g(p)) = 1 - g(p)$ with respect to $p$, which gives $$(-g(p) - p \cdot 
   g'(p)) \phi'(1 - p \cdot g(p)) 
   = -g'(p).$$ Rearranging this expression to isolate $g'(p)$, while using Proposition \[pr:K\], along with the fact that $\phi'(1-x) = \mu-\phi''(1) x
+ o(x)$ as $x\rightarrow 0$, we get $$g'(p) 
   \; = \; \frac{g(p) \phi'(1-p \cdot g(p))}{1-p\cdot \phi'(1-p \cdot g(p))}
   \; = \; \frac{2\mu^3}{\phi''(1)}+o(1)$$ as $p \downarrow p_c$. $\Cox$

Define invasion percolation on an arbitrary tree $T$ as follows. Start with $I^T_0 = \bf 0$ where we recall that $\bf 0$ is the root of $T$. Inductively define $I^T_{n+1}$ to consist of $I^T_n$ along with the vertex of minimal weight $U_v$ adjacent to $I^T_n$. The invasion percolation cluster is defined as $I^T := \bigcup I^T_n$. Note that $I^T$ is measurable with respect to the $U$ variables. Let $I := I^\TT$ \[I-def\] denote the invasion cluster of the random tree $\TT$. By independence of the $U$ variables and $\T$, the conditional distribution of $I$ given $\TT$ agrees with that of invasion percolation.

\[prop:infst\] For any $p > p_c$, $I$ almost surely reaches some vertex $v$ such that $v \conn_p \infty$ in $\TT (v)$.

We consider the following coupling that generates $I$ 