Remember, that an incomplete r.e. set $A$ is cuppable if there is an incomplete r.e. set $B$ such that $A\oplus B \equiv\_T 0'$. It's relatively easy to build a low cuppable set but my question is whether for every cuppable set $A$ there is a low r.e. set $B$ such that $A \oplus B \equiv 0'$


I'm pretty sure this must be false (or at least unsolved but I'm betting false) or people wouldn't have had to prove results like the ZBC theorem (one can think of this as the claim that if S is r.e. in 0' then there is an r.e. set A with $A' \equiv 0' \oplus S$ and a $A$ r.e. set $B$ which is low relative to $A$ and cups to $A'$) but maybe someone can point me to a proof.




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Note that I'm asking about cupping and capping in R (set of re degrees here) not in D to clarify whats going on with the equivalence claim.


Also, the link with ZBC theorem results was mistaken as just because $0' \oplus S \equiv\_T B'$ doesn't guarantee that S itself is re in B. Hence, you can't use low-cuppability to show the ZBC theorem.



Does every cuppable r.e. set cup with a low set?
It seems that a short comment is not sufficient.


***The answer is negative. I.e. there is a cuppable r.e. set which is no low cuppable.***


By the results from Soare's book, low cuppability is equivalent to prompt simplicity which is equivalent to noncappbability.


Now the fact is that there is an r.e. degree which is both cappable and cuppable. The result was proved by Harrington in 1970's in an unpublished handwritten notes. Yang and I have an improvement of this by showing that there is a cappable degree which does not belong to the ideal generated by the union of nonbounding and noncuppable degrees (see <https://www.jstor.org/stable/pdf/27588357.pdf?refreqid=excelsior%3Aa9865b098fcf6a982c2dfeeb4155a184>).


Certainly you may find earlier published papers that imply the result.


