Prove that $L = \small\{ ww | w \in (0, 1)^* \small\}$ is $\textbf{not}$ a context-free language.
Assume $L$ is a context free language.
Let $s = 1^p0^p1^p0^p$ where clearly $|s| \geq p$ and $s \in L\:$and $s =uvxy$
The exhaustive ways to build $vxy$ are:
1) contains $1$'s (first or second half) Considering $|vy|>0$ then either $v$ or $y$ contains at least one 1. Let $i = 0 \:$ $s' = uv^0xy^0z$ will have at least one missing 1. $s'$ is either for first-half $1^{p-1}0^p1^p0^p$ or $1^p0^p1^{p-1}0^p$ for second-half neither of which is $\in L$
2) contains $0$'s (first or second half) Considering $|vy|>0$ then either $v$ or $y$ contains at least one 0. Let $i = 0 \:$ $s' = uv^0xy^0z$ will have at least one missing 0. $s'$ is either for first-half $1^p0^{p-1}1^p0^p$ or $1^p0^p1^p0^{p-1}$ for second-half neither of which is $\in L$
3)contains 1's and 0's (first or second half)
Consider $|vy|>0$ then $v$ or $y$ must contain at least one 1 or 0 or be a mix of them. Let $i = 2 \:$ $s' = uv^2xy^2z$ will have at least one extra 1 or 0.
For first-half:
$s' = 1^{p+x}0^p1^p0^p$ $x > 0$ clearly $s' \notin L$
$s' = 1^p0^{p+x}1^p0^p$ $x > 0$ clearly $s' \notin L$
$s' = 1^{p+x}0^{p+y}1^p0^p$ $x + y > 0$ clearly $s' \notin L$
For second-half:
$s' = 1^p0^p1^{p+x}0^p$ $x > 0$ clearly $s' \notin L$
$s' = 1^p0^p1^p0^{p+x}$ $x > 0$ clearly $s' \notin L$
$s' = 1^p0^p1^{p+x}0^{p+y}$ $x + y > 0$ clearly $s' \notin L$
4)contains $0$'s and $1$'s in the middle Considering $|vy|>0$ then either $v$ or $y$ contains at least one 0 or 1 or is a mix of them. Let $i = 0 \:$ $s' = uv^0xy^0z$ will have at least one 0 or 1 missing.
Cases:
$s' = 1^p0^{p-x}1^p0^p$ $x > 0$ clearly $s' \notin L$
$s' = 1^p0^p0^{p-x}0^p$ $x > 0$ clearly $s' \notin L$
$s' = 1^p0^{p-x}1^{p-y}0^p$ $x + y > 0$ clearly $s' \notin L$ even if $x = y$ so that number of removed $0$'s and $1$'s are same then $num(first\: consecutive \: 1's)$ > $num(second\: consecutive \:1's)$ clearly $s' \notin L$
No matter what decomposition we choose for $1^p0^p1^p0^p$ for each one there is an $i$ such that $s' = uv^ixy^iz \notin L$. Therefore L is not a context-free-language.
Can someone please check whether my proof is correct or whether I've missed any case? Thanks.