non-distributive lattice shown on the right is compatible. We show that there is no distributive lattice that is compatible with $G$.

Assume, towards contradiction, that $G$ has a compatible distributive lattice. By Proposition \[prop:HasseSubgraph\], ${\mathbf{0}}$ and ${\mathbf{1}}$ must be the vertices labelled $0$ and $1$ in the figure. Further ${\mathbf{0}}$ must have unique cover $a$ and ${\mathbf{1}}$ must cover $j$. So ${\bm{\langle} j ]} \cap {[ a \bm{\rangle}}$ is a distributive sublattice with zero $a$ and unit $j$.

As the set $\{d,e\}$ is the intersection of maximal cliques, it is a conservative set, so induces a sublattice. The only $2$ element lattices is the chain, so we may assume, without loss of generality, that $d \leq e$.

The set $\{d,e,h\}$ is also an intersection of maximal cliques, so induces a sublattice of three elements, so must also be a chain. If $h \leq e$ then by $a$ and $h$ are adjacent, so $h \geq e$. Similarly $b \leq d$.

The set $\{b,d,e,f,h\}$ is a maximal clique, so induces a lattice. As $b$ is not adjacent to $a$ or $j$, it follows from that it can neither be above or below $d$ or $e$, so it is incomparable with them. Thus 