Math+150++F09-+Hwk+3+Solutions

Math+150++F09-+Hwk+3+Solutions - SOLUTIONS TO HOMEWORK 3...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

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

Unformatted text preview: SOLUTIONS TO HOMEWORK 3 MATH 150, FALL 09 Problem 1. Section 1.5/ Exercise 12 { , > , } is not complete. To show that, we claim that for every with one sentence symbol A , and connective symbols among , > , , either is a tautology, a contradiction, or A | = . The proof, as usual, is by induction on : (1) If is A , or > , or , then the conclusion is clear. (2) If = > , then is equivalent to . Using the induction hypoth- esis for , we get the conclusion for . (3) If = , then is equivalent to , and so is a contradiction. (4) If = , then we have a the following cases: if both and are tautologies, then so is . if at least one of and is a contradiction, then so is . if A | = and either A | = or is a tautology, then A | = . if A | = and either A | = or is a tautology, then A | = ....
View Full Document

Page1 / 2

Math+150++F09-+Hwk+3+Solutions - SOLUTIONS TO HOMEWORK 3...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online