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

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(C) To determine the nature of the statement $\sim(p \leftrightarrow \sim q)$,we construct a truth table:$1$. The biconditional $p \leftrightarrow \si

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equivalent to $(p \leftrightarrow q)$

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(C) To determine the nature of the statement $\sim(p \leftrightarrow \sim q)$,we construct a truth table:$1$. The biconditional $p \leftrightarrow \sim q$ is true when $p$ and $\sim q$ have the same truth value.$2$. The negation $\sim(p \leftrightarrow \sim q)$ is true when $p \leftrightarrow \sim q$ is false.Truth Table:$p$ | $q$ | $\sim q$ | $p \leftrightarrow \sim q$ | $\sim(p \leftrightarrow \sim q)$$T$ | $T$ | $F$ | $F$ | $T$$T$ | $F$ | $T$ | $T$ | $F$$F$ | $T$ | $F$ | $T$ | $F$$F$ | $F$ | $T$ | $F$ | $T$Comparing this to $p \leftrightarrow q$:$p$ | $q$ | $p \leftrightarrow q$$T$ | $T$ | $T$$T$ | $F$ | $F$$F$ | $T$ | $F$$F$ | $F$ | $T$Since the truth values of $\sim(p \leftrightarrow \sim q)$ and $(p \leftrightarrow q)$ are identical for all combinations of $p$ and $q$,the statement is equivalent to $(p \leftrightarrow q)$.

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

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