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.