1. **State the problem:** We need to verify and explain the set identities: (a) $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ (b) $(A \cap B)' = A' \cup B'$ (De Morgan's Law) (c) $(A \cup B)' = A' \cap B'$ (De Morgan's Law) (d) $A \setminus B = A \cap B'$ 2. **Recall important rules:** - Union ($\cup$) means elements in either set. - Intersection ($\cap$) means elements common to both sets. - Complement ($'$) means elements not in the set. - Set difference ($\setminus$) means elements in the first set but not in the second. - De Morgan's Laws: $$ (A \cap B)' = A' \cup B' $$ $$ (A \cup B)' = A' \cap B' $$ 3. **Verify (a) distributive law:** - Left side: $A \cup (B \cap C)$ means elements in $A$ or in both $B$ and $C$. - Right side: $(A \cup B) \cap (A \cup C)$ means elements that are in $A$ or $B$, and also in $A$ or $C$. 4. **Proof of (a):** - If $x \in A$, then $x$ is in both sides. - If $x \in B \cap C$, then $x \in B$ and $x \in C$, so $x \in A \cup B$ and $x \in A \cup C$. - Hence, both sides contain the same elements. 5. **Verify (b) and (c) De Morgan's Laws:** - $(A \cap B)'$ contains elements not in both $A$ and $B$. - $A' \cup B'$ contains elements not in $A$ or not in $B$. - These are equivalent by De Morgan's Law. - Similarly for $(A \cup B)'$ and $A' \cap B'$. 6. **Verify (d) set difference:** - $A \setminus B$ means elements in $A$ but not in $B$. - $A \cap B'$ means elements in $A$ and not in $B$. - These are the same by definition. 7. **Example with given sets:** - For $S = \{1,2,...,10\}$, $A = \{x | x - 1 \geq 3\} = \{4,5,6,7,8,9,10\}$ - $B = \{x | 8 < 4x < 30\} = \{3,4,5,6,7\}$ - Check $A \cup (B \cap C)$ and $(A \cup B) \cap (A \cup C)$ with $C$ similarly defined. 8. **Summary:** These identities hold true and can be visualized with Venn diagrams showing the overlapping and non-overlapping regions. **Final answer:** All given set identities are correct and follow from the definitions of union, intersection, complement, and set difference, as well as De Morgan's Laws.