Let $$p$$ and $$q$$ be two statements. Then $$\sim(p \wedge (p \to \sim q))$$ is equivalent to

We need to simplify $$\sim(p \wedge (p \to \sim q))$$.

First, recall that $$p \to \sim q \equiv \sim p \vee \sim q$$.

So: $$p \wedge (p \to \sim q) \equiv p \wedge (\sim p \vee \sim q)$$

Using distribution: $$= (p \wedge \sim p) \vee (p \wedge \sim q)$$

Since $$p \wedge \sim p$$ is always false (contradiction):

$$= F \vee (p \wedge \sim q) = p \wedge \sim q$$

Therefore: $$\sim(p \wedge (p \to \sim q)) = \sim(p \wedge \sim q)$$

By De Morgan's law: $$= \sim p \vee q$$

$$= (\sim p) \vee q$$

The correct answer is Option 3: $$(\sim p) \vee q$$.