Let $$E_{TM} = \left \{ \langle M\rangle \mid L(M) = \emptyset \right\}$$

Prove that there are two languages $L_1, L_2$ such that

  1. $L_1, L_2 $ are infinite.
  2. $L_1 \cup L_2 = E_{TM}$
  3. $L_1 \cap L_2 = \emptyset$
  4. $L_1$ is decidable, $L_2$ is not recognizable.

I'm finding it really hard to find two languages that satisfy these conditions.

Especially the second condition, which two language unify to $E_{TM}$?

  • 4
    $\begingroup$ Hint 1, can you construct one $M$ such that $L(M)=\emptyset$? Hint 2, can you construct infinitely many such $M$? Hint 3, let $L_2=E_{TM}\setminus L_1$. $\endgroup$
    – John L.
    Mar 27 '19 at 20:04
  • $\begingroup$ Thank you, I'll try again! $\endgroup$
    – Alan
    Mar 28 '19 at 8:30
  • 3
    $\begingroup$ If you has figured out this problem, please write an answer (yes, you can answer your own question). $\endgroup$
    – xskxzr
    Mar 28 '19 at 8:53
  • $\begingroup$ Certainly, I will $\endgroup$
    – Alan
    Mar 28 '19 at 8:55
  • $\begingroup$ The padding lemma states that there is a total computable function that given any (encoding of) TM $M$, it returns a TM $N$ having a "longer" encoding than the one of $M$, where $L(M)=L(N)$. Essentially, the function adds a few redundant states to $M$ to make its encoding "longer". You could try to apply this function many times, iterating it. $\endgroup$
    – chi
    Mar 28 '19 at 12:30

A slight improvement:

$L_2 = E_{TM} \backslash L_1 $ (since we require that $L_1 \cap L_2 = \emptyset$).

$L_1$ must be a language of turing machines, i.e. $L_1 = \{\langle M\rangle \mid \dots\} $, so that $L_1 \cup L_2 = E_{TM}$.

So let's set $L_1 =$ $\{\langle M\rangle\mid C_0\text{ is a reject state}\}$.

The following holds:

  • $L_1 $ is infinite and decidable (easy to decide given $M$, what the configuration of the first state is).

  • $L_2 $ is infinite and unrecognizable.

  • 1
    $\begingroup$ thank you for your answer! $\endgroup$
    – Alan
    Apr 3 '19 at 18:32

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