If FLT had really been solved then I would know that. (Note- I am not bragging about how much math I knew; I am bragging about how much math I knew of.)Is that a rigorous proof technique? No, but it is useful and often correct.
(All of the math that I refer to below is done rigorously in my writeup here.) As a Grad student I noticed the following:
- Let C(x) be the length of the shortest description of x (the usual Kolmogorov complexity).
- It is easy to show that C ≤T HALT.
- The only proof I had seen that C was not computable was by contradiction: assume C is computable and then use that to create a string x that had no short description along with a short description of x. This proof did not use HALT at all! It was the only proof I knew of that a set was undecidable that did not use HALT (are there others?)
- Hence, from what I knew, it was possible that C was of intermediary Turing degree- neither decidable nor Turing-complete. This would be a natural intermediary Turing degree!! (As opposed to sets that are constructed for the sole purpose of being of intermediary Turing degree.)
If there was a natural intermediary Turing degree then I would know that. (Note- I am not bragging about how much computability theory I knew; I am bragging about how much computability theory I knew of.)Was my intuition correct? Unfortunately yes- one can show that HALT ≤T C. Too bad- I would have preferred to be wrong and have seen a natural intermediary Turing degree. (The writeup I pointed to above has the classic proof that C is undecidable, the proof that C ≤T HALT and the proof that HALT ≤T C. The proof of the latter I got from Peter Gacs.) (NOTE- I had it as Gac but a commenter pointed out that it is Gacs.)