# does recursive (decidable) languages closed under division (Quotient) with any language?

I need to prove or disprove that R languages are closed under divison. I have managed to prove thet CFL are't closed under division. I read in wikipedia that RE languages are closed, but I didn't find any proof and also I didn't find anything about R languages. I would really apreciate any help.

The quotient of one language $$L$$ by another $$R$$ is the set of strings $$x$$ such that $$xy\in L$$ for some $$y\in R$$.
If both $$L$$ and $$R$$ are computably enumerable (what you call RE), then the quotient is clearly enumerable, since we can simply search for all strings $$x$$ and $$y$$ such that $$xy\in L$$ and $$y\in R$$, and when found, output $$x$$. This will enumerate the quotient $$L/R$$.
But in the case of decidable sets (what you call R), it is not true that the quotient is necessarily decidable. To see this, let $$L$$ be the sets of strings consisting of strings of the form $$xy$$, where $$x$$ codes a Turing machine program $$p$$ and input $$n$$ (with a suitable end-of-code marker) and $$y$$ codes the halting computation of $$p$$ on $$n$$, provided that it does halt. And let $$R$$ be the string with just the strings $$y$$ coding the halting computations. These are each decidable, since we can look at a string and easily decide if it codes the information or not.
But the quotient $$L/R$$ will consist of strings coding the halting TM program and input pairs that halt. That is, it is the halting problem, and this is not decidable.