how to prove that L is not context free Given  $\sum_2$ = {$\begin{bmatrix} 0 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$,$\begin{bmatrix} 1 \\ 0 \end{bmatrix}$,$\begin{bmatrix} 1 \\ 1\end{bmatrix}$} , and a language $L$ = {$w$ $\in$ $\sum_2^*$ | $Top(w)$ = $Bott(w)^R$ $\wedge$ $|w|_0$ = $|w|_1$} where
 $Top(w)$ is the top row and $bott(w)$ is the bottom row.
For example, $w$ = $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ then $Top(w)$ = $10$ and $bottom(w)$ = $01$.
I want to prove that $L$ isn't context free language.
I have tried to do so with pumping lemme and could not find a word that would get me to the needed contradiction (I assumed L is indeed context free).
Any help will do.
Thanks in advance.
 A: This answer assumes that $|w|_\sigma$ is the total number of $\sigma$'s in both rows of $w$.
Suppose that $L$ were context-free. Then so would be
$$
L' = L \cap
\left(
\begin{bmatrix} 0 \\ 0 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\right)^*
\left(
\begin{bmatrix} 1 \\ 1 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\right)^*
\left(
\begin{bmatrix} 0 \\ 1 \end{bmatrix}
\begin{bmatrix} 1 \\ 1 \end{bmatrix}
\right)^*
\left(
\begin{bmatrix} 0 \\ 1 \end{bmatrix}
\begin{bmatrix} 0 \\ 1 \end{bmatrix}
\begin{bmatrix} 0 \\ 0 \end{bmatrix}
\right)^* =
\left\{
\left(
\begin{bmatrix} 0 \\ 0 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\right)^a
\left(
\begin{bmatrix} 1 \\ 1 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\right)^b
\left(
\begin{bmatrix} 0 \\ 1 \end{bmatrix}
\begin{bmatrix} 1 \\ 1 \end{bmatrix}
\right)^c
\left(
\begin{bmatrix} 0 \\ 1 \end{bmatrix}
\begin{bmatrix} 0 \\ 1 \end{bmatrix}
\begin{bmatrix} 0 \\ 0 \end{bmatrix}
\right)^d : \\(011)^a(11)^b(01)^c(000)^d = (011)^d (11)^c (01)^b (000)^a, \\
4(a+d) + (b+c) = 2(a+d) + 3(b+c)
\right\}.
$$
The first constraint implies that $a=d$ and $b=c$, simplifying the second constraint to $4a+b=2a+3b$, which in turn implies that $a=b$. Therefore
$$
L' =
\left\{
\left(
\begin{bmatrix} 0 \\ 0 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\right)^n
\left(
\begin{bmatrix} 1 \\ 1 \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\right)^n
\left(
\begin{bmatrix} 0 \\ 1 \end{bmatrix}
\begin{bmatrix} 1 \\ 1 \end{bmatrix}
\right)^n
\left(
\begin{bmatrix} 0 \\ 1 \end{bmatrix}
\begin{bmatrix} 0 \\ 1 \end{bmatrix}
\begin{bmatrix} 0 \\ 0 \end{bmatrix}
\right)^n :
n \geq 0
\right\}.
$$
Let $h$ be the homomorphism which maps $\begin{bmatrix}0\\0\end{bmatrix}$ to $a$, $\begin{bmatrix}1\\1\end{bmatrix}$ to $b$, and the other two letters to $\epsilon$. If $L'$ were context-free then so would be
$$
L'' = h(L') = \{ a^n b^{2n} a^n : n \geq 0 \}.
$$
Let $k\colon \{a,b,c\}^* \to \{a,b\}^*$ be the homomorphism given by $k(a) = k(c) = a$ and $k(b) = bb$. If $L''$ were context-free then so would be
$$
L''' = k^{-1}(L'') \cap a^*b^*c^* = \{a^n b^n c^n : n \geq 0\}.
$$
However, $L'''$ is well-known not to be context-free.
