I have been working to understand the pumping lemma better, but I am quite stuck at proving these two languages is not regular:
\begin{align} L_1 &= \{(ab)^n c^m \mid n\ge 1, m\ge 2n \} \\ L_2 &= \{(ab)^n a^k (ba)^n \mid k<3\} \end{align}
For $L_2$ my approach was:
Let us be given a number $p$. Let $z = (ab)^p a^k (ba)^p$, which satisfies $|z| = 2p > p$, and let $z = uvw$ be its decomposition satisfying $|uv| \leq p$ and $|v| > 0$. This means that $v = (ab)^j$ for some $0 \le j \le p$. We choose $i = 2$ for $uv^iw$, which equals $(ab)^{p+j} a^k (ba)^p$. This word has more $ab$ than $ba$, which means that it doesn't belong to the language. Therefore $L_2$ is not regular.
Mainly I am actually confused with the $(ab)^n$, we should decomposed it so that we can pump $v$, but it is necessary to consider different cases of $v$, or is this sufficient?