# What undecidable language $B$ is reducible to its complement?

I encountered a problem which asks to give an example of an undecidable language $$B$$ such that $$B \leq_m \overline{B}$$...

However, I could find it hard to construct an example ... my difficulty is that given an undecidable but Turing recognizable language, say $$A_{TM}$$, its complement $$\overline{A_{TM}}$$ is not Turing recognizable and loops. If I reduce such a language (say $$x \in A_{TM} \leq_m y \in \overline{A_{TM}}$$, the instance $$y \in \overline{A_{TM}}$$ cannot be recognized by any TM (since by definition, $$\overline{A_{TM}}$$ is looping)...

Any help ?

Let $$H$$ be the language of all Turing machines that halt on empty input. Clearly $$H$$ is undecidable.
Let $$L = \{ (1,T) : T \in H \} \cup \{ (0,T) : T \not\in H \}$$.
Clearly $$L$$ is undecidable. If $$L$$ were decidable, then a Turing machine $$M$$ for $$L$$ would also imply the existence of a Turing machine $$M'$$ that decides $$H$$. $$M'$$ with input $$T$$ simply simulates $$M$$ with input $$(1,T)$$.
Moreover, for a Turing machine $$T$$ and $$x \in \{0,1\}$$ we have: $$(x, T) \in L \iff (1-x, T) \not\in L \iff (1-x, T) \in \overline{L}.$$
This, combined with the fact that we can decide whether a given word encodes a valid Turing machine, shows that $$L$$ is reducible to $$\overline{L}$$.