Statement -1 : sim (p leftrightarrow sim q) i | vedclass

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(C) We know that $p \leftrightarrow q$ is equivalent to $\sim (p \leftrightarrow \sim q)$.Let us verify this using a truth table:$p$ | $q$ | $\sim q$

Vedclass · Accepted Answer

Statement $-1$ is True,Statement $-2$ is False.

Vedclass · Answer

(C) We know that $p \leftrightarrow q$ is equivalent to $\sim (p \leftrightarrow \sim q)$.Let us verify this using a truth table:$p$ | $q$ | $\sim q$ | $p \leftrightarrow \sim q$ | $\sim (p \leftrightarrow \sim q)$ | $p \leftrightarrow q$$T$ | $T$ | $F$ | $F$ | $T$ | $T$$T$ | $F$ | $T$ | $T$ | $F$ | $F$$F$ | $T$ | $F$ | $T$ | $F$ | $F$$F$ | $F$ | $T$ | $F$ | $T$ | $T$From the truth table,the column for $\sim (p \leftrightarrow \sim q)$ is identical to the column for $p \leftrightarrow q$. Thus,Statement $-1$ is True.Since the truth values of $\sim (p \leftrightarrow \sim q)$ depend on the truth values of $p$ and $q$ (it is $T$ when $p$ and $q$ have the same truth value and $F$ otherwise),it is not a tautology. Thus,Statement $-2$ is False.

Statement $-1$ : $\sim (p \leftrightarrow \sim q)$ is equivalent to $p \leftrightarrow q$.
Statement $-2$ : $\sim (p \leftrightarrow \sim q)$ is a tautology.

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Statement $-1$ : $\sim (p \leftrightarrow \sim q)$ is equivalent to $p \leftrightarrow q$.Statement $-2$ : $\sim (p \leftrightarrow \sim q)$ is a tautology.