Boolean algebra

At a Glance

• Frequency: 15 sub-parts across 12 of 13 years (2014, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2023, 2024, 2025)
• Priority tier: T1
• Marks (count): 10 (3), 15 (7), 5 (3), 7.5 (2)
• Average solve time: ~8 min
• Difficulty mix: easy 13, medium 2
• Section: B | Dominant type: computation

Why This Chapter Matters

Boolean algebra appears in 12 of the last 13 years — the most consistent T1 atom in Paper 2 — and the questions are almost always rated easy to medium. The examiner rotates through four tasks: simplify an expression (and draw its gate network), find the canonical DNF (sum of minterms), find the canonical CNF (product of maxterms), or read a Boolean function off a truth table. Each task has a mechanical algorithm. Unlike most Paper 2 topics, no creativity is needed — only a reliable procedure and careful bookkeeping. That makes Boolean algebra one of the most bankable 10–15 marks in the entire paper.

Minimum Theory

Boolean algebra. A Boolean algebra is a set $\{0,1\}$ with operations $+$ (OR), $\cdot$ (AND), and $\overline{\phantom{x}}$ (NOT), satisfying commutativity, associativity, distributivity, identity ($A+0=A$, $A\cdot1=A$), complement ($A+\bar A=1$, $A\cdot\bar A=0$), and idempotence ($A+A=A$, $A\cdot A=A$). The key derived laws that drive simplification are absorption ($A+AB=A$, $A(A+B)=A$), annihilation ($A+1=1$, $A\cdot0=0$), and De Morgan ($\overline{A+B}=\bar A\bar B$, $\overline{AB}=\bar A+\bar B$). Every simplification step should be named (UPSC awards method marks for this).

Canonical forms. A Boolean function of $n$ variables has $2^n$ rows in its truth table. The disjunctive normal form (DNF, sum of minterms) expresses $F$ as the OR of all minterms at rows where $F=1$; a minterm $m_i$ at row $i$ is the AND of all $n$ literals (variable unprimed if its bit is $1$, primed if $0$). The conjunctive normal form (CNF, product of maxterms) expresses $F$ as the AND of all maxterms at rows where $F=0$; a maxterm $M_i$ at row $i$ is the OR of all $n$ literals ($0\!\to\!$ unprimed, $1\!\to\!$ primed — opposite convention from minterms). The mnemonic: minterms live where $F=1$; maxterms live where $F=0$; priming conventions are dual.

Karnaugh maps. A K-map arranges the truth-table rows in a grid whose adjacent cells differ in exactly one variable (Gray-code order for columns). Grouping $2^k$ adjacent 1-cells (DNF) or 0-cells (CNF) eliminates $k$ variables from the corresponding term. Cells wrap around the left/right and top/bottom edges. The minimal SOP or POS is the cover using the fewest, largest groups.

Question Archetypes

Five recurring patterns cover every Boolean algebra question in the corpus.

boolean-simplification (6 question(s); 2016, 2017, 2018, 2019, 2022, 2025)

Recognition Cues

• Question gives a Boolean expression and says “simplify”, “minimise”, “reduce to simplest form”.
• Often appended with “draw a block/logic/gate diagram of the simplified expression.”
• Some questions also ask for the minterm normal form from the truth table.
• A circuit-first variant gives the expression as a hierarchical circuit and asks for the truth table then the simplification (2022).

Solution Template

1. Write out the expression and scan for immediate opportunities: absorption ($A(A+X)=A$ or $A+AX=A$), complementation ($A\bar A=0$, $A+\bar A=1$), and $x\bar y+y=x+y$.
2. Expand any products of sums via the distributive law if no shortcut is visible; then collect and apply laws. Name each rule as you use it.
3. Verify the simplified form against the original using an 8-row (or 4-row) truth table. Write both the original and simplified columns and confirm they match.
4. Draw the gate diagram for the simplified expression: inputs on the left, one logic gate per operator, output on the right. Label every wire and gate.
5. If a minterm normal form is also required: identify the rows where $F=1$ in the truth table and write their full minterms (every variable, primed or unprimed).

Worked Example(s)

2016 Paper 2, 2016-P2-Q8c (15 marks)

Simplify $A\cdot(A+B+C)\cdot(\bar A+B+C)\cdot(A+\bar B+C)\cdot(A+B+\bar C)$ and draw a block diagram.

