I wonder how come that the following language is in $\mathrm R$.
$L_{M_1}=\Bigl\{\langle M_2\rangle \;\Big|\;\; M_2 \text{ is a TM, and } L(M_1)=L(M_2), \text{ and } |\langle M_1\rangle| > | \langle M_2 \rangle| \Bigr\} $
(I know that it's in $\mathrm R$ since there's an answer for this multi-choice question, but without explanation).
I immediately thought that the $L_{M_1} \notin \textrm{co-RE} \cup \textrm{RE}$ since we know that checking if two machines accept the same language is really not decidable, I came to think: is it immediate "False", but it can't be since there's a lot of Turing machines who accepts the same answer and have different codings.
Thanks!