Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Finite Model Theory Unit 2 Dan Suciu Spring 2018 Dan Suciu Finite Model Theory { Unit 2 Spring 2018 1 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types 599c: Finite Model Theory Unit 2: Expressive Power of Logics on Finite Models Dan Suciu Finite Model Theory { Unit 2 Spring 2018 2 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Resources Libkin, Finite Model Theory, Chapt. 3, 4, 11. Gr¨adel,Kolaitis, Libkin, Marx, Spencer, Vardi, Venema, Weinstein: Finite Model Theory and Its Applications, Capt. 2 (Expressive Power of Logics on Finite Models). Dan Suciu Finite Model Theory { Unit 2 Spring 2018 3 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Where Are We Classical model theory is concerned with truth, D ⊧ ', and its implications. Finite model theory is concerned with: ▸ Expressibility: which classes of finite structures can be expressed in a given logic. ▸ Computability: connection between computational complexity and expressibility in that logic. ▸ (Asymptotic) probabilities: study the probability (asymptotic or not) of a sentence. Dan Suciu Finite Model Theory { Unit 2 Spring 2018 4 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Unit 2: Expressibility Ehrenfeuched-Fraisse Games Infinitary logics and Pebble Games Dan Suciu Finite Model Theory { Unit 2 Spring 2018 5 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types The Expressibility Problem Given a property P, can it be expressed in a logic L? Example properties: CONNECTIVITY, EVEN, PLANARITY. Example logics: FO, SO, FO+fixpoint, Datalog. Dan Suciu Finite Model Theory { Unit 2 Spring 2018 6 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Example 1: EVEN Find a sentence ' s.t. G ⊧ ' iff G has an even number of nodes. In class Impossible! µn(') = 0 when n = odd, µn(') = 1 when n = even, violates 0/1-law. Find a sentence ' s.t. G ⊧ ' iff G has an even number of edges. The 0/1 law no longer helps. Dan Suciu Finite Model Theory { Unit 2 Spring 2018 7 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Example 2: CONNECTED 1 ∗ G = (V ; E) is connected if forall a; b ∈ V there exists a path a → b. Find an FO sentence s.t. G ⊧ iff G is connected. ∀x∀yE(x; y)? ∀x∀y∃z(E(x; z) ∧ E(z; y))? ::: Impossible! Let's prove that. 1The correct term is strongly connected. Dan Suciu Finite Model Theory { Unit 2 Spring 2018 8 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Example 2: CONNECTED Suppose G ⊧ iff G is connected. Fix two fresh constants c; d, and, forall n ≥ 1, define: 'n =(¬(∃z1⋯∃zn(E(c; z1) ∧ E(z1; z2) ∧ ⋯ ∧ E(zn; d)))) It says \c; d are not connected by any path of length n". def Σ = { } ∪ {'n S n ≥ 1} is finitely satisfiable why? By Compactness, Σ has a model G On one hand G ⊧ hence it is connected, on the other hand c; d are not connected in G, contradiction. But is Connectivity expressible over finite graphs? This proof does not answer it. Dan Suciu Finite Model Theory { Unit 2 Spring 2018 9 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Isomorphism Assume a relational vocabulary σ = (R1;:::; Rk ; c1;:::; cm) (no functions). A A A A B B B B Fix A = (A; R1 ;:::; Rk ; c1 ;:::; cm), B = (B; R1 ;:::; Rk ; c1 ;:::; cm). Definition An isomorphism f ∶ A → B is a bijection A → B such that: A B Forall R ∈ σ, (a1;:::; ak ) ∈ R iff (f (a1);:::; f (ak )) ∈ R . A B Forall c ∈ σ, f (c ) = c . We write A ≃ B if there exists an isomorphism A → B. Remark: if A ≃ B then for any sentence ' in a \reasonable" logics (like FO, or extensions), A ⊧ ' iff B ⊧ '. Dan Suciu Finite Model Theory { Unit 2 Spring 2018 10 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Elementary Equivalence Definition A and B are elementary equivalent if forall ', A ⊧ ' iff B ⊧ '. We write A ≡ B. Isomorphisms implies elementary equivalence: if A ≃ B then A ≡ B. Over the finite structures, the converse holds too: if A ≡ B, then A ≃ B. We cannot find two finite graphs, one connected and one disconnected, that are elementary equivalent! Dan Suciu Finite Model Theory { Unit 2 Spring 2018 11 