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2 votes
2 votes

if L1 = { anbncn   | n>= 0 } 

and L2 = { anbmck | k,n,m>=0}

L1 is CSL and L2 is regular.

Now  L3 = L1.(L2)*.

Is L3 is regualar or CSL?

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3 Answers

8 votes
8 votes
Regular

L1 = { $\epsilon$, abc, aabbcc, ... }

L2 = a*b*c*

L3 is L1.(L2)*, means $a^nb^nc^n(a^*b^*c^*)^*$

The important thing to notice is that n can be 0. So L3 will be  $(a^*b^*c^*)^*$. Which is regular.
0 votes
0 votes

L1 is CSL

and L2 is regular.

then (L2)* will be reguler (by using closure properties)

and now L1.(L2)*=CSL.Reguler(push up) 

CSL.CSL=CSL

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