← Back to Mathematical Logic, Set Theory and Combinatorics
Course: MAT1MN104 — Mathematical Logic, Set Theory and Combinatorics Scope: Module II §2.1 (The 24 Laws of Sets) — see Module II — Set Theory for theory and Set Laws — Worked Examples for 10 fully-solved model answers.
⚠️ This is instructor-authored practice, not a real past paper. Only one genuine paper exists for this course so far — see the Previous Year Questions page. These questions are modeled on that paper’s Section A/B/C pattern, mark values, and question style (Q4 and Q12 there are set-law questions), so they’re a realistic rehearsal — but they haven’t actually been examined.
Pattern (as in Oct 2024): Section A — 3 marks each, ceiling 24. Section B — 6 marks each, ceiling 36. Section C — 10 marks, answer one of two.
Section A (3 marks each)
A1. State the Idempotent and Domination laws for sets.
A2. Let $A = \{p, q, r, s\}$, $B = \{r, s, t, u\}$, and $U = \{p, q, r, s, t, u, v\}$. Find $A’ \cap B’$ and $(A \cup B)’$. What do you observe?
A3. Write down both forms of the Distributive law for sets.
A4. Simplify $A \cap (A \cup B)$ using the laws of sets, naming the law used at each step.
A5. Write the difference identity and the symmetric difference identity for two sets $A$ and $B$.
A6. Let $|A| = 8$, $|B| = 6$, $|A \cap B| = 3$. Find $|A \cup B|$.
Section B (6 marks each)
B1. Using the laws of sets, simplify $(A \cup B) \cap (A \cup B’)$. Justify every step with the name of the law used.
B2. Prove, using the laws of sets, that $A - (B \cap C) = (A - B) \cup (A - C)$.
B3. Simplify $(A \cap B) \cup (A \cap B’) \cup (A’ \cap B)$ using the laws of sets. Briefly explain your final answer with reference to a Venn diagram.
B4. Using the laws of sets, show that $(A \cup B)’ \cup B = A’ \cup B$.
B5. Let $A = \{2,4,6,8,10\}$, $B = \{4,8,12,16\}$, $C = \{6,8,10,12\}$, $U = \{2,4,6,\ldots,20\}$. Verify that $A - (B \cup C) = (A - B) \cap (A - C)$ for these sets by listing both sides.
B6. Prove, using the laws of sets, that $A \cap (A’ \cup B) = A \cap B$.
Section C (10 marks — answer one of the following)
C1. (a) Prove, using the laws of sets, that $A \oplus B = (A \cap B’) \cup (A’ \cap B)$, starting from the definition $A \oplus B = (A \cup B) - (A \cap B)$. [6] (b) Take $A = \{1,2,3,4,5\}$, $B = \{4,5,6,7\}$. Verify the identity in part (a) by listing both sides explicitly. [4]
OR
C2. (a) State and prove that $A \subseteq B$ if and only if $A \cap B’ = \emptyset$. [6] (b) Let $U = \{1,2,\ldots,12\}$, $A = \{1,2,3,4,5,6\}$, $B = \{4,5,6,7,8,9\}$, $C = \{1,2,4,6,8,10\}$. Find $(A \cup B \cup C)’$ and $A’ \cap B’ \cap C’$, and verify they are equal (generalised De Morgan’s law for three sets). [4]
Answer Key / Hints
Questions whose method matches a fully-solved model appear in Set Laws — Worked Examples — check your law-chaining against that example, then confirm your final answer here.
| Question | Final answer / result | See also |
|---|---|---|
| A1 | $A \cup A = A$, $A \cap A = A$ (Idempotent); $A \cup U = U$, $A \cap \emptyset = \emptyset$ (Domination) | — |
| A2 | $A’ \cap B’ = \{v\}$, $(A \cup B)’ = \{v\}$ — equal, confirming De Morgan’s | Example 8 |
| A3 | $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$; $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ | — |
| A4 | $A$ (Absorption) | Example 2 |
| A5 | $A - B = A \cap B’$; $A \oplus B = (A \cup B) - (A \cap B)$ | Example 10 |
| A6 | $8+6-3 = $ 11 | — |
| B1 | $A$ | Example 3 |
| B2 | — (proof) | Example 5 |
| B3 | $A \cup B$ | Example 7 |
| B4 | $A’ \cup B$ | Example 8 |
| B5 | Both sides = $\{2\}$ | Example 6 method |
| B6 | $A \cap B$ | Example 4 |
| C1(a) | — (proof) | Example 10 |
| C1(b) | $A \oplus B = \{1,2,3,6,7\}$; RHS $= (A \cap B’) \cup (A’ \cap B) = \{1,2,3\} \cup \{6,7\} = \{1,2,3,6,7\}$ — equal | Example 10 |
| C2(a) | — (proof) | Example 9 |
| C2(b) | $A \cup B \cup C = \{1,\ldots,10\}$, so $(A \cup B \cup C)’ = \{11,12\}$. $A’ \cap B’ \cap C’ = \{7,8,9,10,11,12\} \cap \{1,2,3,10,11,12\} \cap \{3,5,7,9,11,12\} = \{11,12\}$ — equal | — |
Working for B5: $B \cup C = \{4,6,8,10,12,16\}$. $A - (B \cup C) = \{2\}$. $A - B = \{2,6,10\}$. $A - C = \{2,4\}$. $(A - B) \cap (A - C) = \{2\}$. Both sides equal $\{2\}$. $\checkmark$