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xxxx0xxxx said:I disagree on the following point.
I overlooked that you are referring to: y∈{x | φ(x)} ↔ φ(y) as though it is a definition, when it is actually not.
y∈{x | φ(x)} ↔ φ(y) is a theorem of naive set theory that restates the axiom of abstraction using the {...|...} operator.
In ZF, this theorem admits Russell's paradox, and cannot be a theorem of ZF.
The correct theorem in ZF is:
[tex]y \in \{x | \phi (x) \} \Rightarrow \phi (y)[/tex]
The equivalence cannot be justified from the separation axiom, and must be an implication to avoid the paradox.
And as you may probably have already guessed, and operation definition for {x | φ(x)} is required in order to eliminate {x | φ(x)} entirely to wff's in the object language:
[tex]\{x | \phi (x) \} = w \Leftrightarrow [ \forall x (x \in w \Leftrightarrow \phi (x)) \wedge w \mbox{ is a set} ] \vee [\neg \exists B \forall x (x \in B \Leftrightarrow \phi (x)) \wedge w = \emptyset][/tex]
This definition along with
[tex]\{ x | \phi (x) \} \not = \emptyset[/tex]
proves the implication, but the converse
[tex]\phi (y) \Rightarrow y \in \{x | \phi (x) \}[/tex]
cannot be shown to follow from separation.
Yes, I understand you completely. But I think you fail to understand me. You're taking the point-of-view of Suppes. This is not standard terminology!
Please, read another book on set theory and you will see what I'm talking about.
I can make suggestion to you that will increase your understanding and point-of-views. But only you can follow them
Good luck on your research!