the component lattices are.

On the other hand, if a disconnected graph has a compatible lattice, then as each component is a subalgebra, and subalgebras are closed under polymorphisms, each component is closed under the lattice operations. Thus each component induces a sublattice, so is compatible with the component by Fact \[fact:sublat\]. If a lattice is distributive, then so is any sublattice.

The following, which does not hold for semilattices, is a huge simplifaction.

\[prop:HasseSubgraph\] For a connected reflexive graph $G$ with a compatible lattice $L$, the Hasse diagram of $L$ is a subgraph of $G$.

It is enough to show for any cover $v {\prec}u$, that $uv$ is an edge of $G$.

Observe first that the upset ${[ v \bm{\rangle}}$ is a connected subgraph of $G$. Indeed as $G$ is connected, for $u_0$ and $u_p$ in ${[ v \bm{\rangle}}$, there is a path $u_0 \sim u_1 \sim \dots \sim u_p$ between them in $G$. So $(v \vee u_0) \sim (v \vee u_1) \sim \dots \sim (v \vee u_p)$ is a walk between them in ${[ v \bm{\rangle}}$.

The same proof in connected ${[ v \bm{\rangle}}$ then shows that the downset ${\bm{\langle} u ]}$ in ${[ v \bm{\rangle}}$ is connected. 