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Let $f(n)$ be the length of the shortest statement whose shortest proof has length $n$ or more.

What are the asymptotics of $f(n)$? With standard symbols and length counted by character.

For any standard theory, such as PA or ZFC.

MJD
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TROLLHUNTER
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    This question does not make sense until you specify what symbols you're using, what axioms you're using, and what proof rules you allow. – Qiaochu Yuan May 31 '12 at 01:12
  • The 20 or so commonly used. Including $\forall$ and $\exists$. – TROLLHUNTER May 31 '12 at 01:14
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    OP said you could pick any standard theory, and offered PA and ZFC, which are quite different, so I presume that means that the answerer will get to decide, as OP is not planning to be fussy about the details. I think it's a fair question. – MJD May 31 '12 at 01:31
  • The question does seem to be well-posed. Take a theory, say PA. Generate all proofs of length 1; take the final step from each; the length of the shortest of these is $f(1)$. Now generate all proofs of length 2; take the final step from each; discard those that appeared in step 1; the length of the shortest one remaining is $f(2)$. Repeat ad lib. – MJD May 31 '12 at 01:35
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    @MarkDominus: If it said "the length of the shortest statement whose shortest proof has length exactly $n$", that would work. But it said "$n$ or more". So there might be a statement of length 2 whose shortest proof has length 76576428414812. You won't know this until you enumerate the proofs of length 76576428414812. And if there is a statement of length 2 that is unprovable, you may never know that it isn't counted in $f(2)$. – Robert Israel May 31 '12 at 01:46
  • Yes, you're right. It's ill-posed as written. Still there is something that the OP is getting at that I think can be reasonably discussed. – MJD May 31 '12 at 01:51
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    It is a well-defined function (for a given formal system), but it might not be computable. – Robert Israel May 31 '12 at 01:53
  • If you take an undecidable statement whose length is $n$ then its shortest proof has "infinite length" so to speak, and i think using Gödel's speedup theorem you can find sentences of length at most $n$ (or at least some other fixed bound) which have proofs of arbitrarily large length -note that in PA or ZF there are infinitely many sentences with proof length 1: the axioms. Though counting symbols instead of modus ponens applications would give a different count, but you also have infinitely many variables to write different formulas. – plm Aug 30 '23 at 04:55
  • I fell on the Ehrenfeucht-Mycielski theorem, which you may like, as much of the rest of this survey by Pudlak: users.math.cas.cz/~pudlak/length.pdf -see section 7. -Note that the editor messed up the typesetting and many characters -commas in particular- appear as "Γ". – plm Aug 30 '23 at 08:29

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In any theory to which Gödel's theorems apply, this is going to grow extremely slowly, in the following very strong sense. Suppose $g(m)$ is a function such that for any positive integer $m$, $f(g(m)) > m$. Then given a statement $S$ of length $\le m$, if $S$ is provable the shortest proof of $S$ must have length at most $g(m)$. But then $g(m)$ can't be a computable function (otherwise we could test whether $S$ is provable by enumerating all proofs of length at most $g(m)$, and we'd have an algorithm for solving the Halting Problem).

Putting it another way, for any computable function $g: {\mathbb N} \to {\mathbb N}$, there is some $m$ such that $f(g(m)) \le m$.

Robert Israel
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