Equivalent Definitions for Finite Tree

Theorem
Let $T$ be a finite tree of order $n$.

The following statements are equivalent:


 * $(1): \quad T$ is connected and has no circuits.


 * $(2): \quad T$ has $n-1$ edges and has no circuits.


 * $(3): \quad T$ is connected and has $n-1$ edges.


 * $(4): \quad T$ is connected, and the removal of any one edge renders $T$ disconnected.


 * $(5): \quad$ Any two vertices of $T$ are connected by exactly one path.


 * $(6): \quad T$ has no circuits, but adding one edge creates a cycle.

Proof
Statement $1$ is the usual definition of a tree.

1 implies 2

 * The fact that $T$ has no circuits is part of statement $1$.


 * The fact that $T$ has $n-1$ edges is proved in Tree has One Less Edge than it has Nodes.

2 implies 3

 * The fact that $T$ has $n-1$ edges is part of statement $2$.


 * The fact that $T$ is connected is proved in Tree has One Less Edge than it has Nodes.

3 implies 1

 * The fact that $T$ is connected is part of statement $3$.


 * Given that $T$ has $n-1$ edges, the fact that it has no circuits is proved during the course of the proof of Tree has One Less Edge than it has Nodes.

1 implies 4
As $T$ has no circuits, then from Condition for an Edge to be a Bridge, every edge is a bridge.

Thus removing any edge of $T$ will disconnect $T$.

4 implies 1
If by removing any one edge between two vertices of $T$ renders it disconnected, then that means each edge must be a bridge.

So by Condition for an Edge to be a Bridge, $T$ has no circuits.

1 implies and is implied by 5
This is proved by Paths in Trees are Unique.

1 implies 6
Suppose $T = \left({V, E}\right)$ is connected and has no circuits.

Let $u, v \in V$ be any two vertices of $T$.

Let $P = \left({u, u_1, u_2, \ldots, u_{n-1}, v}\right)$ be a path from $u$ to $v$.

Let a new edges $\left\{{u, v}\right\}$ be added.

Then $\left({u, u_1, u_2, \ldots, u_{n-1}, v, u}\right)$ is now a cycle, which is by definition also a circuit, in $T$.

Note that this applies even when $P = \left({u, v}\right)$: $\left({u, v, u}\right)$ is still a cycle in $T$, but now $T$ is a multigraph.

6 implies 1
Suppose $T$ has no circuits, but adding one edge creates a cycle, which is by definition also a circuit.

If $T$ were disconnected, then it would be possible to add an edge $e$ to connect two components of $T$.

By definition, $e$ would be a bridge.

From Condition for an Edge to be a Bridge, it follows that $e$ does not lie on a circuit.

So, if the only way to add an edge to $T$ forms a cycle, it follows that $T$ must be connected.

So $T$ is connected and has no circuits.

Thus, all the above can be used as a definition for a finite tree.