zlacker

>>jeremy+(OP)
Here's a philosophical question that's been bothering me for a while. For large enough n, we can construct an n-state Turing machine that attempts to prove a contradiction in our current most powerful axiomatic system (let's say ZFC). We must assume that this machine loops forever, but Gödel's incompleteness theorem implies that we can't prove that.

What does this construction imply about BB(n)? In what sense is the BB sequence even well-defined if we can prove that it can't be determined?

(Edited for clarity.)

>>codefl+Ue
> In what sense is the BB sequence even well-defined if we can prove that it can't be determined?

I think it may help you to understand how the BB sequence can be well-defined if you change your definition of "we can prove".

A proof is a sequence of logical deductions in an axiomatic system, so what "we can prove" depends a lot on which axiom system you're using. For some types of questions, it helps to make that explicit.

For instance, in weak axiom systems, we can't prove that every hydra game terminates. [0] In very strong axiom systems, ones so strong we call them inconsistent, we can prove everything! People argue over intermediate systems sometimes by pointing out things they can prove that are counter-intuitive. [1][2]

For any terminating Turing machine, there is a (possibly very long) proof in a very weak axiom system of that fact: the full trace of the execution. (For non-terminating machines, this is not true.) So, if ZFC, a rather strong axiom system, cannot prove a machine halts, it does not halt.

For another example of this kind of thing, see Terry Tao's discussion of the consequences of the independence of Goldbach's conjecture from ZFC on MathOverflow. [3]

[0] https://en.wikipedia.org/wiki/Goodstein%27s_theorem [1] https://en.wikipedia.org/wiki/Freiling%27s_axiom_of_symmetry [2] https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox [3] http://mathoverflow.net/a/27764

>>jbappl+7i
But if the machine that stops when it has proven ZFC is inconsistent does not halt, then surely it means there is no proof of the inconsistency of ZFC? Hence ZFC is consistent? Which is contradicted by Godel.

I would think instead ZFC can't prove or disprove that this machine halts, which just means it's undecidable wether it halts or not.

This also implies that for sufficiently large n, we won't be able to prove that BB(n) = N, or even BB(n) < N for any N using the ZFC axioms. Of course, although they can't compute or bound its value, they can still easily prove that BB(n) is finite.