Does that mean to prove $p\rightarrow q$ is a true statement? Then since when $p$ is false, $p\rightarrow q$ is vacuously true, do I only have to prove $q$ is true when $p$ is true?
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8You need to show that there is no situation where $p$ is true and $q$ is false. – Rushabh Mehta Jan 23 '20 at 22:17
2 Answers
That is the gist of it, yeah.
There are, in practice, several ways to do this, and here is a short summary. A direct proof uses intermediate, already-known implications chained together like this. $$ p\to p_1\\ p_1\to p_2\\ \vdots\\ p_n\to q $$
A contrapositive proof is a direct proof of the statement $$ \text{not }q\to\text{not }p $$ and a proof by contradiction is a direct proof of the statement $$ (p\text{ and not }q)\to \text{contradiction / absurdity} $$
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In classical logic, $p \implies q$ means only that it is false that both $p$ is true and $q$ is false.
$p \implies q \space \space \equiv \space\space \neg (p \land \neg q)$
We can prove $p\implies q$ by either:
Assuming $p$ is true, and then proving that $q$ must also be true.
Assuming $q$ is false, and then proving that $p$ must also be false.
Assuming $p$ is true and $q$ is false, and then obtainimg a contradiction of the forms $r\land \neg r$ or $r \iff \neg r$
Proving $p$ is false. (Then there is no need to prove anything about $q$.)
Proving $q$ is true. (Then there is no need to prove anything about $p$.)
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