MathematicsMedium115×since 2008Q4309Statement-1 : ∼(p↔∼q)\sim \left( {p \leftrightarrow \sim q} \right)∼(p↔∼q) is equivalent to p↔q{p \leftrightarrow q}p↔q. Statement-2 : ∼(p↔∼q)\sim \left( {p \leftrightarrow \sim q} \right)∼(p↔∼q) is a tautology.AStatement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1BStatement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1CStatement-1 is true, Statement-2 is falseDStatement-1 is false, Statement-2 is trueCheck answerSkip