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 $\set {v', v}$

This procedure directly follows from the theorem:


 * A graph $G$ is $3$-connected (A) 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 $e = x y$ with $\map \deg x, \map \deg y \ge 3$ and $G_i = G_{i + 1} / e$ 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:
 * $G_i = G_{i + 1} / e$

where $G_{i + 1}$ is $2$-connected.

Let $S$ be a vertex cut such that $\card S \le 2$, and $C_1, C_2$ be two components of $G_{i + 1} - S$.

Since $x, y$ are connected, we can assume:
 * $\map V {C_1} \cap \set {x, y} = \O$

Then, $C_2$ cannot contain both $x$ and $y$ nor it can contain any $v \notin \set {x, y}$.

(otherwise $v$ or $u_{x y}$ 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:
 * $\map \deg x, \map \deg y \ge 3$