An equivalent condition for distributivity of a lattice

Question

Let $(L,\land,\lor)$ be a lattice. Show that $L$ is distributive if and only if for all $x,y,z \in L$ holds

$$(x \land y) \lor (x \land z) \lor (y \land z) = (x \lor y) \land (x \lor z) \land (y \lor z)$$

I have already done the "$\Longrightarrow$"-part.

However, I am having trouble with the other direction. I understand that

$$x \lor (y \land z) = x \lor ( (x \land y) \lor (x \land z) \lor (y \land z) )$$

But I do not see how to continue from here to get $(x \lor y) \land (x \lor z)$. Using the premise I see that

$$x \lor ( (x \land y) \lor (x \land z) \lor (y \land z) ) = x \lor ( (x \lor y) \land (x \lor z) \land (y \lor z) ),$$

but I do not see what to do now. Could you help me?

Answer 1

We want to prove that $$(x\wedge y)\vee (y\wedge z)\vee (x\wedge z) = (x\vee y)\wedge (y\vee z)\wedge (x\vee z) \tag{1}$$ implies distributivity. This is Birkhoff's proof, from his book Lattice Theory (Theorem II.6.8).

If we have $x\geq z$, then the left hand side of $(1)$ becomes $$(x\wedge y)\vee (y\wedge z) \vee z = (x\wedge y)\vee\Bigl((y\wedge z)\vee z\Bigr) = (x\wedge y)\vee z;$$ and the right hand side of $(1)$ becomes $$(x\vee y)\wedge (y\vee z) \wedge x = \Bigl((x\vee y)\wedge x\Bigr)\wedge (y\vee z) = x\wedge (y\vee z).$$ Thus, $$x\wedge (y\vee z) = (x\wedge y)\vee z\ \quad\text{ if }x\geq z. \tag{2}$$

Now take arbitrary $x,y,z$. Since $x\geq (x\wedge y)\vee (x\wedge z)$, we have: $$\begin{align*} x\wedge\Bigl( (y\wedge z)\vee (x\wedge y) \vee (x\wedge z)\Bigr)&= x\wedge \Bigl( (y\wedge z)\vee \bigl( (x\wedge y)\vee (x\wedge z)\bigr)\Bigr)\\ &= (x\wedge y\wedge z) \vee \Bigl( (x\wedge y)\vee (x\wedge z)\Bigr)\tag{$\star$}\\ &= (x\wedge y)\vee (x\wedge z)\tag{3} \end{align*}$$ where $(\star)$ holds by $(2)$, and $(3)$ holds because $x\wedge y\wedge z\leq (x\wedge y)$ and $x\wedge y\wedge z\leq x\wedge z$; and $$\begin{align*} x\wedge \Bigl( (x\vee y)\wedge (x\vee z)\wedge (y\vee z)\Bigr) &= \Bigl( x\wedge (x\vee y)\Bigr) \wedge (x\vee z)\wedge (y\vee z)\\ &= x\wedge (x\vee z)\wedge (y\vee z)\\ &= \Bigl( x\wedge (z\vee x)\Bigr) \wedge (y\vee z)\\ &= \Bigl( (x\wedge z)\vee x\Bigr) \wedge (y\vee z)\tag{$\star$}\\ &= x\wedge (y\vee z),\tag{4} \end{align*}$$ where again $(\star)$ holds by (2), since $x\geq x$.

Therefore, if we let $u$ be the left hand side of $(1)$ and we let $v$ be the right hand side of $(1)$, we have $$\begin{align*} x\wedge (y\vee z) &= x\wedge v &\text{(by }(4))\\ &= x\wedge u &\text{(by }(1))\\ &= (x\wedge y)\vee (x\wedge z) &\text{(by }(3)), \end{align*}$$ as desired.

Answer 2

As an alternate answer to Arturo's, I can give you a quick way to see this works, though I suspect it's not what you want. Recall a lattice $L$ is distributive if and only if it does not have a copy of $N_5$ or $M_3$ living inside it. That is, there should be no points $v,w,x,y,z$ in either of the following configurations inside $L$:

But you can check that in both cases, the $3$ middle elements of this configuration do not satisfy your axiom. So any lattice satisfying your axiom must avoid these configurations, and is distributive.

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I hope this helps ^_^