can effectively find a disjunction θ0(x) ∨ · · · ∨ θm−1(x) of formulas from Θn,r, with r ≥ k, which is logically equivalent to ϕ(x).

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Fra¨ ıss´ e-Hintikka Theorem

Proof of (1).

◮ Induction on k (for all n simultaneously). For k = 0, by definition of Θn,0, at most one θ(a) can be true in A. Also by definition, Θn,0 is logically true, so at least one θ(a) must be true in A. This proves (1). ◮ Assume (1) for k (and all n). By inductive hypothesis, for any (n + 1)-tuple (a, c) from A extending a there is a unique θ′ ∈ Θn+1,k such that A | = θ′(a, c). ◮ Each such θ′ is some χi in the standard enumeration of Θn+1,k. Let X be the set of indices of all such θ′. ◮ Put θ =

i∈X ∃xn · χi(a, xn) ∧ ∀xn · i∈X χi(a, xn). Then, A |

= θ(a), and θ ∈ Θn,k+1. ◮ We have thus checked all possible extensions of a to an (n + 1)-tuple. Having a different set Y of indices will either make one of ∃xn · χi(a, xn) false, or the disjunction

• i∈X χi(a, c) false for some c. Therefore, θ is unique, proving (1).

Exercise 2

Prove (3) by induction on k. Give an example showing that if k > 0, then the procedure of finding an unnested equivalent increases quantifier rank.

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Fra¨ ıss´ e-Hintikka Theorem

Proof of (1).

◮ Induction on k (for all n simultaneously). For k = 0, by definition of Θn,0, at most one θ(a) can be true in A. Also by definition, Θn,0 is logically true, so at least one θ(a) must be true in A. This proves (1). ◮ Assume (1) for k (and all n). By inductive hypothesis, for any (n + 1)-tuple (a, c) from A extending a there is a unique θ′ ∈ Θn+1,k such that A | = θ′(a, c). ◮ Each such θ′ is some χi in the standard enumeration of Θn+1,k. Let X be the set of indices of all such θ′. ◮ Put θ =

i∈X ∃xn · χi(a, xn) ∧ ∀xn · i∈X χi(a, xn). Then, A |

= θ(a), and θ ∈ Θn,k+1. ◮ We have thus checked all possible extensions of a to an (n + 1)-tuple. Having a different set Y of indices will either make one of ∃xn · χi(a, xn) false, or the disjunction

• i∈X χi(a, c) false for some c. Therefore, θ is unique, proving (1).

Exercise 2

Prove (3) by induction on k. Give an example showing that if k > 0, then the procedure of finding an unnested equivalent increases quantifier rank.

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Fra¨ ıss´ e-Hintikka Theorem

Proof of (2).

◮ Induction on k (for all n simultaneously). For k = 0, by definition of ≈0, we have (A, b) ≈0 (B, b) iff for every unnested atomic ϕ(x) we have A | = ϕ(a) ⇔ B | = ϕ(b). ◮ This means that precisely the same conjunctions θ ∈ Θn,0 have A | = θ(a) and B | = θ(b). By (1) there is exactly one θ ∈ Θn,0 with A | = θ(a), and then B | = θ(b) as required. ◮ The converse follows immediately by converting ϕ to a disjunction of formulas from Θn,0. ◮ For the inductive step, assume (2) holds for k. Assume (A, a) ≈k+1 (B, b). ◮ By Lemma 14, for every c from A there exist d from B (and vice versa), such that (A, a, c) ≈k (B, b, d). By inductive hypothesis, we get (⋆) For every c from A there is χi ∈ Θn+1,k such that A | = χi(a, c) and B | = χi(b, d) (where d depends on c); and vice versa, for every d from B there is χj ∈ Θn+1,k such that B | = χj(b, d) and A | = χj(a, c) (where c depends on d). ◮ As in the proof of (1) gather into X all indices of formulas χi ∈ Θn+1,k such that A | = χi(a, c); put θ =

i∈X ∃xn · χi(a, xn) ∧ ∀xn · i∈X χi(a, xn). We have A |

= θ(a). ◮ Now (⋆) above implies that for every i ∈ X there is d from B such that B | = χi(b, d) and for every j ∈ X there is d from B such that B | = χj(b, d). ◮ Thus, B | = θ(b) as required. ◮ Conversely, assume there is θ ∈ Θn,k+1 such that A | = θ(a) and B | = θ(b). ◮ By the form of θ it implies that (⋆) holds. Using Lemma 14 backwards, we are done.

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Game equivalences: summary

Theorem 20 (Game equivalences)

Let A and B be similar structures of finite signature. The following hold:

• 1. A ≡k B if and only if A ≈k B.
• 2. A ≡ B if and only if A ≈k B for every k < ω.
• 3. If card(A), card(B) ≤ ω, then A ∼

= B if and only if A ∼ω B. ◮ By Lemma 13 we have that ∼ω implies ≡. ◮ We also have that ≡ is precisely

k<ω ≡k.

Theorem 21 (A hierarchy of equivalences)

∼ = ∼ω ≡ . . . ≡k . . . ≡1 ≡0 where the arrows denote proper inclusions.

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Game equivalences: summary

Theorem 20 (Game equivalences)

Let A and B be similar structures of finite signature. The following hold:

• 1. A ≡k B if and only if A ≈k B.
• 2. A ≡ B if and only if A ≈k B for every k < ω.
• 3. If card(A), card(B) ≤ ω, then A ∼

= B if and only if A ∼ω B. ◮ By Lemma 13 we have that ∼ω implies ≡. ◮ We also have that ≡ is precisely

k<ω ≡k.

Theorem 21 (A hierarchy of equivalences)

∼ = ∼ω ≡ . . . ≡k . . . ≡1 ≡0 where the arrows denote proper inclusions.

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Game equivalences: summary

Theorem 20 (Game equivalences)

Let A and B be similar structures of finite signature. The following hold:

• 1. A ≡k B if and only if A ≈k B.
• 2. A ≡ B if and only if A ≈k B for every k < ω.
• 3. If card(A), card(B) ≤ ω, then A ∼

= B if and only if A ∼ω B. ◮ By Lemma 13 we have that ∼ω implies ≡. ◮ We also have that ≡ is precisely

k<ω ≡k.

Theorem 21 (A hierarchy of equivalences)

∼ = ∼ω ≡ . . . ≡k . . . ≡1 ≡0 where the arrows denote proper inclusions.

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