A very simple question about the relationship between incompleteness and consistency of any theory strong enough to express PA.

We know from Gödel that if such a theory is consistent, then it is incomplete (First incompleteness theorem). We also know that, if such a theory is consistent, it cannot prove its own consistency (Second incompleteness theorem).

But we also know that if a theory is inconsistent, then it is complete. So, why can't we infer from the fact that a given theory T is incomplete that T is consistent? The reasoning is that if T is inconsistent, then we could not find statements that are not provable in it.

It seems this conclusion is false because it contradicts the Second incompleteness theorem. But where's the flaw in the reasoning? Possible explanations I could come up with so far:

  • It is trivially the case that if a theory is inconsistent, then it is complete. But within the inconsistent theory, one could both "prove" that T is both complete and incomplete, i.e. it is false that "we could not find statements that are not provable in it". So really nothing follows from the fact a theory is provably incomplete.

  • The reasoning is sound; the reason this does not work is that it cannot be done in T, but is only an observation about T in a metatheory (which would again be subject to the Incompleteness theorems)?


2 Answers 2


Yes, an incomplete theory is in fact consistent.

But Gödel's theorem is a theorem about $\sf ZFC$, or $\sf PA$, and it is done in the meta-theory. And in the meta-theory, we assume that $T$ is consistent. Otherwise, there's no point in investigating it, is there?

The "fact" that $\sf ZFC$ is incomplete follows from the fact that it is consistent. Namely, the correct fact is "If $\sf ZFC$ is consistent at all, then it is incomplete".

Just a minor remark, though, there is an additional requirement for the incompleteness theorems, that the theory is recursively enumerable. Otherwise, simply take $\operatorname{Th}(\Bbb N)$, and this is a complete theory which is consistent and interprets Peano arithmetic. In fact, this is how you prove that $\operatorname{Th}(\Bbb N)$ is not recursively enumerable.


There is also the important fact that the proof of Godel’s first incompleteness theorem assumes the consistency of PA/ZFC, and so deriving the consistency of such a system from the incompleteness theorem would be circular. Additionally, from a contradiction anything will follow- so proving any statement does not entail in the metatheory that the system is consistent.

  • $\begingroup$ Yes, I did say exactly that in my answer. That the correct "fact" would be "If ZFC is consistent at all, then it is incomplete". $\endgroup$
    – Asaf Karagila
    Mar 12, 2018 at 21:24
  • $\begingroup$ Yes; I was pointing out that not only is your answer correct, but what would be wrong with saying that consistency follows from the incompleteness of ZFC- such an argument would be circular. $\endgroup$ Mar 13, 2018 at 0:13
  • $\begingroup$ No, that's actually incorrect. An incomplete theory is almost by definition a consistent one, as it does not prove contradictions. However it is false to say that the incompleteness theorem implies the consistency of ZFC (or any theory it's applied to), because the incompleteness theorem is a conditional which is vacuously true if the theory is inconsistent. $\endgroup$
    – Asaf Karagila
    Mar 13, 2018 at 6:39
  • $\begingroup$ What you have mentioned is precisely what it means for an argument to be circular. Yes, the fact is a conditional: “If ZFC is consistent, then ZFC is incomplete.” The following is an incorrect proof of ZFC: “If ZFC is consistent , then ZFC is incomplete. Therefore ZFC is consistent.” It is incorrect because, as you say, the antecedent of the first conditional could be false. That is exactly what it means for the argument to fail because it is circular. I believe we are actually on the same page. $\endgroup$ Mar 13, 2018 at 14:37
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    $\begingroup$ No of course you did not. I was not critiquing your post; I was pointing out what was wrong with the OP’s original attempted proof that the incompleteness theorem implies that ZFC is consistent. $\endgroup$ Mar 13, 2018 at 14:45

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