1. **State the problem:** We need to determine if the logical statement $$(p \wedge q) \to p$$ is a tautology. 2. **Recall definitions:** - $p \wedge q$ means "p AND q". - $\to$ means implication: $A \to B$ is false only when $A$ is true and $B$ is false; otherwise, it is true. 3. **Construct the truth table:** | $p$ | $q$ | $p \wedge q$ | $(p \wedge q) \to p$ | |-----|-----|--------------|-----------------------| | T | T | T | T | | T | F | F | T | | F | T | F | T | | F | F | F | T | 4. **Analyze the truth table:** The implication $(p \wedge q) \to p$ is true in all possible truth value combinations of $p$ and $q$. 5. **Conclusion:** Since the statement is true for all truth assignments, $(p \wedge q) \to p$ is a tautology. **Final answer:** $(p \wedge q) \to p$ is a tautology because whenever $p \wedge q$ is true, $p$ must be true, making the implication always true.