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generalized Boolean algebra
A lattice $L$ is called a generalized Boolean algebra if

$L$ is distributive,

$L$ is relatively complemented, and

$L$ has $0$ as the bottom.
Clearly, a Boolean algebra is a generalized Boolean algebra. Conversely, a generalized Boolean algebra $L$ with a top $1$ is a Boolean algebra, since $L=[0,1]$ is a bounded distributive complemented lattice, so each element $a\in L$ has a unique complement $a^{{\prime}}$ by distributivity. So ${}^{{\prime}}$ is a unary operator on $L$ which makes $L$ into a de Morgan algebra. A complemented de Morgan algebra is, as a result, a Boolean algebra.
As an example of a generalized Boolean algebra that is not Boolean, let $A$ be an infinite set and let $B$ be the set of all finite subsets of $A$. Then $B$ is generalized Boolean: order $B$ by inclusion, then $B$ is a distributive as the operation is inherited from $P(A)$, the powerset of $A$. It is also relatively complemented: if $C\in[X,Y]$ where $C,X,Y\in B$, then $(YC)\cup X$ is the relative complement of $C$ in $[X,Y]$. Finally, $\varnothing$ is, as usual, the bottom element in $B$. $B$ is not a Boolean algebra, because the union of all the singletons (all in $B$) is $A$, which is infinite, thus not in $B$.
One property of a generalized Boolean algebra $L$ is the following: if $y$ and $z$ are complements of $x\in[a,b]$, then $y=z$; in other words, relative complements are uniquely determined. This is true because in any distributive lattice, complents are uniquely determined. As $L$ is distributive, so is each lattice interval $[a,b]$ in $L$.
In fact, because of the existence of $0$, we can actually construct the relative complement. Let $bx$ denote the unique complement of $x$ in $[0,b]$. Then $(bx)\vee a$ is the unique complement of $x\in[a,b]$: $x\wedge((bx)\vee a)=(x\wedge(bx))vee(x\wedge a)=0\vee a=a$ and $x\vee((bx)\vee a)=(x\vee(bx))\vee a=b\vee a=b$.
Conversely, if $L$ is a distributive lattice with $0$ such that any lattice interval $[0,a]$ is complemented, then $L$ is a generalized Boolean algebra. Again, $(bx)\vee a$ provides the necessary complement of $x$ in $[a,b]$.
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