Is there any situation where $(p \implies q) \iff (\neg p \lor q)$ isn't true? Looking at the truth table for implication, $p \implies q$ is only false when $p$ is true and $q$ is false.
$\neg p \lor q$ is only false when $p$ is true and $q$ is false as well.
Because of that, I would think that $(p \implies q) \iff (\neg p \lor q)$; that being said, I might be missing something.
If I have missed anything, please let me know, thank you.

The two statements are logically equivalent, as can be checked via truth tables.
For some intuition: "$p \implies q$" means "if p, then q", which you can also think of as "q only if p", or "q, or not p", which is equivalent to $\neg p \lor q$.
