Connected Graph is Tree iff Removal of One Edge makes it Disconnected/Sufficient Condition
Theorem
Let $G = \struct {V, E}$ be a tree.
Then for all edges $e$ of $G$, the edge deletion $G \setminus \set e$ is disconnected.
Proof 1
Let $G$ be a tree.
Then by definition $G$ has no circuits.
From Condition for Edge to be Bridge, every edge of $G$ is a bridge.
Thus by definition of bridge, removing any edge of $G$ will disconnect $G$.
Proof 2
Let $G$ be a tree.
Hence a fortiori $G$ has no cycles.
Let $v, v' \in V$.
Let the edge $\set {v, v'}$ be removed.
Aiming for a contradiction, suppose $G$ is still connected.
Then a priori $v$ and $v'$ are connected.
By If Vertices are Connected then Path Exists between them, there is a path $\tuple {v, v_1, \ldots, v'}$ of length $2$ or more.
Hence $\tuple {v, v_1, \ldots, v', v}$ is a cycle in $G$.
This contradicts the statement that $G$ has no cycles.
The result follows by Proof by Contradiction.