C}(v_r)|}\right\}\\
&< (1-p)(1-q)^{|{\cal C}(v_r)|}\\
&=\Pr\left({\bf X}^*[0,1]\right).\end{aligned}$$ In a summary, $\Pr\left({\bf X}^*([0,t]\right)$ is a none-increasing function of $t\in[1,\infty)$ when $k=0.$

![Example of Lemma \[lem:inequalitytime\][]{data-label="figure:timeinequality"}](figure/timeinequlationlemma.pdf "fig:"){width="200"}\

[**Step 2:**]{} Assume (\[eqn:3\]) holds for $k\leq n,$ and consider $k=n+1.$ Clearly $t\geq n+1\geq 1$ for each ${\bf X}[0,t]$ such that $${\bf F}({\bf X}[0,t])\in {\cal Y}^{n+1}.$$ Furthermore, the set of subtrees ${\cal T}=\{T^{-v_r}_{u}|u \in {\cal C}(v_r)\}$ are divided into two subsets: ${\cal T}^h=\{T^{-v_r}_{u}|u \in {\cal
C}(v_r), {\bf Y}(T^{-v_r}_{u})\cap{\cal I}=\emptyset\}$ and ${\cal
T}^i={\cal T}\backslash {\cal T}^h,$ where ${\bf Y}(T^{-v_r}_{u})$ is the vector of $\bf Y$ restricted to subtree $T^{-v_r}_u.$ In Figure 3, ${\cal
T}^h=\{T^{-v_r}_{v_4}\}$ and ${\cal
T}^i=\{T^{-v_r}_{v_1},\ T^{-v_r}_{v_2},\ T^{-v_r}_{v_3}\}.$ We note that given $t_{v_r}^R,$ the infection processes on the sub-trees are mutually independent.

[**Step 2.a**]{} Recall that ${\cal T}^h$ is the set of subtrees having no infected nodes. Following the argument for the $k=0$ case, we can obtain that if $T_u^{-v_r}\in {\cal T}^h,$ then $$\begin{aligned}
&\Pr({\bf X}^*([0,t],{T}^{-v_r}_u)|t_{v_r}^R)=\max\left\{(1-q)^{t_{v_r}^R},qp(1-q)^{|{\cal
C}(u)|}\right\}\end{aligned}$$ when $t\geq t_{v_r}^R$ and $$\begin{aligned}
&\Pr({\bf X}^*([0,t],{T}^{-v_r}_u)|t_{v_r}^R)= \max\left\{(1-q)^{t},qp(1-q)^{|{\cal
C}(u)|}\right\}\end{aligned}$$ when $t< t_{v_r}^R.$ So $\Pr({\bf X}([0,t],{T}^{-v_r}_u)|t_{v_r}^R)$ is non-increasing in $t$ given any $t_{v_r}^R.$

[**Step 2.b**]{} For $T_u^{-v_r}\in {\cal T}^i$, given the sample path $\tilde{{\bf X}}([0,t+1], T_u^{-v_r}),$ we will construct a sample path ${\bf X}([0,t], T_u^{-v_r})$ which occurs with a higher 