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I know there are multiple different formulations for the axiom schema of replacement, some of which are equivalent and others not or only under other axioms. But all of them look something like this:

If $\phi$ is a well-formed formula where $x$, $y$, $A$, and $\vec w$ are free variables, then $$ \forall A\forall\vec w\quad (\forall x\in A\,\forall y\forall y'\ \phi(x,y,A,\vec w)\wedge\phi(x,y',A,\vec w)\Rightarrow y=y') \Rightarrow\exists B\forall y(y\in B\Leftrightarrow\exists x\in A\ \phi(x,y,A,\vec w)) $$

Specifically, all of the formulations that I have come across have $A$ as a variable in $\phi$.

My question is, why is that? Why is it usefull to do so? And if one were to not put $A$ as a variable in $\phi$, like so

If $\phi$ is a well-formed formula where $\mathbf x$, $\mathbf y$, and $\mathbf {\vec w}$ are free variables, then $$ \forall A\forall\vec w\quad (\forall x\in A\,\forall y\forall y'\ \mathbf {\phi(x,y,\vec w)}\wedge\mathbf {\phi(x,y',\vec w)}\Rightarrow y=y') \Rightarrow\exists B\forall y(y\in B\Leftrightarrow\exists x\in A\ \mathbf{\phi(x,y,\vec w)}) $$

would that formulation not be equivalent to the one above?

I found discussions about this in the answers and comments of this and this question, but ultimately it is never clarified why it is there.

Thomas.M
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  • @BrianM.Scott I've updated the question to include it. – Thomas.M Jul 06 '20 at 17:01
  • Conceptually, you want it there because sometimes you want the (partial) function defined by $\phi$ to depend on the domain to which it's being applied. For instance, the first version of replacement can directly get us, for any $A$, the set of cardinalities of elements of $A$ that are less than $|A|$; the second form cannot. But in this particular example the use of $A$ as a parameter isn't essential; maybe someone else can think of a case where it would be. – Malice Vidrine Jul 06 '20 at 23:51
  • @MaliceVidrine Wait, you said that the second form cannot get the set you described, but're also saying that having $A$ as a parameter isn't needed for that? Wouldn't that mean that the second form can in fact get that set? This seems like a contradiction to me, but maybe I'm missing something. – Thomas.M Jul 07 '20 at 03:39
  • I didn't say you can't get the existence of the same set from the second version-- I said you can't get it directly. In the second case, you can make up for the deficiency in the replacement schema if you have Separation as well, but you don't get the existence of the set I describe solely as an instance of the schema you lay out (at least not in any way I can see). By contrast, you can get the set I describe by applying a single instance of the usual schema. – Malice Vidrine Jul 07 '20 at 04:01
  • Oh okay, I see what you mean. – Thomas.M Jul 07 '20 at 04:37

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