InterviewSolution
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Negate (i) p → (q ∧ r) (ii) q ∨ [~(p ∧ r)] (iii) (p → q)∧( q → p/q) (iv) p→ (q ∧ ~r) |
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Answer» (i) ~ [p → (p ∧ r)] = p ∧~(q ∧ r) ≡ p ∧ (~q ∨ ~r) (ii) ~ [q ∨ (~ (p ∧ r)] ≡ ~q ∧ ~[~ (p ∧ r) ≡ ~q ∧ (p ∧ r) (iii) ~[(p →q) ∧ (q → p)] ≡ ~ (p →q) ∨ ~(q → p) ≡ (p ∧ ~q) ∨ (q ∧ ~p) (iv) ~[p → (q^~r)] = p ∧ ~(q ∧ ~r) = p ∧ [(-q ∨ ~(~r)] = p ∧(~q ∨ r) |
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