This is the question I am trying to answer: Let $G = (V,E)$ be a directed graph and $s,t$ $\epsilon$ $V$ with $s$ $\neq$ $t$, prove that the minimal number of edges in G that must be removed in order to disconnect $s$ and $t$ is equal to the maximum amount of edge disjoint $s-t$ Paths in $G$.
This is my attempt at proving the above statement:
Let $C_1,...,C_n$ be s-t paths in G with $n$ $ \epsilon$ $\Bbb N$ and $n$ maximal
let $k$ denote the minimal number of edges in $G$ that must be removed in order to disconnect $s$ and $t$.
It is clear that $k$ $\geq$ $n$, therefore all that we must show is that there exists a set $F = \{e_1,...,e_n\}$ of $n$ edges such that there is no $s - t$ path in $(V(G), E(G)$ without $F)$
Construct $F$ as follows:
If the path $C_i$ shares its first edge $e_{1_i}$ with another $s-t$ path in G then add $e_{1_i}$ to $F$ otherwise add the last edge in $C_i$ (the edge whose end node is $t$) to $F$.
I believe by removing these edges from $G$ there can no longer be an $s-t$ path in $G$. I don't know how to prove this though, any help would be appreciated!