Tutte's Wheel Theorem

Theorem
Every 3-connected graph can be obtained by the following procedure:
 * Start with $G_0 := K_4$
 * Given $G_i$ pick a vertex $v$
 * Split into $v'$ and $v$ and add edge $\{v',v\}$

This procedure directly follows from the theorem:


 * A graph $G$ is 3-connected (A) iff there exists a sequence $G_0, G_1, . . ., G_n$ of graphs with the following properties (B):
 * $G_0 = K_4$ and $G_n = G$;
 * $G_{i+1}$ has an edge $e = xy$ with $deg \left({x}\right), deg \left({y}\right) \geq 3$ and $G_i = G_{i+1} \thinspace / \thinspace e$ for every $i < n$.

A $\implies$ B
Directly follows from Tutte's Wheel Theorem/Lemma.

B $\implies$ A
If $G_i$ is 3-connected then so is $G_{i+1}$ for every $i < n$.

Suppose not. Then $G_i = G_{i+1} \thinspace / \thinspace e$ where $G_{i+1}$ is 2-connected.

Let $S, \: |S| \leq 2$ be a vertex cut, and $C_1, C_2$ be two components of $G_{i+1}-S$.

Since $x, y$ are connected, we can assume $V\left({C_1}\right) \cap \{x, y\} = \emptyset$.

Then, $C_2$ cannot contain both $x$ and $y$ nor it can contain any $v \notin \{x,y\}$ (otherwise $v$ or $u_{xy}$ would be separated from $C_1$ in $G_i$ by at most two vertices.)

But now $C_2$ contains only one vertex: either $x$ or $y$.

This contradicts $\deg\left({x}\right), \deg\left({y}\right) \geq 3$.