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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 $L_1$.

ViX28
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  • 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 06:55
  • $L_1 = {a^nb^nc^{2n} : n \geq 0}$ 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? – Sarvottamananda Apr 24 '16 at 07:28
  • L1/L2 will be (a*) which is regular. is that right? – ViX28 Apr 24 '16 at 07:36
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    Possible duplicate of Closure against right quotient with a fixed language – Ran G. Apr 24 '16 at 21:41

1 Answers1

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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.

Klaus Draeger
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