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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 ).
Archive for Mathematical Logic – Springer Journals
Published: Jun 7, 2008
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