The supporting lemma: $$ ((p \land q) \Rightarrow r) \Leftrightarrow(q \Rightarrow(\neg p \lor r)) $$
\begin{align} &(p \land q) \Rightarrow r && \text{Start} \\ &p\rightarrow(q\rightarrow r) && \text{given (outside source)} \\ &\neg p \lor (q\rightarrow r) && \text{implication} \\ &\neg p \lor (\neg q \lor r) && \text{implication} \\ &(\neg p \lor \neg q) \lor r && \text{associative} \\ &(\neg q \lor \neg p) \lor r && \text{commutative} \\ &\neg q \lor (\neg p \lor r) && \text{associative} \\ &\neg q \lor (\neg p \lor r) && \text{associative} \\ &q \rightarrow(\neg p \lor r) && \text{implication} \end{align}
Is to be used to prove the distributive property, where the distributive property is written as:
$$ a \land (b \lor c) \Rightarrow (a \land b) \lor (a \land c) $$
$$ a \lor (b \land c) \Rightarrow (a \lor b) \land (a \lor c) $$
I feel super stupid. The start of this proof is to make the correlation between $q\implies(\lnot p\lor r)$ and $a \land (b\lor c) \implies (a\land b)\lor(a\land c)$ where q is $b \lor c$, $\lnot$p is a (where a takes the negative form), and r is $(a\land b)\lor(a\land c)$
This would give us the following equation:
$(b\lor c) \implies (\lnot a \lor(a\land b)\lor(a\land c))$
$a\rightarrow(a\land b)\lor(a\land c)$ Implication
$a$ Implication
$b\lor c$ Implication
The second part of the of the problem can be solved in a similar fashion.