1 — Scan. The leading factor $A$ appears as a bare literal. Three of the four bracketed factors contain a bare $A$; by absorption $A\cdot(A+X)=A$, each collapses: $A\cdot(A+B+C)=A,\quad A\cdot(A+\bar B+C)=A,\quad A\cdot(A+B+\bar C)=A.$ Only $(\bar A+B+C)$ survives: $F=A\cdot(\bar A+B+C).$

2 — Complement elimination. $A(\bar A+B+C)=A\bar A+A(B+C)=0+A(B+C).$ $\boxed{\;F=A(B+C)=AB+AC.\;}$

3 — Truth table. $F=1$ exactly when $A=1$ and at least one of $B,C$ is $1$: rows $(1,0,1),(1,1,0),(1,1,1)$.

4 — Gate diagram.

  B ──┐
      ├─[OR]──(B+C)──┐
  C ──┘               ├─[AND]── F = A·(B+C)
  A ─────────────────┘

2017 Paper 2, 2017-P2-Q5c (10 marks)

Simplify $z(y+z)(x+y+z)$. Name the rules used. Verify by truth tables.

Step 1. $z(y+z)$: expand to $zy+zz=zy+z$ (distributive, idempotent). Then $zy+z=z(y+1)=z\cdot1=z$ (absorption/annihilation). Rule: absorption ($z\cdot(z+y)=z$).

Step 2. $z\cdot(x+y+z)$: by the same absorption law ($z\cdot(z+\cdots)=z$), this collapses to $z$. Rule: absorption.

$\boxed{\;z(y+z)(x+y+z)=z.\;}$

Truth table verification: both expressions produce columns $(0,1,0,1,0,1,0,1)$ over all 8 rows — identical ✓.

2018 Paper 2, 2018-P2-Q8a (15 marks)

Simplify $(a+b)(\bar b+c)+b(\bar a+\bar c)$ and write its minterm normal form.

Expand. $(a+b)(\bar b+c)=a\bar b+ac+b\bar b+bc=a\bar b+ac+bc$.
$b(\bar a+\bar c)=\bar ab+b\bar c$.

Combine. $E=a\bar b+ac+bc+\bar ab+b\bar c$.
Group $bc+b\bar c=b$. Then $\bar ab+b=b$. Then $a\bar b+b=a+b$. Then $a+b+ac=a+b$.

$\boxed{\;E=a+b.\;}$

Truth table. $E=0$ only at rows $(0,0,0)$ and $(0,0,1)$; $E=1$ at all others. Minterm form: $\sum m(2,3,4,5,6,7)$: $\bar ab\bar c+\bar abc+a\bar b\bar c+a\bar bc+ab\bar c+abc.$

2019 Paper 2, 2019-P2-Q8a (15 marks)

$X=AB+ABC+A\bar B\bar C+A\bar C$. Draw the logic diagram, minimise, draw reduced diagram.

Logic diagram (original). Four AND gates ($AB$, $ABC$, $A\bar B\bar C$, $A\bar C$) feed one OR gate; inverters produce $\bar B,\bar C$.

Minimise.

• $AB+ABC=AB(1+C)=AB$ (absorption).
• $A\bar B\bar C+A\bar C=A\bar C(\bar B+1)=A\bar C$ (absorption).

$\boxed{\;X=AB+A\bar C=A(B+\bar C).\;}$

Reduced diagram. NOT gate on $C$ → $\bar C$. OR gate on $B,\bar C$ → $(B+\bar C)$. AND gate on $A,(B+\bar C)$ → $X$. (3 gates vs 5+ originally.)

2022 Paper 2, 2022-P2-Q6b (15 marks)

Find a combinatorial circuit for $f=[x(\bar y+z)]+y$ and write its input/output table.

Circuit (original expression). Gate hierarchy: NOT on $y$ → $\bar y$; OR on $\bar y,z$ → $(\bar y+z)$; AND on $x,(\bar y+z)$ → $x(\bar y+z)$; OR with $y$ → $f$. Components: 1 NOT, 2 OR, 1 AND.

Simplify. $f=x\bar y+xz+y$. Use $x\bar y+y=x+y$ (identity $A\bar B+B=A+B$). Then $x+y+xz=x(1+z)+y=x+y$.

$\boxed{\;f=x+y.\;}$

Truth table ($z$ irrelevant).

