Lecture 4 Notes

But sn is consistent for any n hence s must be

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

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