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.
 A: It is actually not strictly necessary to suppose that $(A \cup B)' \neq \emptyset$ because of a rule of inference called "ex falso sequitur quodlibet", meaning "from falsity, anything follows" (I believe this name may come from the Scholastics), which asserts that anything may be proved from a contradiction (the idea is that if we have lowered our standards of truth to the point that we've let something false be true, then we kind of have to let anything be true).
So, if $(A \cup B)' = \emptyset$ then supposing $x \in (A \cup B)'$ gives a contradiction since $(A \cup B)'$ has no elements, thus $x \in A' \cap B'$ by ex falso. 
A: You are correct that this proof is assuming that $(A\cap B)'$ is nonempty. Luckily, we can quite easily check that it holds in that case, as it means that there are no elements not in both $A$ and $B$, and so $A'\cap B'=\emptyset$. Likewise for the other direction, if $A'\cap B'=\emptyset$ then there are no elements that are in neither $A$ nor $B$, so $(A\cup B)'=\emptyset$. The second part of the theorem is likewise easy to show.
A: 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.
