Prove that (p ∧ q) ∨ (p ∧ r) is equivalent to ¬(p ∧ q) → (p ∧ r)

Question

this is how far I've got: (p ∧ (q ∨ r) ⇔ ¬(p ∧ q) → (p ∧ r)

Answer 1

Let's denote $x$ as $p\wedge q$ and $y$ as $p\wedge r$. Hence, we seek to show that $$x\vee y=\lnot x\to y$$

Note that for arbitrary $s,t$, $s\to t$ is the same as $\lnot s\vee t$. So, $\lnot x\to y$ is the same as $x\vee y$ and we are done.

Answer 2

For simplicity, denote $p\land q $ by $Q$ and $p\land r $ by $R$.

The statement to prove is then $$Q\lor R \iff \lnot Q\implies R$$

Note that the only case that the $RHS$ is false is if both $Q$ and $R$ are false and this is exactly the only case that the $LHS$ is false.

Thus the two sides are equivalent.