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$L1$ and $L2$ are regular languages. $L3$ is CFL and $L4$ is a recursive enumerable language. $L5$ is the reversal of $L4$.

If $L6= \{ (L3/y)$ intersection $L5 \}$ where $y$ is input alphabet, then $L6$ is a:

1. Regular Language
2. CFL
3. Recursive Enumerable Language
4. Non Recursive Enumerable Language

### 1 comment

k belong to input alphabet

L is CFL then L/k is always CFL???

Here L3 is CFL then also L3/y is CFL.
Now l5 is reversal of l4 and REL is closed under reversal.

Eg- a^nb^nc^nd^n after reversal it will still be REL.

So we know REL are closed under intersection so answer is c) Recursive Enumerable Language  .

in one word -  L3/y is cfl and L5 is recursive enumerable language. the intersection will result in recursive enumerable language.
by

L3 is {a^nb(n+1))

and suppose y is any alphabet or a string of single alpabet but not a string of more than 1 alphabet then L3 / y is cfl of language (a^nb^n)

ryt???
yes exactly  @shubhanshu
thanks @Bikram Sir

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