Prove $A \cup A' = U$ and $A \cap A' = \emptyset$
$A \cup A' = U$
set union definition with negation on the second $A$.
$[x: x \in A \lor x \notin A]$
This means that x is in $A$ or x is not in $A$. I wrote that since this is an or statement we can choose one or the other, but apparently that isn't a good reason to demonstrate that $A \cup A' = U$ . So, how do I prove it without set union definition? I don't think substituting $A \cup A' = U$ as $U=U$ is a good idea. That's too easy, and if that's the case, then I know it's wrong.
$A \cap A' = \emptyset$
is really easy to prove.
Using set definition of $A \cap A'$
$[x: x \in A \land x \notin A ]$
which tells me that x is in A and x is not an A. This is a very absurd statement. That's why it's an empty set.