If I want to avoid quantifiers? In mathematics quantifiers always is used with restrictions $\forall x\in A$ etc and mathematicians often write:
$$x\in A\implies p(x)\;\;\;\text{instead of}\;\;\;\forall x\in A:p(x).$$ 
Is there a counterpart expression for $\exists x\in A: p(x)$ without quantifier?
Is writing like this logical unobjectionable? 
I'm interested in trying to replace 'predicate logic' in mathematics with a 'mathematical propositional algebra', that is, only use propositions of the type $x\in A\implies p(x)$. For such propositions $P_i$, there would be a "mathematical ring" with elements
$$\displaystyle
\bigoplus_i\bigwedge_j p_j^{n_i}.
$$
I guess it won't work but in the process I might learn something about logic and semantics.
 A: The fact that the universal quantifier on $x$ is omitted doesn't mean it's not supposed to be there. The correct parsing of $\forall x\in A: p(x)$ is $\forall x(x\in A\rightarrow p(x))$.
For existential quantifier this would be $\exists x(x\in A\land p(x))$. You may feel it is okay to omit the outer quantifier, but you have to remember it's still there.
A: Ther is no way to omit the $\exists$ quantifier in every context.
If we use the convention of omitting initial universal quantifiers, we assume that the sentence :

$\forall x(x > 0)$ 

(which is false in the domain $\mathbb N$ of the natural numbers) can be abbreviated as :

$x > 0$.

Consider now the negation of $\forall x(x > 0)$, i.e. $\lnot \forall x(x > 0)$, which is $\exists x \lnot (x > 0)$ i.e :

$\exists x (x \le 0)$

which is true in $\mathbb N$, as expected.
If we start from $x > 0$ and negate it, we will have $\lnot (x > 0)$, i.e. :

$x \le 0$.

According to our convention, this is an abbreviation for :

$\forall x(x \le 0)$

which is false in $\mathbb N$.
Thus we have that the negation of a false sentence is also false, which is impossible.
A: If you begin a proof with $x \in A$ and demonstrate that $p(x)$ is valid, then its really the same thing as saying $\forall x ( x \in A \Rightarrow p(x))$
( Because if $x$ is not a free variable of $\phi \in \Gamma$ and $\Gamma \vdash \phi$ then $\Gamma \vdash \forall x \phi$, so its logical unobjectionable)
If your question is only about how mathematicians write, sometimes I see a clever avoidance of $\exists$: $p(x)$ for some $x \in A$
A: If you really want to avoid using quantifiers, from a logical point of view (I mean logically, not in the sense of logic theory, I don't know much about that) the statements
$$
\forall x \in A : p(x) \qquad \text{ and } \qquad  x \in A \,\, \Rightarrow \,\, p(x)
$$
are equivalent just as 
$$
\exists x \in A : p(x) \qquad \text{ and } \qquad  x \in A \,\, \nRightarrow \,\, \neg p(x)
$$
are equivalent. I meant if you want to express yourself ; technically speaking the quantifiers are just not explicited but they are always there. Either way, if you want to avoid writing quantifiers, I highly suggest either the use of words or partial mathematical formulas (in the sense that you can put the quantifiers into words and put $p(x)$ in formulas for instance). 
Hope that helps,
A: Considering the other answers, I conclude that it really is possible to exclude quantifiers with limitations of the form $x\in X$ ($\forall x\in X$ and $\exists x\in X$) without loss of information, when properly distinguishes between bounded and free variables. Obviously 
$\forall x\in X:P(x)\quad\Leftrightarrow\quad a\in X\Rightarrow P(a)$
and here there are no problems with the interpretation of the alternatives. Less obvious  is
$\exists x\in X:P(x)\quad\Leftrightarrow\quad a\in X\wedge P(a)$
but successive reversible transformations shows:
$\exists x\in X:P(x)\;\Leftrightarrow\;\neg\forall x\in X:\neg P(x)\;\Leftrightarrow\;
\neg(a\in X\Rightarrow\neg P(a))\;\Leftrightarrow\;\neg(a\notin X\vee\neg P(a))\;\Leftrightarrow\;$$(a\in X\wedge P(a))$.
Not all the steps are equally easy to interpret, but never the less they are logical equivalents and it is more a psychological issue that it may be difficult to interpret a specific step in the chain. But the logical machinery works consistent. If one is interested of the interpretation of the statement one have to transform the statement to a more readable form ($\mathbb N_m=\{x\in\mathbb N|x>m\}$ and $\oplus$ is exclusive or below):
$1\oplus(\varepsilon\in\mathbb R_+)\oplus(\varepsilon\in\mathbb R_+)(m\in\mathbb N)
\oplus(\varepsilon\in\mathbb R_+)(m\in\mathbb N)(n\in\mathbb N_m)\oplus$
$(n\in\mathbb N_m)(|x-x_n|<\varepsilon)\Leftrightarrow$
$(\varepsilon\notin\mathbb R_+)\oplus(\varepsilon\in\mathbb R_+)(m\in\mathbb N)
(1+(n\in\mathbb N_m))\oplus(n\in\mathbb N_m)(|x-x_n|<\varepsilon)\Leftrightarrow$
$(\varepsilon\notin\mathbb R_+)\oplus(\varepsilon\in\mathbb R_+)(m\in\mathbb N)
(n\notin\mathbb N_m)\oplus(n\in\mathbb N_m)(|x-x_n|<\varepsilon)\Leftrightarrow$
$(\varepsilon\notin\mathbb R_+)\vee (m\in\mathbb N\wedge n\notin\mathbb N_m)\vee
(m\in\mathbb N\wedge|x-x_n|<\varepsilon)\Leftrightarrow$
$\varepsilon\in\mathbb R_+\Rightarrow m\in\mathbb N\wedge(n\in\mathbb N_m\Rightarrow |x-x_n|<\varepsilon)\Leftrightarrow$
$\forall\epsilon\in\mathbb R_+\exists k\in\mathbb N:n\in\mathbb N_{k}\Rightarrow |x-x_n|<\epsilon\Leftrightarrow \displaystyle \lim_{i\rightarrow\infty}x_i=x$.
Thus, the logic in (non meta) mathematics is substantially propositional logic.
In general predicate logic it is not always possible to eliminate quantifiers.
