1. **Problem:** Construct the truth table for the compound proposition $ (p \wedge q) \to (\neg q \lor p) $. 2. **Formula and rules:** The implication $A \to B$ is false only when $A$ is true and $B$ is false; otherwise, it is true. 3. **Step-by-step truth table construction:** | $p$ | $q$ | $p \wedge q$ | $\neg q$ | $\neg q \lor p$ | $(p \wedge q) \to (\neg q \lor p)$ | |-----|-----|--------------|----------|-----------------|-------------------------------------| | T | T | T | F | T | T | | T | F | F | T | T | T | | F | T | F | F | F | T | | F | F | F | T | T | T | 4. **Explanation:** - Compute $p \wedge q$ which is true only if both $p$ and $q$ are true. - Compute $\neg q$ which negates $q$. - Compute $\neg q \lor p$ which is true if either $\neg q$ or $p$ is true. - Finally, evaluate the implication $(p \wedge q) \to (\neg q \lor p)$ using the truth values. 5. **Final answer:** The truth table shows the compound proposition is true for all combinations of $p$ and $q$ except when the antecedent is true and the consequent is false, which does not occur here. Thus, the proposition is a tautology.