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It is unknown whether ZFC (or even PA) is consistent, that is, there might be a proof of a statement of form $P \land \neg P$ in ZFC. So it seems natural to try to find lower bounds for this inconsistency, like people do for other problems of same kind like Collatz Conjecture. My question is:

Has anyone done some work on this topic? What are the current known lower bounds for the number of symbols of a proof of an inconsistent statement in ZFC?

(I'm also curious about the analogous question about PA. I'm concerned however whether the question still makes sense if PA was inconsistent, perhaps this should be a separate question.)

Carla is my name
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    For any $n$, PA proves the consistency of the fragment with $n$ symbols. Same with ZFC. – Andrés E. Caicedo Jun 06 '21 at 14:31
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    @AndrésE.Caicedo How can that be proved? I still think my question is valid if we use a weaker system to talk about ZFC (or PA), for example one that can iterate every possible proof and check if it's a contradiction. – Carla is my name Jun 06 '21 at 15:09
  • @Carlaismyname This follows from the fact that PA proves its own finite sub-theories consistent. – Z. A. K. Jun 06 '21 at 15:28
  • :@AndrésE.Caicedo Can you please clarify how your observation relates to the question? It's not clear to me at all. – Z. A. K. Jun 06 '21 at 15:29
  • If I remember correcty, Pudlák showed that for reasonable theory $T$ there is $0<ε_T<1$ and $1<c_T∈ω$ such that for all natural number $n$, the length of the proof of "there is no contradiction in $T$ in less than $n$ steps" is somewhere between $n^{ε_T}$ to $n^{c_T}$ – Holo Jun 06 '21 at 16:52
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    @Z.A.K. If there were a proof of a contradiction using only $n$ symbols, then the axioms used in the proof would have at most $n$ symbols each. So the comment by Andrés Caicedo immediately gives an answer to the question, provided the metatheory is strong enough (PA or ZFC, respectively). – Andreas Blass Jun 06 '21 at 20:56

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