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Answer by Asaf Karagila for Definability and the Separation and Replacement Axiom Schemata

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Henning answered the original question fairly well. Let me address the clarification.

The proof that Replacement implies Separation can be really just put into reverse gear here. If Separation wasn't a consequence of $\sf ZF-S$, then you can find some formula $\varphi(x,p_1,\ldots,p_n)$ whose corresponding separation axiom is false in a model of $\sf ZF-S$.

It's not just "some" consequence. You can immediately say that this "consequence" is one of the separation axioms, and thus find some formula from which that axiom was defined.

Now just repeat the usual proof that Replacement implies Separation applied to that formula. And we will have that the statement we thought is consistently false with $\sf ZF-S$ is really provable from that theory. So it's a contradiction.

But of course, all that really just shows that you should prove this directly. Like the usual proof goes.


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