Lattices are congruence-distributive

Solution 1:

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 $x\,\big((\theta_1\cap\theta_2)\lor(\theta_1\cap\theta_3)\big)\,y$.