Equivalent Definitions for Finite Tree

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

The following statements are equivalent:


 * $$1$$: $$T$$ is connected and has no circuits.


 * $$2$$: $$T$$ has $$n-1$$ edges and has no circuits.


 * $$3$$: $$T$$ is connected and has $$n-1$$ edges.


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


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


 * $$6$$: $$T$$ 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 Number of Edges in Tree.

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 Number of Edges in Tree.

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 Number of Edges in Tree.

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 5
This is proved by Paths in Trees are Unique.

5 implies 1
Suppose $$T$$ had a circuit, say $$\left({u, u_1, u_2, \ldots, u_n, v, u}\right)$$.

Then there are two paths from $$u$$ to $$v$$: $$\left({u, u_1, u_2, \ldots, u_n, v}\right)$$ and $$\left({u, v}\right)$$.

Hence, by Modus Tollendo Tollens, if any two vertices of $$T$$ have only one path between them, $$T$$ must have no circuits.

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.