1. **State the problem:** We want to justify the rule of universal modus tollens, which states 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 $x$, $P(x)$ implies $Q(x)$, and for a specific $a$, $Q(a)$ is false, then $P(a)$ must also be false. 3. **Express the premises:** - Premise 1: $\forall x (P(x) \to Q(x))$ means for every element $x$ in the domain, if $P(x)$ is true, then $Q(x)$ is true. - Premise 2: $\neg Q(a)$ means $Q(a)$ is false for the particular element $a$. 4. **Apply universal instantiation:** From $\forall x (P(x) \to Q(x))$, we can instantiate for the particular element $a$: $$P(a) \to Q(a)$$ 5. **Apply modus tollens:** Given $P(a) \to Q(a)$ and $\neg Q(a)$, by modus tollens, we conclude: $$\neg P(a)$$ 6. **Explanation:** If $P(a)$ were true, then $Q(a)$ would have to be true by the implication. But since $Q(a)$ is false, $P(a)$ cannot be true. Therefore, $\neg P(a)$ must hold. **Final conclusion:** $$\boxed{\neg P(a)}$$