I've heard of many examples of statements that have been proven to be independent of a formal system, meaning that they can't be proven within that formal system (for example, the Continuum Hypothesis is independent of ZFC). I've also heard of examples of proofs showing that certain proof techniques are not powerful enough to prove certain results, such as the result that proofs that relativize cannot resolve P versus NP.

This Stack Overflow question (which might be soon closed or migrated) asks whether we know whether it's possible to resolve P versus NP or not. That got me thinking of whether there are any examples of conjectures where we know that the conjecture can either be proven or disproven within some formal system, but where we don't actually know whether the conjecture is true or not. In other words, are there any conjectures where we can nonconstructively argue that the conjecture is either provable or disprovable, but where we don't know which of these is the case?

I apologize if this is a silly question - I don't have much background in proof theory or model theory.


  • $\begingroup$ It's not at all silly.It'a a good Q. $\endgroup$ Aug 8, 2015 at 23:46
  • $\begingroup$ What happened to the bounty here? $\endgroup$ Aug 11, 2015 at 18:29

3 Answers 3


It's sometimes possible to show existence of proof of a statement in ZF by showing that the statement is true in ZFC, and arguing using absoluteness.

An example would be answer to this question, in which Andreas Blass argues as follows: for any model of ZF, Monsky's theorem holds in an inner model satisfying axiom of choice, and this statement is "simple enough" (involves only finitely many reals) to apply absoluteness, so it held in original model as well. So every model of ZF satisfies Monsky's theorem, hency by Godel's completeness theorem, ZF proves Monsky's theorem.

Note that this argument does not provide us a proof in ZF of Monsky's theorem, it only provides a proof in ZF that there is a proof in ZF on this theorem.

Edit: Actually, I have realized that I haven't answered exactly the question you asked (I thought you are asking for proof of existence of proof without providing one). I think the above proof technique might be applicable for your actual question, so I'll leave the answer here for now.

  • $\begingroup$ Note that proving that ZF proves something is actually much weaker than giving an explicit way to write down the proof in ZF. This is because if ZF is $Σ_1$-unsound it can prove that it proves something even if there is in fact no such proof! In the worst case, such as for $\def\con{\text{Con}}$$\def\zf{\text{ZF}}$$T = \zf + \neg \con(\zf)$, $T$ proves that $T$ itself proves everything! This is despite $T$ being consistent... $\endgroup$
    – user21820
    Jul 28, 2016 at 11:41

Many. Such questions amount to some finite computation, since you can just search for a proof and simultaneously disproof until you find (enumerate all proofs and check if its a proof or disproof). So just pick one which requires a large computation such as; Is there an odd number of pairs of twin primes below 10^999999999999999 ? This is provable, just count them.

I cant prove there isnt some shortcut to this, but there are some statements where you can prove that the proof or disproof is longer then n for any n you choose, see here: https://en.wikipedia.org/wiki/G%C3%B6del%27s_speed-up_theorem

  • $\begingroup$ This is a good point. To make this have more "gravitas," is there a nontrivial conjecture that we've reduced to searching a massive space? $\endgroup$ Aug 3, 2015 at 0:05
  • $\begingroup$ @templatetypedef (sorry for that kind of a reply) There is an MO question precisely answering your question: mathoverflow.net/questions/112097/… $\endgroup$
    – Wojowu
    Aug 12, 2015 at 20:29

Your question asks whether there is a sentence $Q$ such that (within the conventional foundational system $S$ for mathematics) we can prove ( $S$ proves $Q$ or $S$ proves $\neg Q$ ) but as of now can neither prove $Q$ nor prove $\neg Q$. The answer is "yes", and it is easy to construct one even if we restrict $Q$ to be an arithmetical sentence.

Fix a Turing-complete programming language whose programs and input/output are binary strings.

$Q$: Over all pairs $(p,x)$ of binary strings of length at most $5^{10009}$ such that $p$ halts on $x$ within $7^{10013}$ steps, the largest output of $p(x)$ is a multiple of $3$.

We can easily prove that even a weak system like PA$^-$ or TC either proves $Q$ or proves $\neg Q$, because $Q$ is a statement about finitely many programs each run for finitely many steps. However, absolutely nobody has any clue whether $Q$ is true or false!


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