Is every language $L \notin RE$ is $RE$-hard? Similarly, is every language $L \notin RE \cup coRE$ is $RE$-hard and $coRE$-hard?

It seems like a simple question but I can't find an answer. I tried to disprove using a counting argument using the fact that there are $2^{\aleph_0}$ such languages and only $\aleph_0$ reductions, but the reductions are not necessarily onto functions so they do not determine the language (same reduction may work for two different language). I also find that for similar questions, such as the existence of a language in $RE \setminus R$ that is not $RE$-complete, the counter examples may not be natural.

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    $\begingroup$ Are we talking about hardness with respect to Turing reductions, many-to-one reductions, or something else? $\endgroup$ Apr 7 at 15:42
  • $\begingroup$ @AndrejBauer Many-one reudctions (mapping reductions) $\endgroup$ Apr 7 at 19:49

1 Answer 1


Partial answer here: I think it at least depends on the chosen reduction.

For example, consider $H\in \mathsf{RE}$ the halting problem. Then $\overline{H}\notin \mathsf{RE}$, but there is no many-one reduction from $H$ to $\overline{H}$ (otherwise $H$ would be in $\mathsf{R}$). That means that $\overline{H}$ is not $\mathsf{RE}$-hard (for many-one reduction).

  • $\begingroup$ Thanks! what about the second question about language which is not in RE and not in coRE? $\endgroup$ Apr 11 at 6:10

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