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BB(3, 3) is Hard

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Re: BB(3, 3) is Hard

#141

Earlier quoted context omitted.

Excuse me?

I’m saying that all you’ve proven is that if you know a priori whether ZFC is consistent, you can construct a Turing machine that returns this value. If you consider that to be window dressing I don’t know what else I can tell you.

I’m saying that it doesn’t matter if you know if ZFC is consistent; I have proven there exists a TM that halts iff it is.

Re: BB(3, 3) is Hard

#142
post #138

Earlier quoted context omitted.

His point is if you know the value of BB(748) then you don't have to wait forever, just BB(748) steps, as after that the Turing machine is guaranteed not to halt. The problem with his argument is that we don't know the value of BB(748). Not only that, it is incomputable, which resolves the contradiction.

It is categorically false that BB(748) is not computable. On the contrary any particular BB(n) can be computed by some Turing Machine even though there is no Turing Machine that can compute BB(n) for every n.

So then what's stopping you from running the BB(748) machine, getting the number, then running the ZFC machine and proving ZFC consistent or not?

Re: BB(3, 3) is Hard

#143
post #138

Earlier quoted context omitted.

It is categorically false that BB(748) is not computable. On the contrary any particular BB(n) can be computed by some Turing Machine even though there is no Turing Machine that can compute BB(n) for every n.

So then what's stopping you from running the BB(748) machine, getting the number, then running the ZFC machine and proving ZFC consistent or not?

A proof of existence is not the same as a construction, so the fact that we know that there exists a TM that computes BB(748) does not mean that we know which specific TM does it or how to construct such a TM.

Re: BB(3, 3) is Hard

#144
post #98

Earlier quoted context omitted.

> I think the point here is that it was not known whether all 3,3 halting problems are reducible to trivial Collatz-like problems. True, but it still isn't known either way. Nothing changed in that regard.

'trivial' isn't a mathematical concept. If no one has shown why this problem is trivial, it isn't trivial.

This problem isn't trivial, but neither are the ones we could already reduce BB(3, 3) to (via Conway's construction).

Re: BB(3, 3) is Hard

#145
post #137

Earlier quoted context omitted.

Yes. My reasoning applies to a function n=> BB(n) Isn't that what "the function is not computable" is about? Or is the thesis that the value of BB(748) can't be computed?

BB(n) for any particular n is always computable, no exceptions. There simply is no single computable function that can compute BB(n) for every n, but for any particular n there absolutely is a TM that computes it.

Indeed, the value of BB(n) for each n is just a number; and it's easy to construct a TM which outputs some particular number. However, we don't know what those TMs are, for the same reason we don't know what those BB(n) numbers are.

The same applies to more limited settings too, e.g. boolean questions like whether the Collatz conjecture holds: I know that one of the following programs calculates the right answer, but I don't know which:

- PRINT "true"; HALT

- PRINT "false"; HALT

(Perhaps we should also allow `PRINT "independent of the given axioms"; HALT` as well!)

Re: BB(3, 3) is Hard

#146
post #98

Earlier quoted context omitted.

'trivial' isn't a mathematical concept. If no one has shown why this problem is trivial, it isn't trivial.

This problem isn't trivial, but neither are the ones we could already reduce BB(3, 3) to (via Conway's construction).

Some reductions via Conway's method give non-trivial Collatz problems even though other simple techniques show the halting question for those machines to be trivial; the Conway reduction sometimes reduces trivial Turing machines to non-trivial Collatz problems.
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