1 |(A --> B) & ~B assumption
2 |A --> B from 1
3 |~B from 1
4 |~A v B equivalent to 2
5 |~A follows from 3 & 4
6 ((A --> B) & ~B) --> ~A from 1-5
Q.E.D.
2 |A --> B from 1
3 |~B from 1
4 |~A v B equivalent to 2
5 |~A follows from 3 & 4
6 ((A --> B) & ~B) --> ~A from 1-5
Q.E.D.
posted by:
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Cleveland
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Re: proof of modus tollens
Sun, May 9, 2004 - 5:04 PMUri,
What I meant when I said that I don't remember the proofs, is that I don't remember if tollens and ponens are interchangeable. That is, can you derive one from the other.
You'd need to prove either:
(((a -> b) & ~b) -> ~a ) -> (((a -> b) & a) -> b)
Or the opposite direction.
At the moment it seems to me unattainable except by using tollens to prove it (i.e., negating the negation). However, I haven't had much time recently. I'll look at this again when I have some time.
Noam -
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Re: proof of modus tollens
Sun, May 9, 2004 - 5:20 PMbut i proved tollens from no premises, from which it follows that it can be proved from ponens. adding premises can't make a valid argument invalid, at least in classical logic.
i haven't shown that ponens can be proved from no premises, but it's a fact. therefore ponens and tollens follow from each other.
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