I feel like there must be a classical result answering this question (or easily modified to do so) but a quick flip through Soare didn't produce anything so rather than waste time I figured I'd just ask.


Given incomplete r.e.sets $C \nleq\_T B$ must there exist a low r.e. set $A <\_T C$ with $A \nleq\_T B$?


I'm guessing the answer is no and somehow you can put together Robinson low splitting with one of the non-splitting/non-bounding theorems to show this but I'm not seeing it right away.


EDIT: Great answer to the original question below (thanks Ted!) but I realized I should have specified that I wanted the set produced uniformly for the application I wanted. 



Given B,C incomplete, incomparable r.e. sets must C compute low r.e. set avoiding cone below B? (ADDED: Uniformly?)
By Sacks’s splitting theorem, the degree of C is the join of the degrees of two low r.e. sets, both strictly below the degree of C. They can’t both be recursive in B, so at least one of them can serve as A. So, yes, there is such an A. 


