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If language is regular, draw NFA . Otherwise, use pumping lemma to prove it A, {a^mb^n :m,n >=0} B, {a^mb^n : m different n, m,n>=0}
If language is regular, draw NFA . Otherwise, use pumping lemma to prove itA, {a^mb^n :m,n >=0} B, {a^mb^n : m different n, m,n>=0}
A, {a^mb^n :m,n >=0}
B, {a^mb^n : m different n, m,n>=0}
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