# Pumping Lemma to prove that L is not context free

I have the language $L = \{a^ib^jc^k : 0 \leq i \leq j \leq k\}$ and I want to prove that is not context-free.

So I started like this:
$p\in \mathbb{N}_{0}$ is variable.
Choose w = $a^pb^pc^p$

Case 1:
vxy has no c. Choose i = 2
$uv^ixy^iz$ has more a than c or more b than c.
$\Rightarrow uv^ixy^iz\notin L$

Case 2:
vxy has no a.

At case 2 I don't know how to continue, I'm confused since I seem to be not able to prove that there is a case where vxy has no a and the word's b-count or c-count is smaller than the amount of a's.

• The pumping lemma also includes the case $i=0$ (where the $i$ is from the Lemma, not the $i$ from the definition of $L$). – Hendrik Jan May 14 '16 at 17:11
• What are you trying to say? You mean I shouldn't be shadowing those variables? – Leo Pflug May 14 '16 at 17:13
• In Case 2 take $i=0$ in the pumping. That will delete $b$'s and or $c$'s while the $a$'s stay constant. What I added is that there is another $i$ in the definition of $L$. That can be confusing, but with some care you are OK. – Hendrik Jan May 14 '16 at 18:29
• Ah thanks. so those 2 cases are enough, right? – Leo Pflug May 14 '16 at 18:54
• Yes, but you have to argue that, using the length condition on $vxy$. – Hendrik Jan May 14 '16 at 23:08