1. Statement of the problem: We are given the premises $\forall x\, (P(x) \to Q(x))$ and $\neg Q(a)$ for a particular element $a$ in the domain, and we must show $\neg P(a)$. 2. Formula and key rules: We use universal instantiation which allows us to derive $P(a) \to Q(a)$ from $\forall x\, (P(x) \to Q(x))$. We use modus tollens which allows one to infer $\neg P$ from $P \to Q$ and $\neg Q$. 3. Proof by application: From $\forall x\, (P(x) \to Q(x))$ we instantiate to get $P(a) \to Q(a)$. Together with the premise $\neg Q(a)$, applying modus tollens to $P(a) \to Q(a)$ and $\neg Q(a)$ yields $\neg P(a)$. 4. Conclusion: Therefore the premises $\forall x\, (P(x) \to Q(x))$ and $\neg Q(a)$ entail $\neg P(a)$, which is the rule of universal modus tollens. Final answer: $\neg P(a)$.