Turing Incomparability in Scott Sets
逻辑
2007-05-23 v2
摘要
For every Scott set F and every nonrecursive set X in F, there is a Y in F such that X and Y are Turing incomparable.
引用
@article{arxiv.math/0602439,
title = {Turing Incomparability in Scott Sets},
author = {Antonin Kucera and Theodore A. Slaman},
journal= {arXiv preprint arXiv:math/0602439},
year = {2007}
}