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 $i=2$ $z\notin \mathcal{L}$