# $L_1= \{\langle M\rangle \mid$ there exists $x \in \Sigma^*$ such that for every $y \in L(M), xy \notin L(M)\}.$ Is $L_1$ RE or not RE?

I tried to prove $$L_1$$ is not Recursively enumerable via Rice's theorem, however i've been told by a mentor that examples i used are not valid. Can someone point out the mistakes in my understanding?

We can have $$Tyes$$ for $$Σ^*$$ and $$Tno$$ for $$ϕ$$. Hence, $$L_1$$ is no recursive for sure.

$$L(M)$$ is recursive so obviously $$\exists$$ TM for it, so this TM says $$T_{yes}$$ for L(M) and also there exists a TM which says no for $$\Sigma^*$$

Any non-monotonic property of the LANGUAGE recognizable by a Turing machine (recursively enumerable language) is unrecognizable

For a property of recursively enumerable set to be non-monotonic, there should exist at least two recursively enumerable languages (hence two Turing machines), the property holding for one ($$Tyes$$ being its TM) and not holding for the other ($$Tno$$ being its TM) and the property holding set (language of $$Tyes$$) must be a proper subset of the set not having the property (language of $$Tno$$).

we are pretty sure that $$L(M)$$ is a proper subset because it is given in the definition of the $$L_1$$ there exists x ϵ Σ* such that for every y ϵ L(M), xy ∉ L(M).

So there exist a TM which is holding properties of L(M), which can say $$Tyes$$ for L(M), and another TM which says $$Tno$$ for $$\Sigma^*$$

$$L(M) \subset \Sigma^*$$

Property holding set is non monotonic because it is proper subset of of the set $$\Sigma^*$$

Hence $$L_1$$ is non RE