probability. Denote the infection time of $u$ in sample path ${\bf X}([0,t], T_u^{-v_r})$ by $t^I_u$. We let $\tilde{t}^I_{u}$ denote the infection time in sample path $\tilde{{\bf X}}([0,t+1], T_u^{-v_r}).$

If $\tilde{t}^I_{u}>1$, we choose $t^I_{u}=\tilde{t}^I_{u}-1$, i.e., $u$ is infected one time slot later in $\tilde{{\bf X}}[0,
t+1]$ than that in ${\bf X}[0,t].$ Assume the infection processes after $u$ was infected are the same in the two sample paths ${\bf X}([0,t], T_u^{-v_r})$ and $\tilde{{\bf X}}([0,t+1], T_u^{-v_r}).$ Therefore, we have $$\Pr(\tilde{{\bf X}}([0,t+1],T^{-v_r}_{u}))=(1-q)^{\tilde{t}^I_{u}-1}q\Pr(\tilde{{\bf
X}}([0,t+1],T^{-v_r}_{u})|\tilde{t}_{u}^I),$$ and $$\Pr({\bf
X}([0,t],T^{-v_r}_{u}))=(1-q)^{t^I_{u}-1}q\Pr({\bf
X}([0,t],T^{-v_r}_{u})|t^I_{u}).$$ where $\Pr({\bf
X}([0,t],T^{-v_r}_{u})|t^I_{u})$ is the probability of ${\bf
X}([0,t],T^{-v_r}_{u})$ after $u$ was infected. Since the sample paths ${\bf X}([0,t], T_u^{-v_r})$ and $\tilde{{\bf X}}([0,t+1], T_u^{-v_r})$ are the same after $u$ was infected, we obtain $$\Pr({\bf X}([0,t],T^{-v_r}_{u})|t^I_{u})=\Pr(\tilde{{\bf
X}}([0,t+1],T^{-v_r}_{u})|\tilde{t}_{u}^I).$$ Therefore, with $t^I_{u}=\tilde{t}^I_{u}-1$, we get $$\begin{aligned}
&\Pr(\tilde{{\bf X}}([0,t+1],T^{-v_r}_{u}))<\Pr({\bf
X}([0,t],T^{-v_r}_{u}))\end{aligned}$$ If $\tilde{t}^I_{u}=1,$ we set $t^I_{u}=\tilde{t}^I_{u}=1$.[^1] Based on the induction assumption, for $k\leq n$ since ${\bf Y}(T^{-v_r}_{u})\in {\cal Y}^m, m\leq n,$ we have $$\begin{aligned}
&\max_{{\bf X}([0,t],T^{-v_r}_{u})\in {\cal X}(t,T^{-v_r}_{u})}
\Pr({\bf X}([0,t],T^{-v_r}_{u}))\\
&> \max_{\tilde{\bf X}([0,t+1],T^{-v_r}_{u}) \in {\cal
X}(t+1,T^{-v_r}_{u})} \Pr(\tilde{\bf X}([0,t+1],T^{-v_r}_{u})),\end{aligned}$$ where ${\cal
X}(t,T^{-v_r}_{u})=\{{\bf X}([0,t],T^{-v_r}_{u}): {\bf F}({\bf X}([0,t],T^{-v_r}_{u}))={\bf Y}(T^{-v_r}_u) \}.$ Therefore, given any $\tilde{\bf X}([0,t+1],T^{-v_r}_{u}),$ we can always find a corresponding sample path ${\bf X}([0,t],T^{-v_r}_{u}),$ which occurs with a higher probability.

[**Step 2.c**]{} Now we consider 