Is the following language in RE?
$$L = \{\langle M\rangle : M\text{ is a TM that does not accept }010\}$$

I could use Rice's Theorem with the property $P = \{L : 010\text{ is not in }L\}$ to show it isn't in R, but how do I show it is in RE?

  • 1
    $\begingroup$ Hint, is $L$ in co-RE? $\endgroup$ – John L. May 25 '19 at 18:41
  • $\begingroup$ yes, I think it is :) thx $\endgroup$ – Jon Nir May 26 '19 at 9:52

The classic technique of dovetailing can be used to show the complement of $L$, $\{\langle M\rangle : M\text{ is a TM that accept }010\}$ is recursively enumerable. Check this answer for details.

Since $L$ is not decidable as shown by Rice's theorem, $L$ cannot be recursively enumerable.

Exercise. Show that $\{\langle M\rangle : M\text{ is a TM that accept } 010\text{ and }101\}$ is not in RE.


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