1. **Problem statement:** Show that from the premises $\forall x (P(x) \to Q(x))$ and $\neg Q(a)$ for a particular element $a$, we can conclude $\neg P(a)$. 2. **Recall the rule of universal modus tollens:** If for all $x$, $P(x) \to Q(x)$ is true, and for a particular $a$, $Q(a)$ is false, then $P(a)$ must also be false. 3. **Step-by-step justification:** - From $\forall x (P(x) \to Q(x))$, by universal instantiation, we get $P(a) \to Q(a)$. - We are given $\neg Q(a)$. - Modus tollens states that if $P(a) \to Q(a)$ and $\neg Q(a)$, then $\neg P(a)$. 4. **Explanation:** Since $P(a)$ implies $Q(a)$, but $Q(a)$ is false, $P(a)$ cannot be true; otherwise, it would contradict the implication. 5. **Final conclusion:** $$\neg P(a)$$