Introduction to graph theory/Proof of Theorem 4

Statement
For a graph $$G$$, the following are equivalent:
 * $$G$$ is a tree
 * $$G$$ is a minimal connected graph, in the sense that the removal of any edge will leave a disconnected graph.
 * $$G$$ is a maximal acyclic graph, in the sense that the addition of any missing edge will create a cycle.

Corollary of
Corollary 3: A graph is a tree if and only if for every pair of distinct vertices $$u,v$$, there is exactly one $$u,v$$-path.

Proof
Suppose $$G$$ is a tree, and let $$uv$$ be an edge not in $$G$$. By Corollary 3, there is exactly one $$u,v$$-path. If this path is $$ux_1\ldots x_kv$$, then adding the edge $$uv$$ will create the cycle $$ux_1\ldots x_kvu$$. As a tree is defined to be a connected forest and a forest is defined to be acyclic. A tree is therefore a maximal acyclic graph.

Now suppose $$G$$ is a tree, and let $$uv$$ be an edge of $$G$$. By Corollary 3, there is exactly one $$u,v$$-path. This path is clearly just the edge $$uv$$. Thus, in the graph $$G-uv$$, there is no $$u,v$$-path, and hence $$G-uv$$ is disconnected. Since a tree is defined to be a connected forest, a tree is therefore a minimal connected graph.

Now suppose $$G$$ is a maximal acyclic graph, but is not a tree. Since an acyclic graph is called a forest, and a tree is defined to be a connected forest, $$G$$ must be disconnected. Thus there exists vertices $$u,v$$ such that there is no $$u,v$$-path in $$G$$. In particular $$uv$$ is not an edge of $$G$$. As $$G$$ is a maximal acyclic graph, $$G+uv$$ must have a cycle. Since $$G$$ has no cycle, this cycle must include the edge $$uv$$. Let this cycle be $$ux_1\ldots x_kvu$$. Then $$ux_1\ldots x_kv$$ is a $$u,v$$-path in $$G$$. This contradiction shows that maximal acyclic graphs are trees.

Finally, suppose $$G$$ is a minimal connected graph, but is not a tree. As a tree is defined to be a connected forest, it follows that $$G$$ is not a forest, and hence (by definition), $$G$$ must have a cycle. Let $$u,v$$ be adjacent vertices on the cycle, and let the cycle be $$ux_1\ldots x_kvu$$. As $$G$$ is a minimal connected graph, $$G-uv$$ must be disconnected. Consider any vertex $$x$$. In $$G$$, there was a path from $$x$$ to $$v$$. If the path went through the edge $$uv$$, then in $$G-uv$$, there is clearly an $$x,u$$-path. Otherwise, there is clearly an $$x,v$$-path in $$G-uv$$. Either way, the connected component of $$G-uv$$ contains either $$u$$ or $$v$$. Since $$ux_1\ldots x_kv$$ is a $$u,v$$-path in $$G-uv$$, the connected component containing $$u$$ also contains $$v$$. Thus every vertex in $$G-uv$$ is contained in the same connected component, meaning that $$G-uv$$ is connected. This contradiction shows that minimal connected graphs are trees.