are regular languages closed under division

Question

I am trying to solve this question which appeared in previous exam paper

Can someone help me what i am failing to understand

For languages $A$ and $B$ define $A \div B = \{x \in \Sigma^{\ast} : xy \in A \text{ for all } y \in B\}$. If $R$ is regular and $L$ is any language, is it always the case that $R\div L$ is regular? Explain.

I am thinking as regular languages closed under right quotient hoping we can use the same proof here but I understand that in right quotient we prove it for any $y$ that belongs to $B$ but for division it is for all $y$ in $B$ .

Also if I have $L_1$ is regular and $L_2$ is context-free, is $L_1/L_2$ Always regular? Always context-free but not always regular? Or not always context-free? Which one is true?

When $L_1$ is context-free and $L_2$ is regular, is $L_1/L_2$ always regular? Always context-free but not always regular? Or not always context-free?

Any ideas will really help me in preparing for my test.

Answer 1

If $L_1$ is regular and $L_2$ is any language, then $L_1 / L_2$ is regular. In fact, the construction of a DFA for $L_1 / L_2$ is almost trivial (with one caveat): It's the same as the DFA for $L_1$, except that the "final" states are the the states which can reach the original accepting states given a string from $L_2$.

Suppose that $M_1 = (Q, \Sigma, \delta, q_0, F_1)$ is the DFA which accepts $L_1$. Then a DFA which accepts $L_2$ is $M_2 = (Q, \Sigma, \delta, q_0, F_2)$, where:

$$F_2 = \left\{ q \in Q : \exists x \in L_2. \delta(q,x) \in F_1 \right\}$$

Now here's the caveat: this proof sketch is not constructive! To turn this into an algorithm requires some way to enumerate or recognise strings in $L_2$, and remember, $L_2$ can be any language.

If $L_1$ is context free and $L_2$ is regular, $L_1 / L_2$ is always context-free (proof left as an exercise), but it is not necessarily regular. (To see why, consider $L_2 = \{ \epsilon \}$.)

EDIT I've just noticed that the definition in the question is different from the usual definition of "right quotient". The usual definition is:

$$A/B = \left\{x\in \Sigma^{*} : \exists y \in B. xy \in A\right\}$$

The definition in the question is:

$$A\div B = \left\{x\in \Sigma^{*} : \forall y \in B. xy \in A\right\}$$

My answer us for the usual definition, because these are standard results. The answers are similar for the definition in the question, except that the construction is slightly different. (See David Smith's comment.)

I suspect there is a typo in the original question.