The congruence class of a distributive lattice I need to prove the next thing:
For every non-empty ideal $I$ of a lattice $L$ consider the relation $\theta(I)$ defined by:
$ \theta(I) = \{⟨a, b⟩ : ( \exists c \in I) a \vee c = b \vee c\}$
Prove that if $L$ is a distributive lattice, then for every non-empty ideal $I$ of $L$ the relation $\theta (I)$ is a congruence on $L$ and $I$ is one of its equivalence classes.
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 assume $L$ is a distributive lattice, but I don't know how to continue. I want to use the fact that for any ideal, if $a,b \in I$ then $a\vee b \in I$, I have been trying things, but I'm still lost, my idea is to use the fact that if 2 things are not the ideal, the join is there too, but I don't know where to go from there. To prove that something is a congruence, so I need to prove that ny 2 elements are related by it. I will really appreciate any suggestions you have, I have been going in circles for a long time now. 
 A: To prove that $\theta(I)$ is a congruence you need to consequently check the definition you gave.
Hints: 


*

*This relation is obviously reflexive and symmetric. Assume that $a \theta(I) b \theta(I) c$ with elements $d$ and $e$ respectively:
$$a \vee d = b \vee d, \quad b \vee e = c \vee e, $$
for some elements $d, e \in I$. Prove that $a \theta(I) c$ with element $d \vee e \in I$.

*To prove that this relation is compatible with lattice operations assume that $a_1 \theta(I) a_2$ and $b_1 \theta(I) b_2$ with elements $a$ and $b$ respectively. Prove that $(a_1 \vee b_1) \theta(I) (a_2 \vee b_2)$ with element $a \vee b \in I$. To prove that $(a_1 \wedge b_1) \theta(I) (a_2 \wedge b_2)$ consider the following identity:
$$(a_1\vee a)\wedge(b_1\vee b) = (a_2\vee a)\wedge(b_2\vee b) \Longleftrightarrow \{\text{$L$ is distributive}\} \Longleftrightarrow$$
$$(a_1\wedge b_1) \vee (a_1\wedge b) \vee (a\wedge b_1) \vee (a\wedge b) = (a_2\wedge b_2) \vee (a_2\wedge b) \vee (a\wedge b_2) \vee (a\wedge b).$$ Make sure that you understand why $c = (a_1\wedge b) \vee (a\wedge b_1) \vee (a\wedge b) \in I$ and $d = (a_2\wedge b) \vee (a\wedge b_2) \vee (a\wedge b) \in I.$ It is left to prove that $c = d$. To prove this use the following identity: $$c = c \vee (a \wedge b) = d \vee (a \wedge b) = d,$$
where in the middle we use distributivity and that $a_1 \theta(I) a_2$ and $b_1 \theta(I) b_2$ with elements $a$ and $b$ respectively.

*To prove that $I$ is one of its equivalence classes observe that if $a, b \in I$ then $a \theta(I) b$ with $a \vee b$. Conversely, if $a \in I$ and $a \theta(I) b$ with $c$ then $b = b \wedge(b \vee c) = b \wedge (a \vee c) \in I$.
