I asked this question [here](https://math.stackexchange.com/q/1548838/75923) 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.
> 
> 
> 



Is this a sufficient condition for distributivity of a lattice?
Yes that's true. See page 67 of [this book](https://books.google.tk/books?id=Ll0JXd11SW0C&pg=PA67#v=onepage&q&f=false).


