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BusyBeaver(6) Is Quite Large

(scottaaronson.blog)
271 points bdr | 1 comments | 28 Jun 25 16:53 UTC | HN request time: 0.266s | source
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Scarblac ◴[28 Jun 25 17:29 UTC] No.44406478[source]▶
It boggles my mind that a number (an uncomputable number, granted) like BB(748) can be "independent of ZFC". It feels like a category error or something.
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tromp ◴[28 Jun 25 19:20 UTC] No.44407378[source]▶
What makes BB(748) independent of ZFC is not its value, but the fact that one of the 748-state machines (call it TM_ZFC_INC) looks for an inconsistency (proof of FALSE) in ZFC and only halts upon finding one.

Thus, any proof that BB(748) = N must either show that TM_ZF_INC halts within N steps or never halts. By Gödel's famous results, neither of those cases is possible if ZFC is assumed to be consistent.

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Scarblac ◴[28 Jun 25 19:40 UTC] No.44407526[source]▶
I don't understand, surely if we assume ZFC is consistent then it's obvious that it won't halt? Even if its consistency can't be proven, neither can its inconsistency, so it won't halt. Or is that only provable outside of ZFC?

I guess it's also hard when we have an arbitrary Turing machine and have to prove that what it's doing isn't equilavent to trying to prove an undecibable statement.

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1. LegionMammal978 ◴[28 Jun 25 19:52 UTC] No.44407627[source]▶
If you believe that ZF is consistent, then you believe that the machine cannot halt (assuming you trust its construction). But you cannot write a proof in ZF that the machine cannot halt. Such a proof must include a new axiom "ZF is consistent", or some stronger axiom.