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if we will take the complement of L3 it will be regular and regular is closure under complement

1 Answer

+1 vote

L1 = {a1^1 , a2^2, a3^3............} ={a1,a4,a9 ...... } which is not in AP {Arithmetic progression}

L2 = {a1^1, a4^2, a9^3..........} not in AP.

L3 ={a2^1, a3^1, a4^1.........} put n=1 and m>n

regular expression for L3= {aa+}

hence, L3 is regular.

by Loyal (7.7k points)
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but in L3 if we change value of n in every term then it is not in AP. why u put n=1 in every term
0
because {aa+} will generate all lang.
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