$(p \wedge \sim q) \wedge (\sim p \wedge q)$ is

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

If $p \to (q \lor \sim r)$ is false, then the truth values of $(p \leftrightarrow q) \land r$ and $\sim p \to \sim q$ are :

The correct logical equivalences from the following are:
$(I)$ $p \to (q \to r) \equiv (p \land q) \to r$
$(II)$ $(p \to q) \to r \equiv p \to (q \lor r)$
$(III)$ $(p \to q) \to r \equiv (p \to r) \land (\sim q \to r)$
$(IV)$ $p \to (q \to r) \equiv q \to (p \to r)$

If the Boolean expression $(p \wedge q) \circledast (p \otimes q)$ is a tautology,then $\circledast$ and $\otimes$ are respectively given by

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