# Show L is not context free using the CFL pumping lemma

I am trying to use the pumping lemma to show this language is not context free: $L = a^nb^{n+1}c^{2n} : n \ge 0$

So I took $z = a^mb^{m+1}c^{2m}$ where $|z| = 4m+1 > m$. We can decompose $z = a^mb^{m+1}c^{2m}$ to $z = uvwxy$ for $|vwx| \le m$ and $|vx| \ge 1$.

I am confused about what cases to consider to show this is not context free?

• You should revisit how the Pumping lemma works. Hint: Can you prove that $a^nb^nc^n$ is not context-free? – Raphael Apr 10 '14 at 7:34

The idea is to look where the portion $vwx$ of $z$ can be in the string. Since the length of $vwx$ is restricted it makes it so there are not too many cases to consider. First is to consider when $vwx$ will be entirely in one letter. So it may be entirely in $a$, $b$ or $c$. So when you pump up or down you will only be affecting that portion. Then it is possible for it to span over two letters but not all three. If $v$ or $x$ is itself across two letters then pumping up will create an equal amount of letters but will be out of order. Another case is if $v$ is only $a$ and $x$ is only $b$ or similarly for $b$ and $c$. Try coming up with the contradictions for all these cases.