# How can I prove this language is not CFL? [duplicate]

I have a question to find out that $L = \{a^m b^n\mid n>0, m - is prime \}$ is CFL or not. I know that it is not a CFL. But I don't know how to prove that. I know how to prove that $L = \{a^m\mid m - is prime \}$ is not CFL, by taking a random length for $vx$ ($uvwxy$) and prove that the length of the whole word is not prime (I manage to prove that this length is product of two factors). But with this problem I can't do this because of that $n$. At the end it results a non prime number (this product), but I have to decrease the length of $b$'s so I know nothing about that final number. Is there another method to prove that this language is not CFL or am I wrong somewhere?

## marked as duplicate by D.W.♦, André Souza Lemos, David Richerby, Luke Mathieson, JuhoJun 13 '15 at 20:23

The easiest way goes as follows: if $L$ were context-free than so $L \cap a^*$ would be. Now $L' = L \cap a^* = \{ a^m : m \text{ is prime}\}$ is a unary language, so $L'$ is context-free iff it is regular iff it is eventually periodic. If it were eventually periodic then it would be either finite or have finite asymptotic density, but we know that primes are infinite and have zero asymptotic density.
You can also prove that $L'$ is not context-free using the pumping lemma. The pumping lemma shows that for some $p$, every prime $q > p$ satisfies the following: there is $1 \leq t \leq p$ such that $q + (\ell-1)t$ is prime for all $\ell$. Choosing $\ell = q+1$, we get that $q(t+1)$ has to be prime, which it isn't.