De Morgan's Laws proof 
My only problem with this proof is that they seem to be assuming that $(A \cup B)'$ is nonempty (similar idea for $A' \cap B')$. Is that because this holds trivially if $(A \cup B)'$ is empty? In other words, $A \cup B = U$ where $U$ is the universal set.

When it says "Let $x\in (A\cup B)'...\;$", this is a way of saying "For all $x$, if $x\in  (A\cup B)'...$" Similarly for "Suppose $x\in A'\cap B'...$"  in the second part. Without altering the logic the proof can be re-worded as
(1)....$\forall x\;(\; [x\in (A\cup B)']\implies$ $ [x\not \in A\cup B]\implies$ $ [x\not \in A\land x\not \in B]\implies$ $ [x\in A'\land x\in B']\implies$ $ [x\in A'\cap B']\;).$
(2)....$\forall x\;(\;[x\in A'\cap B']\implies$  $[x\in A'\land x\in B']\implies$ $ [x\not \in A\land x\not \in B]\implies$ $ [x\not \in A\cup B]\implies$ $ [x\in (A\cup B)']\;).$
(3)....From ( 1)and (2) we have $$\forall x\;(\;[x\in (A\cup B)]\implies [x\in A'\cap B']\;),$$ $$ \forall x\;(\;[x\in A'\cap B']\implies [x\in (A\cup B)']\;).$$ Therefore $\forall x\;(\;[x\in (A\cup B)']\iff [x\in A'\cap B']\;).$ Therefore $(A\cup B)'=A'\cap B'.$
The proof is about twice as long as necessary.Notice that (2) is (1) backwards. That is, every "implies" in (1) and in (2) can be replaced by "iff".
At no time is it assumed that anything actually does belong to any of the sets that are mentioned.
