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Question
Using rules in logic, prove the following:
p ↔ q ≡ ∼(p ∧ ∼q) ∧ ∼(q ∧ ∼p)
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Solution
(p ↔ q) ≡ (p → q) ∧ (q → p)
p ↔ q ≡ (∼ p ∨ q) ∧ (∼ q ∨ p)] .......(Conditional Law)
p ↔ q ≡ ∼ (p ∧ ∼ q) ∧ ∼ (q ∧ ∼p)
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