# Why Right-Division of regular language with RE\E language is regualr?

I think I can't understand the meaning of language being decidable. The next case makes no sense to me:

Considering I have language L1 which is regular, and language L2 which is in RE\R (in particular, it cannot be decided).

The right-division of any regular language with some language is regular, and in particular in R.

so:

L1/L2={x ∈ Σ* | ∃ y ∈ L2 : xy ∈ L1}

is decidable.

But I can't see why. I can't describe an algorithm that decides L1\L2, so how come this language is decidable? I know how to define the transition function, but does it means that the language is decidable? After all, I can also define a transition function to the accepting problem, it still does not make it decidable.

After all, to check if x belongs to L1 \ L2 I need to go over words and check if there is a word in L2 that completes x to a word in L1, but if there is none, I will run forever.

I think I'm missing something, so I'm very confused. Would appreciate help!

In fact, if $$L_1$$ is regular then $$L_1/L_2$$ is regular for any $$L_2$$.
Indeed, consider a DFA for $$L_1$$ with transition function $$\delta$$ and accepting states $$F$$. We modify the set of accepting states to $$F' = \{ q : \delta(q,w) \in F \text{ for some } w \in L_2 \}.$$
You might object that "$$F'$$ cannot be computed". So what? We never said that you can compute a DFA for $$L_1/L_2$$ given one for $$L_1$$ and some reasonable representation of $$L_2$$. All we claim is that $$L_1/L_2$$ is regular.