# Context-free Language, Pumping lemma [duplicate]

I want to prove that $L = {a^n b^m c^{ \lfloor \frac{n}{m} \rfloor } }$

isn't context free language, so I choose N - constant from lemma

so the word is $w = a^N b^N c$ and $w = uvxyz$

1 case

v and y contains only a $v,y \in a^{*}$

and $w_2 = a^{N+B} b^{N} c$ where $B > 0$ is it correct to say that it doesn't belong to language, because $N + B \neq N$ ?

edit

or second version

$v,y \in a^{*}$ so $w_0 = uv^0xy^0z = uxz$ and $w_0 = a^{N-B}b^Nc$ which doesn't belong to language because $\lfloor \frac{N-B}{N} \rfloor = 0 \neq 1$

and what with case where $v \in a^+$ and $y \in b^+$

• I trust you realize that you left off quite a few other cases. For instance, it might be the case that $vxy = a^tb^s$. Also, in the case you noted, it might be the case that $\lfloor (N+B)/N\rfloor$ might still be 1 (though that can be fixed by more pumping). – Rick Decker Jun 12 '14 at 17:27
• Yes, I know that there are more cases, but I was not sure if this one is correct, but if this one is incorrect, how to prove that ? I added second version – user19334 Jun 12 '14 at 17:35
• Strings of that form are a context free language, so you won't be able to get a contradiction. – vonbrand Jun 12 '14 at 20:03
• maybe on $w = a^{N}b^{N}c$ pumping lemma not work , but language L isn't context free. I proved it to word $w = a^{N^2} b^N c^N$ – user19334 Jun 12 '14 at 20:32