# Existential Quantifier as Predicate?

I was reading this document on axiomatic set theory, and on page 4, it defines $\exists \{x:P\}$ as an alternative notation for $\exists y \forall x P$. Since the first notation looks very much like a predicate, I was wondering if it was possible to treat the existential quantifier as a predicate (at least in set theory?).

For example, suppose $E(s)$ is the predicate "the set $s$ exists". This could be equivalent to $\exists xP(x)$, where $P(x)$ is the property of the set. This in turn could be equivalent to $\exists x \forall y [Q(y) \rightarrow (y \in x)]$, where $Q(y)$ is the property that the elements of the set must have.

To make things concrete:

• $E(\text{positive even integers})$ $\iff$ "the set of positive even integers exists"

• $\exists xP_{\text{positive even integers}}(x)$ $\iff$ "there is a set $x$ such that the statement '$x$ is the set of positive even integers' is true"

• $\exists x \forall y [Q_{\text{positive even integer}}(y) \rightarrow (y \in x)]$ $\iff$ "there is a set $x$ such that for all sets $y$, the statement 'if $y$ is an even positive integer, then $y$ is in $x$' is true"

If the above is possible, then $\exists s E(s)$ would mean "some set exists". Can this be expressed in the other two forms?

I apologize if this is nonsense, but I'm just curious.

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I have never seen this new notation you've provided before, but then again, I don't see why it doesn't make sense. –  Samuel Reid Jan 3 '12 at 19:20
@russell11: I'm afraid the answer has to be no. A predicate operates on variables ranging over the elements of the model, whereas the $P$ in $\exists \{ x : P \}$ is a ‘variable’ ranging over predicates of the language. In order to turn this into something that makes sense one has to internalise the language (i.e. represent formulae as sets) and then internalise the truth predicate. But then you run into significant difficulties... –  Zhen Lin Jan 4 '12 at 2:24