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It is a remarkable fact that the $\in$-induction scheme (i.e. the claim that $\phi(x)$ holds for any $x$, whenever $\phi(\emptyset)$ and $\forall x(\forall y\in x \phi(y)) \Rightarrow \phi(x)$) is equivalent (I mean in ZFC minus foundation) to the axiom of foundation, which is a single axiom not a scheme.

One can ask if similar properties hold for the other axiom schemes in ZFC ; thus :

  • Is there a sentence $\alpha$ such that the replacement scheme is equivalent in ($ZFC$ minus replacement) to $\alpha$ ?

  • Is there a sentence $\alpha$ such that the separation scheme is equivalent in ($ZFC$ minus replacement) to $\alpha$ ? (here I consider $ZFC$ minus replacement instead of $ZFC$ minus separation, because it is well known that replacement implies separation).

The answer to both those questions are NO, as far as I can remember (else set theorists would probably put the non-scheme version of the axiom instead of the scheme, in the definition of ZFC).

Can anyone clarify this ?

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First of all, this depends greatly on your axiomatization. Since Replacement implies Separation, you can reduce the Separation schema to $\exists x(x=x)$ in presence of Replacement.

So we're really just left with Replacement. Now here's the kicker. You cannot have a finite axiomatization of $\sf ZFC$. Unless of course the theory itself is inconsistent. The reason is that $\sf ZFC$ proves the consistency of any finite number of its consequences.

So you can't quite reduce Replacement to a single schema in the presence of $\sf ZF-Rep$.

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