3244.sagews

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

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

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

By remark 1, we may assume that $(A\cup B)\cap(A\cup C)$ is not empty. Let $x \in (A\cup B)\cap(A\cup C)$. That means that we have ($x \in A$ or $B$), and ($x \in A$ or $C$). In the case where $x \in A$, by theorem 4a, $x \in A\cup (B\cap C)$. In the case where $x \not\in A$, which means that $x \in B$ and $x\in C$, we can write $x\in (B\cap C)$ and $x\in A \cup (B\cap C)$ by theorem 4a. Therefore, $(A\cup B)\cap(A\cup C) \subseteq A\cup(B\cap C)$.