is this language context-free? a tricky one

Question

Is this language context-free?

$$L = \{a^nb^nc^{2n} \mid n \ge 0\}$$

It's tricky in my opinion because I know that $a^nb^nc^n$ is not context-free, but can I determine from this that $L$ is not context-free, too?

Thanks in advance

EDIT: I can use these closures: all the closures of regular languages, the closure of intersection of regular language with a context-free language, and closure under homorphishms

Answer 1

So at the first time I end up answering my own question. Hope that I will get more chances such as those at the future to help others

Here it is:

Let's consider that this language IS context-free. We'll define a regular function F such that:

$F(c) = c' + c$

$ F(b) = b$

$F(a) = a$

So $F(L)$ is also context-free from closure to homomorphisms.

Now consider the language $L' = F(L) ∩ \{a^*b^*(cc')^*\}$, that's also context-free due to closure under intersection with regular.

Finally, define a function G such that:

$G(c') = \epsilon$

$G(a) = a$,

$ G(b) = b$,

$ G(c) = c$

So $G(L') = \{a^nb^nc^n\}$ is context-free. But we know it's not - BAM!

Therefore, $L$ is not context-free!

Answer 2

It is not context-free. You can show this using the pumping lemma for context-free languages; the proof is very similar to the one for the language $\{a^nb^nc^n:n>0\}$, which is given in the linked article.

Added: Since you’re restricted to closure properties, perhaps the easiest argument is to use closure under inverse homomorphisms, using the homomorphism $h$ such that $h(a)=a$, $h(b)=b$, and $h(c)=cc$. If $L$ were context-free, $\{a^nb^nc^n:n\ge 0\}$ would also be context-free.