why does ${a^nb^n}$ fit the pumping lemma for context-free languages?

Question

I am writing somthing about Ppumping Lemma. I know that the language $L = \{ a^nb^n| n ≥ 0 \}$ is context-free. But I don't understand how this language satisfies the conditions of pumping lemma (for context-free languages) ?

if we pick the string $s = a^pb^p, |s| > p , |vxy| < p \land |vy| > 0$.

it seems it will be out of the language when we pump it (pump up or down) or there is something I'm missing.

Any explanation would help.

Edit: I am applying pumping lemma to a^nb^n and it fails to stay in the language for all cases. So, why is it Context free?

Answer 1

This language satisfies the conditions of the pumping lemma. (By the way, your question title is wrong: you are not asking why it is context-free; you are asking why it satisfies the conditions of the lemma, which it does.)

Take the string $s=a^mb^m$ and assume $m>0$. Don’t use $p$ or $n$; otherwise it is confusing, because those letters are used in the lemma. Now let $v=a$, $y=b$, $x=\epsilon$ (the empty string), $u=a^{m-1}$, and $z=b^{m-1}$.

You can definitely pump it as much as you want. In other words, you take a string such as $aaaaabbbbb$, then you pick the two middle letters: $aaaa\mathbf{ab}bbbb$. Then you pump them: $aaaa\mathbf{aaabbb}bbbb$ — still in the language.