Reduced circuit: one OR gate ($x$ and $y$ as inputs). The variable $z$ drops out entirely.

2025 Paper 2, 2025-P2-Q6b (15 marks)

Simplify $F(x,y,z)=xyz+x'yz+xy'z+xyz'$ and draw the GATE network.

The four terms are minterms $m_7,m_3,m_5,m_6$, so $F=\sum m(3,5,6,7)$.

Pair adjacently (use $AB+AB'=A$, idempotent to reuse $xyz$): $xyz+x'yz=yz,\quad xyz+xy'z=xz,\quad xyz+xyz'=xy.$

$\boxed{\;F=xy+yz+zx.\;}$

This is the 3-input majority function ($F=1$ iff at least 2 of the 3 inputs are 1).

Gate network. Three 2-input AND gates compute $xy$, $yz$, $zx$; their outputs feed one 3-input OR gate.

  x,y ─►[AND]── xy ──┐
  y,z ─►[AND]── yz ──┼─►[OR]── F
  z,x ─►[AND]── zx ──┘

Common Traps

• Absorption direction: $A\cdot(A+X)=A$ (the lone $A$ absorbs a factor containing $A$); do not confuse with $A+AX=A$ (the OR-absorption direction). All three maxterms containing bare $A$ vanish simultaneously (2016).
• Naming rules: UPSC explicitly asks which law was applied at each step. Writing the manipulation without naming the rule costs partial marks (2017).
• Minterm vs simplified form: “minterm normal form” requires every variable in every term, complemented or unprimed — writing only the simplified $a+b$ is not acceptable as the normal form (2018).
• Reusing a minterm: when pairing minterms on a K-map, the same minterm cell can belong to multiple groups (idempotent law $m+m=m$) — see 2025 where $xyz=m_7$ appears in all three pairs.

normal-form-cnf (4 question(s); 2020, 2021, 2022, 2023)

Recognition Cues

• “Express as product of maxterms”, “find the CNF”, “principal conjunctive normal form.”
• Function is given as an expression (may already be in POS but not canonical), or implicitly via a formula to evaluate.
• The answer must be a product of full sum terms — every clause must contain all $n$ variables.

Solution Template

From a truth table (most reliable):

1. Evaluate $F$ for all $2^n$ input combinations.
2. Identify the rows where $F=0$.
3. For each such row $(b_1,b_2,\ldots,b_n)$: write the maxterm as the OR of all variables, with variable $x_i$ unprimed if $b_i=0$, primed if $b_i=1$ (opposite of minterms!).
4. The CNF is the product (AND) of all these maxterms.

From a partial POS (inflate incomplete clauses):

1. For each sum factor missing variable $v$: replace $(f)$ by $(f+v)(f+\bar v)$ (since $v+\bar v=1$, so $f=(f+v)(f+\bar v)$).
2. Collect distinct maxterms (remove duplicates by idempotence).

Worked Example(s)

2020 Paper 2, 2020-P2-Q5c (10 marks)

$g(w,x,y,z)=(w+x+y)(x+\bar y+z)(w+\bar y)$. Find CNF and express as product of maxterms.

Each factor is a partial sum (not all 4 variables); inflate each:

Factor 1 $(w+x+y)$ — missing $z$: $(w+x+y)=(w+x+y+z)(w+x+y+\bar z).$

Factor 2 $(x+\bar y+z)$ — missing $w$: $(x+\bar y+z)=(w+x+\bar y+z)(\bar w+x+\bar y+z).$

Factor 3 $(w+\bar y)$ — missing $x$ and $z$. First: $(w+\bar y)=(w+x+\bar y)(w+\bar x+\bar y)$. Then add $z$ to each: $(w+\bar y)=(w+x+\bar y+z)(w+x+\bar y+\bar z)(w+\bar x+\bar y+z)(w+\bar x+\bar y+\bar z).$

Collect distinct maxterms (idempotence removes $w+x+\bar y+z$, which appears in factors 2 and 3):

$\boxed{\;g=\underbrace{(w{+}x{+}y{+}z)}_{M_0}\underbrace{(w{+}x{+}y{+}\bar z)}_{M_1}\underbrace{(w{+}x{+}\bar y{+}z)}_{M_2}\underbrace{(w{+}x{+}\bar y{+}\bar z)}_{M_3}\underbrace{(w{+}\bar x{+}\bar y{+}z)}_{M_6}\underbrace{(w{+}\bar x{+}\bar y{+}\bar z)}_{M_7}\underbrace{(\bar w{+}x{+}\bar y{+}z)}_{M_{10}}.\;}$

Product-of-maxterms compact form: $g=\prod M(0,1,2,3,6,7,10)$.

