In a connected and anticonnected graph, a modular partition with at least two parts whose quotient is prime consists of the maximal proper modules

Every part $N\in \mathcal{P}$ is a nonempty module of $G$, and $N\mathrm{\ne }V(G)$ because $\mathcal{P}$ has another part, which is nonempty and disjoint from $N$; so every part is a proper module.

Fix $M_0\in \mathcal{P}$ and $x\in M_0$, and let $M(x)$ be the largest proper module of $G$ containing $x$. Then $M_0\subseteq M(x)$ by step 1.1 and the maximality in [L3].

Let $\mathcal{X}=\{N\in \mathcal{P}:N\cap M(x)\mathrm{\ne }\mathrm{\varnothing }\}$ and let $N\in \mathcal{X}$. Both $N$ and $M(x)$ are proper modules and they meet, so $N\cup M(x)$ is a proper module by [L2]; it contains $x$, so [L3] gives $N\cup M(x)\subseteq M(x)$ and hence $N\subseteq M(x)$.

Every vertex of $M(x)$ lies in a part, and that part meets $M(x)$ and so lies in $\mathcal{X}$; with step 3.1 this gives $M(x)={\bigcup }_{N\in \mathcal{X}}N$.

By [L1] the set $\mathcal{X}$ is therefore a module of $G\mathrm{/}\mathcal{P}$, hence trivial by [F3]. It is not empty, since $M_0\in \mathcal{X}$ by step 2.1; and it is not all of $\mathcal{P}$, since that would give $M(x)=V(G)$, contradicting properness. So $\mathcal{X}$ is a singleton, and by step 2.1 its unique member is $M_0$, whence $M(x)=M_0$.

So each part of $\mathcal{P}$ equals $M(x)$ for each of its vertices $x$, and conversely each $M(v)$ is the part containing $v$ by the same computation; hence $\mathcal{P}$ is exactly the set of maximal proper modules, which by [L4] is a modular partition with at least two parts and a prime quotient, and no other modular partition of $G$ with at least two parts has a prime quotient.