Lemma 3.7

Let $L$ be a complete lattice, and let $a\mapsto a^\perp$ satisfy $a\wedge a^\perp=\bot$, $a\vee a^\perp=\top$, both De Morgan identities $(a\wedge b)^\perp=a^\perp\vee b^\perp$ and $(a\vee b)^\perp=a^\perp\wedge b^\perp$, and $(a^\perp)^\perp=a$. Assume also that $b\leq a$ and $d\leq a^\perp$ imply $a\wedge(b\vee d)=(a\wedge b)\vee(a\wedge d)$ for all $a,b,d\in L$. Let $S\subseteq L$ contain $\bot,\top$, be closed under $a\mapsto a^\perp$, binary meets, and binary joins, satisfy both distributive identities $a\wedge(b\vee d)=(a\wedge b)\vee(a\wedge d)$ and $a\vee(b\wedge d)=(a\vee b)\wedge(a\vee d)$ for all $a,b,d\in S$, and be maximal under inclusion among subsets with these closure and distributivity properties. Let $B\subseteq S$ be nonempty, exclude $\bot$, have a common lower bound in $B$ for every pair of its members, and satisfy $a=(a\wedge b)\vee(a\wedge b^\perp)$ for every $a,b\in B$; assume $B$ is maximal among all subsets of $L$ with these properties. Then

Thus an element belongs to $S$ exactly when it belongs to every subset containing $B,\bot,\top$ and closed under those operations; the subsets $T$ in this intersection are not required to satisfy either distributive identity.