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:Run $M$ on $w$If $M$ rejected get yourself into an inf loop 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 $L\notin RE$?
Is the following language recursively enumerable?
computability;turing machines;reductions;undecidability
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:Run $M$ on $w$ for $|x|$ stepsIf $M$ rejected within $|x|$ steps ...If $M$ accepted within $|x|$ steps ...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.