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Show that the following language is not regular.

$L=$ {$a^nb^k:n>k$} $\cup$ {$a^nb^k:n\neq k-1$}.
in Theory of Computation by Boss (15k points) | 52 views

1 Answer

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In both the given regular expressions, we need to count the number of a's and number of b's to check the given conditions, which cannot be implemented by a DFA/NFA. So they are not regular languages and the Union of these two languages is also not regular.

We need to use PDA.

by Junior (551 points)

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