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Question
Using the algebra of statement, prove that
[p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p] ≡ p
Solution
L.H.S.
= [p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p]
≡ [p ∧ (q ∨ r)] ∨ [(~ r ∧ ~ q)∧ p] ...[Associative Law]
≡ [p ∧ (q ∨ r)] ∨ [(~q ∧ ~r) ∧ p] ....[Commutative Law]
≡ [p ∧ (q ∨ r)] ∨ [~ (q ∨ r) ∧ p] ....[De Morgan’s Law]
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] .....[Commutative Law]
≡ p ∧ [(q ∨ r) ∨ ~(q ∨ r)] ....[Distributive Law]
≡ p ∧ t ......[Complement Law]
≡ p .....[Identity Law]
= R.H.S.
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