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a
(p⇒q)⇔(~q⇒∼p) is a contradiction
b
p∨(~p) is a tautology
c
~(~p)⇔p is a tautology
d
p∧(~p) is a contradiction
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Correct option is A
p⇒q is logically equivalent to ~q⇒∼p, therefore, (p⇒q)⇔(~q⇒∼p) is a tautology but not a contradiction So (a) is false.Similar Questions
The false statement in the following is
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