Completely distributive lattice

Remarks

Example: the power set of a set $X$. A point $x$ lies in $\bigcap_j \bigcup_k x_{j,k}$ if and only if for each $j$ there is $k = f(j)$ with $x \in x_{j, f(j)}$. This is membership in the right side, where the axiom of choice picks $f$. Non-example: the lattice of open subsets of $\R$ is a frame but not completely distributive. Take $J = \N_{\ge 1}$, $K_j = \set{1, 2}$, $x_{j,1} = (-1/j, 1/j)$ and $x_{j,2} = \R \setminus \set{0}$. The left side is $\R$. Each meet on the right is the interior of an intersection and never contains $0$, so the right side is $\R \setminus \set{0}$. Complete distributivity is self-dual: the condition is equivalent to the same identity with $\bigwedge$ and $\bigvee$ exchanged. Every completely distributive lattice is a distributive lattice.