Turing Incomparability in Scott Sets
Abstract
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.
0
Turn this paper into a full lesson
ArcXiv compiles a staged curriculum from this paper: 8-12 lessons across beginner → advanced, synthesised section guides, visuals, flashcards, a quiz, exercises, and on-demand deep dives per section. Grounded in the abstract, never invented.