logical equivalence proof using propositional algebra

Question

i need help to show that (p⇔r)⇒(q⇔r) is equivalent to ∼[(∼p∨r)∧(p∨ ∼r)]∨[(∼q∨r)∧(q∨ ∼r) using propositional algebra. I did it using truth tables but i am struggling with propositional algebra.

Answer 1

1) $p \Rightarrow q$ is true except when $p$ is true and $q$ is false, so it is logically equivalent to $\neg p \lor q$.

2) $p \iff q$ is true except when one is true and the other is false, so it is logically equivalent to $(p \land q) \lor (\neg p \land \neg q)$.

Using these ideas, we have

$$(p \iff r) \implies (q \iff r) \equiv \neg[(p \land r) \lor (\neg p \land \neg r)] \lor [(q \land r) \lor (\neg q \land \neg r)]$$

and so we want to show that

$$\neg[(p \land r) \lor (\neg p \land \neg r)] \lor [(q \land r) \lor (\neg q \land \neg r)]$$

is equivalent to (guessing where your mismatched bracket goes)

$$\neg[(\neg p \lor r)\land(p \lor \neg r)]\lor[(\neg q \lor r) \land (q \lor \neg r)]$$