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Usually, the axiom schemata of separation and replacement are stated for a given formula $\varphi(x,u_1,\dots,u_n)$. Would it be wrong if I just stated them for a formula $\varphi(x)$? Can't we just take $x$ to be an ordered $(n+1)$-tuple, so that the parameters need not be specified explicitly? Maybe it would be better to write $\varphi(x,u)$ and take $u$ to be an $n$-tuple, so that we can quantify over $x$, but still the underlying question remains the same.

Is this line of reasoning wrong? I did find an old conversation about this in the comments of this answer, but the discussion did not come to a clear conclusion and also it was mentioned in the answer that the proof of the equivalence between the versions of replacement with and without parameters is very complex, which my proof using tuples is not. Am I getting this wrong?

Elvis
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  • Tupling gives a simple way to reduce an arbitrary number of parameters to a single parameter, but I don't see any basis to the idea that it can reduce it to no parameters. – spaceisdarkgreen Dec 22 '24 at 02:29
  • @spaceisdarkgreen okay, so the $\varphi(x,u)$ way is the correct one? – Elvis Dec 22 '24 at 02:44
  • Yes, like you can take a separatation instance ${x\in a: \varphi(x,u_0,\ldots, u_n)}$ and define $\varphi'(x, u) := \varphi(x,(u)_0,\ldots, (u)_n)$ and then you have the same set expressed ${x\in a: \varphi'(x, \langle u_0,\ldots, u_n\rangle)}$ as an instance of separation with one parameter. – spaceisdarkgreen Dec 22 '24 at 02:57

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