# Turing Machines – Why {w in {0, 1 } ^ ast mid exists x in {0, 1 } ^ ast colon M_w (x) neq x } n & # Is not recursively enumerable?

Language

$$L_1 = {w in {0, 1 } ^ ast mid exists x in {0, 1 } ^ ast colon M_w (x) = x$$

($$w$$ is an encoding of a DTM, $$M_w$$ is the respective DTM.)

It's not decidable, according to Rice's theorem. However, it is recursively enumerable: a DTM simply chooses a $$x$$ for which the condition is fulfilled in a non-deterministic manner.

Consider very similar language

$$L_2 = {w in {0, 1 } ^ ast mid exists x in {0, 1 } ^ ast colon M_w (x) neq x }$$

Rice's theorem argues that it is undecidable, but it is also not semi-decidable according to my sources.

Why can not we argue $$L_2$$ as we did for $$L_1$$, indicating that a DTM chooses the entry $$x$$ in a non-deterministic way, the condition is fulfilled, $$L_2 in mathsf {RE}$$?

The proof that $$L_2 not in mathsf {RE}$$ performs the reduction $$overline H leq L_2$$, or $$H$$ is the problem of stopping. I understand this proof by reduction, but it seems to me to contradict the argument of recursive enumerability that I give above.

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