Is $\{ 0^p1^q0^r \mid p \neq r \}$ context-free?

Question

Consider the following languages:

$$ \begin{align*} L_1 &= \{ 0^p 1^q 0^r \mid p,q,r \ge 0 \}, \\ L_2 &= \{ 0^p 1^q 0^r \mid p,q,r \ge 0, \; p \neq r \}. \end{align*} $$

Which one of the following statements is false?

A. $L_2$ is context-free.

B. $L_1 \cap L_2$ is context-free.

C. $\overline{L_2}$ is recursive.

D. $\overline{L_1}$ is context-free but not regular.

Doesn't the following show that $L_2$ is context-free, and so option A should be false as well?

Assume $L_2$ is context-free. Consider the string $w = 0100 \in L_2$. Breaking $w$ into $w= uvxyz$, where $u = \epsilon$, $v = 0$, $x = 1$, $y = \epsilon$, $z = 00$, according to the pumping lemma (with $p=r=2$), $uv^2xy^2z = 00100 \in L_2$, which is false.

Answer 1

The mistake in your proof is that the pumping lemma says that such a division (w=uvxyz) exists.
So in order to say $L_2$ is not context free, you have to show that for every division, you get a contradiction.

In your example we can use this division:
$u=01, \, v=00, \, x=\epsilon, \, y=\epsilon , z=\epsilon$
So for every $n \geq 0$:
$uv^nxy^nz=01(00)^n \in L_2$