proving logical equivalence involving biconditional

Question

$$\neg(p\iff q) \equiv p\iff\neg q$$

I am having problem proving this statement. I can prove it by a truth table or a diagram, but I can't prove it by logically (like using symbols like this).

Answer 1

$$\neg (p \leftrightarrow q) \equiv \text{ (use $p \leftrightarrow q \equiv (p \land q) \lor (\neg p \land \neg q)$}$$

$$\neg ((p \land q) \lor (\neg p \land \neg q)) \equiv \text{(DeMorgan)}$$

$$\neg (p \land q) \land \neg (\neg p \land \neg q) \equiv \text{(DeMorgan x 2)}$$

$$(\neg p \lor \neg q) \land (\neg \neg p \lor \neg \neg q) \equiv \text{(Double Negation)}$$

$$(\neg p \lor \neg q) \land (p \lor \neg \neg q) \equiv \text{(Commutation)}$$

$$(\neg p \lor \neg q) \land (\neg \neg q \lor p) \equiv \text{(Implication x 2)}$$

$$(p \rightarrow \neg q) \land (\neg q \rightarrow p) \equiv $$

$$p \leftrightarrow \neg q$$