Let $P=\{L \mid  \exists w\in \Sigma^* s.t. w\in L  \wedge \forall z>w\colon z\in L\}$ . Now denote $L=\{\langle M \rangle \mid L(M)  \in P \}$.That is, $L$ is the set of all TMs $M$ s.t. there exists a $w\in \Sigma^*$ s.t. $w$ is accepted by $M$ and for each $z>w$, $z$ is accepted by $M$ as well. I am trying to find out whether $L$ is in $RE$ or not.I can easily show that $L$'s complement is $\notin RE$ using Rice's Theorem. But here, if using Rice's Theorem, I can only show that $L \notin R$, but I have no guarantee over $RE$. I thought about using reduction - either to prove $L\in RE$ or to disprove, but none came about. I thought about using $L_{acc}$ or its complement, $L_{\Sigma^*}$, but could not come up with an idea that will work for either direction of the reduction.Could someone assist?
Is this language recursively enumerable?
computability;turing machines;reductions;undecidability