Q.Let $A$, $B$ and $C$ be sets. Then show that $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

Concept: Distributive law of intersection over union.

We prove set equality by showing each side is a subset of the other.

($\subseteq$) Let $x \in A \cap (B \cup C)$. Then $x \in A$ and $x \in B \cup C$. The second condition means $x \in B$ or $x \in C$. If $x \in B$, then $x \in A \cap B$; if $x \in C$, then $x \in A \cap C$. Either way, $x \in (A \cap B) \cup (A \cap C)$. …

The distributive law for set operations mirrors the distributive property in algebra, where multiplication distributes over addition. Here, intersection plays the role of "multiplication" and union plays the role of "addition." To prove two sets are equal, we show each is a subset of the other: every element of the left-hand side belongs to the right-hand side, and vice versa.

The strategy is to track what it means for an arbitrary element $x$ to belong to each side, translating set operations into logical statements about membership.

Proof that $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$:

1. Let $x \in A \cap (B \cup C)$ be arbitrary.

2. By definition of intersection, $x \in A$ and $x \in B \cup C$.

3. Since $x \in B \cup C$, by definition of union we have $x \in B$ or $x \in C$ (or both).

4. Case 1: If $x \in B$, then since we already know $x \in A$, we have $x \in A \cap B$. Therefore $x \in (A \cap B) \cup (A \cap C)$.

5. Case 2: If $x \in C$, then since $x \in A$, we have $x \in A \cap C$. Therefore $x \in (A \cap B) \cup (A \cap C)$.

6. In both cases, $x \in (A \cap B) \cup (A \cap C)$.

This establishes the first inclusion.

Proof that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$:

1. Let $x \in (A \cap B) \cup (A \cap C)$ be arbitrary.

2. By definition of union, $x \in A \cap B$ or $x \in A \cap C$ (or both).

3. Case 1: If $x \in A \cap B$, then $x \in A$ and $x \in B$. Since $x \in B$, we have $x \in B \cup C$. Combined with $x \in A$, this gives $x \in A \cap (B \cup C)$. …