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The well-known Axiom Schema of Replacement in ZFC says that for any formula $\varphi$ of the Set Theory with free variables among $w_1,\dots,w_n,A,x,y$ the following holds:

$$\forall w_1,\dots,w_n\;\forall A ([\forall x\in A\;\exists !y\;(\phi(x,y,w_1,\dots,w_n,A)]\Rightarrow \exists B\;[\forall x\in A\;\exists y\in B\;\;\varphi(x,y,w_1,\dots,w_n,A)]),$$ where $B$ is not free for $\varphi$.

I am interested in the meaning of the quantifier $\exists !$ in the above formula.

To illustrate the problem, let us consider the formula $\varphi$ with two free variables $x,y$: $$\mbox{($y$ is a cardinal) $\wedge$ $(|x|\le |y|<|\mathcal P(x)|)$}.$$ This formula defines a function assigning to each set $x$ its cardinal $y=|x|$ if and only if GCH (the Generalized Continuum Hypothesis) holds. So, for this formula the quantifier $\exists !$ holds if and only if GCH is true. This example shows that there are at least two meanings of the Axiom Schema of Replacement: the standard one and the absolute, in which we take into account only the formulas which determine a function in any model.

Question. Does this "absolute" version of the Axiom Schema of Replacement lead to an essentially different Set Theory comparing to the standard ZFC?

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    $\begingroup$ The "absolute" version is what one might call the "replacement rule": from a deduction of $\Gamma \vdash \forall x \in A\,\exists! y\,\ldots$ infer that $\Gamma \vdash \exists B\,\forall x \in A\,\exists y \in B\,\ldots$. A common motto in proof theory is that "rules are weaker than axioms". That said, I never considered this particular instance. $\endgroup$
    – François G. Dorais
    Commented Sep 14, 2022 at 23:10
  • $\begingroup$ There are two versions of this that I can see: the theory consisting of $\mathsf{ZC}$ + replacement for all $\varphi$ that $\mathsf{ZFC}$ prove define functions, or the smallest extension $\Gamma$ of $\mathsf{ZC}$ by replacement instances such that whenever $\Gamma$ proves that $\varphi$ defines a function we have the replacement instance for $\varphi$ in $\Gamma$. (This ignores parameters, but meh.) Which are you asking about? $\endgroup$
    – Noah Schweber
    Commented Sep 14, 2022 at 23:35
  • $\begingroup$ @NoahSchweber Probably the second option is more natural, but also more complicated to formulate. $\endgroup$
    – Taras Banakh
    Commented Sep 15, 2022 at 0:56
  • $\begingroup$ I don't think the unique existential quantifier holds only if GCH is true, without assuming anything about whether there are cardinals between $\vert x\vert$ and $\vert\mathcal P(x)\vert$ we may let $y$ be the cardinal $\vert x\vert$ (using the von Neumann cardinal assignment) and we have $\varphi(x,y)$. $\endgroup$
    – C7X
    Commented Sep 15, 2022 at 1:48
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    $\begingroup$ These are equivalent. Replacement for a formula $\phi(x,y,\dots)$ follows from "absolute replacement" for the formula $\forall y'\,(\phi(x,y',\dots)\leftrightarrow y'=y)\lor(y=\varnothing\land\neg\exists!y'\,\phi(x,y',\dots))$ (i.e., "if there is a unique element satisfying $\phi$, then $y$ is this element, otherwise $y$ is a canonical fixed element"). $\endgroup$
    – Emil Jeřábek
    Commented Sep 15, 2022 at 9:39

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