## Turing Machines – Trying to prove the semi-decidability of undecidable language

We know that the stopping problem is semi-decidable, but the problem that does not stop is not semi-decidable (otherwise the stopping problem would be decidable). Therefore, one way to show that a language is not semi-decidable is to reduce the non-stop problem.

Let $$langle M, x rangle$$ to be an entry the non-stop problem. We are building a new Turing machine $$M$$ who on the inputs $$1,2,3,4.5$$ stop, as input $$6$$ simulated $$M$$ sure $$x$$, and on all other entries never stops. then $$M in S$$ Yes Yes $$M$$ do not stop on $$x$$.

We can also build another Turing machine $$M & # 39;$$ who on the inputs $$1,2,3.4$$ stop, as input $$5$$ simulated $$M$$ sure $$x$$, and on all other entries never stops. then $$M & # 39; notin S$$ Yes Yes $$M$$ do not stop on $$x$$.

These reductions show that neither $$S$$ nor its complement are semi-decidable.