Is there a minimal arithmetic degree $d <_a 0^{\omega}$ such that $0^{\omega}$ is a minimal cover of $d$ in the arithmetic degrees? [1]
While whether or not $0^{\omega}$ is a minimal cover at all (see this question) seems to be an open question surely it can't be true that we can get to $0^{\omega}$ starting from $0$ by taking two minimal covers. Could this really be an open question?
[1]: In other words, if $x$ is comparable to both $d$ and $0^{\omega}$ under arithmetic reducibility then $x$ is arithmetically equivalent to one of $0, d, 0^{\omega}$.
