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$L_1$ =${a^nb^nc^n|n\geq 2}$

$L_2$=${a^nb^mc^k|n,m,k\geq 0}$

$L_3$=$L_1.(L_2)^*$ ,then $L_3$ is

  1. Regular
  2. CFL
  3. Csl
  4. None
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$L_3 = L_1(L_2)^*$

So, We can write $L_3$ in the following way :

$L_3 = L_1(L_2^0 + L_2^1 + L_2^2 + ...)$

$L_3 = L_1(\left \{ \in \right \} + L_2 + L_2.L_2 + L_2.L_2.L_2 + ...)$

$L_3 = ( L_1 + L_1.L_2 + L_1.L_2.L_2 + L_1.L_2.L_2.L_2 + ...)$

Now, Because in $L_1$, $n$ is greater than $1$. $L_1$ can't have null string and which makes all the strings of the language $L_3$ starting with $a^nb^nc^n$ where $n \geq 1$ Which makes $L_3$ CSL because whatever logic we apply for $L_3$, We have to parse   $a^nb^nc^n$ before proceeding further in the string. 

Hence, $L_3$ is $CSL$.


Note that if $L_1$ had contained $\in$ i.e. if $n \geq 0$ then $L_3$ will be Regular. Moreover, in that case $L_3$ would be $\Sigma^*$.

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L2 is regular as it has no restriction. L1 is CSL as it has double comparison, Now, L3=CSL.Regular(push up). L3=CSL.CSL. CSL is closed under concatenation. Hence L3 is CSL.

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