**($\Rightarrow$)** Suppose $\mathcal{V}(\phi_t(x)) \leq \mathcal{V}(x)$ for all $t \geq 0$ and all $x \in E$. Take any $b > a \geq 0$. By the Mean Value Theorem applied to $t \mapsto \mathcal{V}(\phi_t(x))$ on $[a, b]$, there exists $t' \in (a, b)$ such that \begin{align*} \mathcal{V}(\phi_b(x)) - \mathcal{V}(\phi_a(x)) = \dot{\mathcal{V}}(\phi_{t'}(x)) (b - a). \end{align*} Treating $\phi_a(x)$ as the initial condition, the hypothesis gives $\mathcal{V}(\phi_{b-a}(\phi_a(x))) \leq \mathcal{V}(\phi_a(x))$, i.e., $\mathcal{V}(\phi_b(x)) \leq \mathcal{V}(\phi_a(x))$. Hence $\dot{\mathcal{V}}(\phi_{t'}(x)) \leq 0$. Since $a < b$ were arbitrary, $\dot{\mathcal{V}} \leq 0$ everywhere on trajectories. **($\Leftarrow$)** Suppose $\dot{\mathcal{V}}(\phi_{t'}(x)) \leq 0$ for all $t' \geq 0$ and all $x \in E$. Take any $t > 0$ and any $x \in E$. By the Mean Value Theorem, there exists $t' \in (0, t)$ such that \begin{align*} \mathcal{V}(\phi_t(x)) - \mathcal{V}(x) = \dot{\mathcal{V}}(\phi_{t'}(x)) \cdot t \leq 0, \end{align*} where the inequality uses the hypothesis. Hence $\mathcal{V}(\phi_t(x)) \leq \mathcal{V}(x)$ for all $t \geq 0$.