# I think I have a regular expression for a non-regular language

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?

$W_1 = \{a^5 . a^*\}$ 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.
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.