2021 Paper 2, 2021-P2-Q5c-ii (5 marks)

Principal CNF of $(\neg P\to R)\wedge(Q\leftrightarrow P)$.

Simplify. $\neg P\to R\equiv P\vee R$. $Q\leftrightarrow P\equiv (P\wedge Q)\vee(\neg P\wedge\neg Q)$.

Truth table (3 variables, 8 rows). Expression $=(P\vee R)\wedge(Q\leftrightarrow P)$ equals 0 at five rows:

$\boxed{\;(P\vee Q\vee R)\wedge(P\vee\neg Q\vee R)\wedge(P\vee\neg Q\vee\neg R)\wedge(\neg P\vee Q\vee R)\wedge(\neg P\vee Q\vee\neg R).\;}$

Compact: $\prod M(0,2,3,4,5)$.

2022 Paper 2, 2022-P2-Q5c-ii (5 marks)

Express $F(x,y,z)=xy+x'z$ as product of maxterms.

Truth table. $F=0$ at rows $(x,y,z)\in\{(0,0,0),(0,1,0),(1,0,0),(1,0,1)\}$. Maxterms:

• $(0,0,0)$: $x+y+z$
• $(0,1,0)$: $x+\bar y+z$
• $(1,0,0)$: $\bar x+y+z$
• $(1,0,1)$: $\bar x+y+\bar z$

$\boxed{\;F=(x+y+z)(x+\bar y+z)(\bar x+y+z)(\bar x+y+\bar z)=\prod M(0,2,4,5).\;}$

2023 Paper 2, 2023-P2-Q7a-i (7.5 marks)

Find CNF of $f(x,y,z,t)=xyz+\bar xyt+\bar x y\bar z$.

Factor. $f=y[xz+\bar x(t+\bar z)]=y\cdot g$ where $g(x,z,t)=xz+\bar x t+\bar x\bar z$.

Truth table of $g$ (8 rows in $x,z,t$). $g=0$ at $(x,z,t)\in\{(0,1,0),(1,0,0),(1,0,1)\}$.

Maxterms of $g$: $(x+\bar z+t)$, $(\bar x+z+t)$, $(\bar x+z+\bar t)$.

Apply consensus: $(\bar x+z+t)(\bar x+z+\bar t)=\bar x+z$. So minimal CNF of $g$ is $(\bar x+z)(x+\bar z+t)$.

Re-attach $y$:

$\boxed{\;f=y\cdot(\bar x+z)\cdot(x+\bar z+t)\;\text{ (minimal CNF).}\;}$

For the canonical CNF (every clause uses all 4 variables): enumerate all $(x,y,z,t)$ rows where $f=0$ (11 rows) and list their 4-literal maxterms. (Omit if time is short; the minimal form above is usually accepted.)

Common Traps

• Maxterm vs minterm convention: in a maxterm, a variable is primed when its bit is 1 (opposite of minterms where priming happens when bit is 0). A single confusion here flips every term.
• Inflate, don’t simplify: the canonical CNF asks for full $n$-variable clauses; $y$ alone is a valid minimal CNF clause but not a canonical one. Provide both forms if the question says “principal”.
• Idempotence when collecting: if the same maxterm appears from two different inflated factors, count it only once ($M\cdot M=M$); missing this gives the wrong count (2020).

normal-form-dnf (3 question(s); 2015, 2023, 2025)

Recognition Cues

• “Express as sum of minterms”, “find the DNF”, “principal disjunctive normal form.”
• Sometimes accompanied by “is the expression a tautology or contradiction?” — if all rows are 1 (tautology), the DNF is all $2^n$ minterms; if all rows are 0 (contradiction), the DNF is empty.

Solution Template

1. Simplify the expression algebraically if possible (De Morgan, absorption, etc.).
2. Build the truth table for the simplified form over all $2^n$ rows.
3. Identify rows where $F=1$; for each, write its minterm: AND of all $n$ literals, variable unprimed if 1, primed if 0.
4. The DNF is the OR of all these minterms.
5. Report both minimal form (the simplified expression) and canonical/principal form (the explicit minterm sum) — UPSC often asks for both.

