Notes

Outline

Finite Model Theory Lecture 2 Ehrenfeucht-Fraisse Games

Summary of Notions

Infinite v.s. Finite Everything can be defined “in the finite”  f has a finite model  f is valid in the finite if for every finite A, A ² f; notation ²fin f  f is a consequence in the finite of G if for every finite A, A ² G implies A ² f; notation G ²fin f

Completeness Fails “f has a finite model” is r.e.  [why] Trachtenbrot’s theorem: “f has a finite model” is undecidable Hence there is no proof system ` s.t.   ` f iff ²fin f     [why ?]

Compactness Fails There exists an infinite set of formulas G s.t. for every finite subset G0 ½ G, G0 has a finite model, but G does not have a finite model [why ?]

Outline Today’s lecture: Ehrenfeucht-Fraisse games Applications

Equivalences Let A, B 2 STRUCT[s] They are elementary equivalent, A º B, if 8 f, A ² f iff B ² f They are isomorphic, A @ B, if there exists a isomorphism f : A ! B

Equivalences If A @ B, then A º B  [why ?] There are A, B s.t. A º B and not A @ B [give examples in class] If A is finite and A º B then A @ B [why ?]

Partial Isomorphism Let c = (c1, …, cn) be the constants in s Let A, B be two structures in STRUCT[s] Definition A partial isomorphism is given by a = (a1, …, am) and b = (b1, …, bm) s.t.  the structures (a, c) and (b, c) are isomorphic.

Ehrenfeucht-Fraisse Games Given two structures A, B Two players: spoiler and duplicator play k rounds Round i  (i =1, …, k): Spoiler picks a structure A or B Spoiler picks an element in that structure ai 2 A or bi 2 B Duplicator responds by picking an element in the other structure, bi 2 B or ai 2 A Duplicator wins if a, b is a partial isomorphism

Ehrenfeucht-Fraisse Games If duplicator has a winning strategy in k rounds then we write A ºk B Note: if A ºn B and k < n then A ºk B

Quantifier Rank The quantifier rank of a formula f is defined inductively: qr(t1 = t2) = qr(R(t1,…,tn)) = 0 qr(f1 Æ f2) = max(qr(f1), qr(f2)) qr(9 x.f) =1+qr(f) etc FO[k] = all formula f s.t. qr(f) · k

Ehrenfeucht-Fraisse Games Theorem The following two are equivalent: A and B agree on FO[k] A ºk B We will prove this next time. For now, let’s start playing the game !

Games on Sets Let s = ; (i.e. no relation names, no constants), and assume |A| ¸ k, |B| ¸ k. Theorem A ºk B.  [why ?] Corollary EVEN is not expressible in FO when s = ;

Games on Linear Orders Theorem Let k > 0 and L1, L2 be linear orders of length ¸ 2k.  Then L1 ºk L2.

Proof 1 Define: a-1 = min(L1),  a0 = max(L1) b-1 = min(L2),  b0 = max(L2) Let a = (a-1, a0, a1, …, ai),      b = (b-1, b0, b1, …, bi) Duplicator plays such that 8 j, l: If d(aj, al) < 2k-i then d(aj, al) = d(bj, bl) If d(aj, al) ¸ 2k-i then d(bj, bl) ¸ 2k-i aj · al iff bj · bl

Proof 2 Remark: if  L1 ºk L2 then the duplicator has a winning strategy where it responds with min(L2) to min(L1), and with max(L2) to max(L1), and vice versa.  [why ?] Lemma If L1· a ºk L2 · b and L1¸ a ºk L2 ¸ b then (L1 ,a) ºk (L2 ,b) [how does this help us prove the theorem ?]

EVEN Corollary EVEN is not expressible in FO(<) [why ?] Corollary Finite graph connectivity (CONN) is not expressible in FO [proof in class]