I have to prove that the language $L_2:=${$<M>$|$L(M)=\overline{A_TM}$} is (un-)decidable. In a previous assignment we proved that $L_1:=${$<M>$|$L(M)=A_TM$} is undecidable.
I would say that $L_1$ is decidable because we can build a decider S.
S="on input w": (rough outlined)
breadth first search
(a) if no accept state was found, accept
(b) else reject
Am I wrong with my guess?
$\overline{A_TM}$={<M,w>|M is a TM and does not accept w}