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

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?

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The details of the proof are escaping me at the moment, but your suggestion is the general idea. Write out everything you know about $x$, $y$, and $z$ regarding complementarity and distributivity. Bang the formulas up against each other until you get $y = z$. – Austin Mohr Sep 8 '11 at 17:39
Thank you very much. – Eternal_Light Sep 8 '11 at 17:40

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.

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So the way to go is trying to prove it by contradiction? In your answer you are in a distributive lattice so this makes it a boolean algebra. Is it right if I use + and * instead of ∨ and ∧ ? – Eternal_Light Sep 8 '11 at 19:45
Yes, contradiction or contrapositive, depending on how you see it. If a lattice is complemented and distributive, there cannot be 2 complements. Putting it another way, if you do find an $x$ with 2 complements, it must be the case that it is not distributive. Secondly, if it is a boolean algebra, then you can certainly use $+$ and $\cdot$. But it might be weird in this particular case, because ultimately you are proving that the lattice is not a boolean algebra :). (Nothing logically wrong though with using $+$ and $\cdot$.) – Srivatsan Sep 8 '11 at 19:51
Thanks a lot :) – Eternal_Light Sep 8 '11 at 19:57
@Srivatsan what is an "OP's strategy"? – avDec25 Dec 6 '15 at 19:25