I am beginning to study Set Theory, so naturally one might begin with the axioms. When reading the schemata of separation and replacement, my initial thought was, "Why would we need separation if we have replacement? It seems like the sets which exist as a consequence of the separation axioms would also exist as a consequence of replacement." Sure enough, there is a question here on stack exchange asking exactly this. The answer seems to indicate yes, there is a proof indicating that replacement implies separation. Perhaps this is the case, but I'm not buying it yet.
I have a notion in my head that the set of definable functions available to the schema of replacement would be severely limited without already having the separation axioms. Perhaps the definable functions would be so limited that there might exist a set that exists due to separation that would be impossible to show exist due to the replacement axioms without separation. Or perhaps the two build on each other, in a sense that each set that exists due to a separation axiom spawns a set of new replacement axioms, each of which in turn define more sets that spawn more separation axioms, etc etc etc.
I believe working toward a proof of this kind may be very helpful in my understanding of the axioms, but I am having trouble setting up my proof. How do I formally talk about the set of definable functions available to a set theory axiom? There seems to be some weird self-reference going on here.
I would like to work in pure set-theory language: only variable symbols, $\in$, and first order logic symbols. Of course anything like $=$ and $\subset$ that are definable from these are allowed.
The axioms:
Null set: $\exists X \forall y (y \notin x)$
Extensionality: $\forall X \forall Y ( \forall z (z \in X \iff z \in Y) \implies X = Y$
Regularity: $\forall X ( (\exists a : a \in X) \implies \exists b (b \in X \land \neg \exists c (c \in b \land c \in X)))$
Separation: $\forall \vec{w} \forall X \exists Y \forall z (z \in Y \iff (z \in X \land \varphi(z, \vec{w}, X ) )) $
Pairing: $\forall x \forall y \exists z (x \in z \land y \in z)$
Union: $\forall X \exists Y \forall a \forall b ( (b \in a \land a \in X) \implies (b \in Y) ) $
Replacement: $\forall \vec{w} \forall A ( (\forall x \in A \exists ! y \varphi(x,y,\vec{w}, A) \implies (\exists B \forall y (y \in B \iff \exists x \in A \varphi(x,y,\vec{w}, A))))$
Infinity: $ \exists X ( \emptyset \in X \land \forall y (y \in X \implies y \cup \{ y \} \in X))$
Power Set: $\forall x \exists y \forall z (z \subseteq x \implies z \in y) $
Choice, if it makes a difference here.
Further clarification:
If separation follows from replacement, then every model of ZF is a model of ZF without separation (ZF-S).
Assume separation does not follow from replacement. Then there exists a $\psi$ which is a consequence of ZF, but not a consequence of ZF-S. In fact there might be a $\psi$ that is false in every model of ZF-S. When $\psi$ has something to do with the $\varphi$ included in the axioms of replacement, the same axiom will assert the existence of a set with different properties in ZF than it would in ZF-S. Seems reasonable to me. Where could I find a contradiction?
