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If I replace the below axiom in ZFC (or NBG) by the below inference rule, there are any consequence in what can be demonstrated?

Axiom: If two sets (or classes) have the same elements then their are equal. $$ \forall{ }A\forall{ }B(\forall{ }x((x\in A) \Leftrightarrow (x\in B))\Rightarrow A = B) $$

Inference Rule: From $\varphi(A)$ and $\forall{ }x((x\in A) \Leftrightarrow (x\in B))$ infers $\varphi(B)$. $$ \{\varphi(A), \forall{ }x((x\in A) \Leftrightarrow (x\in B))\} \vdash \varphi(B) $$

Comment

This substitution is based on Leibniz Law (but different of this Math.SE question do not quantify about predicates). This substitution have three motives:

  1. Play with logic.
  2. Remove the equality symbol.
  3. Remove the axiom (I like natural deduction, when possible): I imagine that this "formalism" have special significance for classes, since a class is defined by the logical property of their elements then one can imagine that the extension (see this Math.SE answer) is "not a truth" but only a "program" to perform a proof (see this Math.SE answer).
DanielWainfleet
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I.F.F. dos Santos
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    I think it might even be sufficient to use the special case $\forall A \forall B (\forall x (x\in A \leftrightarrow x \in B)) \rightarrow (\forall x (A \in x \leftrightarrow B \in x))$ as a replacement for the axiom. From there, you should be able to prove your inference rule for general $\varphi$ by structural induction (as a metatheorem). – Daniel Schepler Feb 21 '23 at 21:50
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    In both the axiom and the rule, & was presumably nintended to be $\leftrightarrow$. – Andreas Blass Feb 21 '23 at 22:45
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    @MauroALLEGRANZA The OP said in motive #2 that he wants to avoid the equality symbol. – Nico Feb 22 '23 at 23:43
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    I think that in ZF or ZFC you could replace every instance of $x=y$ with $\forall z, (z\in x\iff z\in y)$. It would be very cumbersome but apparently equivalent..... My edit was for spelling (bellow/below). Awl Inglish spelling iz broken and needz reepair. – DanielWainfleet Feb 27 '23 at 19:48

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