Thank you for your help.
I am now trying to understand that the set of complete lattice congruences of a complete lattice is again complete lattice.
Let $\mathcal{L}$ be a complete lattice. An equivalence relation $\theta$ of $\mathcal{L}$ is called a complete lattice congruence if $(x_i, y_i)\in \theta$ for all $i\in I$ ($I$ is not necessary finite) then $(\vee\{x_i\}, \vee\{y_i\})$ and $(\wedge\{x_i\}, \wedge\{y_i\})$ belong to $\theta$.
We denote by $\mathsf{Con^c}(\mathcal{L})$ the set of all complete lattice congruences of $\mathcal{L}$.
Then, the set is $\mathsf{Con^c}(\mathcal{L})$ is a complete lattice.
How to show this?
Clearly $L^2$ is a complete congruence of $L$.
So it's enough to show that $\theta = \bigcap_{I\in I}\theta_i$ is a complete congruence, whenever each of the $\theta_i$ is.
Let $(a_j,b_j) \in \theta$, for all $j \in J$.
Let $a^+ = \bigvee_j a_j$ and $b^+ = \bigvee_j b_j$, and $a_- = \bigwedge_j a_j$ and $b_- = \bigwedge_j b_j$.
Then $(a^+, b^+), (a_-, b_-) \in \theta_i$, for all $i \in I$, because $\theta_i$ is complete.
It follows that $(a^+,b^+), (a_-, b_-) \in \theta$.
Hence $\theta$ is a complete congruence relation of $L$, and the lattice of complete congruences of $L$ is complete.