20

So, one of the things that mathematical logic does is study theorems as abstract objects. There also many theorems about mathematical logic, and these theorems can have connections to other fields.

Additionally, mathematical logic also proven that certain conjectures are not formally decidable within certain formal systems. This means that mathematicians need stronger formal systems to prove them.

However, has a mathematical theorem ever been proven by constructing the proof with mathematical logic? Or proving that a proof exists?

By this, I mean for statement $C$ which is a mathematical conjecture, instead of writing out a proof of $C$, they write either a constructive or nonconstructive proof that proof exists.

For example:

  • Use proof theory to prove there are a sequence of inferences to lead to $C$ (this is implied by the previous two bullets, of course)
  • Use model theory to prove that all models of some theory $T$, where $T$ is a theory that mathematicians typically use for proofs, satisfy $C$
  • Use computer science to construct a program $P$, such that the Curry–Howard correspondence applied to $P$ results in a proof of $C$

What I am not looking for is examples where someone writes a proof of $C$, citing theorems from mathematical logic. In this case, they have actually written down a proof, not simply proved it exists. Like, that is still super cool, and has huge implications for the rest of mathematics, but I am asking about something different.

EDIT: If the conjecture uses a lemma $L$ that satisfies the above, that's fine. However, it must directly use the statement the lemma is stating, not the fact that is has abstractly been proven. That is, the proof of $C$ involves showing that $L$ implies $C$, not that the existence of a proof of $L$ implies $C$.

This is one technical caveat. T can not prove "$\forall C. Provable(C) \implies C$". Moreover, if it does for some $C$, it is vacuously true even within $T$, since one can technically just write a direct proof of $C$ in $T$. However, if $T$ is sound, then Provable(C) does in fact imply C. Since mathematicians usually implicitly assume that the theories they work in are sound (unless they are working in the theory for its meta-mathematical significance instead of its soundness), this should be acceptable. Moreover, proving a proof exists might be more tractable than writing the proof is some instances.

Christopher King
  • 10,365
  • 4
    Not exactly what you are after, but Herbrand's theorem has been used quite successfully to extract explicit bounds from theorems in several areas of analysis whose proofs only establish the existence of bounds, typically via nonconstructive (or nonexplicit) means. – Andrés E. Caicedo Feb 12 '19 at 20:21
  • @PyRulez For more on such uses of Herbrand's theorem and related tools, you want to search for "proof mining". – Andrés E. Caicedo Feb 12 '19 at 20:28
  • Here is a concrete obstacle to what you are after literally, if not in spirit. – Andrés E. Caicedo Feb 12 '19 at 21:34
  • @PyRulez I do not know that Goodstein's theorem is "almost equivalent" to PA. It is an independent sentence at the next level of complexity, and the extra layer of complexity has explicit consequences. For instance, Goodtein's theorem is still unprovable after adding to PA all true $\Pi^0_1$ statements. – Andrés E. Caicedo Feb 12 '19 at 21:36
  • @AndrésE.Caicedo I added an edit addressing the issue. I actually was just realizing the problem when you posted that comment. – Christopher King Feb 12 '19 at 21:43
  • @AndrésE.Caicedo Oh, maybe I was thinking of some other simple statement that was equivalent to Con(PA). – Christopher King Feb 12 '19 at 21:43
  • Just a brief passing comment : if there is a, say ZF(C) (same goes with PA, or any theory to which Gödel's theorem applies) proof of "there is a ZF(C) proof of $\varphi$" (where $\varphi$ is a theorem you'd like to prove), then that doesn't mean there is a ZF(C) proof of $\varphi$; you have to assume the consistency of ZF(C) (or your theory) to get there. This is in essence Gödel's theorem, or if you like, a reformulation of it, known as Löb's theorem, which essentially states : "if there is a proof that the provability of $\varphi$ implies $\varphi$, then there is a proof of $\varphi$" (1/2) – Maxime Ramzi Feb 12 '19 at 21:52
  • In other words, if you can go from knowing that a proof exists to knowing that the theorem is true, well you could have not bothered and just proved the theorem itself – Maxime Ramzi Feb 12 '19 at 21:53
  • No, ZF(C) + "ZF(C) is inconsistent" is not inconsistent (if ZF is consistent), so it cannot prove every statement – Maxime Ramzi Feb 12 '19 at 21:55
  • @Max Oh yes, my bad. I deleted the wrong comment. Also, if consistency sufficient then? I think "ZFC proves p$ could be independent of ZFC + Con(ZFC). – Christopher King Feb 12 '19 at 21:58
  • @Max However, Lob theorem does not say the proofs will be the same length. The meta proof could be shorter than the regular proof. – Christopher King Feb 12 '19 at 21:59
  • Sure, it doesn't say anything about the lengths of the proofs; but what I mean is that conceptually, as long as you don't assume in your proofs "ZF is consistent", then you have no way of proving $\varphi$ from a proof that a proof exists without proving $\varphi$ itself. Of course in terms of length, perhaps there is a shorter proof of "$Prov(\varphi) \land (Prov(\varphi)\implies \varphi)$", but that sounds unrealistic – Maxime Ramzi Feb 12 '19 at 22:02
  • @Max Well, I think "$Con(ZF) \implies p$" is an acceptable result, given that Con(ZF) is widely believed (otherwise we would not be using it). You are write about not being able to formalize this process completely within ZFC or some other theory, though. If you want to be real formal, just use a stronger theory for the metaproof then the proof. If the stronger theory proves the weaker theory is sound, this will then be a valid proof of the statement in the metatheory. If the metatheory is sound, then the weaker theory also proves it. – Christopher King Feb 12 '19 at 22:06
  • Though theoretically I don't have any argument to justify that "unrealistic" bit; and I don't know if there's any research concerning lengths of proofs (the problem with such research is that the length of proof is highly system-dependant, unlike provability) - so you'll have to either find an example where there is a shorter proof or be convinced that it's unrealistic – Maxime Ramzi Feb 12 '19 at 22:07
  • Yes, but the problem with assuming $Con(ZF)$ is that if ZF is consistent, there are models where $\neg Con(ZF)$ holds, so that's a bit of a restriction - I guess your stance on that ultimately depends on your philosophy concerning foundational mathematics. But anyway, at least we agree on the technical stuff, I'm just not sure whether I agree with you on how to interpret it untechnically :) – Maxime Ramzi Feb 12 '19 at 22:08
  • @Max Well, I know that a theoretical example is the existence of Tree(n). But yes, I also agree with everything technical thing you said. Arguing about nontechnical things is tricky. If we continue nontechnical debates, it would probably be better to do in chat. – Christopher King Feb 12 '19 at 22:11
  • 1
    You can prove that something is provable in a logical system without writing down a proof in that system - if you are allowed go outside that system for the proof of the proof. For example, you can prove that a proof of $3425 \cdot 452 = 1548100$ exists in Peano arithmetic by pulling out your pocket calculator, but it would probably be pretty annoying to write down the proof itself. – Jair Taylor Feb 12 '19 at 22:41

