I'm working on the following problem:

Is the following language Turing recognizable (recursively enumerable) ?

$$L = \{ \langle \langle \ M\ \rangle \rangle \ | \ M \ \text{does not > accept}\ 010 \} $$

The way I see it: Suppose that a machine $M$ loops forever on $010$. If a $TM$ recognizes $L$, it should accept $M$ in that case. But that means that it should know if $M$ loops forever or not, which is not possible. So, $L$ is not Turing recognizable.

Is my proof correct, and can it be more formal?

  • $\begingroup$ It's not a proof, its mostly hand waving and (your) intuition. To show $L$ does not belong in a certain class, you must show a specific TM reduction to another TM language that is known not to be in that class. $\endgroup$
    – lox
    May 19, 2019 at 16:23

1 Answer 1


You are presenting an argument which falls short of being a proof. In particular, it is not clear why a Turing machine recognizing $L$ should know whether $M$ loops forever or not; indeed, it is not so clear what do you mean by know in this context.

Here is one way a proof could go. Suppose that $L$ were r.e. The language of Turing machines which do accept $010$ is also r.e. By running both machines in parallel, we can decide whether a given Turing machine accepts $010$, i.e., we could solve the halting problem, which we know is undecidable. Therefore $L$ cannot be r.e.

  • $\begingroup$ First of all, thank you for your answer! Now, I understand the proof you are presenting, but I don't quite get why my argument is not a proof. Could you be more elaborate on that? $\endgroup$
    – Da Mike
    May 19, 2019 at 15:02
  • $\begingroup$ Your claim "But that means that it should know if $M$ loops forever or not" is unsubstantiated. $\endgroup$ May 19, 2019 at 15:03
  • $\begingroup$ So is this specific claim false, or have I not provided a solid proof of it? $\endgroup$
    – Da Mike
    May 19, 2019 at 15:16
  • $\begingroup$ You should explain why "that means that it should know if $M$ loops forever or not", and indeed what "know" means in this context. $\endgroup$ May 19, 2019 at 15:34

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