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

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(D) To evaluate Statement $-1$,we analyze the truth table for $\sim (p \leftrightarrow \sim q)$.Recall that $(p \leftrightarrow \sim q)$ is true when

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Statement $-1$ is true,Statement $-2$ is false.

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(D) To evaluate Statement $-1$,we analyze the truth table for $\sim (p \leftrightarrow \sim q)$.Recall that $(p \leftrightarrow \sim q)$ is true when $p$ and $\sim q$ have the same truth value,which means $p$ and $q$ have opposite truth values.Thus,$(p \leftrightarrow \sim q)$ is equivalent to $\sim (p \leftrightarrow q)$.Therefore,$\sim (p \leftrightarrow \sim q)$ is equivalent to $\sim (\sim (p \leftrightarrow q))$,which simplifies to $(p \leftrightarrow q)$.So,Statement $-1$ is true.To evaluate Statement $-2$,we check if $\sim (p \leftrightarrow \sim q)$ is a tautology.Since $\sim (p \leftrightarrow \sim q) \equiv (p \leftrightarrow q)$,and $(p \leftrightarrow q)$ is not true for all values of $p$ and $q$ (it is false when $p$ and $q$ have different truth values),the expression 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.