Simplifying $(A \cup E) \cup E$. For example
$$(A^c\cap B^c)^c\cup E$$
First of all, De Morgan is definitely a must to simply this to
$$\big((A^c)^c\cup(E^c)^c\big)\cup E$$
Then double negation to remove the double complement to get
$$(A\cup E)\cup E$$
Unfortunately, how do I further simplify this statement?
Logically it seems like an absorption law to make the 3rd statement as
$$A\cup E$$
However, I'm not sure and neither is any of the law matching the 3rd statement to further simplify the statement.

It's very easy:
Since $E \subseteq E \cup A$, and since for any $x \in E \cup (E \cup A)$ we have $x \in E$ or $x \in E \cup A $ and by our observasion, $x$ is always is in $E \cup A$; So we do have $E \cup (E \cup A) = E \cup A$.
And we are done.
