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B
A contradiction
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C
Both a tautology and a contradiction
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D
Neither a tautology nor a contradiction
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Solution
The correct option is B A contradiction (p∧∼q)∧(∼p∧q)=(p∧∼p)∧(∼q∧q) =f∧f=f (By using associative laws and commutative laws) ∴(p∧∼q)∧(∼p∧q) is a contradiction.