2 Answers2

16

I do not know about a conjecture, but I would like to mention the Ax–Grothendieck_theorem.

A very nice way of proof is to show that the (first-order) theory $ACF_0$ of algebraically closed fields with characteristic zero is complete (through quantifier elimination iirc).

Thus, if the statement is false, there is a “somewhat” equivalent (because of the number of variables and degree of polynomials) first-order statement the negation of which can be proved in $ACF_0$.

Since the proof has finite length, there exists some prime $p$ such that the assumption “$p \neq 0$” is not used in the proof. So that first-order statement never holds in any algebraically closed field with characteristic $p$.

It remains to prove that the Ax-Grothendieck theorem holds in the algebraic closure of $\mathbb{F}_p$ for each prime $p$.

In a nutshell, we actually disprove the existence of a general disproof in $ACF_0$ ; since $ACF_0$ is complete, this entails the existence of a proof.

Christopher King
  • 10,365
Aphelli
  • 34,439
  • Yes, that definitely works! I think you can technically translate it to a proof not using mathematical logic, but that's just a consequence of Löb's theorem. – Christopher King Feb 12 '19 at 22:02
  • I was kind of hoping for a more general theory like PA of ZFC, but I did not specify that in the original question. Even the specific theory is interesting anyways. – Christopher King Feb 12 '19 at 22:03
  • 2
    If you've proven something by proving that a lower-order proof exists then you will always be able to find a lower-order proof, because after all one was proven to exist. – Display Name Feb 12 '19 at 22:29
  • 1
    @DisplayName Oh whoops, I meant a "simple-ish" proof. As long as the meta-theory is ω-consistent, then yes, the direct proof will exist. – Christopher King Feb 12 '19 at 22:34
  • 2
    @DisplayName: That's not quite true. You need the system you're working in to be Σ1-sound before you can conclude that if you prove the existence of a proof it really implies the existence of a proof. For all we know, strong extensions of ZFC might be consistent but Σ1-unsound, in which case ZFC may prove the existence of a proof over ZFC but actually there is none. So far though, most theorems do not use anything near the full strength of ZFC, so it is not hard to convert a proof of provability to an actual proof. – user21820 Feb 13 '19 at 09:43
  • 2
    @PyRulez: As per my above comment, you only need Σ1-soundness (which is implied by either ω-consistency or arithmetical-soundness). You may also be interested in this post. – user21820 Feb 13 '19 at 09:47
11

The original proof of the Halpern-Läuchli theorem seems to be the sort of thing you asked for. In their paper, Halpern and Läuchli first set up a formal deductive system and show that a certain formula is deducible in this system. Then they provide a semantics, i.e., meanings for the formulas of their system. They show that the system is sound, i.e., the meanings of deducible formulas are true. And finally, they note that the particular formula whose deducibility they established earlier has, as its meaning, the conclusion that they want to prove.

The MathSciNet citation for the paper is

MR0200172 (34 #71)

Halpern, J. D.; Läuchli, H.

A partition theorem.

Trans. Amer. Math. Soc. 124 1966 360–367.

Andreas Blass
  • 71,833