Is "A and B imply C" equivalent to "For all A such that B, C"? So I mostly study PDE, harmonic analysis, image processing, and so on, but for whatever reason I decided to be a TA for an undergraduate "introduction to proofs" course this semester.  I suppose I wanted a challenge.  The experience has been a rewarding one so far, but I've come across a couple things that I've found difficult to explain in a clear and concise manner.  For example, consider the standard definition of uniform continuity:
"For all $\epsilon>0$ there is a $\delta>0$ such that whenever $x,y\in S$ satisfy $\vert x-y\vert<\delta$, we have $\vert f(x)-f(y)\vert<\epsilon$"
Students are asked to "convert this to logical symbols", and I have seen the following two approaches: 
"$\forall \epsilon>0$ $\exists\delta>0$ s.t. $(x,y\in S \wedge \vert x-y\vert<\delta)\Rightarrow\vert f(x)-f(y)\vert<\epsilon$" 
and what I consider the more standard version: 
"$\forall\epsilon>0$ $\exists\delta>0$ s.t. $\forall x,y\in S$ s.t. $\vert x-y\vert<\delta$, $\vert f(x)-f(y)\vert<\epsilon$."
So my question is:  Are they 100% logically equivalent...?  In particular, it seems that "for all A such that B, C" is equivalent to "A and B imply C".  Is there a way to "prove" this equivalence, or is it just a "functional" equivalence (i.e. they tell you to do the same thing)?  I've been struggling to explain the two approaches/advocate for one over the other, so bonus points if you have simple examples for the two.
Thanks ahead!
 A: Consider the "statement" $(\exists k\in \mathbb{N})(a=2k)$. It doesn't make sense because $a$ isn't quantified (there's no $\forall$ or $\exists$ associated with it). You can't tell wether that "statement" is true or false because it isn't a statement. The following formula is a statement: $(\forall a\in \mathbb{R})(\exists k\in \mathbb{N})(a=2k)$, you can tell wether it's true or false. 
Regarding your question, $x$ and $y$ aren't quantified. I could very well assume the $x,y$ in $$(\forall \epsilon>0) (\exists\delta>0) \bigl((x,y\in S \wedge \vert x-y\vert<\delta)\Rightarrow\vert f(x)-f(y)\vert<\epsilon\bigr) \tag 1$$
are existencial, meaning I could interpret the above pseudo-statement as meaning $$(\forall \epsilon>0) (\exists\delta>0) (\exists x,y)\bigl((x,y\in S \wedge \vert x-y\vert<\delta)\Rightarrow\vert f(x)-f(y)\vert<\epsilon\bigr) \tag 2$$
which isn't what you want at all.
Please note that what I said above was merely an intuitive approach to the issue at hand. I simply cannot decide that $(1)$ means $(2)$, I cannot decide it means anything because it isn't a statement.
The question rises: what is a statement? The answer to that question isn't appropiate to post here, I believe. However @dtldarek has already provided some insight into that.
I know that in the literature you'll find such "definitions" of continuity. Well, they're badly written. I know it's hard to believe, but that's the case.
If any of the variables in a formula isn't quantified, then that formula isn't a statement and has no meaning.
Edit: The following problem is analogous to your question, the difference is the formulae at hand are statements. Hopefully this will be helpful to you:
$$(\forall \epsilon>0) (\exists\delta>0) (\forall x,y)\bigl((x,y\in S \wedge \vert x-y\vert<\delta)\Rightarrow\vert f(x)-f(y)\vert<\epsilon\bigr) \tag i$$
$$(\forall \epsilon>0) (\exists\delta>0) (\forall x,y\in S)\bigl((\vert x-y\vert<\delta)\Rightarrow\vert f(x)-f(y)\vert<\epsilon\bigr) \tag {ii})$$
Are $(i)$ and $(ii)$ equivalent? In fact they are and that equivalence is "very strong". It's more than the fact that you can conclude one from the other. The statement $(ii)$ is just short hand notation for the formally correct $(i)$.
(In the above paragraph I chose to ignore the quantification regarding $\delta$ and $\epsilon$ for the sake of simplicity. The quantification regarding $\delta$ and $\epsilon$ has to be dealt with in a similar manner).
Allow me illustrate this with a simpler version of this problem.
The statement $(\forall x\in \mathbb{R})(x\ge 0)$, (which is obviously false, but that's of no interest to us) is just short for the formally correct $(\forall x)(x\in \mathbb{R} \Longrightarrow x\ge 0)$. The existential quantifier $\exists$ is a bit different. Consider the statement $(\exists n\in \mathbb{N})(n=1337)$. This of course, is just short for $(\exists n)(n\in \mathbb{N} \wedge n=1337)$ and not short for the silly $(\exists n)(n\in \mathbb{N} \Longrightarrow n=1337)$. Well, it's not silly, just not what one would expect $(\exists n\in \mathbb{N})(n=1337)$ to mean.
