150 likes | 163 Views
This lecture provides an outline and proof of the Ehrenfeucht-Fraisse theorem in finite model theory, introducing the concept of rank-k m-types and the back-and-forth property. It also discusses Hanf's Lemma and its applications.
E N D
Finite Model TheoryLecture 3 Ehrenfeucht-Fraisse Games
Outline • Proof of the Ehrenfeucht-Fraisse theorem
Notation If A is a structure over vocabulary sand a1, …, an2Athen (A,a1, …, an) denotes the structure over vocabulary sn = s[ {c1, …, cn} s.t. the interpretation of each ci is ai In particular, (A,a) ' (B,b) means that there is an isomorphism A'B that maps a to b
Types In classical model theory an m-type for m ¸ 0 is a set t of formulas with m free variables x1, …, xm s.t. there exists a structure A and m constants a = (a1, …, am) s.t. t = {f | A²f(a) } In finite model theory this is two strong: (A,a) and (B,b) have the same type iff they are isomorphic (A,a) ' (B,b)
Rank-k m-Types FO[k] = all formulas of quantifier rank · k Definition Let A be a structure and a be an m-tuple in A. The rank-k m-type of a over A istpk(A,a) = {f2 FO[k] with m free vars | A²f(a) } How any distinct rank-k types are there ? [finitely or infinitely many ?]
Rank-k m-Types For m ¸ 0, there are only finitely many formulas up to logical equivalence over m variables x1, …, xm in FO[0] [why ?] For m ¸ 0, there are only finitely many formulas up to logical equivalence over m variables x1, …, xm in FO[k+1] [why ?]
Rank-k m-Types • For each rank-k m-type t there exists a unique rank-k formula f s.t. A²f(a) iff tpk(A,a) = t • In other words, if M = {f1, …, fn} are all formulas in FO[k] with n free variables, then for every subset M0µ M there exists a f2 M s.t. f = (Æy2 M0) y Æ (ÆyÏ M0:y) [WAIT ! Isn’t this a contradiction ?]
The Back-and-Forth Property The k-back-and-forth equivalence relation 'k is defined as follows: • A'0B iff the substructures induced by the constants in A and B are isomorphic • A'k+1B iff the following hold: Forth: 8 a 2A9 b 2B s.t. (A,a) 'k (B,b) Back: 8 b 2B9 a 2A s.t. (A,a) 'k (B,b)
The Back-and-Forth Property • What does A'kB say ? • If we have a partial isomorphism from (A, a1, …, ai) to (B,b1, …, bi), where i < k, and ai+12A, then there exists bi+12B s.t. there exists a partial isomorphism from (A, a1, …, ai, ai+1) to (B, b1, …, bi, bi+1); and vice versa
Ehrenfeucht-Fraisse Games Theorem The following two are equivalent: • A and B agree on FO[k] • AkB • A'kB Proof 2 , 3 is straightforward 1 , 3 in class
. . . . . . . . . . . . More EF Games (informally) Prove, informally, the following: (N,S) (N,S) [ (Z,S) k (Perfectly balanced binary trees are not expressible in FO)
More EF Games (informally) k CONN is not expressible in FO
Hanf’s Lemma • One of several combinatoric methods for proving EF games formally Definition. Let A be a structure. The Gaifman graph G(A) = (A, EA) is s.t.(a,b) 2 EA iff 9 tuple t in A containing both a and b Definition. The r-sphere, for r > 0, is: S(r,a) := {b 2 A | d(a,b) · r}
Hanf’s Lemma Theorem [Hanf’s lemma; simplified form] Let A, B be two structures and there exists m > 0 s.t. 8 n · 3m and for each isomorphism type t of an n-sphere, A and B have the same number of elements of n-sphere type t. Then Am B. Applications: previous examples.
Summary on EF Games • Complexity: examples in class are simple; but in general the proofs get quite complex • Informal arguments: We are all gamblers: • “If you play like this […] you will always win”. We usually accept such statements after thinking about […] • “here is a property not expressible in FO !”. We don’t accept that until we see a formal proof. • Logics v.s. games: Each logic corresponds to a certain kind of game.