my solution is that $L\in co-RE$ by showing that $\overline{L}\in RE$

TM $M$ on input:$\langle M_1,M_2,w\rangle$

  1. Build TM $M_1,M_2$
  2. Simulate $M_1$ on $x\in\Sigma^*$, before that check if $x\neq w$
  3. If $M_1$ accepts some $x$, than Simulate $M_2$ on $x$
  4. If $M_2$ accepts than $M$ accepts. If $M_2$ reject, that we back at $M_1$ on the next $x$

$* $ $M_1$ and $M_2$ will run on each $x$ at the most $|x|$ steps.

Is that correct?



closed as unclear what you're asking by dkaeae, Evil, Discrete lizard Aug 17 at 13:43

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No. It is not correct.

For example, suppose that $x=0$, $w=00$, $L(M_1)\bigcap L(M_2)=\{x,w\}$. If $M_1(x)$ or $M_2(x)$ takes more than $|x|=1$ steps to answer "yes", then your proposed algorithm will failed.


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