In this question, the professor asks us in parts i through iii to prove using truth tables that:
i. $\neg (p \lor q) \Leftrightarrow (\neg p) \land (\neg q)$
ii. $\neg (p \land q) \Leftrightarrow (\neg p) \lor (\neg q)$
iii. $\neg (\neg p) \Leftrightarrow p$
Then in part iv. he changes the game and asks us to "use (i-iii) to derive (by the use of a succession of $\Leftrightarrow$'s)" that $p \lor q \Leftrightarrow (\neg p) \rightarrow q$. That is, a truth table is now specifically not allowed.
I am just stuck. I get that statements i. through iii. are now "tools" that we can use to prove that $p \lor q \Leftrightarrow (\neg p) \rightarrow q$. I understand that $p \lor q$ means one or both must be true, so if one is false (i.e. $\neg p$), then the other must be true. But I don't see how I am supposed to prove that a statement is equivalent to an implication using i. through iii. if none of these statements i. through iii. involve an implication!
I tried submitting this combination of $\Leftrightarrow$ statements and exposition:
"$p \lor q \Leftrightarrow \neg (\neg (p \land q)) \Leftrightarrow \neg((\neg p) \land (\neg q))$ where this last statement means that $p$ and $q$ cannot be simultaneously false, so if $p$ is false, then $q$ must be true,"
However, the professor said there's an easier way, and I should be able to prove the statement using ONLY $\Leftrightarrow$ statements, and no exposition. I feel like I'm just missing something obvious. Or maybe he is assuming we're going to use outside knowledge and not JUST statements i. through iii.?
EDIT: In case anyone ever comes across this, the answer is that you are to use a combination of the rules i. through iii. as described, and also common, outside facts. So you have $(p \lor q) \equiv \neg(\neg p) \lor q \equiv \neg p \rightarrow q$.