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Statement$-I :$ $\sim (p\leftrightarrow q)$ is equivalent to $(p\wedge \sim q)\vee \sim (p\vee \sim q) .$
Statement$-II :$ $p\rightarrow (p\rightarrow q)$ is a tautology.
Statement$-1$ is True, Statement$-2$ is True; Statement$-2$ is a correct explanation for Statement$-1.$
Statement$-1$ is True, Statement$-2$ is True; Statement$-2$ is NOT a correct explanation for Statement$-1.$
Statement$-1$ is True, Statement$-2$ is False.
Statement$-1$ and Statement$-2$ both are False
Solution
$ \sim \left( {p \leftrightarrow q} \right) \equiv \sim \left( {\left( {p \to q} \right) \wedge \left( {q \to p} \right)} \right)$
$ \equiv \sim \left( {p \to q} \right) \vee \sim \left( {q \to p} \right)$
$ \equiv \left( {p \wedge – q} \right) \vee \sim \left( { \sim q \vee p} \right)$
and $p \to \left( {p + q} \right) \equiv p \to \left( { \sim p \vee q} \right)$ is not a tautology.
