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.