If $A \leq B$, $B \leq \overline{A}$, and $B$ is r.e., is $A$ recursive? We know that $A ≤ B$, $B ≤ \overline{A}$ and $B$ is recursively enumerable. Is $A$ recursive or not?
I tried negating whats given, but didn't achieve much. I know that from the very first inequality that $A$ is also recursively enumerable. I assumed whether $B$ is also recursive or not to maybe get to a contradiction of some sort, but I always ended up with &quot;sometimes recursive, sometimes not&quot;.
Notice that $A \leq B \leq \overline{A}$. Since $B$ is r.e., $A$ is r.e., hence $\overline{A}$ is co-r.e. Since $A \leq \overline{A}$, $A$ is also co-r.e. Hence $A$ is recursive.