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Let $L' \subseteq \Sigma^*$ be a fixed, non-empty language.

Define the language:

$$ L = \left\{ \langle M \rangle \;\middle|\; L(M) = L' \right\} $$

That is, $L$ contains the descriptions of Turing machines $M$ whose language is exactly $L'$.

Is it always true that $L \notin R$, i.e., that $L$ is undecidable?

In particular:

  • If $L' \in \text{RE}$, it seems Rice’s Theorem can be applied.
  • But what if $L' \notin \text{RE}$? Rice's Theorem only applies to non-trivial properties of RE languages, so it doesn't directly help here.

Can we still prove that $L \notin R$ in that case — for instance, by reducing from a known undecidable problem?a

csmathstudent8
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1 Answers1

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There are cases in which $L \in R$ and cases in which $L \notin R$.

If $L' \in RE$ we get from Rice theorem, as you mentioned, that $L \in R$.

However, if $L' \notin RE$, then there's no TM that accepts $L'$. Meaning, $L' = \emptyset \in R$.

Yuvi
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