I have read through the outline. Even though it is too sketchy to count as a full proof, I think I can reconstruct enough of the argument to figure out where the error in reasoning is going to be. Basically, in order for Chaitin’s theorem (10) to hold, the Kolmogorov complexity of the consistent theory T has to be less than l. But when one arithmetises (10) at a given rank and level on page 5, the complexity of the associated theory will depend on the complexity of that rank and level; because there are going to be more than 2^l ranks and levels involved in the iterative argument, at some point the complexity must exceed l, at which point Chaitin’s theorem cannot be arithmetised for this value of l.
(One can try to outrun this issue by arithmetising using the full strength of Q_0^*, rather than a restricted version of this language in which the rank and level are bounded; but then one would need the consistency of Q_0^* to be provable inside Q_0^*, which is not possible by the second incompleteness theorem.)
I suppose it is possible that this obstruction could be evaded by a suitably clever trick, but personally I think that the FTL neutrino confirmation will arrive first.
[...] a theory slightly weaker than PA that he calls Q 0 *. Of course, we know that this theory cannot prove its own consistency (if it is consistent), by the second incompleteness theorem
This confuses me. Second incompleteness applies to extensions of PA . I’ll bet Terry Tao knows this too. So what does this remark mean?
More here, including a comment by Terence Tao.
Specifically, Tao’s comment:
He’s such a glorious mathematician. <3
He gave a more detailed comment on the n-Category Café.
This confuses me. Second incompleteness applies to extensions of PA . I’ll bet Terry Tao knows this too. So what does this remark mean?