Abstract

We give an alternative and more informative proof that every incomplete \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\Sigma^{0}_{2}}$$\end{document} -enumeration degree is the meet of two incomparable \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\Sigma^{0}_{2}}$$\end{document} -degrees, which allows us to show the stronger result that for every incomplete \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\Sigma^{0}_{2}}$$\end{document} -enumeration degree a, there exist enumeration degrees x1 and x2 such that a, x1, x2 are incomparable, and for all b ≤ a, b = (b ∨ x1 ) ∧ (b ∨ x2 ).