An equivalent condition for distributivity of a lattice 
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?
 A: 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.
A: 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.

I hope this helps ^_^
