sim (p Rightarrow q) Leftrightarrow sim p vee | vedclass

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(C) The given statement is $\sim (p \Rightarrow q) \Leftrightarrow \sim p \vee \sim q$. By the laws of logic,$\sim (p \Rightarrow q) \equiv p \wedge \

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Neither a tautology nor a contradiction

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(C) The given statement is $\sim (p \Rightarrow q) \Leftrightarrow \sim p \vee \sim q$. By the laws of logic,$\sim (p \Rightarrow q) \equiv p \wedge \sim q$. Thus,the expression becomes $(p \wedge \sim q) \Leftrightarrow (\sim p \vee \sim q)$. Constructing the truth table for the expression: $p$$q$$p \Rightarrow q$$\sim (p \Rightarrow q)$$\sim p$$\sim q$$\sim p \vee \sim q$$\sim (p \Rightarrow q) \Leftrightarrow (\sim p \vee \sim q)$$T$$T$$T$$F$$F$$F$$F$$T$$T$$F$$F$$T$$F$$T$$T$$T$$F$$T$$T$$F$$T$$F$$T$$F$$F$$F$$T$$F$$T$$T$$T$$F$Since the final column contains both $T$ and $F$,the statement is neither a tautology nor a contradiction. Therefore,the correct option is $C$.

$\sim (p \Rightarrow q) \Leftrightarrow \sim p \vee \sim q$ is

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