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1 Finite Model Theory Lecture 13 FO k, L k 1, ,L  1, , and Pebble Games.

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Presentation on theme: "1 Finite Model Theory Lecture 13 FO k, L k 1, ,L  1, , and Pebble Games."— Presentation transcript:

1 1 Finite Model Theory Lecture 13 FO k, L k 1, ,L  1, , and Pebble Games

2 2 Infinitary Logic Allow infinite conjunctions, disjunctions: Now restrict everything to just k variables: x 1, …, x k Finally: Ç i 2 I  i, Æ i 2 I  i L 1,  L k 1,  L  1,  = [  ¸  L k 1, 

3 3 Quantifier Rank qr(  ) is defined as before: In general it can be an ordinal, e.g. 5, 99,  +3, 4  + 7, etc Over finite structures, we will show that it suffices to have qr ·  qr( Ç i 2 I  i ) = qr( Æ i 2 I  i ) = sup i 2 I qr(  i ) qr( 9 x.  ) = qr( 8 x.  ) = qr(  ) + 1 qr( Ç i 2 I  i ) = qr( Æ i 2 I  i ) = sup i 2 I qr(  i ) qr( 9 x.  ) = qr( 8 x.  ) = qr(  ) + 1

4 4 Example Over ordered finite structures, every property can be expressed in L m+1 1, , where m = max i (arity(R i )), R i 2  Proof: First, show that if  = {<} then every property on finite linear orders is in L m+1 1,  Next, generalize to arbitrary 

5 5 Fixpoints Consider transitive closure, expressed as LFP: tc(u,v) = lfp R [E(x,y) Ç 9 z.(E(x,z) Æ R(z,y)](u,v) Can “unfold” it: tc 0 (x,y) = E(x,y) tc n+1 (x,y) = 9 z.(E(x,z) Æ 9 x.(x=z Æ tc n (x,y))) Hence tc(u,v) = Ç n ¸ 0 tc n (u,v) 2 L  1, 

6 6 Fixpoints In general: Theorem LFP, IFP, PFP µ L  1,  Proof [in class]

7 7 Pebble Games There are k pairs of pebbles, (a 1, b 1 ), …, (a k, b k ) Pebble games are played much like Ehrenfeucht- Fraisse games, HOWEVER now pebbles can be moved, after they have been placed on a structure

8 8 Pebble Games Let A, B 2 STRUCT[  ], and k ¸ 0. Round j ¸ 1: –Spoiler picks a pebble, a i (or b i), i = 1, …, k; if a i (b i) was already placed on a structure, then it removes it; spoiler places the pebble a i on the structure A (or b i on the structure B) –Duplicator needs to respond by placing pebble b i on B (or a i on A)

9 9 Pebble Games The game for n rounds: PG n k (A,B) The game forever: PG 1 k (A,B) Duplicator has a winning strategy for PG n (A,B) if 8 j · n, (a 1 ! b 1, …, a k ! b k ) is a partial isomorphism; similarly for PG 1 (A,B) Notation: A  1  k,n B and A  1  k B

10 10 Pebble Games Theorem A, B 2 STRUCT[  ] agree on all sentences of L k 1,  of qr · n iff A  1  k,n B A, B 2 STRUCT[  ] agree on all sentences of L k 1,  iff A  1  k B Proof next time. For today we will assume the theorem to be true

11 11 Example Can you win this with k=2 pebbles ? A=B=

12 12 Example Can you win this with k=4 pebbles ? A= B=

13 13 EVEN Theorem EVEN is not expressible in L  1,  Corollary EVEN is not expressible in LFP, IFP, PFP Corollary LFP & (LFP+<) inv IFP & (IFP+<) inv PFP & (PFP+<) inv

14 14 A Property Assume A, B agree on all sentences in FO k Then, for every n ¸ 0, they agree on all sentences in L k 1,  of qr · n –Why ? Consider Ç i 2 I  i where 8 i. qr(  i ) · n. Then this is in FO k ! Hence, for every n ¸ 0, A  1  k,n B Hence, A  1  k B –Why ? Remember Koenig’s Lemma ? Hence A, B agree on all sentences in L k 1, 

15 15 A Property Theorem The following are equivalent A, B agree on all FO k sentences A, B agree on all L k 1,  sentences

16 16 Definability of Types Definition. Let A 2 STRUCT[  ], and a 2 A. The FO k type of (A, a) is: tp FO k (A,a) = {  (x) 2 FO k | A ²  (a) } One could define L  1,  types as well, but they turn out to be the same as FO k types [ why ?? ]

17 17 Definability of Types The number of FO k types is infinite (since no restriction on quantifier depth) Each FO k type is definable as Ç i 2 I  i, where each  i 2 FO k. However: Theorem. Every FO k type  is defined in FO k. I.e. there exists   (x) s.t. Tp FO k (A, a) = , A ²   (a) Proof: next time. For now assume it holds

18 18 Applications For every structure A there exists a sentence  A in FO k s.t. 8 B: B ²  A iff A  1  k B Every L k 1,  formula is equivalent to Ç i 2 I  i, where each  i 2 FO k


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