Consider the following argument with premise $\forall _x (P(x) \vee Q(x))$ and conclusion $(\forall _x P(x)) \wedge (\forall _x Q(x))$
$\begin{array}{|ll|l|} \hline (A) & \forall _x (P(x) \vee Q(x)) & \text{Premise} \\ \hline (B) & P(c) \vee Q(c) & \text{Universal instantiation from (A)} \\ \hline (C) & P(c) & \text{Simplification from (B)} \\ \hline (D) & \forall _x P(x) & \text{Universal Generalization of (C)} \\ \hline (E) & Q(c) & \text{Simplification from (B)} \\ \hline (F) & \forall _x Q(x) & \text{Universal Generalization of (E)} \\ \hline (G) & (\forall _x P(x)) \wedge (\forall _xQ(x)) & \text{Conjuction of (D) and (F)} \\ \hline \end{array}$
64.3k questions
77.9k answers
244k comments
80.0k users