[proofplan]
We prove both identities by the Extensionality of Sets: for each proposed equality, it is enough to prove both inclusions. Each inclusion is proved by taking an arbitrary object $x$ in the left-hand side and unpacking the definitions of union and intersection. The only delicate point is the reverse inclusion for the second identity, where the alternatives $x \in A$ and $x \notin A$ separate the argument.
[/proofplan]
[step:Prove that $A \cap (B \cup C)$ is contained in $(A \cap B) \cup (A \cap C)$]
Let $x$ be an arbitrary object such that $x \in A \cap (B \cup C)$. By the definition of intersection, $x \in A$ and $x \in B \cup C$. By the definition of union, either $x \in B$ or $x \in C$.
If $x \in B$, then $x \in A \cap B$, and therefore $x \in (A \cap B) \cup (A \cap C)$. If $x \in C$, then $x \in A \cap C$, and therefore $x \in (A \cap B) \cup (A \cap C)$. Hence, in every case,
\begin{align*}
x \in (A \cap B) \cup (A \cap C).
\end{align*}
Since $x$ was arbitrary,
\begin{align*}
A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C).
\end{align*}
[/step]
[step:Prove that $(A \cap B) \cup (A \cap C)$ is contained in $A \cap (B \cup C)$]
Let $x$ be an arbitrary object such that $x \in (A \cap B) \cup (A \cap C)$. By the definition of union, either $x \in A \cap B$ or $x \in A \cap C$.
If $x \in A \cap B$, then $x \in A$ and $x \in B$. Since $x \in B$ implies $x \in B \cup C$, we get $x \in A \cap (B \cup C)$. If $x \in A \cap C$, then $x \in A$ and $x \in C$. Since $x \in C$ implies $x \in B \cup C$, we again get $x \in A \cap (B \cup C)$. Hence
\begin{align*}
x \in A \cap (B \cup C).
\end{align*}
Since $x$ was arbitrary,
\begin{align*}
(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C).
\end{align*}
[/step]
[step:Conclude that intersection distributes over union]
The previous two steps prove
\begin{align*}
A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)
\end{align*}
and
\begin{align*}
(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C).
\end{align*}
By the Extensionality of Sets, the two [sets](/page/Set) are equal:
\begin{align*}
A \cap (B \cup C) = (A \cap B) \cup (A \cap C).
\end{align*}
[/step]
[step:Prove that $A \cup (B \cap C)$ is contained in $(A \cup B) \cap (A \cup C)$]
Let $x$ be an arbitrary object such that $x \in A \cup (B \cap C)$. By the definition of union, either $x \in A$ or $x \in B \cap C$.
If $x \in A$, then $x \in A \cup B$ and $x \in A \cup C$, so $x \in (A \cup B) \cap (A \cup C)$. If $x \in B \cap C$, then $x \in B$ and $x \in C$. Therefore $x \in A \cup B$ because $x \in B$, and $x \in A \cup C$ because $x \in C$, so $x \in (A \cup B) \cap (A \cup C)$. Hence
\begin{align*}
x \in (A \cup B) \cap (A \cup C).
\end{align*}
Since $x$ was arbitrary,
\begin{align*}
A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C).
\end{align*}
[/step]
[step:Prove that $(A \cup B) \cap (A \cup C)$ is contained in $A \cup (B \cap C)$]
Let $x$ be an arbitrary object such that $x \in (A \cup B) \cap (A \cup C)$. By the definition of intersection,
\begin{align*}
x \in A \cup B
\end{align*}
and
\begin{align*}
x \in A \cup C.
\end{align*}
By the definition of union, the first membership says that $x \in A$ or $x \in B$, and the second says that $x \in A$ or $x \in C$.
If $x \in A$, then $x \in A \cup (B \cap C)$. If $x \notin A$, then the membership $x \in A \cup B$ forces $x \in B$, and the membership $x \in A \cup C$ forces $x \in C$. Thus $x \in B \cap C$, and hence $x \in A \cup (B \cap C)$. In all cases,
\begin{align*}
x \in A \cup (B \cap C).
\end{align*}
Since $x$ was arbitrary,
\begin{align*}
(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C).
\end{align*}
[guided]
We start with an arbitrary object $x$ satisfying
\begin{align*}
x \in (A \cup B) \cap (A \cup C).
\end{align*}
To prove the desired inclusion, we must show that this same object $x$ belongs to $A \cup (B \cap C)$. By the definition of intersection, membership in $(A \cup B) \cap (A \cup C)$ gives both conditions
\begin{align*}
x \in A \cup B
\end{align*}
and
\begin{align*}
x \in A \cup C.
\end{align*}
The goal $x \in A \cup (B \cap C)$ can happen in either of two ways: either $x \in A$, or $x \in B \cap C$. We therefore separate the argument according to whether $x$ lies in $A$.
If $x \in A$, then the definition of union immediately gives
\begin{align*}
x \in A \cup (B \cap C).
\end{align*}
Now suppose $x \notin A$. Since $x \in A \cup B$, the definition of union says that $x \in A$ or $x \in B$. The alternative $x \in A$ is excluded by the assumption $x \notin A$, so $x \in B$. Similarly, since $x \in A \cup C$, the definition of union says that $x \in A$ or $x \in C$, and again $x \notin A$ forces $x \in C$. Hence $x \in B$ and $x \in C$, so by the definition of intersection,
\begin{align*}
x \in B \cap C.
\end{align*}
Therefore, by the definition of union,
\begin{align*}
x \in A \cup (B \cap C).
\end{align*}
Thus every object $x$ in $(A \cup B) \cap (A \cup C)$ lies in $A \cup (B \cap C)$. Since $x$ was arbitrary,
\begin{align*}
(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C).
\end{align*}
[/guided]
[/step]
[step:Conclude that union distributes over intersection]
The previous two steps prove
\begin{align*}
A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)
\end{align*}
and
\begin{align*}
(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C).
\end{align*}
By the Extensionality of Sets, the two sets are equal:
\begin{align*}
A \cup (B \cap C) = (A \cup B) \cap (A \cup C).
\end{align*}
Together with the first distributive identity proved above, this proves both stated distributive laws for the sets $A$, $B$, and $C$.
[/step]
