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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 $<M,w>$:

  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. $<M,w> \in \overline{A_{TM}}$, then $M_1$ accepts any input word, i.e. $M_1 \in L$.

Am I missing something here?

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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.

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