Generalized Boolean algebra

Remarks

Example: the finite subsets of $\N$, ordered by inclusion, form a generalized Boolean algebra. For finite sets $a$ and $b$, the set $x = a \setminus b$ is finite, $(a \cap b) \cup x = a$, and $(a \cap b) \cap x = \emptyset$. This lattice has no greatest element, so it is not a Boolean algebra. Non-example: the chain $0 < 1 < 2$ is a distributive lattice with least element $0$, but for $a = 2$ and $b = 1$ there is no $x$ with $1 \vee x = 2$ and $1 \wedge x = 0$. Indeed $1 \vee x = 2$ forces $x = 2$, and then $1 \wedge x = 1 \ne 0$. Equivalently, a distributive lattice with $0$ is a generalized Boolean algebra if and only if each interval $[0, a]$ is a Boolean algebra. A generalized Boolean algebra with a greatest element is the same as a Boolean algebra.