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Question
Using rules in logic, prove the following:
∼ (p ∨ q) ∨ (∼p ∧ q) ≡ ∼p
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Solution
∼ (p ∨ q) ∨ (∼p ∧ q)
≡ (∼p ∧ ∼q) ∨ (∼p ∧ q) ..........(Negation of disjunction)
≡ ∼p ∧ (∼q ∨ q) ........(Distributive Law)
≡ ∼p ∧ T ............(Complement Law)
≡ ∼p ............(Identity Law)
∴ ∼ (p ∨ q) ∨ (∼p ∧ q) ≡ ∼p
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