The statement (p⇒q)⇔(~p∧q) is a
Tautology
contradiction
Neither A nor B
none of these
Explanation for correct option:
Given statement, (p⇒q)⇔(~p∧q)
Make a truth table
pq~pp⇒q~p∧q(p⇒q)⇔(~p∧q)TTFTFFTFFFFTFTTTTTFFTTFF
(p⇒q)⇔(~p∧q) is neither a contradiction nor tautology.
Hence, option (C) is the correct answer.
Choose the correct option.
Statement 1:(p ∧∼ q)∧(∼ p∧q) is a fallacy.
Statement 2:(p⇒q)⇔(∼ q⇒∼ p) is a tautology.