### A note on distributive and modular lattices

I’ve been working a little with an old friend, modal logic, this weekend. This digressed into a study of lattices and Boolean algebras, so I thought I would write up some of my observations for archival. Also, I just like to say “lattice”.

If you know anything about lattices, you will probably find this stuff trivial. If you don’t know anything about lattices, a free and much better introduction can be found in A Course in Universal Algebra.

Definition: A distributive lattice is a lattice which satisfies the distributive laws:

$D1: x \land (y \lor z) \simeq (x \land y) \lor (x \land z)$
$D2: x \lor (y \land z) \simeq (x \lor y) \land (x \lor z)$

Actually, it can be shown that a lattice satisfies D1 iff it satisfies D2.

Definition: A lattice L is said to be modular if the modular law holds:

$M : x \leq y \Rightarrow x \lor (y \land z) \simeq y \land (x \lor z)$

Lemma: The modular law for lattices is equivalent to the identity

$(x \land y) \lor (y \land z) \simeq y \land ((x \land y) \lor z)$

Proof
Assume that the modular law holds and consider the expression $y \land ((x \land y) \lor z)$. Since $x \land y \leq y$ we can apply the modular law to infer that $y \land ((x \land y) \lor z) \simeq (x \land y) \lor (y \land z)$.

Now, assume that the identity holds and that $x \leq y$ for some $x, y, z \in L$. Then $x = x \land y$ so

$x \lor (y \land z) \simeq$
$(x \land y) \lor (y \land z) \stackrel{\dagger}{\simeq}$
$y \land ((x \land y) \lor z) \simeq$
$y \land (x \lor z)$,

where the identity was used in $\dagger$. Thus, the modular law holds.$\Box$

Theorem: Every distributive lattice is modular.

Proof
Assume that $x \leq y$ for some $x, y, z \in L$ where $L$ is a distributive lattice. Then $y = x \lor y$ so
$x \lor (y \land z) \simeq (x \lor y) \land (x \lor z) \simeq y \land (x \lor z)$

where we have made use of D2. The desired result now follows from the lemma.