The purpose of this short paper is to clarify and correct the proof of the main result contained in [1]: namely, that there exists no low maximal d. c. e. degree. There we gave a simple proof obtained as an immediate corollary of the following posited extension of the Robinson Splitting Theorem ([1, Theorem 1.7]: For any c. e. set A,Denis Hirschfeldt (private communication) was the first to notice a problem with the particular application of the Recursion Theorem in the proof of this result, one which does not occur in the original Robinson proof [5].We present below the following reformulation of the use of the Recursion Theorem sufficient to correct the proof of our main result (the non-existence of a low maximal d. c. e. 1) ), via a degree-theoretic extension of the Robinson Splitting Theorem:Theorem For any d. c. e. degree l, any c. e. degree a, if l is low and l < a, then there are d. c. e. degrees a 0 , a 1 such that l < a 0 , a 1 < a and a 0 ∪ a 1 = a. P r o o f. Let L be a d. c. e. set of low degree. Let L = L 0 − L 1 for some c. e. sets L 0 , L 1 such that L 0 ⊃ L 1 . Let f be a 1 − 1 computable function such that L 0 = {f (x) : x ∈ ω}, and M = f −1 (L 1 ). Then M is c. e. and M ≤ T L. (M is called Lachlan's set for L.)Given a c. e. set A, and a d. c. e. set L, assume that L < T A and L has low degree. First we construct ω-c. e. sets A 0 , A 1 to satisfy the following requirements:where i = 0, 1, e ∈ ω, and {Φ e : e ∈ ω} is a standard list of all partial computable (p. c.) functionals Φ. Let a = deg T (A), l = deg T (L) and a i = deg T (A i ⊕ L) for i = 0, 1. Then a i is an ω-c. e. degree, for each i = 0, 1. By the R-requirement, a 0 ∪ a 1 = a, and by the S-requirements, l ≤ a i < a, i ∈ {0, 1}. Then by the S-and R-requirements we will have l < a i .We will construct ω-c. e. sets so that they are c. e. in Lachlan's set M for L. The theorem follows from the following result from [2].Proposition 1 (Arslanov, LaForte and Slaman [2]) . Let A and C be sets such that C is c. e., A is c. e. in C, C ≤ T A, and A is ω-c. e. Then deg(A) is d.