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Question
Show that the simplified form of (p ∧ q ∧ ∼ r) ∨ (r ∧ p ∧ q) ∨ (∼ p ∨ q) is q ∨ ∼ p.
Sum
Solution
(p ∧ q ∧ ∼ r) ∨ (r ∧ p ∧ q) ∨ (∼ p ∨ q)
= (p ∧ q ∧ ∼ r) ∨ (p ∧ q ∧ r) ∨ (∼ p ∨ q) ...(Associative law)
= (p ∧ q) ∧ [∼ r ∨ r] ∨ (∼ p ∨ q) ...(Distributive law)
= [(p ∧ q) ∧ T] ∨ (∼ p ∨ q) ...(Complement law)
= (p ∧ q) ∨ (∼ p ∨ q) ...(Identity law)
= [(p ∧ q) ∨ ∼ p] ∨ q ...(Associative law)
= [(p ∨ ∼ p) ∧ (q ∨ ∼ p)] ∨ q ...(Distribution law)
= [T ∧ (q ∨ ∼ p)] ∨ q ...(Complement law)
= (q ∨ ∼ p) ∨ q ...(Identity law)
= (q ∨ q) ∨ ∼ p ....(Associative law)
= q ∨ ∼ p
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