Worked Example(s)

2015 Paper 2, 2015-P2-Q5c (10 marks)

Find the principal DNF of $((p\wedge q)\to r)\vee((p\wedge q)\to\neg r)$. Tautology or contradiction?

Simplify. Let $A=p\wedge q$. Expression $=(A\to r)\vee(A\to\neg r)=(\neg A\vee r)\vee(\neg A\vee\neg r)=\neg A\vee r\vee\neg r=\neg A\vee\mathbf{1}=\mathbf{1}$.

The expression is a tautology — true for all 8 valuations of $(p,q,r)$.

Principal DNF. A tautology’s DNF is the disjunction of all $2^3=8$ minterms: $\boxed{\;(\neg p\wedge\neg q\wedge\neg r)\vee(\neg p\wedge\neg q\wedge r)\vee\cdots\vee(p\wedge q\wedge r).\;}$

(All 8 three-literal minterms in lexicographic order.)

2023 Paper 2, 2023-P2-Q7a-ii (7.5 marks)

Express $f(x,y,z)=x+\overline{(\bar x\bar y+\bar x z)}+z$ in DNF and construct its truth table.

Simplify. Apply De Morgan to the complemented bracket: $\overline{\bar x\bar y+\bar x z}=\overline{\bar x(\bar y+z)}=x+(y\bar z).$ So $f=x+(x+y\bar z)+z=x+y\bar z+z$.

Now $y\bar z+z=y+z$ (identity $A\bar B+B=A+B$). Hence $\boxed{\;f=x+y+z.\;}$

Truth table. $f=0$ only at $(0,0,0)$; $f=1$ at the other 7 rows.

Canonical DNF (7 minterms): $\bar x\bar y z\,\vee\,\bar x y\bar z\,\vee\,\bar x yz\,\vee\,x\bar y\bar z\,\vee\,x\bar y z\,\vee\,xy\bar z\,\vee\,xyz = \sum m(1,2,3,4,5,6,7)$.

2025 Paper 2, 2025-P2-Q5c-ii (5 marks)

Truth table and full DNF of $F(x,y,z)=(x+y+z')(x'+y')$.

Evaluate. $(x'+y')=0$ only when $x=y=1$; $(x+y+z')=0$ only when $x=0,y=0,z=1$.

$F=1$ at rows $0,2,3,4,5$.

$\boxed{\;F=x'y'z'+x'yz'+x'yz+xy'z'+xy'z=\sum m(0,2,3,4,5).\;}$

Common Traps

• Tautology DNF: when the formula simplifies to $\mathbf{1}$, the principal DNF is the OR of all $2^n$ minterms (2015). Students often write only the non-trivial ones.
• Minimal vs canonical: $x+y+z$ is the minimal DNF but not the principal (canonical) DNF — the principal form has every variable in every minterm (2023).
• De Morgan direction: $\overline{A+B}=\bar A\bar B$ and $\overline{AB}=\bar A+\bar B$. Mixing these up when stripping the outer complement inverts the whole expression.

boolean-from-truth-table (2 question(s); 2021, 2024)

Recognition Cues

• A truth table is given (all $2^n$ rows filled, $F$ column given); derive the Boolean function.
• Often followed by “simplify” and “draw the GATE circuit.”
• A bit-sequence variant (2024): input and output sequences are given as $n$-bit strings; evaluate the formula bit by bit.

Solution Template

1. Write the SOP (sum of minterms): identify rows where $F=1$ and write their minterms.
2. K-map or absorption: for 3 variables, draw the 3-variable K-map, place the 1-values, and find the minimal groups. For algebraic simplification, look for adjacent minterms differing in one variable.
3. Simplified expression from the K-map groups.
4. Draw the gate circuit for the simplified form.
5. Bit-sequence evaluation (2024 variant): evaluate the simplified expression bit-by-bit across the input columns.

Worked Example(s)

2021 Paper 2, 2021-P2-Q6b (15 marks)

Boolean function from truth table; simplify; GATE network.

Given: $F=1$ at rows $(x,y,z)\in\{(1,1,1),(1,1,0),(1,0,1),(0,1,1)\}$.

SOP. $F=xyz+xy\bar z+x\bar y z+\bar x y z$.

K-map simplification (3 variables):

Group $yz$-column (cells $m_3,m_7$): $yz$. Group $x=1$ row cells $m_5,m_7$: $xz$. Group $x=1$ row cells $m_6,m_7$: $xy$.

