Pick any path $v$ in $S(t_0+\tau, u(t_0 + \tau))$ and consider the spliced path $w = u \underset{t_0+\tau}{\bowtie} v$, which, by property [**S4**]{}, belongs to the funnel $S(t_0, a)$. Since $u$ maximizes $\zeta$ over $S(t_0, a)$, $\zeta(u) \ge \zeta(w)$, i.e., $$\int_0^\infty e^{-\lambda t} \vp(u(t_0 + t))\,dt \ge \int_0^\infty e^{-\lambda t} \vp(w(t_0 + t))\,dt\,.$$ But $$\begin{aligned}
& \int_0^\infty e^{-\lambda t} \vp(w(t_0 + t))\,dt = \int_0^\tau e^{-\lambda t} \vp(u(t_0 + t))\,dt + 
\int_\tau^\infty e^{-\lambda t} \vp(v(t_0 + t))\,dt = \\ 
& \int_0^\tau e^{-\lambda t} \vp(u(t_0 + t))\,dt + e^{-\lambda \tau}\,\int_0^\infty e^{-\lambda t} \vp(v(t_0 + \tau + t))\,dt\,,
\end{aligned}$$ while $$\int_0^\infty e^{-\lambda t} \vp(u(t_0 + t))\,dt = \int_0^\tau e^{-\lambda t} \vp(u(t_0 + t))\,dt + e^{-\lambda \tau}\,\int_0^\infty e^{-\lambda t} \vp(u(t_0 + \tau + t))\,dt\,.$$ Hence, $$\int_0^\infty e^{-\lambda t} \vp(u(t_0 + \tau + t))\,dt \ge \int_0^\infty e^{-\lambda t} \vp(v(t_0 + \tau + t))\,dt\,,$$ which means $\sigma_\tau u$ is a maximizer in $S(t_0+\tau, u(t_0 + \tau))$. To check property [**S4**]{} for $V_\zeta[S]$, pick $u\in V_\zeta[S(t_0, a)]$ and $v\in V_\zeta[S(t_0+\tau, u(t_0+\tau))]$ and consider the spliced path $w = u \underset{t_0+\tau}{\bowtie} v$. We have to show that $w$ maximizes $\zeta$ over $S(t_0, a)$ and hence belongs to $V_\zeta[S(t_0, a)]$. This follows from a simple calculation that takes into 