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?
I think I have a regular expression for a non-regular language
regular languages;automata;regular expressions;pumping lemma
$W_1 = \{a^5 . a^*\}$ 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.