Lecture 4 Notes

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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.)...
View Full Document

Ask a homework question - tutors are online