Axiom of empty set In axiomatic set theory, we have Axiom of empty set: $\exists \varnothing \forall x ( x\notin \varnothing)$. Is there any equivalent statement without the use of quantifiers? For example, $ \exists x\in y (P (x)) $ is equivalent to $ x\in y\: \And\: P (x) $ and $\forall x\in y (P (x) $ is equivalent to $ x\in y\Rightarrow P (x) $ (or are they?). Any try to clear things up is appreciated.
 A: 1) '$\exists \varnothing \forall x ( x\notin \varnothing)$' is ill-formed. '$\varnothing$' is a constant, not a variable. 
2) Extensionality tells you that any sets $a$ and $b$ which lack members, if such sets exist, are the same set. So adopt the axiom that there is a set which lacks members: $\exists y \forall x ( x\notin y)$. It will now follow, given that consequence of extensionality, that $\exists! y \forall x ( x\notin y)$ -- there is a unique empty set. So that justifies introducing a constant `$\varnothing$' to denote the unique empty set.
3) Note, in introducing the empty set that way, you do need an existential axiom.
4) You could, I suppose, instead build set theory using a classical first-order language with the signature $\{\in, \varnothing\}$ from the start, and then just make do with the universal axiom $\forall x\, x \notin \varnothing$: but that would be a bit cheaty -- as you'd be presupposing the useful but not compulsory convention that all constants denote.  
5) Re: 'A statement must include some form of (implicit or explicit) quantifier.' Really? '$\neg (\varnothing \in \varnothing)$' is a perfectly good quantifier free statement if $\varnothing$ is introduced -- as is often the case -- as a constant. [The justification for the introduction is something quantificational, but what is introduced is a constant.]
A: What you wrote about bounded quantifiers is very wrong.
$\forall x\in y\varphi(x)$ is not $x\in y\rightarrow\varphi(x)$. It's an abbreviation for $\forall x(x\in y\rightarrow\varphi(x))$. Similarly $\exists x\in y\varphi(x)$ is abbreviation for $\exists x(x\in y\land\varphi(x))$.
The axiom asserts existence, so you cannot avoid $\exists x(\ldots)$ in its form (unless you prefer $\lnot\forall x(\ldots)$ instead). 
A: Deduced from your given example, one could form
$$ x \text{ is an Element} \Rightarrow x\notin\emptyset$$
but that actually just omits $\forall \text{Elements } x\ : x\notin\emptyset$. As said in the Comments, a statement must include some form of (implicit or explicit) quantifier.
A: Any statement without quantifiers, e.g. $x\cdot x=x^2$ is at most equivalent to a corresponding statement with allquantors (e.g. $\forall x\colon x\cdot x=x^2$) because the usual rules of inference include both $\phi(x)\vdash \forall x\colon \phi(x)$ and $\forall x\colon \phi(x)\vdash \phi(x)$, but cannot guarantee the existence of an object. 
The main purpose of an Axiom of Empty set is the existence of a set at all, for once you have any set $a$ (which may also follw from other axioms such as the Axiom of Infinity, if included) you can define $\emptyset := \{\,x\in a\mid \neg (x=x)\,\}$ using the other axioms. But you can never get existence from a formula introduced with only allquantors (or a formula without quantors, having implicit allqantors)
