Let p, q and r be statements. Use the Laws of Logical Equivalence and the equivalence of → to a disjunction to show that: ∼ ((p ∨ (q →∼ r)) ∧ (r → (p∨ ∼ q))) ≡ (∼ p ∧ q) ∧ r.

Let us show that $∼ ((p ∨ (q →∼ r)) ∧ (r → (p∨ ∼ q))) ≡ (∼ p ∧ q) ∧ r:$

$∼ ((p ∨ (q →∼ r)) ∧ (r → (p∨ ∼ q))) \\ ≡ ∼ ((p ∨ (\sim q \lor ∼ r)) ∧ (\sim r \lor (p∨ ∼ q)))\\ ≡ ∼ ((p ∨ (∼ r \lor \sim q)) ∧ (\sim r \lor (p∨ ∼ q)))\\ ≡ ∼ ((p ∨ ∼ r) \lor \sim q) ∧ (\sim r \lor (p∨ ∼ q)))\\ ≡ ∼ ((∼ r ∨ p) \lor \sim q) ∧ (\sim r \lor (p∨ ∼ q)))\\ ≡ ∼ ((∼ r ∨( p \lor \sim q)) ∧ (\sim r \lor (p∨ ∼ q)))\\ ≡ ∼ (∼ r ∨( p \lor \sim q)) \\ ≡ ∼ (∼ r )\land \sim( p \lor \sim q)) \\ ≡ r \land( \sim p \land\sim( \sim q))) \\ ≡ r \land( \sim p \land q)) \\ ≡ ( \sim p \land q)) \land r.$