Modular lattice

Remarks

The inequality $a \vee (x \wedge b) \le (a \vee x) \wedge b$ holds in every lattice when $a \le b$, so only the reverse inequality is a condition. Equivalently, $L$ satisfies the identity $((a \wedge b) \vee x) \wedge b = (a \wedge b) \vee (x \wedge b)$ for all $a, b, x \in L$, so the modular lattices form a variety.

Example: the lattice of subspaces of a vector space $V$ is modular. If $A \subseteq B$ and $X$ are subspaces, and $b = a + x \in (A + X) \cap B$ with $a \in A$, $x \in X$, then $x = b - a \in B$, so $b \in A + (X \cap B)$.

Non-example: the pentagon lattice N5 with $0 < a < c < 1$ and $b$ incomparable to $a$ and $c$. Here $a \le c$, but $a \vee (b \wedge c) = a \vee 0 = a$ while $(a \vee b) \wedge c = 1 \wedge c = c$.

Every distributive lattice is modular. The diamond lattice M3 is modular but not distributive.