0

Our teacher gave us a truth table regarding when p implies q but didn't give one for when q implies p. So, it was a little confusing when the teacher asked for q implies p.

"If 1+1=2, then a square is a circle." Obviously, p is true and q is false. And with the table given, p implies q is evidently false. What would then q implies p be?

In the teacher's method, they first converted it to true and false then basically determined the answer by "true implies false" and "false implies true." If p is true and q is false, then p implies q is false. But if p is false and q is true, then p implies q is true. We now do the converse. If p is true and q is false, then what is q implies p? And if p is false and q is true, then what is q implies p?

Would the order of p and q really won't matter as long as they are converted to their truth values? Like p implies q and q implies p will not matter as long as the first ones are of the same truth value and as such of the second ones. Why even consider p is the hypothesis and q is the conclusion in this matter if they are only to be made no use of and is completely ignored?

Ralph Louie Gatan
  • 1
  • The truth table for "p implies q" covers ALL four cases: specifically, the case F-T will result T. Thus, nothing change with "q implies p". – Mauro ALLEGRANZA Aug 30 '23 at 14:14
  • $(p \implies q) \iff [(\neg p) \vee q].$ – user2661923 Aug 30 '23 at 14:58
  • 1
    Once you have the truth-table for $p \to q$, then the truth-table for $q \to p$ immediately follows. That is, the truth-table for $p \to q$ tells you the entries for $True \to True$, $True \to False$, $False \to True$, and $False \to False$. So with those, you can fill in the table for $q \to p$ as well, since there you are dealing with those very 4 cases as well ... you just need to make sure you put them in the right order, e.g in the row where $p$ is True and $q$ is False, $p \to q$ is evaluated as $True \to False$, while $q \to p$ would be evaluated as $False \to True$ – Bram28 Aug 30 '23 at 18:16

0 Answers0