**Definitions:**


A set $X$ is *of PA degree relative to a set $Y$* if every infinite $Y$-computable binary tree has an infinite $X$-computable path.
A set $X$ is *low* if $X'$ is computable from $\emptyset'$.


**Easy facts:**


By the relativized Low basis theorem and using the fact that a low relative to a low is still low, $\emptyset'$ is of PA degree relative to every low set.
Of course, if $\emptyset'$ is of PA degree relative to a set $X$, then $X$ is $\emptyset'$-computable.


**Question:**



> 
> Are there non-low set $Y$ such that $0'$ is of PA degree relative to $Y$ ?
> 
> 
> 



Is 0' of PA degree relative to a non-low set?
No, by the Arslanov completeness criterion $0'$ is only DNC (Diagonally non-computable) relative to low sets. And PA implies DNC.


