Find an undecidable language that is mapping-reducible to its complement As the title suggests. Also, such a language must satisfy that neither it nor its complement are semi-decidable. I already know that $All_{TM}, EQ_{TM}, T$ (that is the set of all deciders) satisfy this property. But I tried reducing these to their complements directly, and via some sort of intermediate language, but to no avail. Can anyone help?
Try the following construction. Given a language $L$, use
$$ \{0x : x \notin L\} \cup \{1x : x \in L\}. $$