Why are the following not logically equivalent? Why are the following statements not logically equivalent? 
$$((\forall x\in D,P(x))\vee(\forall x\in D, Q(x))$$
and $$\forall x\in D, (P(x)\vee Q(x)$$
I have thought for several hours about this and I can't think of any reason why they shouldn't be equivalent. May someone help me?
 A: For example, let $D=\mathbb R$ and $P(x)$ be $x\ge0$ and $Q(x)$ be $x<0$.  Then your first statement is not true, because both $\forall x\in D,P(x)$ and $\forall x\in D, Q(x)$ are false, but your second statement is true.
A: Because if, for example, each $x$ satisfies $P$ or $Q$ but not both, with both cases occurring, the first statement is false but the second one is true.
A: One can find a simple countermodel showing that these are not equivalent using this tree proof generator.  The countermodel is for the following implication:
$$\forall x(Px \lor Qx) \to (\forall xPx \lor \forall xQx)$$
The model specifies a domain $\{0,1\}$ and an interpretation that $P$ is true only for $x=1$ and $Q$ is true only for $x=0$. With that interpretation, for all $x$ the antecedent is true because for any $x$ in the domain $\{0,1\}$ either $P$ is true or $Q$ is true. However, since $P$ is false for $x=0$ and $Q$ is false for $x=1$, the consequent is false.
To try to do this manually, start with a one element domain, say, $\{0\}$. Interpret $Q(0)$ as true and $P(0)$ as false. This would make the antecedent true, but it would also make the consequent true. Since we suspect there exists a model showing the conditional false, try adding another element to the domain making it $\{0,1\}$ and interpret $P(1)$ to be true and $Q(1)$ to be false. This gives us the same countermodel presented by the software showing the antecedent is true while the consequent is false.
