I am learning for the first time the Pumping Lemma for CFL, and I thought I understood how it works until I came across this example:
"Show that $L = \{a^m b^m c^n \mid m \leq n\}$ is not a CFL."
My problem is that I find a scenario that the above language can be CFL. I am showing my proof below:
- Opponent picks $p$
- We pick $z = a^p b^p c^{2p}$ where $|z| \geq p$
- Opponent divides the string in $z = uvwxy$ where $|vwx| \leq p$ and $vx$ different than $\epsilon$
- We have 5 scenarios to consider:
-1- $vwx$ is all $a$'s
-2- $vwx$ is all $b$'s
-3- $vwx$ is all $c$'s
-4- $vwx$ is a combination of $a$'s and $b$'s
-5- $vwx$ is a combination of $b$'s and $c$'s
For scenario -3-, note that if we pump $v$ and $x$, we still get a word that is part of our given language (the number of $c$'s increases, and that does not affect $a$ or $b$). Am I doing something wrong, or is there a flaw in my logic?