$\boxed{\;F=xy+yz+zx.\;}$

This is the 3-input majority function (1 iff at least 2 inputs are 1).

Gate circuit. 3 AND gates ($xy$, $yz$, $zx$) feeding a 3-input OR gate.

2024 Paper 2, 2024-P2-Q6b (15 marks)

Logical circuit and truth table for $Y=AB\bar C+B\bar C+\bar A B$; evaluate for $A=10001111$, $B=00111100$, $C=11000100$.

Simplify. $AB\bar C+B\bar C=B\bar C(A+1)=B\bar C$. So $Y=B\bar C+\bar A B=B(\bar A+\bar C)$.

$\boxed{\;Y=\bar A B+B\bar C.\;}$

Gate circuit. NOT on $A$ → $\bar A$. NOT on $C$ → $\bar C$. AND on $\bar A,B$ → $\bar A B$. AND on $B,\bar C$ → $B\bar C$. OR → $Y$.

Bit-by-bit evaluation (left to right):

$\boxed{\;Y\text{-sequence}=00111000.\;}$

Common Traps

• K-map cell adjacency: columns are in Gray-code order (00, 01, 11, 10); column “11” is adjacent to “01” and “10”, not to “00”. Reading the K-map in natural binary order misses valid groupings.
• Bit-sequence direction: read the bit sequences left-to-right and pair positions consistently — mixing up which bit is “position 1” flips all outputs (2024).
• The 3-input majority function (2021, 2025) has the symmetric form $xy+yz+zx$; recognising it avoids algebra.

boolean-identity-proof (1 question(s); 2014)

Recognition Cues

• “Show/prove that [expression] = [expression] for Boolean variables.”
• The identity itself is given; you must justify it from the axioms.
• Accept all three proof styles: algebraic, truth table, logical argument.

Solution Template

1. Algebraic proof (primary): manipulate the LHS using named axioms until it equals the RHS. Name every law used.
2. Truth table (secondary): build the table for both LHS and RHS; confirm columns match for all $2^n$ rows.
3. Logical argument (optional): explain in words why the identity holds.

For full marks on a 15-mark identity question, provide at least two independent proofs.

Worked Example(s)

2014 Paper 2, 2014-P2-Q8b (15 marks)

Show that $x+xy=x$ for Boolean variables $x,y$.

Proof 1 — Algebraic. $x+xy=x\cdot1+x\cdot y=x(1+y).$ Since $1+y=1$ (annihilation/null law): $x(1+y)=x\cdot1=x$. $\blacksquare$

Proof 2 — Truth table.

Last column equals first column ($x$) in all 4 rows. $\blacksquare$

Proof 3 — Logical argument. "$x+xy$" is true iff $x$ is true, or both $x$ and $y$ are true. In either case $x$ must be true; and if $x$ is already true, the expression is true regardless of $y$. So the expression equals $x$ exactly. $\blacksquare$

$\boxed{\;x+xy=x.\;}$

Common Traps

• $1+y=1$ is the asymmetry between Boolean algebra and ordinary arithmetic ($1+y\neq1$ in $\mathbb Z$). Name this law explicitly (“null law” or “annihilation”).
• The dual identity $x(x+y)=x$ (AND-absorption) is equally fundamental — expect it to appear in a multi-part question.
• For a 15-mark proof, one-line algebraic manipulation without naming laws or a truth-table check will not earn full marks.

Marks-Aware Writing

5-mark questions (2021-ii, 2022-ii, 2025-ii): These are from the compulsory Section B short-answer slots. One method is sufficient — truth table with maxterms/minterms, compact notation, boxed answer. No gate diagram required.

7.5-mark questions (2023): Give both the minimal form and the canonical form (clearly labelled). Truth table for the sub-function. For CNF questions also briefly verify one maxterm evaluates to 0 at its row.

10-mark questions (2017, 2020, 2022 Q5c-ii): For simplification: named steps + truth table + boxed answer. For normal forms: full truth table + all maxterms/minterms listed. No gate diagram unless asked.

15-mark questions (the majority): Named algebraic steps + full truth table verification + gate diagram + boxed answer. For identity proofs (2014): at least two independent proof methods. For expressions with gate diagrams (2019, 2021, 2024): draw both the original and the reduced circuit.

Practice Set