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This is part of Exercise III.1.16(c) of Bourbaki's set theory.

Let $E$ be a lattice. The condition

$(x\vee y)\wedge(z\vee(x\wedge y))=(x\wedge y)\vee(y\wedge z)\vee(z\wedge x)$

for all $x,y,z\in E$ implies that $E$ is distributive.

I already know that

$(x\vee y)\wedge(y\vee z)\wedge(z\vee x)=(x\wedge y)\vee(y\wedge z)\vee(z\wedge x)$

implies distributivity, so it suffices to prove

$(x\vee y)\wedge(z\vee(x\wedge y))=(x\vee y)\wedge(y\vee z)\wedge(z\vee x)$.

I can't find a proof of this last statement. Either it is true in any lattice, then the proof should be so simple I must be blind not to see it, or I have to use the condition above again, but I can't imagine in what way.

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To see the last equality does not hold for arbitrary lattices, take the 5-element $01$-lattice with $x,y,z$ pairwise distinct, and with all three pairwise joins among them equal to $1$, and all three pairwise meets equal to $0$ (the diamond lattice $M_3$). Then the left hand side evaluates to $z$, the right hand side evaluates to $1$. – Arturo Magidin Dec 18 '10 at 22:14
up vote 2 down vote accepted

$(x\vee y)\wedge(z\vee(x\wedge y))=(x\wedge y)\vee(y\wedge z)\vee(z\wedge x)$


$x\vee((x\vee y)\wedge(z\vee(x\wedge y)))=x\vee((x\wedge y)\vee(y\wedge z)\vee(z\wedge x)).$

The right side equals $x\vee(y\wedge z)$.

For the left side, we use the condition above again, substituting $x\vee y$ for $y$ and $z\vee(x\wedge y)$ for $z$, and get

$x\vee((x\vee y)\wedge(z\vee(x\wedge y)))=(x\vee y)\wedge((z\vee(x\wedge y))\vee x)=(x\wedge y)\vee(x\wedge z).$

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