The negation of $p \wedge (q \rightarrow \sim r)$ is

Let $a: \sim(p \wedge \sim r) \vee(\sim q \vee s)$ and $b: (p \vee s) \leftrightarrow(q \wedge r)$. If the truth values of $p$ and $q$ are $T$ and that of $r$ and $s$ are $F$,then the truth values of $a$ and $b$ are respectively...

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

The truth values of $p \rightarrow r$ is $F$ and $p \leftrightarrow q$ is $F$. Then the truth values of $(\sim p \vee q) \rightarrow (p \vee \sim q)$ and $(p \wedge \sim q) \rightarrow (\sim p \wedge q)$ are respectively:

If $(p \wedge \sim q) \wedge r \to \sim r$ is $F$,then the truth value of $r$ is:

The dual of the converse of the inverse of the logical statement $p \to (q \to r)$ is equivalent to...