# Prove a language is not regular without pumping lemma [duplicate]

How can you prove that $$L=\{a^n b^{2n} \}$$ is not regular without the use of pumping lemma?

• This language is finite hence regular. Did you mean $\{a^nb^{2n}\mid n\in \mathbb{N}\}$? – Nathaniel May 18 at 17:23
• cs.stackexchange.com/q/1031/755 – D.W. May 18 at 18:59

• you can use closure properties of regular languages. For example, if $$L_1$$ and $$L_2$$ are regular then $$L = L_1 \cap L_2$$ must be regular. That means that if $$L$$ is not regular, then either $$L_1$$ or $$L_2$$ is not regular;
• you can suppose a language is recognized by a DFA, and conclude to a contradiction (the DFA must have at least $$n$$ states for all $$n\in\mathbb{N}$$, it must recognize words not in the language, …)