# Why is this language Turing recognizable and not not-Turing recognizable

I read that the following language is r.e. but not not-Turing recognizable

$$L$$: On input $$M$$ (where $$M$$ is a Turing Machine), $$M$$ accepts at least 20 inputs

I am not sure why it is not not-Turing recognizable., since I could perhaps make the following reduction from $$\overline{A_{TM}}$$ to $$L$$ given this procedure $$R$$ namely:

$$R$$: On input $$$$:

1. Construct TM $$M_1$$, where on input $$x$$, if $$x=1$$, accept
2. If input $$x$$ is not equal to $$1$$, run $$M$$ on input $$w$$ for $$|x|$$ steps. If after $$|x|$$ steps, $$M$$ does not accept $$w$$, then accept $$x$$

From this reduction, if $$M$$ does not accept $$w$$, i.e. $$ \in \overline{A_{TM}}$$, then $$M_1$$ accepts any input word, i.e. $$M_1 \in L$$.

Am I missing something here?

What you are missing is that if $$\langle M, w\rangle \notin \overline{A_{\mathrm{TM}}}$$, i.e. if $$M$$ halts on input $$w$$, you don't know whether or $$M_1 \notin L$$. If $$M$$ does halt on $$w$$, but this takes longer than $$20$$ steps, it would also hold that $$M_1 \in L$$. Thus, you don't have a reduction here.
That the language $$L$$ cannot be co-RE is an immediate consequence of Rice's theorem.