Finite Tree has Leaf Nodes/Proof 2

Theorem
Every non-edgeless finite tree has at least two leaf nodes.

Proof
Let the proposition we are proving in the case of a graph of order $k$ be denoted with $P \left({k}\right)$.

Basis of the induction
That $P \left({2}\right)$ is true follows as the Unique Tree of Order 2 contains exactly two leaves (viz. its two nodes).

Induction hypothesis
Suppose $P\left({i}\right)$ is true for all $i$ such that $2 \le i \le k$.

Induction step
Let $T_{k + 1}$ be a tree of order $k + 1$.

Pick an edge $e = \left\{{v_1, v_2}\right\}$ of $T_{k + 1}$, and consider the subgraph $T_0$ obtained by removing $e$.

By Edge of Tree is Bridge, $e$ is a bridge and removing it results in a subgraph with two components.

By Connected Subgraph of Tree is Tree each such component is a tree.

Let $T_i$ be the component containing $v_i$, and denote its order by $k_i$, where $i \in \left\{{1, 2}\right\}$

Since no nodes have been removed from $T_{k+1}$, $k_1 + k_2 = k + 1$.

There are two cases:

Case 1
Neither $T_1$ nor $T_2$ is of order $1$.

Then $2 \le k_i \le k - 1, i = 1, 2$.

By the induction hypothesis, both $T_1$ and $T_2$ have at least two leaves.

Either $v_1$ and $v_2$ are both leaves of the respective trees, or one of them is not.

Suppose that $v_1$ and $v_2$ are both leaves of the respective trees.

Then by adding $e$ back to the graph and thus recreating $T_{k + 1}$, $v_1$ and $v_2$ cease to be leaves.

However, at least one other leaf from each subgraph $T_i, i = 1, 2$ survives the addition, becoming a leaf of $T_{k + 1}$.

Therefore $T_{k + 1}$ has at least two leaves.

Suppose that either $v_1$ or $v_2$ is not a leaf of its respective tree.

Then $T_{k + 1}$ has at least three leaves.

Case 2
$T_1$ or $T_2$ is of order $1$.

Assume WLOG that $T_1$ is of order $1$, i.e. $k_1 = 1$.

This implies that $k_2 = k$.

Then, by the induction hypothesis, $T_2$ has at least two leaf nodes.

Also $k_1 = 1$ implies that $v_1$ is a leaf of $T_{k + 1}$.

Suppose that $v_2$ is one of the leaves of $T_2$.

Then by adding $e$ back and thus recreating $T_{k + 1}$, $v_2$ is no longer a leaf, though at least one other leaf of $T_2$ survives the addition, becoming a leaf of $T_{k + 1}$.

Therefore $T_{k + 1}$ has at least two leaves.

Suppose that $v_2$ is not one of the leaves of $T_2$.

Then $T_{k + 1}$ has at least three leaves.

In both cases, the thesis follows by the Second Principle of Mathematical Induction.