Let $\mathcal{R},\mathcal{S}$ be lattices. Let $\theta $ be a congruence on $ \mathcal{R}$ and $\phi$ a congruence $\mathcal{S}.$ Let $T$ be the relation on $\mathcal{R}\times \mathcal{S}$ given by $$ ((a_1,b_1),(a_2,b_2)) \in T \Leftrightarrow ((a_1,a_2)\in \theta \land (b_1,b_2) \in \phi ).$$
It's not too difficult to see that $T$ is a congruence. I want to show the reciprocal of this. I want to show that any congruence on $\mathcal{R}\times \mathcal{S}$ can be defined on a similar fashion to $T.$ How should I go about this? Any help would be greatly appreciated.
HINT: Let $\tau$ be a congruence on $\mathcal{R}\times\mathcal{S}$. For each $r\in\mathcal{R}$ define a relation $\tau_r$ on $\mathcal{S}$ by $s_0\mathrel{\tau_r}s_1$ iff $\langle r,s_0\rangle\mathrel{\tau}\langle r,s_1\rangle$. Suppose that $r,r'\in\mathcal{R}$, $s_0,s_1\in\mathcal{S}$, and $s_0\mathrel{\tau_r}s_1$. Then $\langle r,s_0\rangle\mathrel{\tau}\langle r,s_1\rangle$, so
$$\langle r\land r',s_0\rangle\mathrel{\tau}\langle r\land r',s_0\land s_1\rangle\mathrel{\tau}\langle r\land r',s_1\rangle$$
and
$$\langle r\lor r',s_0\rangle\mathrel{\tau}\langle r\lor r',s_0\lor s_1\rangle\mathrel{\tau}\langle r\lor r',s_1\rangle\;.$$
But then
$$\langle r',s_0\rangle=\langle(r\lor r')\land r',s_0\rangle\mathrel{\tau}\langle(r\lor r')\land r',s_1\rangle=\langle r',s_1\rangle\;,$$
so $s_0\mathrel{\tau_{r'}}s_1$, and it follows that $\tau_r=\tau_{r'}$. Define a relation $\varphi$ on $\mathcal{S}$ by $s_0\mathrel{\varphi}s_1$ iff $s_0\mathrel{\tau_r}s_1$ for some (any) $r\in\mathcal{R}$; $\varphi$ is a congruence on $\mathcal{S}$.
In similar fashion define a relation $\theta$ on $\mathcal{R}$ by $r_0\mathrel{\theta}r_1$ iff $\langle r_0,s\rangle\mathrel{\tau}\langle r_1,s\rangle$ for some (any) $s\in\mathcal{S}$, and verify that for any $\langle r_0,s_0\rangle,\langle r_1,s_1\rangle\in\mathcal{R}\times\mathcal{S}$, $\langle r_0,s_0\rangle\mathrel{\tau}\langle r_1,s_1\rangle$ iff $r_0\mathrel{\theta}r_1$ and $s_0\mathrel{\varphi}s_1$.