How to prove that any x in a complemented distributive lattice cannot have two complements?

Question

How can I prove the following statement?

In a complemented lattice, if there exist two complements for any x then the lattice is not distributive.

I thought of showing that, in a complemented and distributive lattice, if y and z are both complements of x then y = z so they are the same thing. Would this make any sense or am I really far away from where I should be?

Answer 1

We will follow the OP's strategy and prove the following contrapositive form of the statement:

If a lattice is complemented and distributive, then every element of the lattice has a unique complement.

Convince yourself that this is equivalent to the claim in the question.

A complemented and distributive lattice is a boolean algebra, so we will use $+$ and $\cdot$ in place of $\vee$ and $\wedge$ respectively. Now, of course, every element does have a complement (by definition); the real task is to show uniqueness.

Let $x$ be an arbitrary element, and let $y$ and $z$ be its complements. We want to show that $y = z$. We start from $$ y = y \cdot 1, $$ and replace $1$ by $x+z$. Then applying distributivity and the fact that $yx = 0$, we get $$ y = y(x+z) = yx + yz = 0 + yz = yz. \tag{1} $$ Repeating this argument after switching $y$ and $z$, we get $$ z = zy. \tag{2} $$ Comparing $(1)$ and $(2)$, we are done.