Q.$(p \wedge q) \to r$ is logically equivalent to ________.
A) $p \to (q \to r)$
B) $(p \wedge q) \to \sim r$
C) $(\sim p \vee \sim q) \to \sim r$
D) $(p \vee q) \to r$

$(p \wedge q) \to r \equiv \sim(p \wedge q) \vee r \equiv \sim p \vee \sim q \vee r$ (implication law then De Morgan's law).

Option A: $p \to (q \to r) \equiv \sim p \vee (q \to r) \equiv \sim p \vee (\sim q \vee r) \equiv \sim p \vee \sim q \vee r$. This matches exactly.

Quick checks on the other options show they don't match: option B gives $\sim p \vee \sim q \vee \sim r$ (has $\sim r$, not $r$); option C gives $(p \wedge q) \vee \sim r$; option D gives $(\sim p \wedge \sim q) \vee r$ — none of these equal $\sim p \vee \sim q \vee r$ in general.

Convert both sides to $\sim$/$\vee$ form using the implication law and compare (this identity is the standard exportation law).

Confusing this with the converse or contrapositive form, or stopping after expanding only one side.