TITLE: Factoring out universal quantifier in combination with an implication
QUESTION [6 upvotes]: I just began studying maths and so far everything made sense after tinkering around with it a little bit (e.g. $ \lnot(\forall x \in M : A(x)) = \exists x \in M : \lnot A(x) $ thinking "not all math students are dumb means that there's at least one who is not dumb")
I already prooved this statement right (sry I'm not a native speaker) $( \forall x:P(x))\Rightarrow q = \exists x:(P(x)\Rightarrow q)$, but is there any example (like the one given above) for this as well? I just cannot understand why factoring out the universal quantifier makes any difference... (although I know it has to be right)

REPLY [2 votes]: Brian & Henning have given excellent, precise answers above.
Here is an imprecise attempt at providing some intuitive 'feel'.
I think of $x$ as being a representative element of some countable set $\{x_1,x_2,...\}$.
I think of $(\forall x P(x))$ as $(P(x_1) \land P(x_2) \land ...)$, and I think of $(\exists x R(x))$ as $(R(x_1) \lor R(x_2) \lor...)$.
We have $A \Rightarrow B$ is the same as $\lnot A \lor B$.
Then $( \forall x:P(x))\Rightarrow q$ becomes $\lnot (P(x_1) \land P(x_2) \land ...) \lor q = (\lnot P(x_1) \lor \lnot P(x_2) \lor ...) \lor q$,
The statement $\exists x:(P(x)\Rightarrow q)$ becomes $((\lnot P(x_1) \lor q) \lor (\lnot P(x_1) \lor q) \lor...)$ which is logically the same as the statement above.