$j\in B^0_i$, then $$\begin{aligned}
\int_0^s\frac{|(v^n_j-v^n_i)\cdot(x^n_j-x^n_i)|}{|x^n_j-x^n_i|}\psi(|x^n_j-x^n_i|)d\theta= \Psi(|x^n_j(s)-x^n_i(s)|),\end{aligned}$$ where $\Psi(s)=\frac{1}{1-\alpha}s^{1-\alpha}$ is a primitive of $\psi$. Indeed, we have $$\begin{aligned}
\Psi(|x^n_j(s)-x^n_i(s)|)=\int_0^s\Psi(|x^n_j-x^n_i|)^{'}d\theta\leq \int_0^s\psi(|x^n_j-x^n_i|)\frac{|(x^n_j-x^n_i)(v^n_j-v^n_i)|}{|x^n_j-x^n_i|}d\theta\end{aligned}$$ and since $\psi\geq 0$ we can substitute the above inequality with an equality provided that on $(0,t_n]$ the function $\xi(s):=(x^n_j(s)-x^n_i(s))(v^n_j(s)-v^n_i(s))$ has a constant sign. To prove that $\xi$ has a constant sign it suffices to show that $|\xi|>0$ in $(0,t_n]$, which is an immediate consequence of Lemma \[equitech\]. Thus we proved that [$$\begin{aligned}
I\leq\frac{4}{N}\sum_{j\in B^0_i}\Psi(|x^n_j(s)-x^n_i(s)|)= \frac{4}{N(1-\alpha)}\sum_{j\in B^0_i}\left|(x^n_j(s)-x^n_i(s))\right|^{1-\alpha}\leq \frac{4M^{1-\alpha}}{N(1-\alpha)}\sum_{j\in B^0_i}|s|^{1-\alpha}\leq
\frac{4M^{1-\alpha}}{1-\alpha}|s|^{1-\alpha},\end{aligned}$$ ]{} where we use inequality $|x^n_j(s)-x^n_i(s)|\leq M|s|$ that follows from property 4 from Proposition \[properties\]. To estimate $II$ we first notice that since for all $j\in B^+_i$, we have $|x^n_j(0)-x^n_i(0)|>0$ then there exists $\delta>0$ such that $|x^n_j(0)-x^n_i(0)|>\delta$ for all $j\in B^+_i$. Then, by property 4 from Proposition \[properties\] there exists an $n$ independent interval $[0,t_0]$ on which $|x^n_j-x^n_i|>\delta$ for all $j\in B^+_i$. On this interval $$\begin{aligned}
\psi_n(|x^n_j(s)-x^n_i(s)|)\leq \delta^{-\alpha}.\end{aligned}$$ Therefore $$\begin{aligned}
II\leq \frac{1}{N}\sum_{j\in B^+_i} 2|s| M\delta^{-\alpha}\leq 2 t_0^\alpha M\delta^{-\alpha}|s|^{1-\alpha}\end{aligned}$$ and adding our estimations of $I$ and $II$ we get $$\begin{aligned}
|v_i^n(s)-v_i^n(0)|\leq L |s|^{1-\alpha}\end{aligned}$$ with $L=\frac{4M^{1-\alpha}}{1-\alpha}+2t_0^\alpha M\delta^{-\alpha}$ on interval $[0,t_n]\cap[0,t_0]$. For simplicity let us denote $\min\{t_n,t_0\}$ again by $t_n$. This finishes step 1.\
[Step 2.]{} In step 1 we proved that for each 