TITLE: Is a distributive lattice planar iff it admits no B3 sublattice?
QUESTION [11 upvotes]: A finite lattice is planar if it  admits a Hasse diagram which is a planar graph (we want the Hasse diagram to be represented in the plane so that the $y$-coordinate of each element respects the order).    
Remark:  See the paper Planar lattices and planar graphs (1976) by C.R Platt :
It is shown that a finite lattice is planar if and only if the (undirected) graph obtained from its (Hasse) diagram by adding an edge between its least and greatest elements is a planar graph.
The $B_3$ lattice (below) is not planar.
 
Question: Is a finite distributive lattice planar iff it admits no sublattice isomorphic to $B_3$?

REPLY [9 votes]: It's already been answered positively, but here's another argument that shows something a little stronger: every finite distributive lattice either contains a B3 (and is not planar) or it can be drawn as a planar grid graph.
By Birkhoff's representation theorem, every finite distributive lattice is isomorphic to the lattice of lower sets of a finite partially ordered set. If this partial order has width three or more, (that is, if it has an antichain of three elements), then they generate a B3 and the lattice is not planar. And if the partial order has width two, then by Dilworth's theorem it can be decomposed into two chains. This decomposition can be used to embed the lattice as a grid graph, by setting the two coordinates of each lower set to be the numbers of elements it has in each of the two chains (and then rotating the whole thing by 45 degrees to make it an upward drawing).