How to prove that $P \rightarrow Q$ is equivalent with $\neg P \lor Q $?

Layout:

To prove that $\neg P\lor Q$ is a formal consequence of $P\to Q$, start by assuming $P\to Q$ and further suppose that $\neg (\neg P\lor Q)$ holds. At this point you should prove $P\lor \neg P$ and perform $\lor$-$\text{Elim}$ on this disjunction. It's easy to find contradictions on both cases yielding $\neg \neg (\neg P\lor Q)$.

For the other direction, naturally start by assuming that $\neg P\lor Q$ holds and then assume $P$. Now start yet another subproof (within the assumption that $P$ holds) assuming that $\neg Q$ holds. At this point perform $\lor$-$\text{Elim}$ on the assumption $\neg P\lor Q$. It's easy to find a contradiction in this last subproof.

1. $P → Q$ (given)
2. $Q \lor \neg Q$ (tautology)
3. $\neg Q → \neg P$ (contrapositive of (1) )
4. from (2) and (3): $Q \lor \neg P$ (modus ponens)