What type of statement is sim (p rightarrow q | vedclass

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(C) Let us construct the truth table for the statement $\sim (p \rightarrow q) \Leftrightarrow (\sim p \vee \sim q)$.$1$. For $p = T, q = T$: $\sim (T

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Neither a tautology nor a contradiction

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(C) Let us construct the truth table for the statement $\sim (p \rightarrow q) \Leftrightarrow (\sim p \vee \sim q)$.$1$. For $p = T, q = T$: $\sim (T$ $\rightarrow T) \Leftrightarrow (\sim T \vee \sim T) \implies \sim (T) \Leftrightarrow (F \vee F) \implies F \Leftrightarrow F$,which is $T$.$2$. For $p = T, q = F$: $\sim (T$ $\rightarrow F) \Leftrightarrow (\sim T \vee \sim F) \implies \sim (F) \Leftrightarrow (F \vee T) \implies T \Leftrightarrow T$,which is $T$.$3$. For $p = F, q = T$: $\sim (F$ $\rightarrow T) \Leftrightarrow (\sim F \vee \sim T) \implies \sim (T) \Leftrightarrow (T \vee F) \implies F \Leftrightarrow T$,which is $F$.$4$. For $p = F, q = F$: $\sim (F$ $\rightarrow F) \Leftrightarrow (\sim F \vee \sim F) \implies \sim (T) \Leftrightarrow (T \vee T) \implies F \Leftrightarrow T$,which is $F$.Since the truth values of the statement are not all $T$ (not a tautology) and not all $F$ (not a contradiction),it is neither a tautology nor a contradiction.

What type of statement is $\sim (p \rightarrow q) \Leftrightarrow (\sim p \vee \sim q)$?

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