Regular Graph is Tree iff Complete Graph of Order 2

Theorem
Let $G$ be a non-edgeless regular graph.

Then $G$ is a tree $G$ is $K_2$, the complete graph of order $2$.

Necessary Condition
Let $G$ be a non-edgeless regular graph which is also a tree.

From Finite Tree has Leaf Nodes it follows that $G$ has at least two vertices of degree $1$.

Therefore, for $G$ to be regular it need to be $1$-regular.

Suppose $G$ has $3$ or more vertices.

Let $u, v, w$ be such vertices of $G$.

Suppose each of $\set {u, v}$, $\set {v, w}$ and $\set {u, w}$ were adjacent.

Then there would be a cycle in $G$: $\tuple {u, v, w, u}$ and so $G$ by definition would not be a tree.

So at least one pair $\set {u, v}$, $\set {v, w}$ and $\set {u, w}$ is not adjacent.

suppose this pair is $\set {u, v}$.

By definition, a tree is connected.

Therefore there exists a path $P$ between $u$ and $v$.

As $u$ and $v$ are not adjacent, there exists another vertex $y$ of $G$ through which $P$ passes.

In order for this to be the case, the degree of $y$ needs to be greater than $1$.

It follows that $G$ can not be $1$-regular.

So to be regular, $G$ can contain only $2$ vertices.

By Size of Tree is One Less than Order it follows that $G$ contains $2$ vertices connected together by a single edge.

By definition, this is the complete graph of order $2$.

Sufficient Condition
$K_2$, the complete graph of order $2$, consists of $2$ vertices connected together by a single edge.

From Complete Graph is Regular it follows that $K_2$ is a non-edgeless regular graph.

From Circuit has Three Edges or More it follows that $K_2$ contains no circuits.

It follows by definition that $K_2$ is a tree.