Let $X$ be a set and $A,B$ are subsets of $X$ .

1. $(A^{\complement})^\complement=A$ .
   Proof. $a\in (A^{\complement})^\complement$ iff $a\notin A^{\complement}$ iff $a\in A$ .
2. $\emptyset^\complement = X$ .
   Proof. $a\in \emptyset^\complement$ iff $a\notin \emptyset$ iff $a\in X$ .
3. $X^\complement = \emptyset$ .
   Proof. $a\in X^\complement$ iff $a\notin X$ iff $a\in \emptyset$ .
4. $A\cup A^\complement = X$ .
   Proof. $a\in A\cup A^\complement$ iff $a\in A$ or $a\in A^\complement$ iff $a\in A$ or $a\notin A$ iff $a\in X$ .
5. $A\cap A^\complement =\emptyset$ .
   Proof. $a\in A\cap A^\complement$ iff $a\in A$ and $a\in A^\complement$ iff $a\in A$ and $a\notin A$ iff $a\in \emptyset$ .
6. $A\subseteq B$ iff $B^\complement\subseteq A^\complement$ .
   Proof. Suppose $A\subseteq B$ . If $a\in B^\complement$ , then $a\notin B$ , so $a\notin A$ , or $a\in A^\complement$ . This shows that $B^\complement\subseteq A^\complement$ . On the other hand, if $B^\complement\subseteq A^\complement$ , then by applying what's just been proved, $A=(A^\complement)^\complement \subseteq (B^\complement)^\complement =B$ .
7. $A\cap B=\emptyset$ iff $A\subseteq B^\complement$ .
   Proof. Suppose $A\cap B=\emptyset$ . If $a\in A$ , then $a\in B^\complement$ , or $a\notin B$ , which implies that $A\cap B=\emptyset$ . Suppose next that $A\subseteq B^\complement$ . If there is $a\in A\cap B$ , then $a\in B$ and $a\in A$ . But the second containment implies that $a\in B^\complement$ , which contradicts the first containment.
8. $A\setminus B = A\cap B^\complement$ , where the complement is taken in $X$ .
   Proof. $a\in A\setminus B$ iff $a\in A$ and $a\notin B$ iff $a\in A$ and $a\in B^\complement$ iff $a\in A\cap B^\complement$ .
9. (de Morgan's laws) $(A \cup B)^\complement = A^\complement \cap B^\complement$ and $(A \cap B)^\complement = A^\complement \cup B^\complement$ .
   Proof. See here.