Logic: Lecture 15, 01 October 2015 ---------------------------------- Expressiveness of First-Order Logic Proofs of inexpressiveness using Lowenheim-Skolem theorem usually produce infinite counterexamples. What if we are interested in finite models? In general, for finite models, we assume our language L = (R,F,C) is "relational" --- that is, F is empty. We can encode a k-ary function f by a (k+1)-ary relation r_f such that f(x1,...,xk) = y iff (x1,x2,...,y) in r_f ---------------------------------------------------------------------- Compactness fails over finite models (Libkin, Proposition 3.2) Proposition: There is an infinite set of sentences that is not satisfiable over finite models, such that every finite subset is satisfiable over finite models ---------------------------------------------------------------------- Inexpressibility of evenness (Libkin, Proposition 3.3) Proposition: Assume that L is empty. Then EVEN is not FO-definable. Note: Prop 3.3 fails if we have R = {<}, for instance. How do we extend this result to a richer class of finite models? ---------------------------------------------------------------------- The "natural" approach to show that a property P is not satisfiable over finite models is to exhibit two models A and B that satisfy the same formulas such that A has property P and B does not. Unfortunately, this strategy fails because no two non-isomorphic models satisfy the same FO formulas. Characteristic formula for finite structures (Libkin, Lemma 3.4) Lemma: For every finite structure A, there is a sentence Phi_A such that B |= Phi_A iff B is isomorphic to A. ---------------------------------------------------------------------- A more sophisticated strategy: - Stratify formulas by structural complexity. More formally, define quantifier rank of a formula as maximum depth of nested quantifiers. Let FO[k] denote all formulas of quantifier rank at most k. - Find a family of structures A_k and B_k, one for each k, that agree on all of FO[k] where A has property P and B does not. - Suppose Psi captures property P. Psi is a finite sentence and has a fixed quantifier rank m. Consider the structures A_m and B_m from the previous step. These agree on FO[m], hence agree on Psi, but B does not have property P. This is a contradiction ---------------------------------------------------------------------- Ehrenfeucht games (Libkin, Section 3.2) Partial isomorphism of structures (Libkin, Definition 3.5) - A and B structures with {a1,...,am} and {b1,...,bm} m-tuples from A and B, respectively - The substructures defined by {a1,...,am} and {b1,...,bm} are isomorphic The game: - Spoiler and Duplicator - In each move j - Spoiler picks aj in A or bj in B - Duplicator picks a matching element in the other structure - After k rounds - Duplicator wins if {a1,...,ak} and {b1,...,bk} are partially - isomorphic ---------------------------------------------------------------------- Examples: 1. Games on Sets Let A, B be sets such that |A|, |B| are both at least m. Then A and B are m-equivalent wrt Ehrenfeucht games. 2. Games on Linear Orders Let (A,<), (B,<) be linear orders such that |A|, |B| are both at least 2^m. Then A and B are m-equivalent wrt Ehrenfeucht games. ======================================================================