How to prove $\{\langle M\rangle: L(M)=\emptyset\}$ is undecidable?

Question

Consider the language $$E_{T M}=\{\langle M\rangle: L(M)=\emptyset\}.$$ Prove that $E_{T M} \in \text{coRE} \backslash\text{R}.$

I proved that $$E_{T M} \in\text{coRE}$$ using Turing machine I built which recognizes the completement of the language. What left is to show that it is not in $$ R$$

Any suggestions how can I do that?

This question should have been answered somewhere on this site. — John L., May 28 '23 at 14:17
Does this answer your question? How to show that a function is not computable? How to show a language is not computably enumerable? — Nathaniel, May 28 '23 at 22:16

score 1 · Answer 1 · answered Dec 25 '23 at 06:34

One way to prove that is to reduce complement of A TM to E TM.

For Turing machine M and w as its input. We construct M' as follows:

For input w':

Run M on w:

If M rejected w, reject w'.

Otherwise, accept w'.

It is easy to see that M accepts w if and only if M' belongs to E TM.

From this reduction, we can deduct that E TM beings to coRE. And it doesn't belong to RE. If it does, Then E TM would be computable which contradicts the fact that A TM's complement is reducible to E TM.

score 1 · Answer 2 · answered Dec 28 '23 at 19:40

$\overline{HALT}_{TM} \leq_m E_{TM}$. Define a machine $H$ on input $\langle M, w \rangle$ as follows:

H(<M, w>):
    Construct a machine M' defined as follows:
    M'(x):
        Run M on w for |x| steps
          if (M halts on w after |x| steps)
             Accept x
          else
             Reject x
    Output <M'>

If $\langle M, w \rangle \in \overline{HALT}_{TM}$, then $M$ never halts on $w$, which means $L(M') = \emptyset$ and $\langle M' \rangle \in E_{TM}$. If $\langle M, w \rangle \notin \overline{HALT}_{TM}$, then $M$ halts on $w$ after some finite number of steps, $k$, thus, on inputs string $x$ such that $|x| \geq k$, $M'$ accepts $x$, which means $L(M') \neq \emptyset$ and $\langle M' \rangle \notin E_{TM}$.

Since we can reduce $\overline{HALT}_{TM}$ to $E_{TM}$, if $E_{TM}$ is recognizable, then $\overline{HALT}_{TM}$ is recognizable and the halting problem is decidable (since we know $HALT_{TM}$ is recognizable), a contradiction. Thus, $E_{TM} \notin RE$, which means $E_{TM} \notin R$.

How to prove $\{\langle M\rangle: L(M)=\emptyset\}$ is undecidable?

2 Answers2