# Lattices are congruence-distributive

$\newcommand{\r}[1]{\mathrel{#1}}$ First, a few definitions. Given a lattice $L$, a congruence on $L$ is an equivalence relation $\theta$, compatible with the lattice operations, i.e. if $x_1\r{\theta}x_2$ and $y_1\r{\theta}y_2$, then $x_1\wedge x_2\r{\theta}y_1\wedge y_2$ and $x_1\vee x_2\r{\theta}y_1\vee y_2$. The congruences on $L$ form a lattice, with meet being intersection and join being the equivalence envelope.

I'm trying to prove that this lattice of congruences is distributive. So let's take three congruences $\theta_1,\theta_2,\theta_3$ and attempt to prove that $$\theta_1\cap(\theta_2\vee\theta_3)\subseteq (\theta_1\cap\theta_2)\vee(\theta_1\cap\theta_3)$$

For a start, let's try to prove something easier which should lead me to the general idea. Take elements $x,y,z$ such that $x\r{\theta_1}y,x\r{\theta_2}z$ and $z\r{\theta_3}y$. We want to prove that $x(\r{(\theta_1\cap\theta_2)\vee(\theta_1\cap\theta_3)})y$.

Start with $(x\wedge z)\r{\theta_1}(y\wedge z)$, which gives $x\r{\theta_1}((y\wedge z)\vee x)$. Similarly, $(x\wedge y)\r{\theta_2}(z\wedge y)$ and $x\r{\theta_2}((y\wedge z)\vee x)$. This gives $$x\r{(\theta_1\cap\theta_2)}((y\wedge z)\vee x)$$

I expect a similar manipulation should now give $$((y\wedge z)\vee x)\r{\theta_1\cap\theta_3} y$$ but I can't see how to manage this. Is this even in the right direction? I suppose something other than $((y\wedge z)\vee x$ could be the middle link, but nothing obviously better comes to mind.

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I would use the fact that if $\theta_1$ and $\theta_2$ are congruences on $L$, and $x,y\in L$, then $x\,(\theta_1\lor\theta_2)\,y$ iff there is a chain $$x\land y=z_0\le z_1\le\dots\le z_n=x\lor y$$ such that whenever $0\le k<n$, $z_k\,\theta_2\,z_{k+1}$ or $z_k\,\theta_3\,z_{k+1}$.
Suppose that $x\,(\theta_1\cap(\theta_2\lor\theta_3))\,y$. Then $x\,(\theta_2\lor\theta_3)\,y$, so there is a chain $$x\land y=z_0\le z_1\le\dots\le z_n=x\lor y$$ such that whenever $0\le k<n$, $z_k\,\theta_2\,z_{k+1}$ or $z_k\,\theta_3\,z_{k+1}$. You also have $x\,\theta_1\,y$, so $(x\land y)\,\theta_1(x\lor y)$, and therefore $(x\land y)\,\theta_1\,z_k\,\theta_1(x\lor y)$ for $0\le k<n$. It follows that for $0\le k<n$, $z_k\,(\theta_1\cap\theta_2)\,z_{k+1}$ or $z_k\,(\theta_1\cap\theta_3)\,z_{k+1}$ and hence that $x\,\big((\theta_1\cap\theta_2)\lor(\theta_1\cap\theta_3)\big)\,y$.