# Use the pumping lemma to prove that $L = \{a^kb^mca^nb^m : k < n \space \wedge \space k,m > 0\}$ is not context free.

Heres what I got but I am not sure if ive overlooked something or got anything wrong.

$$L = \{a^kb^mca^nb^m : k < n \space \wedge \space k,m > 0\}$$

Assume that $$L$$ is a CFL. Let $$p$$ be the pumping length given by the pumping lemma. We choose the string $$s = a^{p-1}b^pca^pb^p$$. We know by the pumping lemma that $$s$$ can be divided into the string $$s = > uvxyz$$ where $$uv^1xy^1z \in L$$ for each $$i \geq 0$$, $$|vy| < 0$$, and $$|vxy| \leq p$$.

Consider the three cases:

1. $$vy$$ contains a $$c$$: The string $$uv^2xy^2z$$ then has two $$c$$'s and thus is not in $$L$$.
2. $$vy$$ contains atleast one $$a$$: We pump upwards to $$uv^2xy^2z$$ if $$vy$$ is on the left of the $$c$$. We pump downwards to $$uv^0xy^0z$$ if $$vy$$ is on the right side of the $$c$$. I both cases, $$k \geq n$$ and the string is thus not in $$L$$.
3. $$vy$$ contains atleast one $$b$$: The string $$uv^2xy^2z$$ will make the substrings consinsting of only $$b$$'s not equals. Then the string cant be in $$L$$

Each case results in a contradiction. There for the pumping lemma does not hold and $$L$$ is not context free.