Negation of expression Suppose I want to 'negate' the following expression:
$\exists m$ s.t. $\forall n > 0$, ...
Would the negation be of the form:
(a) $\not \exists m$ s.t. ...
or 
(b) $\forall m \exists$ ...
or 
(c) Does it depend on your definition of negation
And if the answer (as I suspect) is (b), can you explain why it is (b) rather than (a)?
 A: It depends on how far you take negation into the interior of the statement. 
(a) is fine, provided "everything after  s.t." is identical to the initial statement's "everything after s.t.".  
(b) is fine, provided whatever follows $\forall m \exists n  ...$ is the negation of what follows after the "$\exists m \forall n ... $" in the initial statement.
Without knowing your quantified statement in it's entirety, we can't really say which is the correct negation.  But negation is negation, and the first and second forms are equivalent, provided pushing the negation inward proceeds correctly in the case of (b).
For example
Suppose we have for $m, n \in \mathbb Z$, $$\exists m \forall n (m+n = n)\tag{1} $$
It's negation can be written: $$\lnot \exists m \forall n (m+n= n)\tag{$\lnot 1_1$}$$
The negation can also be written, and is equivalent to the negation immediately above, as follows: $$\forall m \lnot\big(\forall n (m+n=n)\big)\tag{$\lnot 1_2$}$$
Which is equivalent to: $$\forall m \exists n \Big(\lnot(m+n = n)\Big)\tag{$\lnot 1_3$}$$
Which is equivalent to: $$\forall m\exists n\big(m+n\neq n\big) \tag{$\lnot 1_4$}$$
A: Let's say (instead of "...") that what follows is some statement dependent on $m$ and $n,$ which we'll denote by $\Phi(m,n).$ Thus, what we want to negate is $$\exists m:\forall n>0,\Phi(m,n).$$
It is certainly fair to say that the negation of this is $$\not\exists m:\forall n>0,\Phi(m,n).$$ This is effectively the same as $$\neg\bigl(\exists m:\forall n>0,\Phi(m,n)\bigr),$$ where $\neg$ indicates negation.
But what does that mean? Well, it means that, no matter what $m$ we choose, it isn't true that $\forall n>0,\Phi(m,n).$ Therefore, there must be some $n>0$ such that $\neg\Phi(m,n).$ Thus, we can instead say that the negation is $$\forall m,\exists n>0:\neg\Phi(m,n).$$
The upshot is that the answer is: "Both, if we're being careful."
