I have a simple question on a Lipschitz flow. Recall that a flow on a non-empty set $X$ is a map $\Phi: \mathbb R \times X \to X$ satisfying
$\Phi(0,x) = x \text{ and } \Phi(s, \Phi(t, x)) = \Phi(s + t, x)$
Now let $X$ be a Banach space and $F$ a Lipschitz-function on $X$. Now it is well known that there exists for every $x \in X$ a unique function $u_x:\mathbb R \to X$ which is continuously strongly differentiable and satisfies:
$u_x'(t) = F(u_x(t)) \text{ where $x \in \mathbb R$ and } u_x(0) = x.$ Where u' is the strong derivative in the Banach space.
Now we can set $\Phi(t,x) = u_x(t)$. This should be a flow, why is that? The first property is easy to verify, but the second one requires us to show that $u_{u_x(t)}(s) = u_x(s + t)$. Can someone give a hint how I obtain this? I'm probably missing something simple.