1. **Stating the problem:** Determine the nature (tautology, contradiction, or contingency) of the logical statement $$ (\neg p \wedge (p \to q)) \to q $$. 2. **Recall the definitions and formulas:** - Negation: $$ \neg p $$ means "not p". - Conjunction: $$ p \wedge q $$ means "p and q". - Implication: $$ p \to q $$ means "if p then q", which is logically equivalent to $$ \neg p \lor q $$. - A statement is a **tautology** if it is always true regardless of truth values of its components. - A **contradiction** is always false. - A **contingency** is sometimes true and sometimes false. 3. **Rewrite the statement using equivalences:** $$ (\neg p \wedge (p \to q)) \to q $$ Using $$ p \to q \equiv \neg p \lor q $$, we get: $$ (\neg p \wedge (\neg p \lor q)) \to q $$ 4. **Simplify the conjunction inside:** $$ \neg p \wedge (\neg p \lor q) = (\neg p \wedge \neg p) \lor (\neg p \wedge q) = \neg p \lor (\neg p \wedge q) $$ Since $$ \neg p \wedge q $$ is a subset of $$ \neg p $$, this simplifies to: $$ \neg p $$ 5. **So the statement reduces to:** $$ (\neg p) \to q $$ Using implication equivalence again: $$ \neg (\neg p) \lor q = p \lor q $$ 6. **Analyze the simplified statement:** $$ p \lor q $$ is true except when both $$ p $$ and $$ q $$ are false. 7. **Conclusion:** The original statement is false only when $$ p = \text{false} $$ and $$ q = \text{false} $$. Therefore, it is not always true (not a tautology), not always false (not a contradiction), so it is a **contingency**. **Final answer:** The statement $$ (\neg p \wedge (p \to q)) \to q $$ is a contingency.