Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If we show that a claim is equivalent to a tautology (which is stronger than showing the claim implies a tautology), how come that isn't a valid method of proof?

share|improve this question
2  
Showing that the claim implies a tautology doesn't tell you anything, since a tautology should be true anyhow. It is sufficient to show that your claim follows from a tautology, so if you show that your claim is equivalent to a tautology, you have shown that it is true. –  Brett Frankel Nov 4 '12 at 3:21
    
What text or class did this come from? –  Doug Spoonwood Nov 4 '12 at 3:29
add comment

2 Answers

If $T$ is a tautology, $(P\Rightarrow Q)\Leftrightarrow T$ is enough to prove $P\Rightarrow Q$, but it's overkill. All you need is $(P\Rightarrow Q)\Leftarrow T$. $(P\Rightarrow Q)\Rightarrow T$ is always true because $T$ is a tautology - it holds whether $P\Rightarrow Q$ is true or not, so it is a tautology in and of itself. On the other hand, $(P\Rightarrow Q)\Leftarrow T$ is only true when $P\Rightarrow Q$ is true, and in fact is equivalent to $P\Rightarrow Q$.

share|improve this answer
add comment

Who says it isn't valid? In what context? For what purpose?

In some (actually most) mathematical contexts, for some (actually most) mathematical purposes, demonstrating that a formula is an instance of a tautology is a perfectly cromulent method of proving it.

Among the exceptions is if the context is a course in formal logic and the purpose is to gain or show familiarity with a particular formal proof system and how it works. In that (fairly narrow) circumstance, appealing to tautology obviously misses the point, at least until you have formally proved in general that every tautology has a proof in the proof system at hand.

share|improve this answer
1  
+1 for apparently producing the first-ever instance of "cromulent" on this site. –  Rick Decker Nov 4 '12 at 19:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.