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Let $W = \{a^n b^m \mid n\ge m+5,m\le 5\}$, where $\Sigma=\{a, b\}$.

I have proved that this language is irregular through pumping Lemma. But through regular expression it is proving that the language is regular. Can anyone please tell me that what should I consider it as. And does that happens?

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$W_1 = \{a^5 . a^*\}$ is a regular language.

$W_2 = \{a^m b^m \mid m\le 5\}$ is a regular (finite) language.

Thus $W = W_1.W_2$ is a regular language.

Something is wrong in your proof.

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The key result of finite automata is that deterministic finite automata, nondeterministic finite automata and regular expressions all define exactly the same set of langauges: the regular languages.

So there must be a mistake somewhere. In this case, your pumping lemma proof is wrong because the language is regular. Hint: to accept, you only need to consider whether the number of $a$s is 0, 1, ..., 9, or "10 or more" and whether the number of $b$s is 0, 1, ..., or 5.

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