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L1 = ∑* = {∈, a,aa,aaa,..... b,bb,bbb,.........c,cc,ccc,..... , ab,aabb,ba,bbaa,bbab,aaba,..... cb,ccbb,bc,ccbcc,... ac,aacc,acacb,....abc,cba,cab,......} ===> it have regular expression = (a+b+c)* ====> it is RL

L2 = { an . bn . c | n > 0 } ===> it doesn't have Regular expression ===> it is not RL

But L1 ⊃ L2

 

Note that :-

1) Subset of Regular language is need not to be Regular

2) Superset of Non-Regular language is need not to be a Non-Regular.

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L1 = ∑* is RL
Since any no of a's, b's, c's exist there. We don't need to compare the no of a's with b's or a's with c's etc...

L2 = { an . bn . c | n > 0 }  is not RL
Since we have to compare the no of a's with b's and a's with c's or b's with c's etc etc and check whether its equal or not. In RL we don't have memory so can't compare. If it was L2 = { am . bn . c | m > 0, n > 0, p > 0 }  then It would be in RL since we don't have to compare a's with b's or c's. 

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