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The question is simple: do we know if there are non-trivial axiom collections stronger than (imply but are not implied by) $ZFC$?

To clarify what I mean: Do we know a way of replacing the axioms in ZFC by other axioms that do not resemble ZFC and such that, together, imply ZFC, but none of whom are implied by ZFC?

To be even clearer of what constitutes an answer for me, suppose we could make up 5 new axioms such that from these we can prove the 8 axioms of ZFC, but starting with the axioms of ZFC we can`t prove any of these 5 new axioms.

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    $\begingroup$ Presumably, you mean something other than ZFC with some known independent axiom(s) added (like the Continuum Hypothesis.) $\endgroup$ – Thomas Andrews Jan 7 '16 at 15:54
  • $\begingroup$ Yes, I mean other than ZFC. But I do not mean to add new axioms, What I mean is replacing the axioms in ZFC by other axioms that, together, imply ZFC, but none of whom are implied by ZFC. $\endgroup$ – Henrique Tyrrell Jan 7 '16 at 16:00
  • $\begingroup$ You can at least drop the axiom of choice and take the generalised continuum hypothesis. Then ZF + GCH imply AC. To make a guess: I think it is possible. $\endgroup$ – Paul K Jan 7 '16 at 16:08
  • $\begingroup$ Beside the point, but to mentioned: ZFC contains an infinite number of axioms. SUB and SEP are a schemes of axioms. $\endgroup$ – drhab Jan 7 '16 at 16:10
  • $\begingroup$ The idea behind this question is basically just a curiosity. Informally speaking, we can use ZFC to prove the propositions in set theory and then construct the rest of mathematics. So I thought, can we invert this idea and find the step that comes "before" ZFC? And then maybe find the step before that... $\endgroup$ – Henrique Tyrrell Jan 7 '16 at 16:51
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I believe Morse-Kelly set theory may satisfy your requirements. While it shares some axioms with ZFC, each theory contains axioms that the other does not, so MK is not simply ZFC + some other axiom, and MK is a second-order theory (sets and classes) while ZFC is only first-order (just sets). Most importantly MK is a proper extension of ZFC, in other words $MK \vdash Con(ZFC)$ but not the other way around.

I've had little experience with MK myself so I'm sure others can chime in about more details, but this article has a very well-written overview of the above statement and some consequences. For example, it points out that ZFC + "there exists an inaccessible cardinal" $\vdash Con(MK)$, so MK is not "much" stronger than ZFC.

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Let P be a proposition which is independent of ZFC, and for each axiom A of ZFC, replace it with "A and P".

It's not what you want, but I don't think you're going to be able to formulate a very well-defined question for what you want.

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    $\begingroup$ @downvoter Thank you for providing an explanation for me. Much appreciated. $\endgroup$ – Dustan Levenstein Jan 7 '16 at 18:40

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