About (P=>Q) = ((not P) or Q)

Asked May 19 '21 at 07:03

Active May 19 '21 at 07:34

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Is (P=>Q) = ((not P) or Q) a definition or a convention or a theorem?

If it's a theorem, how to prove it without the truth table

Can accept the following reasoning?

let's show that from P => Q we can deduce ((not P) or Q)

Suppose P => Q, By the absurd, if not ((not P) or Q). We, therefore, have P and (not Q). Then, because P and (P => Q), implies Q. We get both Q and not Q, ie a contradiction.

Reciprocally,

let's show that from ((not P) or Q) we can deduce P => Q

suppose Q. Then P => (P and Q) => Q, so P => Q
suppose not P. Then (not Q) => ((not Q) and (not P)) => (not P), and finally P => Q

Finally from ((not P) or Q), we can deduce (P => Q)

edited May 19 '21 at 07:34

Mauro ALLEGRANZA

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asked May 19 '21 at 07:03

Pascal

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https://math.stackexchange.com/q/3583323 – Jean Marie May 19 '21 at 07:18
Your proof is correct, but note that this is just a more cumbersome way of stating the same you would see in a truth table. – David May 19 '21 at 07:25
David I wanted to avoid the truth table because I had to be convinced of why does false imply everything – Pascal May 19 '21 at 07:32
can be a theorem or a definition – Mauro ALLEGRANZA May 19 '21 at 07:34
see also this post as well as this one – Mauro ALLEGRANZA May 19 '21 at 07:36
See Material Implication. There is a simple proof of it – Mauro ALLEGRANZA May 19 '21 at 10:18
Thank's for all. I read the links, there are those who take it for a definition and others for a theorem – Pascal May 19 '21 at 12:49

About (P=>Q) = ((not P) or Q)

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