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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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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 $$ (p\text{ and not }q)\to \text{contradiction / absurdity} $$

Arthur
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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:

  1. Assuming $p$ is true, and then proving that $q$ must also be true.

  2. Assuming $q$ is false, and then proving that $p$ must also be false.

  3. Assuming $p$ is true and $q$ is false, and then obtainimg a contradiction of the forms $r\land \neg r$ or $r \iff \neg r$

  4. Proving $p$ is false. (Then there is no need to prove anything about $q$.)

  5. Proving $q$ is true. (Then there is no need to prove anything about $p$.)