A degree is $\alpha$-generic if it has representative that is $\alpha$-generic. Are the $\omega$-generic arithmetic degrees (i.e. the degree structure induced by arithmetic reproducibility) downward closed (excluding the 0 degree)?
For context:
In the Turing degrees a result of Jockusch tells us that every 2-generic degree bounds a degree that doesn't contain a 1-generic. This proceeds by using the fact that every 2-generic $G$ is properly r.e. in some set $B$ such that $B \oplus 0'$ doesn't compute $G$ ($B'$ must compute $G$ but if $B$ was of 1-generic degree then $B' \equiv_T B \oplus 0'$.)
However, in the arithmetic degrees the analog of this claim fails. Indeed, if $G$ is $\omega$-generic and $G$ is $\omega$-REA in $B$ then $B \equiv_a G$. Thus, one might hope that the $\omega$-generic arithmetic degrees are downward closed.
