According to ProofWiki, the principle of explosion says that:
$$p \implies (\lnot p \implies q)$$
but sometimes it's also stated as
$$(p \land \lnot p) \implies q$$
Can one be converted to the other? How do you prove or show these?
If we know that $(p \implies q) \equiv \lnot p \lor q$ by definition, then substituting the 2nd representation:
$$((p \land \lnot p) \implies q) \equiv \lnot(p \land \lnot p) \lor q \equiv \lnot p \lor p \lor q$$
Is this just true due to law of excluded middle? Like either $p$ or $\lnot p$ is true, i.e. $p \lor \lnot p \equiv \text{T}$ before we even get to whatever $q$ is. Is that why we're able to say it?
Then the first representation is weird too:
$$(p \implies (\lnot p \implies q)) \equiv (\lnot p \lor (\lnot p \implies q)) \equiv (\lnot p \lor (p \lor q)) \equiv \lnot p \lor p \lor q$$
I mean am I even on the right track here? How do you normally get these without "working backwards" and showing that they happen to be equivalent? Is there a more intuitive derivation of both of these?