Vtgyjfy

Given $sum_2$ = {$begin{bmatrix} 0 \ 0 end{bmatrix}$, $begin{bmatrix} 0 \ 1 end{bmatrix}$,$begin{bmatrix} 1 \ 0 end{bmatrix}$,$begin{bmatrix} 1 \ 1end{bmatrix}$} , 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.

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}.
$$
Let $h$ be the homomorphism which maps $begin{bmatrix}0\0end{bmatrix}$ to $a$, $begin{bmatrix}1\1end{bmatrix}$ to $b$, and the other two letters to $epsilon$. If $L'$ were context-free then so would be
$$
L'' = h(L') = { a^n b^{2n} a^n : n geq 0 }.
$$
Let $kcolon {a,b,c}^* to {a,b}^*$ be the homomorphism given by $k(a) = k(c) = a$ and $k(b) = bb$. If $L''$ were context-free then so would be
$$
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.