I thought about these languages: $$L_{1} = A_{TM} = \big\{ \langle M, w \rangle \mid M \text{ is TM and }M \text{ accepts } w \big\}$$ $$L_{2} = \overline{HALT_{TM}} = \big\{ \langle M, w \rangle \mid M \text{ is TM and }M \text{ doesn't halt on input } w \big\}$$

Their intersection: $$L_{1}\cap L_{2} = \big\{ \langle M, w \rangle \mid M \text{ is TM and }M \text{ accepts and doesn't halt on input } w \big\}$$

If I assume that the intersection is decidable by TM $T$ so I can use $T$ to decide $A_{TM}$ and that's a contradiction.

Is it true?


1 Answer 1


Your proof is wrong. Note if $M$ accepts $w$, it must halt on $w$, so $L_1\cap L_2$ is the empty set, which is of course decidable.

To disprove the statement, you can set $L_2$ to be $\Sigma^*$, i.e. the language containing all strings. Since $\emptyset$ is Turing recognizable, $L_2$ is co-Turing recognizable. However, $L_1\cap L_2=L_1$ is undecidable.

  • 1
    $\begingroup$ Nitpick: we need to explicitly set $L_1$ so to be undecidable but recognizable, e.g. the halting problem. Otherwise, only knowing that $L_1$ is recognizable does not necessarily imply that it is undecidable. $\endgroup$
    – chi
    Jul 27, 2018 at 10:12
  • $\begingroup$ @chi, right, I just described the change to the proof in OP. $\endgroup$
    – xskxzr
    Jul 27, 2018 at 10:13
  • $\begingroup$ I don't understand your disproving. Did you use these languages: $$E_{TM} = \big\{ \langle M \rangle \mid M \text{ is TM and }L(M)= \emptyset \big\}$$ $$ALL_{TM} = \big\{ \langle M \rangle \mid M \text{ is TM and }L(M)= \Sigma^{*} \big\}$$ What is $L_{1}$ and what is $L_{2}$? $\endgroup$
    – Asaf
    Jul 27, 2018 at 12:37
  • $\begingroup$ @Asaf $L_1=A_{TM}$, $L_2=\Sigma^*=\{s\mid s\text{ can be any string }\}$. $\endgroup$
    – xskxzr
    Jul 27, 2018 at 12:55
  • $\begingroup$ Why $L_{2}$ is co-Turing recognizable? I can use automata that accepts all strings. $\endgroup$
    – Asaf
    Jul 27, 2018 at 13:26

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