I'm trying to prove the following: Show that a lattice is distributive if and only if it does not contain a sublattice isomorphic to either of the two lattices below.
I was able to prove that neither of these two lattices are distributive. Then showed the forward direction, which is as follows (proved by the contrapositive):
Let $D$ be a lattice. Suppose there is some $L\subseteq D$ such that $L\cong A$ or $L\cong B$. Then by our previous argument (above), $D$ would contain at least three elements that do not satisfy the distributive property. Hence, $D$ is not a distributive lattice.
I can't figure out how to get the converse! I started with (contrapositive approach): Suppose $D$ is a lattice, but it is not distributive. Then there exist $x,y,z\in D$ such that either $x\wedge (y\vee z)>(x\wedge y)\vee (x\wedge z)$ or $x\wedge (y\vee z)<(x\wedge y)\vee (x\wedge z)$.
I can't figure out how to show that $D$ will have a sublattice isomorphic to one of the lattices above. Any help is appreciated. Thanks!