I would like to prove that there are two sets $A,B\subset \mathbb{N}$ such that


* $A |\_T B$
* $\emptyset' \equiv\_T A\oplus B$
* for every $e$, if $\{e\}^{A\oplus B}=\emptyset'$ then the map sending $(B,n)$ to the prefix of $B$ used in the computation $\{e\}^{A\oplus B}(n)$ is not computable.


I believe the existence of $A,B$ as above can be proved with a not-so-hard (but maybe not entirely trivial) finite extension argument. However, since these arguments are often tedious to read, and even more to write, I was wondering whether the existence of $A,B$ follows more easily by some result in classical computability.



Computing the halting problem with no computable bound on the use function
The answer is yes.


Since Chaitin's $\Omega$ is $wtt$-reducible to $\emptyset'$, we may replace $\emptyset'$ with $\Omega$.


Now let $A<\_T \emptyset'$ be a $K$-trivial but promptly simple set. Then there is an incomplete c.e. set $B$ so that $\emptyset'\equiv\_T A \oplus B$.


Suppose that there is such an $e$. Since $B$ is c.e. but incomplete, it must be non-$DNC$. So there is a partial computable function $g$ so that



> 
> **$ \exists^{\infty}n(g(n)\mbox{ is the prefix as required})$.**
> 
> 
> 


(Note that $B$ can be low and so non-high. Then we may assume that $g$ is totally computable.)


Therefore for any such $n$, $K(\Omega \upharpoonright n)\leq K(A\upharpoonright |g(n)|)+K(g(n))\leq 2K(n)$ up to a constant. This is a contradiction.


