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a^nb^2n where n > 1 , is regular  language or not    how to proof using pumping  lemma
in Theory of Computation by (13 points) | 11 views
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Let assume "m" is the minimum pumping lemma.

Let take the string a$^m$.b$^m$, so only all a's part is covered, now you take any string in that part as y.. You repeat it, you find the strings which are not in the language. Hence given language is not regular

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