1. **State the problem:** The argument tries to prove that from $\exists x P(x) \wedge \exists x Q(x)$, it follows that $\exists x (P(x) \wedge Q(x))$. 2. **Recall the logical rules:** - $\exists x P(x)$ means "there exists some element $x$ such that $P(x)$ is true." - $\exists x P(x) \wedge \exists x Q(x)$ means there is some element satisfying $P$ and some (possibly different) element satisfying $Q$. - $\exists x (P(x) \wedge Q(x))$ means there is a single element satisfying both $P$ and $Q$ simultaneously. 3. **Analyze the argument steps:** - Step 1 incorrectly uses $\lor$ (or) instead of $\wedge$ (and) in the premise. - Steps 2 and 4 simplify from $\exists x P(x) \lor \exists x Q(x)$, which is not the original premise. - Existential instantiation in steps 3 and 5 introduces the same constant $c$ for both $P$ and $Q$, assuming the same element satisfies both. 4. **Identify the error:** - The key error is assuming the same element $c$ satisfies both $P$ and $Q$ without justification. - From $\exists x P(x)$ and $\exists x Q(x)$ separately, the elements satisfying $P$ and $Q$ can be different. - Therefore, $\exists x (P(x) \wedge Q(x))$ does not necessarily follow. 5. **Conclusion:** - The argument is invalid because it incorrectly assumes the same witness $c$ for both existential statements. - The correct inference is that $\exists x P(x) \wedge \exists x Q(x)$ does not imply $\exists x (P(x) \wedge Q(x))$. **Final answer:** The error is in existential instantiation assuming the same element $c$ satisfies both $P$ and $Q$ without justification, and the misuse of $\lor$ instead of $\wedge$ in the premise.