The total computable maps
satisfy
Hence and .
Let be the diagonal halting set and put
If were a computably enumerable set, then would many-one reduce to , making the complement of computably enumerable. Together with the computable enumerability of , that would make a computable set, a contradiction.
Moreover
If this complement were computably enumerable, would again make computably enumerable. Thus neither nor its complement is computably enumerable.
Let total computable and witness and . Define
Parity testing and integer division by two are computable, so is total computable. For every ,
Therefore .

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