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L1={WWʳ |w€(a+b)*}

L2= Reversal(L1)

What is L1.L2?

What is the answer to this question please don't apply closure property I understood through closure property. My doubt is L1.L2 =W.Wʳ.Wʳ.W and this  isn't accepted by any pda so shouldn't the answer be not cfl ??
ago in Theory of Computation by (6 points) | 29 views
See Here L1 and L2 are same.

L1.L2 will be $(L1)^2$ only.

Say if L1=abba

L2 also be abba.

So, ans will be either $(L1)^2$ or $(L2)^2$

yes, it will be CFL, as CFL under concatenation operation is closed.
L1 and L2 are not same assume L1 = aabbaa then L2 will be baaaab.
How L2 could be baaaab? How a palindrome generate different string in forward and backward?

@Aman kumar jha

as per you, $L_1.L_2 = \{\color{red}{W}\color{green}{W^r}.\color{green}{W^r}\color{red}{W} |\; W∈ (a+b)^* \}$ ?

it is wrong.

$L_1.L_2 = \{\color{red}{W}\color{green}{W^r}.\color{green}{X^r}\color{red}{X} |\; W∈ (a+b)^*,\; X∈ (a+b)^* \}$ ?

if you doesn''t understand, just try with small example.

Let $L_1$={abba,aabbba,abbaabba} then what is $L_2$, $L_1.$$L_2$.

I understood thanks this was perfect
Because I was taking w= aab then wr=baa. So w.wr= aabbaa and wr.w=baaaab


Let L1={abba,aabbaa,abbaabba} then what is L2, L1.L2.

 L1 = {abba ,aabbaa,abbaabba}     

L2= {abba,aabbaa,abbaabba}

L1.L2  also contains abbaaabbaa.So how can this be cfl??

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