Is a finite lattice where each element has exactly one complement distributive? Why or why not? While reading the paper LATTICES WITH UNIQUE COMPLEMENTS by R. P. DILWORTH, I get to know that any number of weak additional restrictions are sufficient for a lattice with unique complement to be a boolean algebra, including properties like modular, etc.
But I'm wondering whether the restriction like "finite" be sufficient enough. However, it's so difficult for me to prove or give a counterexample.
 A: It is immediate that every finite lattice is complete and atomic, i.e., every element is above some atom.
So the following result yields that a finite uniquely complemented lattice is Boolean.

Theorem.[Theorem 16 in Chapter X of Birkhoff's Lattice Theory, 1948, page 170]
Let $\mathbf L$ be any complete atomic lattice with unique complements.
Then $\mathbf L$ is isomorphic to the Boolean algebra of the subsets of its atoms.

Notice that it's not even asked in the hypothesis that $\mathbf L$ is atomistic (which is stronger than being atomic).
We will see by the end of the proof that that will follow from the hypothesis.
Proof:
Let us denote by $\mathcal At(\mathbf L)$ the set of atoms of $\mathbf L$.
To $S \subseteq \mathcal At(\mathbf L)$, let
$$\bigvee S = \bigvee \{ x : x \in S \}$$
and
$$\bigwedge S' = \bigwedge \{ x' : x \in S \}.$$
It follows that
$$\bigvee (S \cup T) = \bigvee S \vee \bigvee T$$
and
$$\bigwedge(S \cup T)' = \bigwedge S' \wedge \bigwedge T'.$$
For each atom $x$ of $\mathbf L$, we have that $x' \prec 1$ ($x'$ is covered by $1$, that is, $x'<1$ and if $x'\leq y \leq 1$ then $y=x'$ or $y=1$).
Indeed, if $x' \leq y \leq 1$, then either $x \leq y$ or $x \nleq y$;
in the former case, $1 = x \vee x' \leq y$, whence $y=1$; in the later, $x \wedge y = 0$, and $x \vee y \geq x \vee x' = 1$, whence $y = x'$.
It follows that for $x \neq y$ in $\mathcal At(\mathbf L)$ we have $x \leq y'$, for otherwise $x \wedge y' = 0$ and $x \vee y' = 1$, yielding $x=y$ by the unique complementation.
Hence, if $S,T \subseteq \mathcal At(\mathbf L)$ are such that $S \cap T  = \varnothing$, then
$$\bigvee S \leq \bigwedge T'.$$
Thus, denoting by $S^c$ the complement of $S$ in $\mathcal At(\mathbf L)$,
\begin{align}
\bigvee S \wedge \bigvee S^c
&\leq \bigwedge(S^c)' \wedge \bigwedge(S^{cc})'\\
&= \bigwedge(S^c)' \wedge \bigwedge S'\\
&= \bigwedge(S^c \cup S)'\\
&= \bigwedge(\mathcal At(\mathbf L))' \\
&= 0
\tag{$\dagger$}
\end{align}
Thus, $\mathcal At(\mathbf L) \cap {\downarrow}\bigvee S = S$ and so $\bigvee S \neq \bigvee T$, whenever $S \neq T$, and therefore the poset whose elements are the family $\{ \bigvee S : S \subseteq \mathcal At(\mathbf L) \}$, with the order inherited from $\mathbf L$ is isomorphic to the power-set $\wp(\mathcal At(\mathbf L))$, which is clearly a Boolean algebra.
It remains to show that $x = \bigvee S$ for some $S \subseteq \mathcal At(\mathbf L)$ and each $x \in L$.
Let
$$S_x = \{ a \in \mathcal At(\mathbf L) : a \leq x \}.$$
We will show that $x = \bigvee S_x$ (i.e., $\mathbf L$ is atomistic).
It is clear that the only atoms below $x \wedge \bigvee S_x^c$ are those which are in $S_x \cap S_x^c = \varnothing$, and so $x \wedge \bigvee S_x^c = 0$.
On the other hand
\begin{align}
x \vee \bigvee S_x^c
&\geq \bigvee S_x \vee \bigvee S_x^c\\
&= \bigvee (S_x \cup S_x^c)\\
&= \bigvee \mathcal At(\mathbf L)\\
&= 1.
\end{align}
Thus $x$ is the (unique) complement of $\bigvee S_x^c$.
From ($\dagger$) and
$$\bigvee S_x \vee \bigvee S_x^c
= \bigvee (S_x \cup S_x^c)
= \bigvee \mathcal At(\mathbf L) = 1,$$
it follows that $x = \bigvee S_x$.
