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प्रश्न
Without using truth table prove that
[(p ∧ q ∧ ∼ p) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ≡ (p ∨ q) ∧ r
उत्तर
[(p ∧ q ∧ ∼ p) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r)
= [(p ∧ ∼ p) ∧ q) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ...(Associative law)
= (F ∧ q) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ...(Complement law)
= F ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ...(Identity law)
= (∼ p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ...(Identity law)
= [∼ p ∧ (q ∧ r)] ∨ [p ∧ (q ∧ r)] ∨ (p ∧ ∼ q ∧ r) ...(Associative law)
= [(∼ p ∨ p) ∧ (q ∧ r)] ∨ (p ∧ ∼ q ∧ r) ...(Distributive law)
= [T ∧ (q ∧ r)] ∨ (p ∧ ∼ q ∧ r) ...(Complement law)
= (q ∧ r) ∨ (p ∧ ∼ q ∧ r) ...(Identity law)
= [q ∨ (p ∧ ∼ q)] ∧ r ...(Distributive law)
= [(q ∨ p) ∧ (q ∨ ∼ q)] ∧ r ...(Distributive law)
= [(q ∨ p) ∧ T] ∧ r ...(Complement law)
= (q ∨ p) ∧ r ...(Identity law)
= (p ∨ q) ∧ r ...(Commutative law)
