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प्रश्न
Without using truth table prove that:
∼ [(p ∨ ∼ q) → (p ∧ ∼ q)] ≡ (p ∨ ∼ q) ∧ (∼ p ∨ q)
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उत्तर
L.H.S. = ∼ [(p ∨ ∼ q) → (p ∧ ∼ q)]
≡ (p ∨ ∼ q) → (p ∧ ∼ q) ...(Apply Negation of Implication Law)
≡ (p ∨ ∼ q) ∧ [∼ p ∨ ∼ (∼ q)] ...(Apply De Morgan’s Law)
≡ (p ∨ ∼ q) ∧ (∼ p ∨ q) ...(Apply Negation of Negation Law)
= R.H.S.
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