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BusyBeaver(6) Is Quite Large

(scottaaronson.blog)
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Scarblac ◴[] 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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ChadNauseam ◴[] No.44406574[source]
The number itself is not independent of ZFC. (Every integer can be expressed in ZFC.) What's independent of ZFC is the process of computing BB(748).
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Straw ◴[] No.44406611[source]
Sure, if someone just gives you the number, ZFC can represent it. But ZFC cannot prove that the value is correct, so how do you know you have the right number? Use a stronger proof system? Go a bit bigger and same issue.
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ajkjk ◴[] No.44406771[source]
Not an expert, but I've read about this a bit because it bothered me also and I think this is the answer:

Most of these 'uncomputable' problems are uncomputable in the sense of the halting problem: you can write down an algorithm that should compute them, but it might never halt. That's the sense in which BB(x) is uncomputable: you won't know if you're done ever, because you can't distinguish a machine that never halts from one that just hasn't halted yet (since it has an infinite number of states, you can't just wait for a loop).

So presumably the independence of a number from ZFC is like that also: you can't prove it's the value of BB(745) because you won't know if you've proved it; the only way to prove it is essentially to run those Turing machines until they stop and you'll never know if you're done.

I'm guessing that for the very small Turing machines there is not enough structure possible to encode whatever infinitely complex states end up being impossible to deduce halting from, so they end up being Collatz-like and then you can go prove things about them using math. As you add states the possible iteration steps go wild and eventually do stuff that is beyond ZFC to analyze.

So the finite value 745 isn't really where the infinity/uncomputability comes from-it comes from the infinite tape that can produce arbitrarily complex functions. (I wonder if over a certain number of states it becomes possible to encoding a larger Turing machine in the tape somehow, causing a sort of divergence to infinite complexity?)

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Scarblac ◴[] No.44407071{3}[source]
And also, if BB were computable, then it could be used to solve the halting problem: run the Turing machine of size n for BB(n) steps, and if it hasn't halted yet, it never will. So the BB function is clearly not computable.

But to me as a layman that seems true regardless of formal axioms chosen, but I guess I need to read that linked thesis.

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1. ajkjk ◴[] No.44407086{4}[source]
That is the standard argument for why BB is uncomputable for general n, but it's not the same as why BB(n) would be independent of ZFC for fixed n.