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Outline

Finite Model Theory Lecture 17 0/1 Laws

Extension Axioms Revisited Lemma Forall t, s0: m(t t,s0) = limn mn(t t,s0) = 1 This is if and only if Lemma Forall s0: m(ss0) = 1 Proof:  Let t1, t2, …, tm be all k-types Then:  ss0 = tt1,s0 Æ … Æ ttm,s0

Extension Axioms Revisited Moreover, this is iff Lemma Forall s0: m(ss0 ¹) = 1 Proof:  Every ss0 is the conjunction of several expressions ss1¹, where s1 states s0 over fewer variables, by repeating some of them.

Extension Axioms Revisited Now let’s see the proof from last lecture that m[t] = 1…

Examples What is m(f) in each case below ? (These are from the book) “There are no isolated nodes”:f =  8 x.9 y.R(x,y) “R is connected” “The domain has even cardinality”:  EVEN “R has even cardinality”:  PARITY

0/1 Law for Lw1w Let qk be the conjunction of all extension axioms with · k variables I.e. tt,s where t is a type over x = (x1, …, xk) Lemma.  Let A, B be such that A ² qk, B ² qk . Then A =1wk B Proof: qk says precisely that the duplicator has a winning strategy with k pebbles.

0/1 Law for Lw1w Theorem Lw1w has a 0/1 law. Proof Let f 2 Lk1w Case 1. There exists a model A for both f and qk .  Then, for every model B of qk we have B ² f, hence m(f) = 1 Case 2. There is no such model.  Then : f and qk have a common model and we prove m(f) = 0.

0/1 Law for Lw1w Proof via a Transfer Theorem [Kolaitis and Vardi] Lemma.  8 k > 0 there is yk 2 FOk such that:  R ² yk  For f in Lk1w there is f’ in FOk such that: yk ² f $ f’

Proof of the Lemma Claim: there are only finitely many inequivalent FOk formulas over R [Proof: in class ?? ] a1, a2, …, am Hence: R ² 8 x(p(x) $ ai(x)) R ² 8 x(: ai $ : aj) R ² 8 x.(9 xj.ai $ al) Let yk be their conjunction; clearly R ² fk

Proof of the Lemma Claim 2: for any f 2 Lk1w there exists f’ 2 FOk s.t. yk ² f $ f’ This by induction on f.

Proof of 0/1 Law for Lk1w Let f be in Lk1w and yk, f’ be as in the Lemma Then m(phik) = 1 and one can check that m(f) = m(f’)

Discussion How does the “random graph” R look like ? Does second order logic SO have a 0/1 law ? “Almost everywhere equivalence”.  Two logics L, L’ are a.e. equivalent if forall f 2 L 9 f’ in L’ s.t. m(f $ f’) = 1. The following are a.e. equivalent: FO, Lw1w, LFP, IFP, PFP