[dandanua]

> if it halts, we have proven by Thm 1 that ZFC is inconsistent. If not, we have similarly proven that ZFC is consistent.

The second part is wrong. We can't physically check that a program runs forever - this requires an infinite amount of time.

[modeless]

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.

[dandanua]

Right, we don't know the value BB(748), and moreover, "knowing" it is not enough, we would still need a proof that a certain number matches BB(748). And such a proof is not easier than a direct proof of ZFC consistency, I presume.

BTW, a value can't be uncomputable, only a function can.

[Kranar]

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.

[modeless]

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?

[Kranar]

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.