Simple Proof Methods

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