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