In general, mathematical conjectures are resolved by proof, disproof, or proof that they are neither provable nor disprovable.

Is it possible that some open conjectures cannot be settled in any of these ways? That is, are there statements that are independent of, say, $\mathrm{ZFC}$, but not provably independent?

  • $\begingroup$ Is this related to the notion: can it be proven that it can be proven that this theorem can/cannot be proven true or false? $\endgroup$ Jun 12, 2013 at 1:14
  • 3
    $\begingroup$ Provable from what assumptions? It's already true that ZFC cannot prove that the continuum hypothesis is independent of ZFC (since this statement implies that ZFC is consistent); I believe the theory that proves that the continuum hypothesis is independent is ZFC + Con(ZFC). If you want an example of a statement independent of ZFC but such that ZFC + Con(ZC) can't prove it, consider Con(ZFC + Con(ZFC)) (this assumes that ZFC + Con(ZFC) is in fact consistent). $\endgroup$ Jun 12, 2013 at 1:49
  • $\begingroup$ You're right. I should be more careful about that.Does this get the job done?: Can it be proven in ZFC+Con(ZFC)+Con(ZFC+Con(ZFC)) that there are statements independent of ZFC that cannot be proven so in ZFC+Con(ZFC)? $\endgroup$
    – dfeuer
    Jun 12, 2013 at 1:56

1 Answer 1


There is always in increase in strength, relative to a fixed theory, to prove that something is unprovable in that theory. It follows from the incompleteness theorems that for a sufficiently strong effective theory $T$ (say ZFC), if $T$ proves "$\phi$ is not provable in $T$" for one sentence $\phi$, then $T$ proves $\text{Con}(T)$ and $T$ is inconsistent.

So whenever we want to prove "$\phi$ is not provable in $T$", we have to do so by means not formalizable in $T$. We can add an assumption such as $\text{Con}(T)$, or we can extend $T$ in other ways, for example if $T$ is Peano arithmetic we might prove the independence result in ZFC.

Thus the way that we know a sentence to be independent of ZFC is by proving that independence in some stronger theory. So, while there may be statements independent of ZFC which we have not yet proved are independent of ZFC, we will not know they are independent, so we will not be able to give them as concrete examples for the question being asked.

It is possible to give plenty of examples of particular theories $S$ extending ZFC, and sentences $\phi$ independent of ZFC, so that $S$ cannot prove that $\phi$ is independent of ZFC. The easiest way is to take $\phi$ to be $\text{Con}(S)$ for some theory $S$ whose consistency is not provable in ZFC, e.g. letting $S$ be ZFC plus the existence of a measurable cardinal. I would be interested to see whether other answerers can give more natural examples.

  • $\begingroup$ Does Con(T) mean that T is consistent? $\endgroup$ Jun 12, 2013 at 13:36
  • $\begingroup$ Yes: Con(T) is the canonical arithmetical sentence asserting the consistency of $T$, as in the incompleteness theorems $\endgroup$ Jun 12, 2013 at 15:21
  • $\begingroup$ I wasn't looking for a concrete example, but more a sense of whether there must be such a pathological case when only considering the "natural" extension of a consistency axiom. $\endgroup$
    – dfeuer
    Jun 12, 2013 at 15:51
  • $\begingroup$ This canonical arithmetical sentence sounds little scary. Could you explain or point some reference which says what is canonical arithmetical sentence? Otherwise I understand your answer. $\endgroup$ Jun 13, 2013 at 9:42
  • $\begingroup$ This reminds me of Boolos simplification of the incompleteness theorem. $\endgroup$
    – Asaf Karagila
    Jul 20, 2013 at 0:11

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