[Solved]: Reducing a non-RE language to its complement

Such a undecidable language $L$ can not be in $RE$, since if there was a computable reduction from $L$ to $overline{L}$, then $L$ would be in $R$. Let $x’$ be the value to which $x$ reduces. $x’ in L$ if and only if $xnot in L$. We could then decide $xin L$ by enumerating $L$, and stopping to report that $xin L$ as soon as we encounter $x$, and report that $xnot in L$ when we encounter $x’$. That would make $L$ decidable, which is a contradiction. So there can be no reduction from a language in $RE$ to its complement. By a similar argument, such a language can not be in $co-RE$ either. There does exist an undecidable language $L$ that can be reduced to its complement. To find one, we must consider a language that is neither in $RE$ nor in $co-RE$. Let $Lin RE$ be an undecidable language. Then $L’={0w|win L} cup {1w|win overline{L}}$ is not recursive enumerable nor is its complement (see here). But there clearly is a reduction from $L’$ to $overline{L’}$.