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B
(p∧∼q)∨∼p
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C
p∧∼(q∨∼p)
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D
None of these
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Solution
The correct option is B(p∧∼q)∨∼p Dual of the statement is obtained by replacing ∨ with ∧, ∧ with ∨, each T with F and each F with T but no need to change the negation.