1. **Problem:** Show that the proposition $ (p \Rightarrow q) \Rightarrow \neg (p \wedge \neg q) $ is a tautology. 2. **Recall the definitions:** - Implication: $p \Rightarrow q$ is equivalent to $\neg p \vee q$. - Negation: $\neg (p \wedge \neg q)$ means "not both $p$ and not $q$". 3. **Rewrite the proposition using implication equivalence:** $$ (p \Rightarrow q) \Rightarrow \neg (p \wedge \neg q) = (\neg p \vee q) \Rightarrow \neg (p \wedge \neg q) $$ 4. **Rewrite the outer implication:** $$ (\neg p \vee q) \Rightarrow \neg (p \wedge \neg q) = \neg (\neg p \vee q) \vee \neg (p \wedge \neg q) $$ 5. **Simplify $\neg (\neg p \vee q)$ using De Morgan's law:** $$ \neg (\neg p \vee q) = p \wedge \neg q $$ 6. **Substitute back:** $$ (p \wedge \neg q) \vee \neg (p \wedge \neg q) $$ 7. **This is a disjunction of a statement and its negation, which is always true:** $$ (p \wedge \neg q) \vee \neg (p \wedge \neg q) = \text{True} $$ 8. **Conclusion:** The proposition is a tautology because it is always true regardless of the truth values of $p$ and $q$. Final answer: **The proposition $ (p \Rightarrow q) \Rightarrow \neg (p \wedge \neg q) $ is a tautology.**