The negation of the expression $q \vee ((\sim q) \wedge p)$ is equivalent to

Let $r \in \{p, q, \sim p, \sim q\}$ be such that the logical statement $r \vee (\sim p) \Rightarrow (p \wedge q) \vee r$ is a tautology. Then $r$ is equal to

Consider
Statement-$1$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a fallacy.
Statement-$2$: $(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$ is a tautology.

The negation of the inverse of $\sim p \rightarrow q$ is

If the statements $p, q$ and $r$ have the truth values $F, T, F$ respectively,then the truth values of the statement patterns $(p \wedge \sim q) \rightarrow r$ and $(p \vee q) \rightarrow r$ are respectively

If the truth value of the compound statement $[(p \vee q) \wedge (q \to r) \wedge (\sim r)] \to (p \wedge q)$ is $False$, then the truth values of $p \to q$ and $q \to p$ are respectively......