Simple Proof Methods

4. Theorem. For any sets $A$, $B$, and $C$, the following holds:$$A \cup (B \cap C) = (A \cup B) \cap (A \cup C).$$ Proof. Let $A$, $B$, and $C$ be any sets. To show that $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$, we need to show that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ and that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$:
   • [Showing that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$] Let $x$ be any element in the set $A \cup (B \cap C)$. This means that $x$ is in the union of $A$ and $B \cap C$. This means that $x \in A$ or that $x \in B$ and $x \in C$. In other words, it means that $x \in A$ or $x \in B$, and that $x \in A$ or $x \in C$. This means that $x$ is in the intersection of $A \cup B$ and $A \cup C$, so $x \in (A \cup B) \cap (A \cup C)$. This implies $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
   • [Showing that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$] Let $x$ be any element in the set $(A \cup B) \cap (A \cup C)$. This means that $x$ is in the intersection of $A \cup B$ and $A \cup C$. This means that $x \in A$ or $x \in B$, and that $x \in A$ or $x \in C$. In other words, it means that $x \in A$ or that $x \in B$ and $x \in C$. This means that $x$ is in the union of $A$ and $B \cap C$, so $x \in A \cup (B \cap C)$. This implies $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.
   Therefore, we conclude that $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ by the definition of set equality.◼