If, one fine day, someone found a contradiction in ZFC (or even ZF), what implications would such an event have for mathematicians? Is there currently any backup axiomatic system on par with ZFC that would enable mathematicians to continue pursuing mathematics, should the unthinkable occur?

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    $\begingroup$ Most mathematicians will not even notice that ;-) since most mathematicians very rarely use ZFC. IMHO, most likely — if ZFC turns out to be inconsistent — it would be possible to slightly restrict the replacement axiom and get a “consistent” axiom system in which all mathematics can be formalized (of course, one cannot be sure that the new axiom system is consistent). $\endgroup$
    – Yury
    Nov 29, 2012 at 5:05
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    $\begingroup$ For most mathematicians it would have little or no practical effect, and no backup axiomatic system would actually be required save as a security blanket. Those who much prefer a set-theoretic foundation to any other (e.g., category-theoretic) sort would doubtless attempt to find a reasonably natural weakening of ZFC that avoided the contradiction and preserved as much of ZFC as possible. $\endgroup$ Nov 29, 2012 at 5:31
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    $\begingroup$ The sets one needs in almost all of mathematics use only a small fragment of the power of ZFC. $\endgroup$ Nov 29, 2012 at 5:53
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    $\begingroup$ Some friends of mine and I like to imagine that a small group of mathematicians has found an inconsistency in ZFC, and that they use it in very subtle ways (so that the inconsistency will remain hidden) to prove some great theorems. $\endgroup$ Jan 5, 2014 at 7:04
  • $\begingroup$ There is a whole hierarchy of foundational systems that we can fall back upon. Currently almost all logicians believe that nobody will ever find an inconsistency in PA, and predicativists mostly believe that ACA and even ATR are consistent. Some logicians are not confident in the consistency of systems beyond that point, starting with $Π_1^1$-CA and going up to full second-order arithmetic Z2. I happen to be of the opinion that not only Z2 but even full higher-order arithmetic HOA is consistent, and that almost every real-world application of mathematics can be proven in HOA. So... $\endgroup$
    – user21820
    Sep 13, 2018 at 14:36

4 Answers 4


The way I see it there are two possible outcomes:

  1. The contradiction will come from a very concrete and identifiable part of the system, e.g. power set axiom and some replacement, which we can remove and replace by milder variants -- much like we did with Frege's comprehension schema. We will then go through the proofs and verify which of the proofs cannot be modified, and hopefully most of the big proofs will go through, in which case there will be some change, but not too much.

  2. The contradiction will come from a much weaker theory, e.g. Peano Axioms or some other mathematically useful theory. In this case it's more than set theory which has a problem. It would mean that many people will now begin to look for alternatives, possibly in several fields. I suppose that set theorists will also try to find themselves an improved theory.

In either cases there will be a lot of work to see which fragments of ZFC we can rescue, and how many of the proofs would go through. Note that Russell's paradox, at the time, wasn't considered as a big deal by non-logicians. So while the second option would affect many mathematicians, the first option would be largely ignored by most (or it will be used as a source to kick set theory around).

If at some point we conclude that we cannot rescue a lot of the reasonable proofs (e.g. some inner model theory, forcing, etc.) then it is possible that ZFC will be just dropped and a new basis will be sought.

Of course it is possible that just something about large cardinals would be inconsistent, and then we will be able to draw a lot of conclusions about the universe of sets (for example, all propositions which imply the consistency of this cardinal would be immediately false. If you have found a contradiction in a measurable cardinal then you can prove that AD is inconsistent with ZF). But that's still quite far from showing that ZFC itself is inconsistent.

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    $\begingroup$ I could also see some people who hate set theory and prefer categories gloating, and returning to use their TG+Universes theory, only to find it is inconsistent. :-) $\endgroup$
    – Asaf Karagila
    Nov 29, 2012 at 8:29
  • $\begingroup$ Haha... This is pretty amusing. $\endgroup$ Nov 29, 2012 at 15:30
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    $\begingroup$ Just for the record, I have not problem with ZF. $\endgroup$ Nov 30, 2012 at 8:06

There was a quite relevant post at here:


and the original lecture is worth watching.

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    $\begingroup$ In particular I thought the last part of Timothy Chow's comment was pretty much spot-on: "For most mathematicians, "ZFC" is just an arbitrary trigraph that is cited when the need arises to specify a particular foundation for mathematics. I daresay many people who toss the trigraph around couldn't even state all the axioms of ZFC precisely. If we scale back to some other system that goes by some other trigraph, it won't take much retraining to learn the new trigraph. For most researchers, that will be the only impact on their day-to-day work." $\endgroup$ Nov 29, 2012 at 9:08

I'm personally a fan of ETCS. See, for example, Todd Trimble's blog post on the subject.

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    $\begingroup$ Aren't most "AST"'s equiconsistent with ZFC (or stronger)? If so it would be like saying "Oh, ZFC? I don't care about that. I only want to use ZF" :-) $\endgroup$
    – Asaf Karagila
    Nov 29, 2012 at 8:15
  • $\begingroup$ The nLab (ncatlab.org/nlab/show/ZFC) claims that ETCS is equivalent to BZC, which appears to be weaker than ZFC. $\endgroup$ Nov 29, 2012 at 8:32
  • $\begingroup$ ETCS is a somewhat weak set theory, as I noted in my recent question. The category of sets in $V_{\omega + \omega}$ gives a (small!) model of ETCS. I haven't studied algebraic set theory yet, but if I understand what I have heard correctly, it should turn out to be equivalent in strength to a classful theory such as NBG or MK. $\endgroup$
    – Zhen Lin
    Nov 29, 2012 at 9:01

Note that an inconsistency implies that any theorem and its negation can be proven. The fact such a thing hasn't happened (no one has ever proven the negation of a theorem in mainstream math) means that wherever ZFC fails, it is almost certain that a more restricted set of axioms can be found where most math can still be done.

To see an example of what could happen you have to look no farther than Russell's Paradox. It shows that naive set theory makes no sense, but the example is far removed from the regular sets (with "actual elements") that we all usually use, and axioms where set theory works were found.


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