Logic: Lecture 10, 3 September 2015 -------------------------------- ---------------------------------------------------------------------- FO satisfiability Theorem: Let X be a set of sentences over L. The following are equivalent. 1. There is an L-structure M = (S,i) that satisfies X 2. There is an L_H structure M' = (S,i') that satisfies X 3. X U Phi_H U Phi_Q U Phi_Eq is propositonally satisfiable (3 => 2) We have a valuation v over all prime formulas in L_H such that v |= X U Phi_H U Phi_Q U Phi_Eq Need to define an L_H structure I = (S,i) that satisfies X First S: Define an equivalence relation on closed terms t ~ t' iff v |= t = t' v |= Phi_Eq guarantees that ~ is an equivalence relation Elements of S are equivalence classes [t] of closed terms Define i as follows: For C : i(c) = [c] For R : ([c1],...,[cn]) in i(r) iff v |= r(c1,...,cn) For F : For closed terms (t1,...,tn), need to identify [t] such that i(f)([t1],...,[tn]) = [t] Choose t = f(t1,...,tn), itself a closed term, so i(f)([t1],...,[tn]) = [f(t1,...,tn)] The fact that interpretations of F and R are compatible with ~ follows from Phi_Eq Claim: For every closed term t, there is a constant c such that [t] = [c] --- i.e. v(t = c) is true Proof: Consider the formula phi(x) = t = x with one free variable. C_H contains a constant c_phi for the formula Ex phi(x). Substituting t for x, we get phi(t) to be t = t. - By the equality axioms v(phi(t)) = v(t = t) = tt. - By the quantifier axiom v(phi(t) -> Ex phi(x)) = tt. Hence, v(Ex phi(x)) = tt. - By the Henkin axiom v(Ex phi(x) -> v(phi(c_phi)) = tt. Hence v(phi(c_phi)) = v(t = c_phi) = tt Thus, [t] = [c_phi]. Claim: For all formulas phi in L_H, I |= phi iff v |= phi Proof: Construction of S ensures that I |= phi iff v |= phi for all atomic formulas. Boolean connectives are easy. phi = Ex psi I |= Ex psi => I |= psi(a) for some element a of s. Every a in S is [t] for some closed term t, and hence (by the previous claim) [c] for some constant c. Hence v |= psi(c). From quantifier axiom, v |= Ex psi v |= Ex psi => v |= psi(c_phi) by Henkin axiom, so I |= psi(c_phi) so I |= Ex psi. From the claim, it follows that I |= X ---------------------------------------------------------------------- Lemma: Finite satisfiability X a set of FO sentences over L. X is satisfiable iff every finite subset Y of X is satisfiable (<==) Every finite subset Y of X is FO satisfiable. To show that X is FO satisfiable. Sufficient to show that X U Phi_H U Phi_Q U Phi_Eq is propositionally satisfiable. Sufficient to show that every finite subset of X U Phi_H U Phi_Q U Phi_Eq is propositionally satisfiable If Y is FO satisfiable, then Y U Phi_H U Phi_Q U Phi_Eq is propositionally satisfiable Each finite subset of X U Phi_H U Phi_Q U Phi_Eq is contained in some Y U Phi_H U Phi_Q U Phi_Eq, hence satisfiable. So the result follows. ---------------------------------------------------------------------- Theorem: Compactness X |= phi iff Y |= phi for some finite Y subset of X X U {~phi} is not satisfiable, so there is finite Y such that Y U {~phi} is not satisfiable, so Y |= A ---------------------------------------------------------------------- Theorem: Lowenheim-Skolem (downward) 1. If L is finite or countable, then X is satisfiable iff X is satisfiable in a countable structure. 2. If L is not countable, X is satisfiable in a structure of cardinality bounded by cardinality of L Proof: 1. From the size of the witnessing expansion Countable union of countable sets is countable. 2. Similar argument for larger cardinalities ---------------------------------------------------------------------- Corollary: No finite/countable language can axiomatize the real numbers (upto isomorphism). i.e., any countable set of axioms that is satisfied over the reals will also be satisfied in some countable structure. ---------------------------------------------------------------------- Finiteness of structures cannot be captured in FO. Theorem: If X has arbitrarily large finite models (for all n in N there is a model of at least size n), then X has an infinite model. Proof Let Y = X U {phi_>=n | n >= 2}. Every finite subset Y_0 of Y is satisfiable by assumption. By finite satisfiability, Y has a model. This must be infinite, so X has an infinite model. ---------------------------------------------------------------------- Theorem: Upward Lowenheim-Skolem If X has an infinite model, then for any set A, X has a model of cardinality at least the size of A (i.e., there exists an injection from A to the underlying set of the structure.) Proof Add constants c_a for a in A Y = X U { ~(c_a = c_b) | a =/= b} Every finite subset Y_0 of Y is satisfiable. Y_0 has a finite set of inequalities ~(c_a = c_b). Map these constants to distinct elements in the structure and map the remaining constants arbitrarily. By finite satisfiability, Y is satifiable. Each constant c_a, a in A, is mapped to a distinct element of the underlying structure for Y. The interpreretation for the constants c_a, a in A, defines an injection from A to the underlying set. ======================================================================