Proof writing involving propositional logic: (p∧ (¬q) ) ↔ ((¬p) ∧ q) ≡ p ↔ q

Question

Prove by using propositional logic:

(p∧ (¬q) ) ↔ ((¬p) ∧ q) ≡ p ↔ q

Is this possible? I tried solving but i get stuck.

LS:

= (p∧ (¬q) ) ↔ ((¬p) ∧ q)

= ((p∧ (¬q) ) → ((¬p) ∧ q) ) ∧ ( ((¬p) ∧ q) → (p∧ (¬q) ))

= (¬(p∧ (¬q) ) ∨ ((¬p) ∧ q)) ∧ ¬((¬p) ∧ q) ∨ (p∧ (¬q) )

= ((¬p) ∨ q) ∨ ((¬p) ∧ q)) ∧ ((p ∨ (¬q)) ∨ (p∧ (¬q) ))

= (¬p ∨ ((¬p) ∧ q)) ∨ (q ∨ ((¬p) ∧ q))) ∧ ((p ∨ (p∧ (¬q) )) ∨ ((¬q) ∨ (p∧ (¬q)))

= (T ∧ (¬p ∨ q) ∨ (q ∨ ¬p) ∧ q ) ∧ (p ∧ (p ∨ (¬q) ) ∨ (q v p) ∧ T)

= ((¬p ∨ q) ∨ ((q ∨ ¬p) ∧ q) ) ∧ ( (p ∧ (p ∨ (¬q) )) ∨ (q v p))

=...

I assume this would be true since ( p ∨ q) ↔ ( p ∧ q ) ≡ p ↔ q is true. Proved that already. What you guys think? Is it possible to keep breaking this statement down till p ↔ q ? am i missing some identities?

Edit: worked a truth table so both statements are equivalent. However, I need to prove this by propositional logic. Any help?

Answer 1

\begin{align} LHS&\equiv((p∧ ¬q)\to(¬p ∧ q)) \land ((¬p ∧ q)\to(p∧ ¬q)) \\ &\equiv ((\neg p\lor q)\lor(¬p ∧ q)) \land ((p \lor \neg q)\lor(p∧ ¬q)) \\ &\equiv (((\neg p\lor q)\lor¬p)\land ((\neg p\lor q)\lor q)) \land (((p \lor \neg q)\lor p)\land ((p \lor \neg q)\lor ¬q)) \\ &\equiv ((\neg p\lor q)\land (\neg p\lor q)) \land ((p \lor \neg q)\land ((p \lor \neg q)) \\ &\equiv (\neg p\lor q) \land (p \lor \neg q) \\ &\equiv (p\to q) \land (q\to p) \\ &\equiv (p↔ q) \end{align}