Let $"t"$ and $"s"$ be a words we will say that two words are "completly different" if for all $1\leqslant i\leqslant |t|$ the $i$ letter in $t$ diffrent from the $i$ letter in $s$.
Prove that the language $\mathcal{L}=\{ts|t,s\in \{0,1\}^*,|t|=|s|,t,s \text{ completly different} \}$ is not a free-context-language
Attempt :
Applaying the pumping lemma for free-contex-language:
Suppose that $\mathcal L$ is regular so exists a word '$z=uxvyw$' with length of at least $n$ such that:
$(1)\,\,\,|xvy|\leqslant n$
$(2)\,\,\,|xy|\geqslant 1$
$(3)\,\,\,ux^ivy^iw \in \mathcal L\,\,\,\,\,\,\,\,\,i\geqslant 0$
Now, let's choose the word $\color{blue}{z=0^n1^n}$ it is obvious that $|z|\geqslant n$ so we can use $(1)-(3)$
$z=0^{\alpha}0^{\beta}0^{\gamma}0^{\lambda}1^n$
So $\alpha+\beta+\gamma+\lambda=n$
I am stuck here.
EDIT: After using @Renato's answer:
Consider $z=0^p1^p0^p1^p0^p1^p\in \mathcal{L}$ since $|z|>p$, there are $u,v,w,x,y$ such that $z=uvwxy,|vwx|\leqslant p, |vx|>0$ and $uv^iwx^iy\in \mathcal{L}$
$vwx$ must straddle the midpoint of $z$ there are fore possibilities:
$vwx$ is in $0^p$ part.
$vwx$ is in $1^p$ part.
$vwx$ is in $1^p0^p$ part.
$vwx$ is in $0^p1^p$ part.
Thus, it is not of the form that we want
For $i=2$ $z\notin \mathcal{L}$
