3243.sagews

Suppose each of $A$, $B$, and $C$ is a set. Then prove that $A\cap(B\cup  C)=(A\cap B)\cup(A\cap C)$.
\begin {enumerate}
\item $A\cap(B\cup C)\subseteq (A\cap B)\cup(A\cap C)$

By remark 1, we may assume $A\cap(B\cup  C)$ is not empty. Let $x\in A\cap(B\cup C)$. That means that we have $x \in A$ and $x \in B$, or, $x \in A$ and $x \in C$. The first case, $x \in A$ and $x \in B$ implies that $x \in (A \cap B)$. The second case, $x \in A$ and $x \in C$ implies that $x \in (A \cap C)$. Since $x \in (A \cap B)$ and $x \in (A \cap C)$, then $x \in (A\cap B)\cup(A\cap C)$. Therefore, $A\cap(B\cup C)\subseteq (A\cap B)\cup(A\cap C)$.

\item $(A\cap B)\cup(A\cap C) \subseteq A\cap(B\cup C)$

By remark 1, we may assume that $(A\cap B)\cup(A\cap C)$ is not empty. Let $x \in (A\cap B)\cup(A\cap C)$. That means that we have $x \in A$ and $B$, or, $x \in A$ and $C$. In both cases, $x$ always $\in A$, and whether $x \in B$ or $x \in C$, by theorem 4a, $x \in (B\cup C)$. Then $x \in A\cap(B\cup C)$. Therefore $(A\cap B)\cup(A\cap C) \subseteq A\cap(B\cup C)$.