By Gödel's incompleteness theorem, we know that $ZFC+\neg\text{Con}(ZFC)$ is consistent (if $ZFC$ is consistent). However, as $ZFC$ is not finitely axiomatizable, what a model in $M$ of $ZFC+\neg\text{Con}(ZFC)$ thinks is $ZFC$ is not what $V$ thinks is $ZFC$. (For reference: Noah Schweber's answer here.)

However, there are finitely axiomatizable conservative extensions to $ZFC$, like von Neumann-Bernays-Gödel set theory ($NBG$). By Gödel's theorem, we should have that a model in the intended model (say $A$) of $NBG+\neg\text{Con}(NBG)$ is consistent (if $A$ is consistent). But how could this be if now the model $A$ and the model $NBG+\neg\text{Con}(NBG)$ agree on what $NBG$ is, as they both have finitely many axioms?

Also, does the result hold that every model of $NBG$ (including $NBG+\neg\text{Con}(NBG)$) has an element that is a model of $NBG$? It does for $ZFC$. (For reference: Joel David Hamkins's answer here.)

  • 1
    $\begingroup$ "[..] we know that $ZFC+\neg\text{Con}(ZFC)$ is consistent." If only that were true... $\endgroup$ Aug 28, 2017 at 18:56
  • 3
    $\begingroup$ @StefanMesken I presume that this question is assuming, as we do most of the time in logic, that ZFC is consistent. $\endgroup$ Aug 28, 2017 at 18:58
  • $\begingroup$ Fair point; edited. $\endgroup$
    – user474153
    Aug 28, 2017 at 18:58
  • 1
    $\begingroup$ @NoahSchweber I'm aware. But I liked it better as stated ;-) $\endgroup$ Aug 28, 2017 at 19:04
  • 2
    $\begingroup$ The problem is that in non-standard theories, you can have non-standard natural numbers. The proof of $0=1$ from a theory can therefore be of nonstandard length, according to the theory. $\endgroup$
    – PyRulez
    Feb 15, 2018 at 5:53

1 Answer 1


Remember where "nonstandard axioms" come from: nonstandard natural numbers which code first-order sentences which could be axioms. Similarly, natural numbers code proofs, and so different models will disagree about what sentences are provable.

The non-finite-axiomatizability of ZFC is relevant because of the reflection principle (I didn't do a good job of explaining this in my linked answer, but it's implicit in my reference to Hamkins' result), and if it were finitely axiomatizable this would contradict Godel's incompleteness theorem via Godel's completeness theorem - there have to be models of ZFC which see nonstandard axioms of ZFC in order to not have every model satisfy "every finite subset of ZFC is consistent." However, the general issue of models of a theory proving the inconsistency of that theory should be thought of as primarily about nonstandard proofs, not nonstandard axioms, since this applies to all appropriate theories.

Now, what about your question of whether, analogously to the ZFC situation, every model of NBG contains a model of NBG as an element? I believe the answer is no. First, note that the proof that the fact holds for ZFC makes crucial use of the reflection principle, which as remarked above fails for NBG (or indeed any finitely axiomatizable theory). So the only reason I know for that fact to be true doesn't hold in the NBG setting.

Now what about a negative answer to your question? Well, the point is that first-order satisfaction is absolute: if $\mathcal{A}$ is a structure in a model $M$ of NBG (or any appropriately strong theory) which classically satisfies $\varphi$, then $M\models$ "$\mathcal{A}\models\varphi$." So letting $\varphi$ be the conjunction of the finitely many NBG axioms, if there is a model of NBG contained in $M$ then $M$ knows that NBG is consistent; so any $M\models\neg$Con(NBG) cannot contain a model of NBG as an element.

It's worth pointing out that the absoluteness of satisfaction is a subtle point, and reasonable-sounding claims along those lines turn out to be false - this is examined in a paper of Hamkins and Yang. The precise statement I quoted above, however, is true, and its saving grace is the fact that $\varphi$ is assumed to be an actual first-ordersentence.

  • $\begingroup$ Oh, so a model of NBG+ ¬Con(NBG) will disagree on what is a valid proof, though they agree on the axioms? But I still don't understand whether every model of NBG will have an element that is a model of NBG. $\endgroup$
    – user474153
    Aug 28, 2017 at 18:57
  • 1
    $\begingroup$ @Zeno That's right. Offhand I don't have an answer to your further question, though, and I'm editing to explain why. EDIT: I think I got it, and the answer is "no." $\endgroup$ Aug 28, 2017 at 19:06
  • $\begingroup$ This makes sense. Thanks. $\endgroup$
    – user474153
    Aug 28, 2017 at 22:18

You must log in to answer this question.