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Question Number 46475 by peter frank last updated on 27/Oct/18

Commented by Joel578 last updated on 27/Oct/18

pls recheck question (b)

plsrecheckquestion(b)

Commented by peter frank last updated on 28/Oct/18

b)[(p→q)∧(q⇔p)∧∼p]→∼q

b)[(pq)(qp)p]→∼q

Answered by Joel578 last updated on 27/Oct/18

(p → q) ∧ (q → ∼r) ∧ r  ≡ (∼p ∨ q) ∧ [(∼q ∨ ∼r) ∧ r]  ≡ (∼p ∨ q) ∧ [(r ∧ ∼q) ∨ (r ∧ ∼r)]  ≡ (∼p ∨ q) ∧ [(r ∧ ∼q) ∨ F]  ≡ (∼p ∨ q) ∧ (r ∧ ∼q)  ≡ [(∼p ∨ q) ∧ ∼q] ∧ r  ≡ [(∼q ∧ ∼p) ∨ (∼q ∧ q)] ∧ r  ≡ [(∼q ∧ ∼p) ∨ F] ∧ r  ≡ (∼q ∧ ∼p) ∧ r    [(∼q ∧ ∼p) ∧ r] → ∼p  ≡ ∼[(∼q ∧ ∼p) ∧ r] ∨ ∼p  ≡ [(q ∨ p) ∨ ∼r] ∨ ∼p  ≡ q ∨ (p ∨ ∼p) ∨ ∼r  ≡ q ∨ T ∨ ∼r  ≡ T

(pq)(qr)r(pq)[(qr)r](pq)[(rq)(rr)](pq)[(rq)F](pq)(rq)[(pq)q]r[(qp)(qq)]r[(qp)F]r(qp)r[(qp)r]p[(qp)r]p[(qp)r]pq(pp)rqTrT

Commented by peter frank last updated on 28/Oct/18

thanks

thanks

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