Alternative characterization of distributive lattice

Question

Let $(X,{\leq},{\wedge},{\vee})$ be a lattice. The lattice is called distributive if for all $x,y,z\in X$ both distributive laws hold: $$ x \wedge (y \vee z) = (x \wedge z) \vee (y \wedge z) \quad\text{and}\quad x \vee (y \wedge z) = (x \vee z) \wedge (y \wedge z). $$

According to this Wikipedia article, the lattice $X$ is distributive if and only if for all $x,y,z\in X$, $$(x \wedge y) \vee (y \wedge z) \vee (z\wedge x) = (x\vee y) \wedge (y \vee z) \wedge (z\vee x).$$

I can do one direction: If $X$ is distributive, then \begin{align} (x \wedge y) \vee (y \wedge z) \vee (z\wedge x) &= (y \wedge (x \vee z)) \vee (z\wedge x) \\ &= (y \vee (z\wedge x)) \wedge ((x\vee z) \vee (z\wedge x)) \\ &= ((y \vee z) \wedge (y \vee x)) \wedge (((x\vee z)\vee z) \wedge ((x \vee z)\vee x)) \\ &= (y \vee z) \wedge (y \vee x) \wedge (x\vee z) \wedge (x \vee z) \\ &= (x\vee y) \wedge (y \vee z) \wedge (z \vee x). \end{align}

But I'm lost on the reverse implication. Can anyone show me how to do it?

Answer 1

The straightforward way to prove it is to use the $M_3-N_5$ theorem:

A lattice is distributive iff it doesn't have a copy of either $M_3$ or $N_5$ as a sub-lattice.

The Hasse diagrams of $M_3$ and $N_5$ are in the image here for reference.

You already showed that distributive lattices satisfy the identity \begin{equation} (x\wedge y) \vee (y \wedge z) \vee (z \wedge x) = (x\vee y) \wedge (y \vee z) \wedge (z \vee x). \label{eq1}\tag{$\dagger$} \end{equation} So the result is proven if we show that every non-distributive lattice fails to satisfy \eqref{eq1}, and for that, using the $M_3-N_5$ theorem, it's enough to see that both $M_3$ and $N_5$ fail to satisfy \eqref{eq1}.

Indeed, and using the same labels in the image, in $M_3$ that would give $0=1$ and in $N_5$, we would have $x=z$.

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Perhaps you were really interested in an equational proof.
Something closer to that is as follows.
(This is very close to a proof in Grätzer, General Lattice Theory, 2nd ed., Theorem III.2.4, page 185.)

Let us start to prove that a lattice satisfying \eqref{eq1} is modular, that is, it satisfies the Modular law: \begin{equation} x \leq y \implies x \vee (y \wedge z) = y \wedge (x \vee z) \tag{M}\label{eq2} \end{equation}

Indeed, if $x \leq y$, then $x = x \wedge y$ and $(y \wedge z) \vee (z \wedge x) = y \wedge z$, yielding \begin{align} x \vee (y \wedge z) &= (x \wedge y) \vee (y \wedge z) \vee (z \wedge x)\\ &= (x \vee y) \wedge (y \vee z) \wedge (z \vee x) \tag{$\because$\eqref{eq1}}\\ &= y \wedge (x \vee z). \end{align}

Now, the distributivity: \begin{align} x \vee (y \wedge z) &= x \vee (x \wedge y) \vee (y \wedge z) \vee (z \wedge x) \tag{$\because$ Absorption}\\ &= x \vee ((x \vee y) \wedge (y \vee z) \wedge (z \vee x)) \tag{$\because$ \eqref{eq1}}\\ &= (x \vee y) \wedge (z \vee x) \wedge (x \vee y \vee z) \tag{$\because$ \eqref{eq2}}\\ &= (x \vee y) \wedge (x \vee z), \end{align} where the application of Modular law above, follows from $x \leq (x \vee y) \wedge (z \vee x)$ (thanks to @azimut for noticing I had used Modular law twice needlessly).