Is this a sufficient condition for distributivity of a lattice?

Question

I asked this question here on Math.SE but uptil now it was not answered. So I decided to give it a try here. Thank you in advance.

If a lattice $L$ is distributive then it can be shown that for $a,b,c\in L$: $$[a\wedge b=a\wedge c\text{ and }a\vee b=a\vee c]\implies b=c$$

So for fixed $a,u,v\in L$ there is at most one $b$ such that $u=a\wedge b$ and $v=a\vee b$.

Is the converse of this true? More precisely, is the following statement true?

If for each triple $a,u,v$ in a lattice there is at most one $b\in L$ such that $u=a\wedge b$ and $v=a\vee b$ then the lattice is distributive.

Minimus Heximus · Accepted Answer · 2015-12-01 11:52:25Z

3

Yes that's true. See page 67 of this book.

answered Dec 1, 2015 at 11:52

Minimus Heximus

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$\begingroup$ Your answer was downvoted. Probably because it only provides a link. But it is enough for me. Thank you. $\endgroup$
– drhab
Commented Dec 1, 2015 at 13:52

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