Distributive law of intersection over union

We prove set equality by showing both inclusions, using an arbitrary element $x_0$ (element-chasing).

($\subseteq$) Suppose $x_0 \in A \cap (B \cup C)$. By the definition of intersection this means $x_0 \in A$ and $x_0 \in B \cup C$. By the definition of union, $x_0 \in B \cup C$ means $x_0 \in B$ or $x_0 \in C$. Consider the two cases separately. Case 1: $x_0 \in B$. Combined with $x_0 \in A$, this gives $x_0 \in A \cap B$. Case 2: $x_0 \in C$. Combined with $x_0 \in A$, this gives $x_0 \in A \cap C$. In either case $x_0$ lies in $x_0 \in A \cap B$ or in $x_0 \in A \cap C$, so $x_0 \in (A \cap B) \cup (A \cap C)$.

($\supseteq$) Suppose $x_0 \in (A \cap B) \cup (A \cap C)$. By the definition of union this means $x_0 \in A \cap B$ or $x_0 \in A \cap C$. If $x_0 \in A \cap B$, then $x_0 \in A$ and $x_0 \in B$, hence $x_0 \in B \cup C$ (since $x_0$ is in $B$, it is in $B$ or $C$), so $x_0 \in A \cap (B \cup C)$. If instead $x_0 \in A \cap C$, then $x_0 \in A$ and $x_0 \in C$, hence $x_0 \in B \cup C$, so again $x_0 \in A \cap (B \cup C)$.

Since every element of the left-hand side lies in the right-hand side and vice versa, the two sets contain exactly the same elements, so $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. This finishes the proof.