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For $A,B\subseteq \Sigma ^{*},$ define

$A/B=\{x\in\Sigma^{*}\mid \exists y\in B,xy\in A\}$

If $L$ is a $\text{CFL}$ and $R$ is $\text{Regular},$ then $L/R$ is$?$

  1. Regular
  2. CFL but not Regular
  3. Recursive but not CFL
  4. None of the above
in Theory of Computation edited by
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pls confirm the answer .....
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1 Answer

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Consider L as any CFL [say wwr] and choose R = {ϵ} [as R is given Regular in question]

So Now for all x in L, there exist y i.e ϵ such that xy ϵ L [as xy = xϵ = x and x is in L only]. So L/R is nothing but L itself. 

Hence L/R is CFL and hence recursive too [so option c is false]

Only correct option is B.

4 Comments

CFL quetionent with RL = CFL

( Note that every RL is CFL )
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@Shaik Masthan

it might be regular too but as CFL already contain Regular so B can be option?

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yes
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