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To show that L is not Turing-recognizable, we can use a reduction from the complement of the ATM problem (ATM'). However, I'm not sure about how we would prove that the complement of L is not Turing-recognizable. I'm wondering if we could just use the same from as for L but instead of rejecting if ATM' accepts and vice versa, but I don't understand these well enough to understand if that's feasible.

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  • $\begingroup$ Hint: Are you familiar with the halting problem? $\endgroup$
    – codeR
    Commented Apr 25 at 9:13

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