# How to prove that $L=\{a^kb^mc^{m-k}|m\ge k\ge0, m-k\ge k\}$ is not context-free language?

Prove that $$L=\{a^kb^mc^{m-k}|m\ge k\ge0, m-k\ge k\}$$ is not context-free language.

We can suppose by contradiction that $$L$$ is context-free and choose $$Z=a^kb^{2k}c^k$$.

Using pumping lemma, $$vwx$$ can't have both $$a$$'s and $$c$$'s because $$|vwx|\le k$$. There're $$3$$ cases:

1) $$a\notin vx, |vx|\ge 1$$ and $$b\in vx$$. For $$i=0:$$ $$Z_0=uv^0wx^0y=a^kb^{2k-l}c^{k-t}\\l\ge 1\\t\ge 0$$ In this case $$2k-l\le 2k\implies Z_0\notin L$$.

2) $$a\notin vx, |vx|\ge 1$$ and $$b\notin vx$$ and $$c\in vx, |vx|\ge 1$$. For $$i=0:$$ $$i=0: Z_0= uv^0wx^0y=a^kb^{2k-t}c^{k-l}\\l\ge 1\\t\ge 0$$ In this case $$k-l

3) $$a\in vx, |vwx|\le k, c\notin vx$$. For $$i=0:$$ $$Z_0=uv^0wx^0y=a^{k-l}2^{2k-t}c^k$$ In this case $$k-l

I'm not sure that my proof good. If not what are my mistakes?