In propositional language 𝑃 ↔ 𝑄 is equivalent to (where ∼ denotes NOT) :

Derivation: Start from the biconditional and rewrite implications.

P↔Q ≡ (P→Q) ∧ (Q→P).

Replace each implication using P→Q ≡ ¬P ∨ Q and Q→P ≡ ¬Q ∨ P to get:

(¬P ∨ Q) ∧ (¬Q ∨ P)

Therefore the biconditional is equivalent to (¬P ∨ Q) ∧ (¬Q ∨ P).

Why the other expressions are not equivalent:

• The expression ¬(P∨Q) ∧ ¬(Q∨P) simplifies to ¬P ∧ ¬Q, which is true only when both P and Q are false.

• The expression (P∨Q) ∧ (Q∨P) simplifies to P∨Q, which is true when at least one of P or Q is true; it does not require P and Q to have the same truth value.

• The implication ¬(P∨Q) → ¬(Q∨P) has identical antecedent and consequent, so it is always true (a tautology); the biconditional is not a tautology.