“…P r o o f . For the convenience of the reader, we recall the argument in[5], adapting it to our context and notations. We look for a computable function f (D, e, x) such that if ϕ e (x) ↓, and ϕ e (x) = A d for some d ∈ D then f (D, e, x) = A ϕ e (x).Let p be a productive function for the pair (U 0 , U 1 ): it is well known that we may assume that p is total.Let {u d,D,e,x , vd,D,e,x : D is a finite subset of ω, d ∈ D, e, x ∈ ω} be a computable set of indices we control by the Recursion Theorem.…”