Lecture 4 Notes

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

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: 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...
View Full Document

Ask a homework question - tutors are online