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BusyBeaver(6) Is Quite Large

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Re: BusyBeaver(6) Is Quite Large

#231

Earlier quoted context omitted.

> 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…

> (2) ZFC is consistent with two different statements, "BB(x) = a" and "BB(x) = b" for two different a, b. This means that a disproof of either statement cannot exist. > This, in turn, means that there is no observation you could ever make that would distinguish between the values a and b (for the identity of BB(x)). No matter what you believe the value of BB(x) might secretly be, there are no consequences; nothing a…

The trick is that if none halt on `b` steps, you don't know that BB(x)<b. Specifically, if you have one TM that keeps going, you don't know whether that TM halts eventually or keeps going forever.

Re: BusyBeaver(6) Is Quite Large

#232

Earlier quoted context omitted.

> (2) ZFC is consistent with two different statements, "BB(x) = a" and "BB(x) = b" for two different a, b. This means that a disproof of either statement cannot exist. > This, in turn, means that there is no observation you could ever make that would distinguish between the values a and b (for the identity of BB(x)). No matter what you believe the value of BB(x) might secretly be, there are no consequences; nothing a…

The trick is that if none halt on `b` steps, you don't know that BB(x)<b. Specifically, if you have one TM that keeps going, you don't know whether that TM halts eventually or keeps going forever.

Sure, you don't know whether BB(x) b for that reason, but wouldn't you still know BB(x) ≠ b, and isn't that good enough?
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