Orthomodular lattice

Remarks

An equivalent form is: if $a \le b$ and $a^\perp \wedge b = 0$, then $a = b$.

Example: the lattice of closed subspaces of a Hilbert space $H$. If $M \subseteq N$ are closed subspaces, every $x \in N$ is $x = Px + (x - Px)$, where $P$ is the orthogonal projection onto $M$; then $x - Px \in M^\perp \cap N$. So $N$ is the closed span of $M$ and $M^\perp \cap N$. When $H$ is infinite-dimensional, this lattice is orthomodular but not modular.

Non-example: the hexagon ortholattice $O_6$ with elements $0, a, b, b^\perp, a^\perp, 1$ and order generated by $0 < a < b < 1$ and $0 < b^\perp < a^\perp < 1$. The map exchanging $a$ with $a^\perp$, $b$ with $b^\perp$ and $0$ with $1$ is an orthocomplementation. Here $a \le b$, but $a^\perp \wedge b = 0$, so $a \vee (a^\perp \wedge b) = a \ne b$.

Every Boolean algebra is orthomodular, and every modular ortholattice is orthomodular.