$T_u^{-v}\cap{\cal I}=\emptyset,$ all infected nodes are in $T_v^{-u},$ so it is obvious $t^*_u=t^*_v+1.$

[**Step 2:**]{} In this step, we will prove that $t^I_v=1$ on the sample path ${\bf X}_u^*[0, t^*_u].$ If $t^I_v>1$ on ${\bf X}_u^*([0,t_u^*]),$ then $$t^*_u-t^I_v=t^*_v+1-t^I_v<t^*_v.$$ Note that according to the definition of $t_u^*$ and $t_v^I,$ within $t_u^*-t_v^I$ time slots, node $v$ can infect all infected nodes on $T_v^{-u}.$ Since $t^*_u=t^*_v+1,$ the infected node farthest from node $u$ must be on $T_v^{-u},$ which implies that there exists a node $a \in
T^{-u}_v$ such that $d(u,a)=t^*_u=t^*_v+1$ and $d(v,a)=t^*_v.$ So node $v$ cannot reach $a$ within $t^*_u-t^I_v$ time slots, which contradicts the fact that the infection can spread from node $v$ to $a$ within $t_u^*-t_v^I$ time slots along the sample path $X_u^*[0,
t_u^*].$ Therefore, $t^I_v=1.$

[**Step 3:**]{} Now given sample path ${\bf X}^*_u[0,t^*_u]$, we construct ${\bf X}_v[0,t^*_v]$ which occurs with a higher probability. We divide the sample path ${\bf X}^*_u[0,t^*_u]$ into two parts along subtrees $T_u^{-v}$ and $T_v^{-u}.$ Since $t_v^I=1,$ we have $$\begin{aligned}
&\Pr({\bf X}^*_u[0,t^*_u])\\
&=q\Pr\left({\bf
X}^*_u\left([0,t^*_u],T^{-u}_v\right)\Big|t^I_v=1\right)\Pr\left({\bf
X}^*_u\left([0,t^*_u],T^{-v}_u\right)\right),\end{aligned}$$ where $q$ is the probability that $v$ is infected at the first time slot. Suppose in ${\bf X}_v[0,t^*_v]$, node $u$ was infected at the first time slot, then $$\begin{aligned}
&\Pr({\bf X}_v[0,t^*_v])=\\
&q\Pr\left({\bf
X}_v\left([0,t^*_v],T^{-u}_v\right)\right)\Pr\left({\bf
