Context-free pumping lemma of $a^nb^n$

Question

I know $a^nb^n$ with $n\geq0$ is considered a context-free language, but if I try:

Using pumping length $p = 3$

$n = p$, thus we have $aaabbb$

$u =aa$ and $y = bb$

$v = a$, $w = b$ and $x=λ$, then $|vwx|=2\leq p=3$ and $|vx| = 1 \geq 1$

$uv^iwx^iy \notin L$, for instance, with $i=2$ we have: $$aaaabbb$$

I know I'm wrong in some part of the process, that's I'm attempting to 'break' the lemma, to fully understand it.

Answer 1

You picked a wrong decomposition. Similarly to the pumping lemma for regular languages, the pumping lemma for context-free languages states that for every context-free language $L$, there exists some legal decomposition for every string $s$ with $|s| \geq p$ where $p$ is a pumping length of $L$.

By legal decomposition, I mean dividing $s$ into partitions $uvwxy$ such that

• $|vx| \geq 1$
• $|uvw| \leq p$
• $uv^nwx^ny \in L$ for all $n \in \mathbb{N}_0$

In the case of the string $aaabbb$ with $p = 3$, a legal decomposition would be eg. $u, w, y = \epsilon$, $v = aaa$, $x = bbb$.