Working in first order languge $\mathcal L(\in, W)$, where $W$ is a constant symbol.

Reflection: if $\varphi$ is a formula in $\mathcal L(\in)$, in which $x$ is free, and $\vec{p}$ is the string of all of its parameters, and if $\psi$ is a formula in which $z$ is free and $y$ not free, then all closures of: $$\vec{p} \in W \wedge \exists x (\varphi) \to \exists x [\varphi\wedge \exists y \in W \ \forall z (z \in y \leftrightarrow z \in x \wedge \psi) ] $$; are axioms.

Now with Extensionality this would interpret all axioms of $\text{ZFC}$ since it would trivially interpret Harvey Friedman $\text{K(W)}$ theory (page 3). Although I'm not sure but I think even Extensionality is interpretable using the above scheme only.

Has there been prior attempts to non-trivially shortly axiomatize ZFC with a single comprehension schema? and had this schema been studied before specially in absence of Extensionality?

  • $\begingroup$ Well, the Replacement scheme implies the Separation scheme, so the answer to your first question is immediately "yes." $\endgroup$ – Noah Schweber Jan 25 at 18:51
  • $\begingroup$ To me all of the axioms of set union, power and infinity, all are individual comprehension axioms. What is not a comprehension axiom is Extensionality, Foundation and Choice. And by the way Replacement doesn't interpret all the others $\endgroup$ – Zuhair Jan 25 at 19:10

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