How can I prove that $\forall x \in D (P(x)\implies Q(X))$ is not equivalent to $(\forall x \in D, P(x))\implies (\forall x \in D, Q(X))$?
Would I prove the negation of this?
$\big(\forall x \in D ,(P(x)\implies Q(x))\big) \iff \big((\forall x \in D, P(x))\implies (\forall x \in D, Q(x))\big)$
If so, how would I do that?
On
Left to right is true; assume that P always implies Q. Then if P is universally true, then Q must also be, trivially. Right?
But right to left is false. Say D has two elements, A and B, and P is true only of A, while Q is true nowhere.
It is false that "P is universally true" so the implication "if P is always true, then Q is always true" holds. However, the left hand side is false: it requires "everything which satisfies P must also satisfy Q," so which is false (for A).
In general finite domains (usually very small!) are enough to find counterexamples for this sort of thing. In fact, properly stated, there is a theorem to this effect (since you're only talking about unary relationships) but I don't know any constructive bounds off the top of my head.
But to approach these problems, if you think two statements are inequivalent, just try different small finite domains until you find a counterexample (or accidentally convince yourself such an example is impossible).
Hope this helps!
(sorry for lack of TeX, I'm on mobile at the moment)
It is true that, if all people are left-handed, then all people are Chinese (because the hypothesis, "all people are left-handed," is false and therefore implies every statement). But it is not true that all left-handed people are Chinese.