1. **Stating the problem:** We are given the logical statements: - (P \wedge Q) \to R - \neg P \lor \neg Q and we need to prove \neg R. 2. **Recall the implication rule:** An implication $A \to B$ is false only if $A$ is true and $B$ is false. 3. **Analyze the premises:** - From $\neg P \lor \neg Q$, at least one of $P$ or $Q$ is false. - Therefore, $P \wedge Q$ is false. 4. **Evaluate the implication:** Since $P \wedge Q$ is false, the implication $(P \wedge Q) \to R$ is true regardless of $R$. 5. **To prove $\neg R$:** Assume for contradiction that $R$ is true. Then $(P \wedge Q) \to R$ holds, but since $P \wedge Q$ is false, $R$ can be either true or false. 6. **However, from the premises, since $P \wedge Q$ is false, the only way for $R$ to be true is if the implication holds, which it does. 7. **But the problem states to conclude $\neg R$, which is not logically guaranteed from the premises given.** **Hence, the conclusion $\neg R$ does not logically follow from the premises.** --- **For part c:** 1. **Stating the problem:** Given: - $\neg (P \wedge \neg Q) \to \neg R$ - $\neg R \to \neg P \lor Q$ Prove: - $\neg (P \wedge \neg Q) \to \neg P \lor Q$ 2. **Use hypothetical syllogism:** If $A \to B$ and $B \to C$, then $A \to C$. 3. **Identify:** - $A = \neg (P \wedge \neg Q)$ - $B = \neg R$ - $C = \neg P \lor Q$ 4. **Apply the rule:** From $\neg (P \wedge \neg Q) \to \neg R$ and $\neg R \to \neg P \lor Q$, we get $$\neg (P \wedge \neg Q) \to \neg P \lor Q$$ which is the conclusion. **Final answers:** - Part b: The conclusion $\neg R$ does not logically follow. - Part c: The conclusion $\neg (P \wedge \neg Q) \to \neg P \lor Q$ is valid by hypothetical syllogism.