Properties of distributive lattices and congruences. Let $L$ be a lattice and let $a,b,c,d \in L$. Show that: 


*

*$\theta(a,b) \subseteq \theta(c,d)$ iff $\langle a,b\rangle \in \theta(c,d)$

*$\theta(a,b)=\theta(a \wedge b, a \vee b)$
Where $\theta$ is a congruence. Definition: given a lattice $L$, a congruence $\theta$ on $L$ is an equivalence relation, compatible with the lattice operations, i.e. if $x_1\theta x_2$ and $y_1 \theta y_2$, then $x_1 \wedge x_2 \theta y_1 \wedge y_2$ and $x_1 \vee x_2 \theta y_1 \vee y_2$.
I specially have a problem with the first part, how to prove that it is a subset. The part of $\Leftarrow$ seems "evident", I have 2 elements, $a,b$, I assume they belong to the congruence, so they have to be inside, but I don't get the more formal way to explain this. For the $\Rightarrow$ part, I assume that $\theta (a,b) \subseteq \theta (c,d)$, so I think I have to argue that all the elements of a subset have a certain characteristic, so in particular $a,b$ have it. Again, if this is right, I am having trouble translating it into a proper proof.
About the second one, I can only think of proving it in a way similar to let $a= a\wedge b$ so $b = a\vee b$ (I have a lemma that $a= a\wedge b$ iff $b = a\vee b$). But it seems like there is something missing. I would very much appreciate your help to sort this out, and your input on how I can write this properly. 
 A: First of all, lattice $L$ in your question is not supposed to be distributive (but the title uses this word).
If $\theta(a, b)$ means the congruence generated by $\langle a, b \rangle$ then:


*

*If $\theta(a, b) \subseteq \theta(c, d)$ then (since $\theta(a, b)$ is generated by $\langle a, b \rangle$) we have $$\langle a, b \rangle \in \theta(a, b) \subseteq \theta(c, d) \Rightarrow \langle a, b \rangle \in \theta(c, d).$$
Conversely, if $\langle a, b \rangle \in \theta(c, d)$ then since $\theta(a, b)$ is the smallest congruence containing pair $\langle a, b \rangle$ and $\theta(c, d)$ is some congruence containing this pair we have $\theta(a, b) \subseteq \theta(c, d)$.

*We want to show two inclusions, which are equivalent (by the first part) to showing that generating pair belongs to congruence. To show that $\langle a, b \rangle \in \theta(a \wedge b, a \vee b)$ observe that:
$$a = a \vee (a \wedge b) \theta(a \wedge b, a \vee b) a \vee (a \vee b) = a \vee b,$$
$$b = b \vee (a \wedge b) \theta(a \wedge b, a \vee b) b \vee (a \vee b) = a \vee b,$$
which means that $a \theta(a \wedge b, a \vee b) b$ and hence $\langle a, b \rangle \in \theta(a \wedge b, a \vee b)$.
To show that $\langle a \wedge b, a \vee b \rangle \in \theta(a, b)$ observe that:
$$ a \wedge b \theta(a, b) a \wedge a = a = a \vee a \theta(a, b) a \vee b,$$
which means that $a \wedge b\theta(a, b)a \vee b$, and hence $\langle a \wedge b, a \vee b \rangle \in \theta(a, b)$. By the first part we have both inclusions and hence $\theta(a, b) = \theta(a \wedge b, a \vee b).$
