[Solved]: Is there a simple example of sets such that $A leq_T B$ but not $A leq_m B$?

For example sets $H = {x , | $ Turing machine with index $x$ halts on input $x}$ and $overline{H} = {x , | $ Turing machine with index $x$ doesn’t halt on input $x}$. Because if $overline{H} leq_m H$, then $overline{H}$ would be recursively enumerable and therefore $H$ would be recursive, which is contradiction. On the other hand $overline{H} leq_T H$, because there is Turing machine with oracle $H$ recognizing $overline{H}$. This machine for input $x$ just checks $x in H$ and negates the answer.