Lecture 4 Notes

But sn is consistent for any n hence s must be

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: mber of axioms of S∞. This means they are also Sn-proofs, where Sn is the member of the sequence that has as its axioms those that are used in these proofs. Thus: Sn A and Sn (∼A). But Sn is consistent, for any n. Hence S∞ must be consistent. Since S∞ is consistent, it has a consistent complete extension, call it T (Prop. 4.39).* Recall from the proof of Prop. 4.39 that T is constructed by again enumerating wfs and constructing a sequence of extensions. In this case, however, we enumerate all wfs of L (not just those with one free variable). And the sequence of extensions begins, in this case, with S . We then go down the list of wfs, checking to see if each is a theorem of S . If it is, we do nothing, if it isn't, we add its negation as a new axiom and get a new member of the sequence, and continue checking the list of wfs for theoremhood in the new extension, repeating * + ∞ ∞ PL 3014 - Metalogic Part II. Use T to define an interpretation I of L+ as follows: 3 1. D I = {closed...
View Full Document

This document was uploaded on 03/25/2014 for the course PL 3014 at NYU Poly.

Ask a homework question - tutors are online