Abstract

We give an alternative and more informative proof that every incomplete ${\Sigma^{0}_{2}}$ -enumeration degree is the meet of two incomparable ${\Sigma^{0}_{2}}$ -degrees, which allows us to show the stronger result that for every incomplete ${\Sigma^{0}_{2}}$ -enumeration degree a, there exist enumeration degrees x 1 and x 2 such that a, x 1, x 2 are incomparable, and for all b ≤ a, b = (b ∨ x 1 ) ∧ (b ∨ x 2 )