Lecture 4 Notes

# But t is consistent thus it must be that t a 2 suppose

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Unformatted text preview: t;. Suppose I A. Now: T (∀xim)Fm(xim) → Fm(cm). (K5, cm is free for xim in Fm(xim), since cm doesn't occur in Fm(xim).) So: I (∀xim)Fm(xim) → Fm(cm). (Prop. 4.4. - axioms are logically valid.) Hence: I Fm(cm). (Prop. 3.26.) Thus: T Fm(cm). (Inductive Hypothesis.) Now: Suppose T A. Then: T ∼A. (T is complete.) Or T ∼(∀xim)Fm(xim). But: T ∼(∀xim)Fm(xim) → ∼Fm(cm). (Gm is an axiom of T.) So: T ∼Fm(cm). But T is consistent. Thus it must be that T A. 2. "⇒". Suppose T A. Now suppose I A. Then: There's a valuation in I that doesn't satisfy A. So: There's a valuation v that doesn't satisfy Fm(xim). Now: v (x im) = d, for some closed term d in DI. And: v(d) = d. (Valuations map constants to constants; hence closed terms to closed terms.) So: v (x im) = v(d). Now: We have the following: 1. Fm(xim) is a wf with xim free. 2. d is a (closed) term free for xim in Fm(xim). 3. v (x im) = v(d). 4. v is i-equivalent to itself. Thus: v satisfies Fm(d) iff v satisfies Fm(xim). (Prop. 3.23.) Hence: v does not satisfy Fm(d). Thus: I Fm(d) . Now: T (∀xim)Fm(xim). (assumption T A.) So: T Fm(d). (K5, d is free for xim in Fm(xim), and MP.) Hence: I Fm(d) . (Inductive Hypothesis.) So it must be that I A. Part III. Lemma 5: For any (open or closed) wf B of L, if S B, then I B. Proof: Suppose S B, for some wf B of L. If B is closed, then T B, hence I B. (Lemma 4: If B is a closed wf of L, it is also a closed wf of L+ .) Suppose B is open. Then: S B'. (Prop. 4.19, B' is the universal closure of B.) Hence: T B '. Thus: I B'. (Lemma 4.) Hence: I B. (Cor. 3.28.)...
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