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Partial Isomorphism Fix a relational vocabulary σ: relations Ri , constants cj . Let A; B be two σ-structures. Definition k A partial isomorphism is a pair a; b, where a = (a1;;:::; ak ) ∈ A , k a b = (b1;:::; bk ) ∈ B s.t. the substructures ASa, BSb are isomorphic via: A B ∀i; ai ↦bi ∀j; cj ↦ cj a A A ASa consists of the universe {a1;:::; ak ; c1 ;:::; cm}. We write a ≃ b. In other words: Forall i; j, ai = aj iff bi = bj . (Equality is preserved.) A B Forall i; j, ai = cj iff bi = cj . (Constants are preserved.) A A (t1;:::; tn) ∈ R where each ti is either some aj or cj , iff ′ ′ B ′ B (t1;:::; tn) ∈ R where ti is bj or cj respectively. Dan Suciu Finite Model Theory { Unit 2 Spring 2018 12 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Ehrenfeucht-Fraisse Games There are two players, spoiler and duplicator. They play on two structures A; B in k rounds, i = 1;:::; k. Round i: Spoiler places his pebble i on an element ai ∈ A or bi ∈ B. Duplicator places her pebble i on an element bi ∈ B or ai ∈ A. Let a = (a1;:::; ak ), b = (b1;:::; bk ) be the pebbles at the end of the game. Duplicator wins if a; b forms a partial isomorphism; otherwise Spoiler wins. Definition We write A ∼k B if the duplicator has a winning strategy for k rounds. Dan Suciu Finite Model Theory { Unit 2 Spring 2018 13 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Ehrenfeucht-Fraisse Games: Main Result The quantifier rank of a formula ' is defined inductively2: qr(F ) =qr(t1 = t2) = qr(R(t1;:::; tm)) = 0 qr(' → ) = max(qr('); qr( )) qr(∀x(')) =1 + qr(') def FO[k] = FO restricted to formulas with qr ≤ k. Theorem (Ehrenfeucht-Fraisse) A ≡k B (meaning: they agree on FO[k]) iff A ∼k B. We will prove it later. First, let's see examples. 2The number of quantifiers can be exponentially larger than qr(') why? Dan Suciu Finite Model Theory { Unit 2 Spring 2018 14 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Ehrenfeucht-Fraisse on Total Orders Let Lk = ({1; 2;:::; k}; <). Play the Ehrenfeucht-Fraisse game on L6; L7 using k = 2 pebbles: L6 ∼2 L7 Play the Ehrenfeucht-Fraisse game on L6; L7 using k = 3 pebbles: L6 ∼~3 L7 Find ' ∈ FO[3] s.t. L6 ⊧ '; L7 ⊧~ ' b b b b b b <x1 1 1 2 1 2 1 ∀x1∀x2(x2 < x1 → ~~L6 is small L6 L6 L6 L6 L6 L6 (∀x3¬(x3 < x2) ∨ ∀x3¬(x2 < x3 < x1))) >x1 ∨∀x2(x2 > x1 → ~~L6 is small (∀x3¬(x3 > x2) L L L L L L 7 ∨ ∀x3¬(7x1 < x3 < x2))) 7 7 7 7 a1 a1 a2 a1 a2 a1 a3 a2 a1 Dan Suciu Finite Model Theory { Unit 2 Spring 2018 15 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Ehrenfeucht-Fraisse on Total Orders Let Lm = ({1; 2;:::; m}; <). <a def >a def Lm = {x ∈ Lm S x < a} Lm = {x ∈ Lm S x < a} Lemma <a <b >a >b If Lm ∼k Ln and Lm ∼k Ln (duplicator wins), then Lm ∼k Ln. Proof. <a <b If spoiler places pebble in Lm then duplicator answers in Ln . >a >b If spoiler places pebble in Lm then duplicator answers in Ln . If spoiler places pebble on a then duplicator places pebble on b. If spoiler plays in the other structure, duplicator answers similarly. <a <b >a >b If Lm Sc ≃ Ln Sd and Lm Sc ≃ Ln Sd (partial isomorphisms), then c ≃ d Dan Suciu Finite Model Theory { Unit 2 Spring 2018 16 / 66 Introduction Games for FO Games for ∃MSO Games for Recursion FOk Types Ehrenfeucht-Fraisse on Total Orders Theorem Let m; n; k be positive integers. The following are equivalent: Lm ∼k Ln k k m = n or both m ≥ 2 − 1 and n ≥ 2 − 1 k Proof. If m; n ≥ 2 − 1, duplicator has winning strategy. Spoiler plays a ∈ Lm. <a k−1 Case 1: SLm S < 2 − 1 (What do we do?) <a <b Duplicator chooses b s.t. Lm ≃ Ln (i.e. isomorphic). Then: >a >b k−1 >a >b SLm S; SLn S > 2 − 1 (why?), Lm ∼k−1 Ln (why?), Lm ∼k Ln (lemma). >a k−1 Case 2: SLm S < 2 − 1 Symmetric: >a >b Duplicator chooses b s.t. Lm ≃ Ln (i.e. isomorphic). Then: <a <b k−1 <a <b SLm S; SLn S > 2 − 1, Lm ∼k−1 Ln , hence Lm ∼k Ln (lemma). <a >a k−1 Case 3: both SLm S; SLm S ≥ 2 − 1 (Is this possible?) <b >b k−1 Duplicator chooses any b s.t. SLn S; SLn S ≥ 2 − 1. Then: <a <b >a >b k−1 <a <b >a >b SLm S; SLn S; SLm S; SLn S ≥ 2 − 1; Lm ∼k−1 Ln ; Lm ∼k−1 Ln ; Lm ∼k Ln.