Question #171390

Use the properties to verify the logical equivalences in the following. Supply a reason for each

step.

a. (p ∧∼ q) ∨ p ≡ p

b. p ∧ (∼ q ∨ p) ≡ p

c. ∼ (p ∨∼ q) ∨ (∼ p ∧∼ q) ≡ ∼ p

d. ∼ ((∼ p ∧ q) ∨ (∼ p ∧∼ q)) ∨ (p ∧ q) ≡ p

e. (p ∧ (∼ (∼ p ∨ q))) ∨ (p ∧ q) ≡ p


1
Expert's answer
2021-03-16T07:35:11-0400

a. (pq)p(p \land \sim q) \lor p \equiv [By the Absorption law] (pq)(p(pq))\equiv (p \land \sim q) \lor (p \lor (p \land q))\equiv [By the Associative and Commutative laws] ((pq)(pq))p)\equiv ((p \land \sim q) \lor (p \land q)) \lor p ) \equiv [By the Distributive law] ((p(qq))p)\equiv ((p \land (\sim q \lor q)) \lor p) \equiv [By the Negation law] ((pt)p)\equiv ((p \land t) \lor p) \equiv [By the Identity law] pp\equiv p \lor p \equiv [By the Idempotent law] p\equiv p

b. p(qp)p \land (\sim q \lor p) \equiv [By the Distributive law] (pq)(pp)\equiv (p \land \sim q)\lor (p \land p) \equiv [By the Idempotent law] (pq)p\equiv (p \land \sim q)\lor p \equiv [Equivalent to expression a.] p\equiv p

c. (pq)(pq)\sim (p \lor \sim q) \lor (\sim p \land \sim q) \equiv [By the De Morgan's law] (pq)(pq)\equiv \sim (p \lor \sim q) \lor \sim (p \lor q) \equiv [By the De Morgan's law] ((pq) (pq))\equiv \sim ((p \lor \sim q) \land \ (p \lor q)) \equiv [By the Distributive law] (p(qq))\equiv \sim (p \lor (\sim q \land q)) \equiv [By the Negation law] (pc)\equiv \sim (p \lor c) \equiv [By the Identity law] p\equiv \sim p

d. ((pq)(pq))(pq)\sim ((\sim p \land q) \lor (\sim p \land \sim q)) \lor (p \land q) \equiv [By the Distributive law] \equiv (p(qq))(pq)\sim (\sim p \land (q \lor \sim q)) \lor (p \land q) \equiv [By the Negation law] (pt)(pq)\equiv \sim (\sim p \land t) \lor (p \land q) \equiv [By the Identity law] p(pq)\equiv p \lor (p \land q) \equiv [By the Absorption law] p\equiv p

e. (p((pq)))(pq)( p \land (\sim (\sim p \lor q))) \lor (p \land q) \equiv [By the De Morgan's law] (p(pq))(pq)\equiv ( p \land ( p \land \sim q)) \lor (p \land q) \equiv [By the Associative law] \equiv ((pp)q)(pq)(( p \land p) \land \sim q) \lor (p \land q) \equiv [By the Idempotent law] (pq)(pq)\equiv ( p \land \sim q) \lor (p \land q) \equiv [By the Distributive law] p(qq)\equiv p \land (\sim q \lor q) \equiv [By the Negation law] pt\equiv p \land t \equiv [By the Identity law] p\equiv p


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS