I'm sure there are easy ways of proving things using, well... any other method besides this! But still, I'm curious to know whether it would be acceptable/if it has been done before?

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    Sure. For certain statements, you can even prove them by showing that there is no proof of their negation. – Andrés E. Caicedo Jan 14 '13 at 06:52
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    @AndresCaicedo not true. If you know you can't disprove something, then it's consistent, not proven. AC and ~AC are both consistent with ZF – John Dvorak Jan 14 '13 at 06:54
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    I'm wondering how would you _non-constructively_ prove that a proof exists. The proof of a proof would then count as a proof of the original concept. – John Dvorak Jan 14 '13 at 06:57
  • @Jan Dvorak Let $P$ be a statement that one might want to prove. Now I can think about proving that a proof for statment $S$ exists. Where $S$ is the statement that a proof for $P$ exists – Amr Jan 14 '13 at 07:00
  • @Amr I mean, in practice. The theory is perfectly clear. – John Dvorak Jan 14 '13 at 07:01
  • @AndresCaicedo note there is a difference between "no proof of negation" and "provably no case where it doesn't hold". – John Dvorak Jan 14 '13 at 07:02
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    @Jan Dvorak I understand your point The interesting question is "Are there any known theorems that use this proof-strategy in their proof"? – Amr Jan 14 '13 at 07:03
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    @JanDvorak I am well aware of these issues, of course. The statement I wrote can be made precise, and the precise versions are true. For example, it is a theorem of ZF that any $\Pi^0_1$ statement about the natural numbers that is not refutable in PA is true. – Andrés E. Caicedo Jan 14 '13 at 07:05
  • @Amr if $P = NP$ is proven (big if), then the proof will most likely be an indirect proof (since the current research is into proving $P \neq NP$ – John Dvorak Jan 14 '13 at 07:05
  • @AndresCaicedo isn't that a theorem of PA, that every consistent statement is true (AKA every statement is decidable)? – John Dvorak Jan 14 '13 at 07:08
  • Consistent with what? – Andrés E. Caicedo Jan 14 '13 at 07:09
  • Consistent with PA – John Dvorak Jan 14 '13 at 07:10
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    Hmm... Maybe you are thinking of Löb's theorem, http://en.wikipedia.org/wiki/L%C3%B6b's_theorem. But no, not every consistent statement is true, and PA certainly does not prove that. Anyway, this is getting off topic. – Andrés E. Caicedo Jan 14 '13 at 07:11
  • Possible Duplicate: http://math.stackexchange.com/questions/18088/existence-proofs – picakhu Jan 18 '13 at 21:08
  • @picakhu Ha! I had even commented on your question. (My memory is going...) – Andrés E. Caicedo Jan 18 '13 at 21:19
  • There is a similar question in computer science: http://cs.stackexchange.com/questions/32325/is-there-an-algorithm-that-provably-exists-although-we-dont-know-what-it-is – Erel Segal-Halevi Nov 18 '14 at 17:39

2 Answers2


There is a disappointing way of answering your question affirmatively: If $\phi$ is a statement such that First order Peano Arithmetic $\mathsf{PA}$ proves "$\phi$ is provable", then in fact $\mathsf{PA}$ also proves $\phi$. You can replace here $\mathsf{PA}$ with $\mathsf{ZF}$ (Zermelo Fraenkel set theory) or your usual or favorite first order formalization of mathematics. In a sense, this is exactly what you were asking: If we can prove that there is a proof, then there is a proof. On the other hand, this is actually unsatisfactory because there are no known natural examples of statements $\phi$ for which it is actually easier to prove that there is a proof rather than actually finding it.

(The above has a neat formal counterpart, Löb's theorem, that states that if $\mathsf{PA}$ can prove "If $\phi$ is provable, then $\phi$", then in fact $\mathsf{PA}$ can prove $\phi$.)

There are other ways of answering affirmatively your question. For example, it is a theorem of $\mathsf{ZF}$ that if $\phi$ is a $\Pi^0_1$ statement and $\mathsf{PA}$ does not prove its negation, then $\phi$ is true. To be $\Pi^0_1$ means that $\phi$ is of the form "For all natural numbers $n$, $R(n)$", where $R$ is a recursive statement (that is, there is an algorithm that, for each input $n$, returns in a finite amount of time whether $R(n)$ is true or false). Many natural and interesting statements are $\Pi^0_1$: The Riemann hypothesis, the Goldbach conjecture, etc. It would be fantastic to verify some such $\phi$ this way. On the other hand, there is no scenario for achieving anything like this.

The key to the results above is that $\mathsf{PA}$, and $\mathsf{ZF}$, and any reasonable formalization of mathematics, are arithmetically sound, meaning that their theorems about natural numbers are actually true in the standard model of arithmetic. The first paragraph is a consequence of arithmetic soundness. The third paragraph is a consequence of the fact that $\mathsf{PA}$ proves all true $\Sigma^0_1$-statements. (Much less than $\mathsf{PA}$ suffices here, usually one refers to Robinson's arithmetic $Q$.) I do not recall whether this property has a standard name.

Here are two related posts on MO:

  1. $\mathrm{Provable}(P)\Rightarrow \mathrm{provable}(\mathrm{provable}(P))$?
  2. When does $ZFC \vdash\ ' ZFC \vdash \varphi\ '$ imply $ZFC \vdash \varphi$?
Andrés E. Caicedo
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  • Good answer! Also, if we were to name this proof technique, what do you think would be appropriate? – chubbycantorset Jan 14 '13 at 18:04
  • (I've moved a comment answering the follow-up question above to the body of the answer, and added some references.) – Andrés E. Caicedo Sep 24 '13 at 15:41
  • A sort of 'flip' of this, of course (and one catch with the purported approach to e.g. Goldbach, which Andres is certainly well aware of), is that there is (almost certainly) no statement $\phi$ for which we can prove that e.g. PA _doesn't_ prove $\phi$! This is because if PA is inconsistent then it proves _everything_, so proving that there's a statement that PA doesn't prove is tantamount to a proof of the consistency of PA, and as such (by Godel) is impossible within PA itself unless the theory is inconsistent. (Note: this doesn't rule out proofs from _outside_ PA,a la Goodstein...) – Steven Stadnicki Sep 24 '13 at 16:04
  • @StevenStadnicki: Hmm you should have said "there is (almost certainly) no statement $φ$ for which we can prove **within PA** that PA doesn't prove $φ$" just so that people don't misunderstand. – user21820 Feb 08 '16 at 03:39

I'd say the model-theoretic proof of the Ax-Grothendieck theorem falls into this category. There may be other ways of proving it, but this is the only proof I saw in grad school, and it's pretty natural if you know model theory.

The theorem states that for any polynomial map $f:\mathbb{C}^n \to\mathbb{C}^n$, if $f$ is injective (one-to-one), then it is surjective (onto). The theorem uses several results in model theory, and the argument goes roughly as follows.

Let $ACL_p$ denote the theory of algebraically closed fields of characteristic $p$. $ACL_0$ is axiomatized by the axioms of an algebraically closed field and the axiom scheme $\psi_2, \psi_3, \psi_4,\ldots$, where $\psi_k$ is the statement "for all $x \neq 0$, $k x \neq 0$". Note that all $\psi_k$ are also proved by $ACL_p$, if $p$ does not divide $k$.

  1. The theorem is true in $ACL_p$, $p>0$. This can be easily shown by contradiction: assume a counter example, then take the finite field generated by the elements in the counter-example, call that finite field $F_0$. Since $F_0^n\subseteq F^n$ is finite, and the map is injective, it must be surjective as well.
  2. The theory of algebraically closed fields in characteristic $p$ is complete (i.e. the standard axioms prove or disprove all statements expressible in the first order language of rings).
  3. For each degree $d$ and dimension $n$, restrict Ax-Grothendieck to a statement $\phi_{d,n}$, which is expressible as a statement in the first order language of rings. Then $\phi_{d,n}$ is provable in $ACL_p$ for all characteristics $p > 0$.
  4. Assume the $\phi_{d,n}$ is false for $p=0$. Then by completeness, there is a proof $P$ of $\neg \phi_{d,n}$ in $ALC_0$. By the finiteness of proofs, there exists a finite subset of axioms for $ACL_0$ which are used in this proof. If none of the $\psi_k$ are used $P$, then $\neg \phi_{d,n}$ is true of all algebraically closed fields, which cannot be the case by (2). Let $k_0,\ldots, k_m$ be the collection of indices of $\psi_k$ used in $P$. Pick a prime $p_0$ which does not divide any of $k_0,\ldots,k_m$. Then all of the axioms used in $P$ are also true of $ACL_{p_0}$. Thus $ACL_{p_0}$ also proves $\neg \phi_{d,n}$, also contradicting (2). Contradiction. Therefore there is a proof of $\phi_{d,n}$ in $ACL_0$.

So the proof is actually along the lines of "for each degree $d$ and dimension $n$ there is a proof of the Ax-Grothendieck theorem restricted to that degree and dimension." What any of those proofs are, I have no clue.

Lucas Wiman
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  • Hi. Do you see how to extend the argument to prove that the inverse should also be polynomial? – Andrés E. Caicedo Jan 18 '13 at 21:09
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    See also: http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/ – Andrés E. Caicedo Jan 18 '13 at 21:10
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    Not off the top of my head. I'm guessing it goes something like this: Since every function on finite fields is a polynomial function, there should be an upper bound $U(n, d, p)$ on the degree of the inverse for every $n$. If that function can be made independent of $p$, then just use "$\phi_{d,n}$ AND there is a polynomial of degree at most $U(n,d)$ which is an inverse of $f$" instead of *just* $\phi_{d,n}$. The proof would go the same. I don't know how to make the upper bound on the degree independent of $p$, however. (Is it possible?) – Lucas Wiman Jan 18 '13 at 23:44
  • Rudin's proof seems more powerful than the model-theory proof above, and it elucidates better *why* the theorem is true. Still, this is a standard proof which directly addresses the original question. – Lucas Wiman Jan 18 '13 at 23:51
  • @RecursivelyIronic: I've always heard it the other way around, that the logical nature is the key reason for the theorem. Can you point out what exactly in Rudin's proof seems more fundamental than the facts used in this proof? – user21820 Feb 08 '16 at 03:44
  • It's a matter of taste. I guess you could view this as the more fundamental viewpoint that, that any statement true for all sufficiently high characteristic algebraically closed fields _must_ be true of the complex numbers as well. I don't know much about complex analysis or algebraic geometry, so the model theoretic proof is certainly more intuitive _to me_. However, I think the nondegeneracy of the Jacobian of all injective holomorphic maps is a better geometric explanation, and nonobvious for polynomial maps in the positive-characteristic case. – Lucas Wiman Feb 15 '16 at 22:56