Show that if A, B, and C are sets, then A ∩ B ∩ C = A ∪ B ∪ C by showing each side is a subset of the other side. using a membership table.

To show that $\overline{A\cap B\cap C}=\overline{A}\cup \overline{B}\cup \overline{C}$

Let $x \in \overline{A\cup B\cup C}$

$x \in U \text{ and } x \notin A \cap B \cap C\\ x \in U \text{ and } x \notin A \text{ or } x \notin B \text{ or } x \notin C\\ x \in U \text{ and } x \notin A \text{ or } x \in U \text{ and } x \notin B \text{ or } x \in U \text{ and } x \notin C\\ x\in \overline{A} \text{ or } x\in \overline{B} \text{ or } x\in \overline{C} \\ x\in \overline{A} \cup \overline{B} \cup \overline{C}\\ \implies \overline{A\cup B\cup C} \subseteq \overline{A} \cup \overline{B} \cup \overline{C}$

Conversely,

Let $x \in \overline{A} \cup \overline{B} \cup \overline{C}$

$x \in \overline{A} \text{ or } x \in \overline{B} \text{ or } x \in \overline{C} \\ x \in U \text{ and } x \notin A \text{ or } x \in U \text{ and } x \notin B \text{ or } x \in U \text{ and } x \notin C\\ x \in U \text{ and } x \notin A \text{ or } x \notin B \text{ or } x \notin C\\ x \in U \text{ and } x \notin A \cap B \cap C\\ x \in \overline{A\cup B\cup C}\\ \implies \overline{A} \cup \overline{B} \cup \overline{C}\subseteq \overline{A\cup B\cup C}$

Hence,

$\overline{A\cap B\cap C}=\overline{A}\cup \overline{B}\cup \overline{C}$