Question #227156

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.


1
Expert's answer
2021-08-18T16:10:32-0400

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


((p(qr))(r(pq)))((p(qr))(r(pq)))((p(rq))(r(pq)))((pr)q)(r(pq)))((rp)q)(r(pq)))((r(pq))(r(pq)))(r(pq))(r)(pq))r(p(q)))r(pq))(pq))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.



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