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# [Solved]: Is the following language recursively enumerable?

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Problem Detail:

Let $L =\{ <M> |$ the amount of words $w\in\Sigma^*$ that $M$ does not halt on is finite $\}$.

I would like to prove that $L\notin RE$.

I can show that $\overline{L}\notin RE$ that is $L\notin CoRE$.

I do this with showing a reduction: $\overline{L_{acc}} \leq \overline{L}$ and since $\overline{L_{acc}}\notin RE$ then $\overline{L}\notin RE$.

The reduction goes as follows:

• input: $<M>,<w>$
• output: $<M_w >$

where $<M_w>$ is a TM that, on an input $x$ ignores it and performs:

1. Run $M$ on $w$
2. If $M$ rejected get yourself into an inf loop
3. If $M$ accepted then accepts as well.

Now if $<M><w>\in\overline{L_{acc}}$ then $w\notin L(M)$, thus $M$ either rejects $w$ or does not end its calculation on it, but in both cases $M_w$ will not end its calculation on all inputs. Otherwise, if $<M><w>\notin \overline{L_{acc}}$ then $w\in L(M)$ and then $M_w$ will halt (and accept) its calculation on every input $x$.

So I have that $\overline{L_{acc}} \leq \overline{L}$ and thus $\overline{L}\notin RE$, that is $L\notin CoRE$.

But I am at a loss on how I can prove $L\notin RE$? I tried a reduction from $\overline{L_{acc}}$, but this idea does not work because $M$ might not halt on $w$. I also tried from $L_{\Sigma^*}$ or from $L_d$, but could not find a reduction that will prove me that $L\notin RE$.

Does someone have an idea for such a reduction? Or perhaps another method to prove $L\notin RE$?

#### Answered By : chi

I'll provide some hint. Try building a reduction $\overline{L_{acc}} \leq L$ changing your argument as follows:

... where $<M_w>$ is a TM that, on an input $x$ it performs:

1. Run $M$ on $w$ for $|x|$ steps
2. If $M$ rejected within $|x|$ steps ...
3. If $M$ accepted within $|x|$ steps ...
4. If $M$ did not halt within $|x|$ steps ...

When $M$ does not accept $w$, you want the above to diverge only on finitely many $x$s.

When $M$ accepts $w$, you want the above to diverge on infinitely many $x$s.

Alternatively, you can try to exploit the Rice-Shapiro theorem.

###### Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/52440

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