Q.Let $A = \{1, 2, 3, 4\}$, $B = \{2, 3, 5\}$ and $C = \{3, 4, 6\}$. Verify the distributive property $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

Work out the left-hand side and right-hand side independently, respecting brackets, then compare.

Computing each side from its own innermost operation and comparing confirms the identity.

Left-hand side: $A \cap (B \cup C)$.

• $B \cup C = \{2,3,5\} \cup \{3,4,6\} = \{2,3,4,5,6\}$.
• $A \cap (B\cup C) = \{1,2,3,4\} \cap \{2,3,4,5,6\} = \{2,3,4\}$ (elements common to both).

Right-hand side: $(A \cap B) \cup (A \cap C)$.

• $A \cap B = \{1,2,3,4\} \cap \{2,3,5\} = \{2,3\}$.
• $A \cap C = \{1,2,3,4\} \cap \{3,4,6\} = \{3,4\}$.
• $(A\cap B) \cup (A\cap C) = \{2,3\} \cup \{3,4\} = \{2,3,4\}$.

Compare. Both sides give $\{2,3,4\}$ — the distributive property is verified for this choice of sets.

Mixing this identity up with the OTHER distributive law (union distributing over intersection) and computing $A \cup (B \cap C)$ instead of $A \cap (B \cup C)$ by mistake — the two distributive properties look similar but test different operation pairs.