In a connected and anticonnected graph with at least two vertices, each vertex lies in a largest proper module, and two such modules are equal or disjoint

The singleton $\{v\}$ is a module of $G$, and $\{v\}\mathrm{\ne }V(G)$ because $\mathrm{\mid }V(G)\mathrm{\mid }\ge 2$, so $\{v\}\in \mathcal{M}(v)$ and $\mathcal{M}(v)$ is nonempty.

Every member of $\mathcal{M}(v)$ is a subset of the finite set $V(G)$, so its cardinality is a natural number at most $\mathrm{\mid }V(G)\mathrm{\mid }$.

The set $\{\mathrm{\mid }V(G)\mathrm{\mid }-\mathrm{\mid }M\mathrm{\mid }:M\in \mathcal{M}(v)\}$ is a nonempty subset of $\mathbb{N}$ by steps 1.1 and 1.2, so it has a least element; a member $M^{\ast }\in \mathcal{M}(v)$ attaining it has $\mathrm{\mid }M\mathrm{\mid }\le \mathrm{\mid }{M}^{\ast }\mathrm{\mid }$ for every $M\in \mathcal{M}(v)$.

Let $M\in \mathcal{M}(v)$. Both $M$ and $M^{\ast }$ are proper modules containing $v$, so they meet, and [L1] makes $M\cup M^{\ast }$ a proper module; it contains $v$, so it lies in $\mathcal{M}(v)$ and step 2.1 gives $\mathrm{\mid }M\cup M^{\ast }\mathrm{\mid }\le \mathrm{\mid }{M}^{\ast }\mathrm{\mid }$.

Since $M^{\ast }\subseteq M\cup M^{\ast }$ and the two have equal cardinality by step 3.1 and [L2], they are equal, so $M\subseteq M^{\ast }$. Writing $M(v)=M^{\ast }$, this is a proper module containing $v$ and containing every proper module that contains $v$.

If $M(u)\cap M(v)\mathrm{\ne }\mathrm{\varnothing }$ then $M(u)\cup M(v)$ is a proper module by [L1]; it contains $u$, so step 4.1 gives $M(u)\cup M(v)\subseteq M(u)$ and hence $M(v)\subseteq M(u)$, and by symmetry $M(u)\subseteq M(v)$, so $M(u)=M(v)$. Otherwise the two are disjoint, and every vertex $v$ lies in $M(v)$.