idependent of $n$, while in step 2 we establish a lower bound on $t_n$ that is independent of $n$.\
[Step 1.]{} It suffices to show the assertion separately for all particles, thus let us fix $i=1,...,N$ and consider $x_i$. By Remark \[setb\] for all $s$, we have $$\begin{aligned}
|v^n_i(s)-v^n_i(0)|=\left|\int_0^s\dot{v}^n_i(\theta)d\theta\right|\leq 
\frac{1}{N}\sum_{k\in B_i(0)}\int_0^s|v^n_k-v^n_i|\psi_n(|x^n_k-x^n_i|)d\theta = \\
\frac{1}{N}\sum_{k\in B^0_i}\int_0^s|v^n_k-v^n_i|\psi_n(|x^n_k-x^n_i|)d\theta + \frac{1}{N}\sum_{k\in B^+_i}\int_0^s|v^n_k-v^n_i|\psi_n(|x^n_k-x^n_i|)d\theta =: I+II,\end{aligned}$$ where $$\begin{aligned}
B^0_i:=\{j\in B_i(0): |x_j(0)-x_i(0)|=0\},\ \ \ B^+_i:=\{j\in B_i(0): |x_j(0)-x_i(0)|>0\}\end{aligned}$$ and $B_i(0)$ is the defined by (\[bi\]) set of all particles that have different trajectories than $x_i$. Thus $B^0_i$ consists of all particles that start from the same position as $x_i$ but with different velocities, while $B^+_i$ consists of all particles that start from a different position than $x_i$. We may assume that $B_i(0)$ is not empty as otherwise all $x^n$ are constantly equal $x^n(0)$ and the assertion holds. Thus at least one of sets $B^0_i$ or $B^+_i$ is nonempty. Now we estimate $I$ and $II$ separately starting with $I$. For $j\in B^0_i$, we have $|v^n_j(0)-v^n_i(0)|>0$ and by its continuity there exists $t_n$ such that $|x^n_j(s)-x^n_i(s)|>0$ and consequently $\psi_n(|x^n_j(s)-x^n_i(s)|)\leq \psi(|x^n_j(s)-x^n_i(s)|)$ in $(0,t_n]$. Together with Lemma \[equitech\] it implies that $$\begin{aligned}
I\leq \frac{4}{N}\sum_{j\in B^0_i}\int_0^s\frac{|(v^n_j-v^n_i)\cdot(x^n_j-x^n_i)|}{|x^n_j-x^n_i|}\psi(|x^n_j-x^n_i|)d\theta.\end{aligned}$$ We claim that, since $\Psi(|x^n_i(0)-x^n_j(0)|)=0$ for all 