using truth table, prove that: ~(pv q)=~ p^~q
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p q p↔q p∧q ∼p ∼q ∼p∧∼q (p∧q)∨(∼p∧∼q)
T T T T F F F T
T F F F F T F F
F T F F T F F F
F F T F T T T T
From column 3 and 8 we get p↔q≡(p∧q)∨(∼p∧∼q)
Hence proved.
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