Statement-I: sim (p leftrightarrow q) is equi | vedclass

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(C) For Statement-$I$: $\sim (p \leftrightarrow q) \equiv \sim ((p$ $\rightarrow q) \wedge (q$ $\rightarrow p))$$\equiv \sim (p$ $\rightarrow q) \vee

Vedclass · Accepted Answer

Statement-$I$ is True,Statement-$II$ is False.

Vedclass · Answer

(C) For Statement-$I$: $\sim (p \leftrightarrow q) \equiv \sim ((p$ $\rightarrow q) \wedge (q$ $\rightarrow p))$$\equiv \sim (p$ $\rightarrow q) \vee \sim (q$ $\rightarrow p)$$\equiv (p \wedge \sim q) \vee (q \wedge \sim p)$.Thus,Statement-$I$ is True.For Statement-$II$: $p$ $\rightarrow (p$ $\rightarrow q) \equiv \sim p \vee (\sim p \vee q) \equiv (\sim p \vee \sim p) \vee q \equiv \sim p \vee q$.This is not a tautology (it depends on the truth values of $p$ and $q$).Thus,Statement-$II$ is False.

Statement-$I$: $\sim (p \leftrightarrow q)$ is equivalent to $(p \wedge \sim q) \vee (q \wedge \sim p)$.
Statement-$II$: $p$ $\rightarrow (p$ $\rightarrow q)$ is a tautology.

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Statement-$I$: $\sim (p \leftrightarrow q)$ is equivalent to $(p \wedge \sim q) \vee (q \wedge \sim p)$.Statement-$II$: $p$ $\rightarrow (p$ $\rightarrow q)$ is a tautology.