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  • $\begingroup$ but this theory proves existence of finite von Neumann ordinals? $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 23, 2018 at 11:10
  • 2
    $\begingroup$ No, it doesn't. It doesn't even prove the existence of the empty set. $\endgroup$
    – Joel David Hamkins
    Commented Jun 23, 2018 at 11:11
  • 2
    $\begingroup$ You can't separate on a set unless it is definable. In my model, every "set" contains all the ordinals, since there aren't any. But you are right, it doesn't have a set whose elements are exactly the finite ordinals. Tell me how you want to define "von Neumann ordinal" since the various usual formulations are likely no longer equivalent in your theory. $\endgroup$
    – Joel David Hamkins
    Commented Jun 23, 2018 at 11:20
  • 2
    $\begingroup$ Yes, your new infinity axiom is not true in my integer structure. And indeed, once you get infinity by your axiom, then you get the empty set and also many finite ordinals and also $P(\omega)$ and so on. $\endgroup$
    – Joel David Hamkins
    Commented Jun 23, 2018 at 11:52
  • 1
    $\begingroup$ OK, I have edited. $\endgroup$
    – Joel David Hamkins
    Commented Jun 23, 2018 at 12:32