We are given two Turing machines $M_1$ and $M_2$ and we wish to decide whether the union of the language $L(M_1)$ accepted by $M_1$ with the language $L(M_2)$ accepted by $M_2$ coincides with $\Sigma^*$.

Is this problem undecidable? In other words, is the language $\{ (M_1, M_2) \mid L(M_1) \cup L(M_2) = \Sigma^*\} $ undecidable?

I'm thinking about doing a proof by contradiction and somehow reducing to $E_{TM}$, but not sure where to start.

  • 1
    $\begingroup$ What are $M_1$ and $M_2$? The question is unclear, please define precisely "the resulting language". And if this is an exercise, please transliterate the exact wording. $\endgroup$ Nov 23, 2021 at 19:34
  • $\begingroup$ I have edited your question. Please check if it matches what you intended to ask. Also it is still unclear what $E_{TM}$ is and how reducing to (and not from) another language would help to prove that your language is undecidable. $\endgroup$
    – Steven
    Nov 23, 2021 at 20:43

1 Answer 1


Yes, the language $L = \{ (M_1, M_2) \mid L(M_1) \cup L(M_2) = \Sigma^* \}$, where $M_1$ and $M_2$ are Turing machines, is undecidable.

If $L$ were decidable then you would be able to decide if the language accepted by a Turing machine $T$ is $\Sigma^*$ by simply checking whether $(T,T) \in L$.

To see that the problem of deciding whether a given Turing machine $T$ satisfies $L(T) = \Sigma^*$ is undecidable you can reduce from the well-known undecidable problem of deciding whether a given Turing machine $M$ halts on empty input. To do so, it suffices to construct $T$ from $M$ by replacing each transition that halts the machine and rejects with a transition that halts the machine and accepts. Then $M$ halts if and only if $L(T)=\Sigma^*$.

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    $\begingroup$ Good job on understanding what the question was! $\endgroup$
    – Nathaniel
    Nov 23, 2021 at 20:21

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