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प्रश्न
Without using the truth table, show that p ∧ [(∼p ∨ q) ∨ ∼q] ≡ p
योग
उत्तर
p ∧ [(∼p ∨ q) ∨ ∼q]
≡ p ∧ (∼ p ∨ q) ∨ ( p ∧ ∼ q) .......(By distributive property)
≡ (p ∧ ∼ p) ∨ (p ∧ q) ∨ (p ∧ ∼ q) .......(By identity law)
≡ ( p ∧ q) ∨ (p ∧ ∼ q) ......(By distributive law)
≡ p ∧ (q ∨ ∼ q)
≡ p ∧ t (By complement law)
≡ p
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Indefinite Integration - Definition of an Integral
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