Ambiguity of ∃x∀y logical syntax? I often see statements such as $\exists x \forall y (P(x,y))$.  If I read this literally, it says “There exists an x for every y, such that (Something about $x$ and $y$).”  However, I get the impression that it’s supposed to be read “There exists an $x$ such that for every $y$, stuff...”.
Is this juxtaposition of $\exists x \forall y$ a standard way of writing statements where a “...such that...” is meant to be inserted between the terms?  Is there a less ambiguous way of writing it?
The converse is not true (I don’t think?). That is, $\forall y \exists x (P(x,y))$ is read “For all $y$ there exists an $x$ such that... (stuff in parentheses)”.  It seems odd that the same syntax would have different meanings with the terms switched.
 A: The notation's being perfectly nice; the fault lies with natural language. "$\mathsf{Q_0}a_0\mathsf{Q_1}a_1$" is always read so that $a_0$ "occurs before" $a_1$ (and this can in fact be made precise, e.g. by talking about game semantics); the fact that the natural language expressions "there is an $a_0$ for every $a_1$" and "for every $a_1$ there is an $a_0$" have the same meaning is a sign that natural language isn't great in this context. (Or, we could make our translations wordier: read "$\forall x$" as "For every $x$ it is the case that" and "$\exists x$" as "There is an $x$ such that.")
A: Yes, $\exists x\forall y\bigl(P(x,y)\bigr)$ means that there is a $x$ such that, for every $y$, $P(x,y)$ holds.
There is nothing peculiar here. The existential quantifier should always be read as if there was a “such that” after it. For instance, $\exists x\bigl(P(x)\rightarrow Q(x)\bigr)$ means that there is a $x$ such that $P(x)$ implies $Q(x)$.
As far as I can see, this is only redundant when there are several existential quantifiers: if, say, you have $\exists x\exists y\bigl(P(x,y)\bigr)$, there is no need to read it as “there is a $x$ such that there is a $y$ such that $P(x,y)$ holds”. It will be enough to read it as “there are a $x$ and a $y$ such that $P(x,y)$ holds”.
A: *

*OP: “Would you also vote then that the expression $∃x{\in}X$, is really
shorthand for $∃x \;(x∈X)$ ?”
Indeed. And $$\exists x{\in} S\;P(x)\\\text{There exists
   an $x$ in $S$ such that $P(x)$}\\\text{For some $x$ in $S,\,$
   $P(x)$}$$ is just the conventional notational shorthand for $$\exists x\;\big(\,x\in
   S\ \land P(x)\,\big)\\\text{There exists an $x$ such that $[x$ is in
   $S]$ and $P(x)$}\\\text{For some $x,\;[x$ is in $S]$ and $P(x)$}.$$


*However, the abbreviation is structured differently for the universal
quantifier: $$\forall x{\in} S\;P(x) \\\text{For each $x$
   in $S,\,$ $P(x)$}$$ is the conventional notational shorthand for  $$\forall
   x\;\big(\,x\in S\to P(x) \,\big)\\\text{For each $x,\;$ if $x$ is in
   $S,$ then $P(x)$}.$$
