# Are these languages are regular?

Consider the languages $L_1, L_2 \subseteq \sum^*$, where $\sum=\{a,b,c\}$. Define

$$L_1/L_2 = \{x : \exists y \in L_2\ such\ that\ xy \in L_1 \}$$

Let $L_1 = \{a^nb^nc^{2n}: n \ge 0\}$ and $L_2 = \{b^nc^{2n}: n \ge 0\}$.

Justify whether $L_1$ and $L_1/L_2$ are regular.

$L_1$ will not be CFL also as it needs more than one stack to count. $L_1/L_2$ gives concatenation and the result will be $a^{n} b^{2n} c^{4n}$ which is again non regular. Am I right? I am little bit confused for $L_1/L_2$ case as for some $y$, $xy$ belongs to $L_1$.

• You haven't understood what $L_1/L_2$ means. You seem to think it's concatenation (in which case what you wrote is still false), but it isn't. Try again. Use the definition. – Yuval Filmus Apr 24 '16 at 6:55
• $L_1 = \{a^nb^nc^{2n} : n \geq 0\}$ is definitely not a CFL so it is not regular, but $L_1/L_2$ seems to be regular. How are you going to prove that? – Shreesh Apr 24 '16 at 7:28
• L1/L2 will be (a*) which is regular. is that right? – ViX28 Apr 24 '16 at 7:36
• We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. – D.W. Apr 24 '16 at 9:18
• Possible duplicate of Closure against right quotient with a fixed language – Ran G. Apr 24 '16 at 21:41

Since this looks like an exercise, I'll just hint at the solution for now; some of this has already been nudged at in comments.

What we have:

• $L_1 = \{a^nb^nc^{2n}\ |\ n\ge 0\}$
• $L_2 = \{b^nc^{2n}\ |\ n\ge 0\}$
• $L_1/L_2 = \{x\ |\ xy\in L_1 \mbox{ for some } y\in L_2\}$.

Plugging in the actual $L_1$ and $L_2$, we get

$\begin{array}{ll} L_1/L_2 &= \{x\ |\ xb^nc^{2n}\in L_1 \mbox{ for some }n\ge 0\}\\ &= \{x\ |\ xb^nc^{2n}=a^mb^mc^{2m} \mbox{ for some }m,n\ge 0\} \end{array}$

Then it is straightforward to show that $m,n$ must actually be equal, and the rest is easy.

• Thanks a lot :) got it. It's from ISI entrance exam not exercise problem. – ViX28 Apr 24 '16 at 14:59