# 40.12 - A More Straightforward Proof of Metatheorem 40.12...

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Unformatted text preview: A More Straightforward Proof of Metatheorem 40.12 Michael Caie 03/20/07 40.12: Let I be an arbitrary interpretation with domain D . Let A be an arbitrary wff. Let s and s be two sequences such that, for each free variable v in A , if v is the k th variable in the fixed enumeration of the variables of Q, then s and s have the same member of D for their k th terms. Then s satisfies A iff s does. Proof. By induction on the complexity of A , i.e. , on the number n of connectives and quantifiers in A . Basis Step: n = 0. There are two possible cases. 1. A is a propositional symbol p . This is trivial. Given any interpretation I and a propositional symbol p either every sequence satisfies p or every sequence fails to satisfy p . 2. A is of the form Ft 1 ,...t n . We know that s satisfies A iff h t 1 ?s,...,t n ?s i ∈ I(F) and s satisfies A iff h t 1 ?s ,...,t n ?s i ∈ I(F) . We prove that for each t i in A , that t i ?s = t i ?s , and therefore that s satisfies A iff s satisfies A . The proof....
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