1. **State the problem:** Given the premises: 1. $\forall x [P(x) \Rightarrow Q(x)]$ 2. $\exists x [P(x) \wedge R(x)]$ 3. $\forall x [(R(x) \wedge Q(x)) \Rightarrow S(x)]$ Prove: $\exists x [P(x) \wedge S(x)]$ 2. **Recall relevant rules:** - Universal quantifier ($\forall$) means the statement holds for all elements. - Existential quantifier ($\exists$) means there is at least one element for which the statement holds. - Modus ponens: From $P \Rightarrow Q$ and $P$, infer $Q$. - Conjunction introduction and elimination allow combining and separating statements. 3. **Proof steps:** - From premise 2, there exists some element $a$ such that $P(a) \wedge R(a)$ is true. - From premise 1, since $\forall x [P(x) \Rightarrow Q(x)]$, then $P(a) \Rightarrow Q(a)$. - Since $P(a)$ is true (from step above), by modus ponens, $Q(a)$ is true. - From premise 3, $\forall x [(R(x) \wedge Q(x)) \Rightarrow S(x)]$, so $(R(a) \wedge Q(a)) \Rightarrow S(a)$. - We know $R(a)$ and $Q(a)$ are true, so by modus ponens, $S(a)$ is true. - We have $P(a)$ and $S(a)$ both true, so $P(a) \wedge S(a)$ is true. - Therefore, $\exists x [P(x) \wedge S(x)]$ is true. 4. **Conclusion:** We have shown that from the given premises, there exists an element $a$ such that $P(a)$ and $S(a)$ hold, proving $\exists x [P(x) \wedge S(x)]$. This completes the proof.