Lecture 4 Notes

note sn a a a same reasoning as in above note so sn

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Unformatted text preview: plication of Gen on a free variable in Gn occurs in the deduction {Gn} Sn (∼Gn).) Note: Sn ((A → ∼A) → ∼A). (Same reasoning as in above note.) So: Sn (∼Gn). In other words, Sn ∼(∼(∀xin)Fn(xin) → ∼Fn(cn)). Note: Sn (∼(∼A → ∼B) → ∼A) and Sn (∼(∼A → ∼B) → B). (Same reasoning as in first note.) So: Sn ∼(∀xin)Fn(xin) and Sn Fn(cn). Now: In the proof of Fn(cn), we can replace all occurrences of cn with some variable y that doesn't occur in the proof. Since cn doesn't appear in any of the axioms of Sn used to derive Fn(cn), we get a proof in Sn of Fn(y)). So: Sn Fn(y). Thus: Sn (∀y)Fn(y). (Gen on y.) So: Sn (∀xin)Fn(xin). (Prop. 4.18.) But Sn was assumed consistent. Hence Sn+1 must be consistent. Lemma 3: S∞ is consistent, for S∞ the extension of S+ that has as axioms all axioms of S0, S1, ... . Proof: Suppose S∞ is not consistent. Then: There's a wf A of L+ such that S∞ A and S∞ (∼A). Note: These S∞-proofs are finite; so they use only a finite nu...
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