The Wikipedia page on the distributive property claims one should be able to distribute implication over implication (Distribution of implication):
$$ P \rightarrow (Q \rightarrow R) \equiv (P \rightarrow Q) \rightarrow (P \rightarrow R) $$
I verified it using a truth table, but I am unable to prove it algebraically. Here's what I have so far:
$$ \begin{align} P \rightarrow (Q \rightarrow R) &\equiv P \rightarrow (\lnot Q \lor R) \tag{Material Implication} \\ &\equiv \lnot P \lor (\lnot Q \lor R) \tag{Material Implication} \\ &\equiv (\lnot P \lor \lnot P) \lor (\lnot Q \lor R) \tag{Idempotence} \\ &\equiv (\lnot P \lor \lnot Q) \lor (\lnot P \lor R) \tag{Associativity} \\ &\equiv \lnot (P \land Q) \lor (\lnot P \lor R) \tag{DeMorgan's} \\ &\equiv \lnot (P \land Q) \lor (P \rightarrow R) \tag{Material Implication} \\ &\equiv (P \land Q) \rightarrow (P \rightarrow R) \tag{Material Implication} \\ \end{align} $$
I don't know the correct way to state the Idempotence step, but I'm using $P \equiv P \lor P$. I'm trying to avoid using Distribution of disjunction over disjunction, also advertised on the same Wikipedia page, but the effect is the same.
Have I made any mistakes? How do I proceed from here?
It's probably easier to start with the right hand side:
$(P \to Q) \to (P \to R) \equiv$
$\neg (\neg P \lor Q) \lor \neg P \lor R \equiv$
$(P \land \neg Q) \lor \neg P \lor R \equiv$
$((P \lor \neg P) \land (\neg Q \lor \neg P)) \lor R \equiv$
$(\top \land (\neg Q \lor \neg P)) \lor R \equiv$
$\neg Q \lor \neg P \lor R \equiv$
$\neg P \lor \neg Q \lor R \equiv$
$P \to (Q \to R)$