For the statements $$p$$ and $$q$$, consider the following compound statements:
$$(a)$$ $$(\sim q \wedge (p \to q)) \to \sim p$$
$$(b)$$ $$((p \vee q) \wedge \sim p) \to q$$
Then which of the following statements is correct?

We need to determine whether statements (a) $$(\sim q \wedge (p \to q)) \to \sim p$$ and (b) $$((p \vee q) \wedge \sim p) \to q$$ are tautologies.

For statement (a), recall that $$p \to q \equiv \sim p \vee q$$. So $$\sim q \wedge (p \to q) = \sim q \wedge (\sim p \vee q) = (\sim q \wedge \sim p) \vee (\sim q \wedge q) = \sim p \wedge \sim q$$, since $$\sim q \wedge q$$ is always false. The statement becomes $$(\sim p \wedge \sim q) \to \sim p$$. Since the hypothesis $$\sim p \wedge \sim q$$ contains $$\sim p$$, the implication is always true. So (a) is a tautology.

For statement (b), we simplify $$(p \vee q) \wedge \sim p = (\sim p \wedge p) \vee (\sim p \wedge q) = F \vee (\sim p \wedge q) = \sim p \wedge q$$. The statement becomes $$(\sim p \wedge q) \to q$$. Since the hypothesis contains $$q$$, the implication is always true. So (b) is also a tautology.

Therefore, both (a) and (b) are tautologies.