My proof of $\{a^p|p\mathbb{isprime}\}$ is not a CFL.

Question

I want to show that the language $L=\{a^p~|~p~~\mathbb{is~~prime}\}$ is not a CFL.

Assuming towards a contradiction that $L$ is a CFL.

Let $p$ be the number from the Pumping lemma for context-free languages, thus $p$ is a prime.

Choose $w=a^p$, then we can decompose $w=uvxyz$.

So $w_{p+1}=uv^{p+1}xy^{p+1}z$, so $|w_{p+1}|=p+p+p=3p$ which is not a prime number, contradiction.

Is that correct?

Thanks!

Now, I am not an expert in CFL, but I would say “let $p$ be a prime at least as big as the ‘pumping length’ from the Pumping lemma formulation”. Also, $|w_{p+1}|=p+|vy|p$, which is divisible by $p$ but bigger than $p$ (as $|vy|\ge 1$), so I think your idea is good but the write-up has a few flaws. — , Jun 04 '22 at 18:08
I have also linked this question to a similar question: How do I show language of prime number of xs not context-free? — , Jun 04 '22 at 18:09
@StinkingBishop Yes I think you are correct, that was helpful, thanks a lot! — Math4Me, Jun 04 '22 at 18:27

score 1 · Answer 1 · answered Jun 04 '22 at 18:46

The pumping length need not be a prime number. Let $q$ be the pumping length of the language. And let $p$ be any prime number that is greater than the pumping length. Now your argument follows (note that to show that it is not a CFL, you need to have a universal quantification over the subdivision, and not an existential one, so change "we can decompose $w=uvxyz$" to "for any decomposition $w=uvxyz$ of w").

Also as pointed out by @StinkingBishop $|w_{p+1}| = p + |vy|p $.

My proof of $\{a^p|p~~\mathbb{is~~prime}\}$ is not a CFL.

1 Answers1

My proof of $\{a^p|p\mathbb{isprime}\}$ is not a CFL.