I'm trying to understand proof of the following theorem:
A nontrivial connected digraph $D$ is Eulerian if and only if outdegree and indegree is equal for every vertex $v$ of $D$.
This is an arguement regarding implication from right to left (so we're assuming that outdegree is equal to indegree of every vertex). We discover that is there is a longest trail from any vertex $v$ to some another vertex $u$ than it is a circuit (so $u$ = $v$). So the only thing left to prove is that this circuit is indeed Eulerian. Let's call the circuit $C$.
The proof in the book goes as follows: "We claim that $C$ contains all of the arcs of $D$ and that $C$ is therefore an Eulerian circuit. Assume,to the contrary, that $C$ does not contain all of the arcs of $D$. Then, since $D$ is connected, there is a vertex $w$ on $C$ that is incident with arcs not on $C$. (...)".
I intuitively understand why the last sentence is true, but could someone provide a formal arguement as to why such $w$ must exist on the circuit?
Take an edge that is not in $C$ and let $u$ be one of its terminal vertices. If $u\in C$ we are done, if not we know that $D$ is connected so pick a vertex $c\in C$ and there must be a path from $u$ to $v$ in $C$, Let $l$ be the first vertex in that path that belongs to $C$ and suppose the path we took looks like this:
$u\dots j,l\dots,v$ , then $jl$ is an edge that does not belong to $C$ and is adjacent to a vertex in $C$ (If $jl$ belonged to $C$ then $j$ would also belong to $C$ and $l$ would not be the first vertex in the path belonging to $C$).