Graphs with no $K_4$ minor have a vertex of degree at most $2$

Note — non-blind read-back. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statements, with full knowledge of the intended meaning. It is not independent or blind testimony, and no reviewer should treat it as such; an independent audit is still needed.

Let $V$ be a finite, nonempty type and $G$ a simple graph on $V$. If $G$ has no $K_4$ minor, then there is a vertex $v\in V$ whose set of neighbours has at most $2$ elements. Here, for graphs $H$ on a vertex type $W$ and $G$ on a vertex type $V$, “$H$ is a minor of $G$” means: there are sets $B_w\subseteq V$ ($w\in W$) such that (i) each $B_w$ induces a connected subgraph of $G$ (in particular $B_w\neq\emptyset$), (ii) $B_{w_1}\cap B_{w_2}=\emptyset$ whenever $w_1\neq w_2$, and (iii) whenever $w_1w_2$ is an edge of $H$, some vertex of $B_{w_1}$ is adjacent in $G$ to some vertex of $B_{w_2}$. “$G$ has a $K_t$ minor” means the complete graph on the $t$-element set $\{0,\dots,t-1\}$ is a minor of $G$ in this sense (so: $t$ pairwise disjoint connected nonempty vertex sets, every two joined by an edge of $G$; for $t=0$ this holds for every graph). “$k$-colourable” means there is a map from $V$ to a $k$-element set of colours giving adjacent vertices different colours (for $k=0$ this forces $V=\emptyset$). All graphs are simple (no loops, no multiple edges).