The statement sim(p leftrightarrow sim q) is | vedclass

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(A) To determine the nature of the statement $\sim(p \leftrightarrow \sim q)$,we construct a truth table:| $p$ | $q$ | $\sim q$ | $p \leftrightarrow \

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equivalent to $p \leftrightarrow q$

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(A) To determine the nature of the statement $\sim(p \leftrightarrow \sim q)$,we construct a truth table:| $p$ | $q$ | $\sim q$ | $p \leftrightarrow \sim q$ | $\sim(p \leftrightarrow \sim q)$ | $p \leftrightarrow q$ || :--- | :--- | :--- | :--- | :--- | :--- || $T$ | $T$ | $F$ | $F$ | $T$ | $T$ || $T$ | $F$ | $T$ | $T$ | $F$ | $F$ || $F$ | $T$ | $F$ | $T$ | $F$ | $F$ || $F$ | $F$ | $T$ | $F$ | $T$ | $T$ |Comparing column $5$ and column $6$,we observe that the truth values are identical for all combinations of $p$ and $q$.Therefore,$\sim(p \leftrightarrow \sim q) \equiv p \leftrightarrow q$.

The statement $\sim(p \leftrightarrow \sim q)$ is

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