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The statement  ~(p↔∼q) is

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By Expert Faculty of Sri Chaitanya
a
equivalent to  (~p→q)
b
a tautology
c
a fallacy
d
equivalent to  p↔q

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detailed solution

Correct option is D

~(p↔∼q)≡∼((p→∼q)∧(~q→p)≡∼((~p∨~q)∧(q∨p))≡∼(~q∨~p)∨(~(q∨p))≡(~(q)∧~(~p))∨(~q∧~p)≡(q∧p)∨(~q∧~p)≡[(q∧p)∨(~q)]∧[(q∧p)∨(~p)]≡[(q∨(~q))∧(p∨(~q))]∧[(q∨(~p)) ∧(p∨(~p))]≡[t∧(~q∨p)]∧[((~p)∨q))∧t]≡(~q∨p)∧(~p∨q)≡(q→p)∧(p→q)≡p↔qAlternative Solution. Use the following table. From the last two colums, we get t~(p↔(~q))≡p↔q


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