# Is a finite lattice where each element has exactly one complement distributive? Why or why not

discrete mathematicslattice-orders

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.

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 $$x = \bigvee S_x$$.