=== 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: :$w = \tuple {r, v}$ or: :$\exists m \in V: w = \tuple {r, \ldots, m, v}$ Thus $v$ is the final vertex of the arc $r v$ or the arc $m v$. {{AimForCont}} 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. {{qed|lemma}} {{AimForCont}} 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. {{qed|lemma}} === 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 $\tuple r$ is a directed walk from $r$ to $v$. Since $r$ is not the final vertex of any arc, $\tuple r$ 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 $\tuple {r, v}$. But $r$ is not the final vertex of any arc. So in fact: :$z = \tuple {r, v}$ If $x \ne r$, then: :$z = \tuple {r, \ldots, x, v}$ Continuing inductively from $x$ proves that $z = w$. {{explain|set up the induction more formally, or less formally, or something. This is messy.}} 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. {{qed|lemma}} === 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 $T = \struct {V, E}$ be the simple graph corresponding to $G$. That is, for $x, y \in V$, let $\set {x, y} \in E$ {{iff}} $\tuple {x, y} \in A$ or $\tuple {y, x} \in A$. 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 $\tuple {x, y} \in A$ and $\tuple {y, x} \in A$. 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. {{AimForCont}} 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 $w + \tuple {x_1, \dots, x_n}$ is another directed walk from $r$ to $x_0$, contradicting uniqueness. Thus $G$ has no directed cycles. {{AimForCont}} that $T$ has a cycle $\tuple {x_0, x_1, \dots, x_n}$, 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. {{explain}} 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. {{qed|lemma}} === 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 $v$. {{AimForCont}} 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 $w = \tuple {x_0, \ldots, x_n}$ and $z = \tuple {x_0, \ldots, x_n, \ldots, x_m}$. 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 $\tuple {w_{k - 1}, w_k, \ldots, w_n, z_{m - 1}, \dots, z_k, z_{k - 1} }$ forms a cycle, contradicting the fact that $T$ is a tree. {{explain}} {{qed}} Category:Arborescences 3313oj1kghruvj5qbb8ynvtui1llsm8