Equivalence of Definitions of Arborescence

Theorem
Let $G = \left({V, A}\right)$ be a directed graph.

Let $r \in V$.

Definition 1 implies Definition 3
Let $G$ be an $r$-arborescence by definition 1.

Let $v \in V$ such that $v \ne r$.

Then there is exactly one directed walk $w$ from $r$ to $v$.

Since $v \ne r$, either:
 * $w = \left({r, v}\right)$

or:
 * $\exists m \in V: w = \left({r, \ldots, m, v}\right)$

Thus $v$ is the final vertex of the arc $r v$ or the arc $m v$.

Suppose for the sake of contradiction that $v$ is the final vertex of distinct arc $x v$ and $y v$.

Then there exist directed walks $w_1$ and $w_2$ from $r$ to $x$ and $r$ to $y$ respectively.

But appending $v$ to $w_1$ and to $w_2$ yields distinct directed walks from $r$ to $v$.

This contradicts the fact that there is exactly one such directed walk.

Thus $v$ is the final vertex of exactly one arc.

Suppose for the sake of contradiction that $r$ is the final vertex of an arc $x r$.

By Definition 1, there is a directed walk $w$ from $r$ to $x$.

But then $w$ appended to $w$ is a directed walk from $r$ to $x$ which is not equal to $w$.

This contradicts definition 1.

Thus we conclude that $r$ is not the final vertex of any arc.

It follows immediately from definition 1 that there is a directed walk from $r$ to each vertex $v \ne r$.

Thus $G$ is an $r$-arborescence by Definition 3.

Definition 3 implies Definition 1
Suppose that $G$ is an $r$-arborescence by Definition 3.

Let $v \in V$.

We must show that there is a unique directed walk from $r$ to $v$.

If $v = r$, then $\left({r}\right)$ is a directed walk from $r$ to $v$.

Since $r$ is not the final vertex of any arc, $\left({r}\right)$ is the only such directed walk.

If $v \ne r$, then there exists some directed walk $w$ from $r$ to $v$.

Suppose that $z$ is a directed walk from $r$ to $v$.

Since $v \ne r$, $v$ is the final vertex of exactly one arc $x v$.

If $x = r$, then $z$ must end with $\left({r, v}\right)$.

But $r$ is not the final vertex of any arc.

So in fact:
 * $z = \left({r, v}\right)$

If $x \ne r$, then:
 * $z = \left({r, \ldots, x, v}\right)$

Continuing inductively from $x$ proves that $z = w$.

Thus for each $v \in V$ there is exactly one directed walk from $r$ to $v$.

So $G$ is an $r$-arborescence by Definition 1.

Definitions 1 and 3 imply Definition 2
Let $G$ be an $r$-arborescence by Definition 1.

From the above, $G$ is then also an $r$-arborescence by Definition 3.

Let $T = \left({V, E}\right)$ be the simple graph corresponding to $G$.

That is, for $x, y \in V$, let $\left\{{x, y}\right\} \in E$ $\left({x, y}\right) \in A$ or $\left({y, x}\right) \in A$.

We will show that $G$ is an orientation of $T$ and that $T$ is a tree.

By Definition 1, there is exactly one directed walk from $r$ to each vertex $v$.

Let $x, y \in V$ and suppose for the sake of contradiction that $\left({x, y}\right) \in A$ and $\left({y, x}\right) \in A$.

Let $w_x$ be the directed walk from $r$ to $x$.

Then appending $y$ to $w_x$ yields a directed walk $w_x + y$ from $r$ to $y$.

But then appending $x$ to $w_x + y$ yields a directed walk $w_x + y + x$ from $r$ to $x$, contradicting the fact that $w_x$ is unique.

Thus $A$ is asymmetric.

So $G$ is an orientation of $T$.

We must now show that $T$ is a tree.

By Equivalence of Definitions of Reachable, each vertex $v$ of $G$ is reachable from $r$.

Let $x, y \in V$.

By Definition 1, there is exactly one directed walk from $r$ to $x$ and exactly one directed walk from $r$ to $y$.

Thus there is a walk $w_x$ in $T$ from $r$ to $x$ and a walk $w_y$ from $r$ to $y$.

Reversing $w_x$ and then appending $w_y$ to it (eliding the duplicate $r$) yields a walk from $x$ to $y$.

Thus $T$ is connected.

Next we show that $G$ has no directed cycle.

Suppose for the sake of contradiction that $x_0, x_1, \dots, x_n$ is a directed walk with $n \ge 2$ and $x_0 = x_n$.

By Definition 1, there is a unique directed walk $w$ from $r$ to $x_0$.

But then $w + \left({x_1, \dots, x_n}\right)$ is another directed walk from $r$ to $x_0$, contradicting uniqueness.

Thus $G$ has no directed cycles.

Suppose for the sake of contradiction that $T$ has a cycle $\left({x_0, x_1, \dots, x_n}\right)$, where $x_0 = x_n$ and $n \ge 2$.

Since $G$ has no directed cycles, the arcs corresponding to the edges in this cycle cannot all go in the same direction around the cycle.

But this implies that some vertex of $x_0, x_1, \dots, x_n$ is the final vertex of two different arcs, contradicting Definition 3.

Thus $T$ has no cycles.

As $T$ is connected, it is a tree.

Thus we have shown that $G$ is an orientation of $T$ and that $T$ is a tree.

Definition 2 implies Definition 1
Let $G$ be an $r$-arborescence by Definition 2.

That is, let $G$ be an orientation of a tree $T$ and that every vertex of $G$ is reachable from $r$.

Let $v \in V$.

By the definition of reachable, there exists a directed walk $w$ from $r$ to $v$.

We must show that $w$ is the only directed walk from $r$ to $v$.

Suppose for the sake of contradiction that $z$ is a directed walk from $r$ to $v$ and $z \ne v$.

Then either $z$ extends $w$, $w$ extends $z$, or there is some $k$ such that $w_k \ne z_k$.

First suppose that $z$ extends $w$.

So $w = \left({x_0, \ldots, x_n}\right)$ and $z = \left({x_0, \ldots, x_n, \ldots, x_m}\right)$.

Then $p = \left({x_n, \dots, x_m}\right)$ is a directed walk from $v$ to $v$ with more than one vertex.

By Directed Circuit in Simple Digraph forms Circuit, the vertices of $p$ form a circuit, contradicting the fact that $T$ is a tree.

Thus there exists $k$ such that $w_k \ne z_k$.

It follows from the Well-Ordering Principle that there must be some smallest $k$ such that $w_k \ne z_k$.

Since $w$ and $z$ both end at $v$, it follows from the Well-Ordering Principle that:
 * there must be some smallest $n > k$ such that there exists $m > k$ such that $w_n = z_m$
 * and that there exists a smallest such $m$.

Then $\left({w_{k-1}, w_k, \ldots, w_n, z_{m-1}, \dots, z_k, z_{k-1} }\right)$ forms a cycle, contradicting the fact that $T$ is a tree.