In probability you combined events with "or," "and" and "not." Events are sets of outcomes, so those were set operations all along. Here they get their general names and notation, building on sets and subsets.
Key idea
For sets inside a universe $U$: the union $A \cup B$ is everything in $A$ or $B$ (or both); the intersection $A \cap B$ is everything in both; the difference $A \setminus B$ is what is in $A$ but not $B$; the complement $A^{c} = U \setminus A$ is everything in $U$ not in $A$. De Morgan's laws hold for sets: $(A \cup B)^{c} = A^{c} \cap B^{c}$ and $(A \cap B)^{c} = A^{c} \cup B^{c}$.
Why it works
Each operation is a logical connective applied to membership. $x \in A \cup B$ means $x \in A \vee x \in B$; $x \in A \cap B$ means $x \in A \wedge x \in B$; $x \in A^{c}$ means $\neg(x \in A)$. So every logical equivalence becomes a set identity. De Morgan's laws for sets are the same laws you proved with truth tables, with "or" and "and" replaced by $\cup$ and $\cap$. (Probability books often write $A'$ for the complement; it means the same as $A^{c}$.)
Worked examples
Worked example: Computing with a universe
Let $U = \{1, 2, \ldots, 10\}$, $A = \{2, 4, 6, 8, 10\}$ (the even numbers) and $B = \{3, 6, 9\}$ (the multiples of $3$). Find $A \cup B$, $A \cap B$, $A \setminus B$, $A^{c}$ and $(A \cup B)^{c}$.
- $A \cup B = \{2, 3, 4, 6, 8, 9, 10\}$ (each element once).
- $A \cap B = \{6\}$, and $A \setminus B = \{2, 4, 8, 10\}$.
- $A^{c} = \{1, 3, 5, 7, 9\}$, the odd numbers.
- $(A \cup B)^{c} = \{1, 5, 7\}$: in neither set. Check De Morgan: $$A^{c} \cap B^{c} = \{1, 3, 5, 7, 9\} \cap \{1, 2, 4, 5, 7, 8, 10\} = \{1, 5, 7\}.$$
Answer: $\{2, 3, 4, 6, 8, 9, 10\}$, $\{6\}$, $\{2, 4, 8, 10\}$, $\{1, 3, 5, 7, 9\}$, $\{1, 5, 7\}$
Worked example: Sizes with De Morgan
In a universe of $40$ people, $|A| = 18$, $|B| = 15$ and $|A \cap B| = 6$. How many people are in $A^{c} \cup B^{c}$?
- De Morgan: $A^{c} \cup B^{c} = (A \cap B)^{c}$.
- So the count is everyone except the $6$ people in both: $40 - 6 = 34$.
Answer: $34$
Common mistake
Complementing each piece without switching the operation. $(A \cap B)^{c}$ is not $A^{c} \cap B^{c}$. To be outside "both $A$ and $B$," you only need to miss one of them, so $(A \cap B)^{c} = A^{c} \cup B^{c}$. Check with one element: in the example above, $3$ is in $(A \cap B)^{c}$ (it is not in both), but $3$ is not in $A^{c} \cap B^{c}$ (it is in $B$).