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Detailed notes on Boolean logic for Cambridge IGCSE Computer Science, covering key concepts, explanations, examples, and exam-focused revision points.
Three equivalent representations: a DESCRIPTION in plain English, a logic EXPRESSION in Boolean algebra, and a CIRCUIT diagram with gates. Translating between them; building truth tables; reading expressions back from circuits.
Mapped to the Cambridge IGCSE 0478 syllabus (2026-2028).
Description, expression, circuit — all say the same thing.
Cambridge tests your ability to translate between three representations.
1. Description. Plain English: 'The output is 1 when A is 1 and B is 1, OR when C is 0.'
2. Expression. Boolean algebra: X = (A AND B) OR (NOT C).
3. Circuit. Gate diagram: A and B feed an AND gate; C feeds a NOT gate; the AND output and the NOT output feed an OR gate; OR's output is X.
All three say the same thing. The skill is moving smoothly between them.
Cambridge tip. Mark scheme typically shows one form and asks you to derive the others. Practice translating IN BOTH DIRECTIONS — description → expression → circuit AND circuit → expression → description.
Read inside-out; each operator becomes a gate.
Given an expression like X = (A AND B) OR (NOT C), draw the circuit:
Step 1. Identify the operators (gates): AND, NOT, OR.
Step 2. Find the INNERMOST operations first. Here, both A AND B and NOT C are inside their own brackets — they happen first.
Step 3. Draw inputs on the LEFT (A, B, C entering from left).
Step 4. Draw the inner gates.
Step 5. Draw the outer gate.
Convention. Inputs LEFT, output RIGHT. Signals flow LEFT → RIGHT.
Cambridge tip. Label intermediate signals clearly. Mark scheme awards method marks for showing the intermediate steps even if the final layout has errors.
See the full worked example for logic expressions and circuits →
Trace from inputs to output; build the expression as you go.
Given a circuit (or a description of one), derive the expression:
Step 1. Trace from INPUTS forward through gates. At each gate, write the operation in terms of its inputs.
Step 2. Combine intermediate signals as you reach later gates.
Step 3. End with the OUTPUT expression.
Example. Circuit description: 'A and B go into an AND gate. The AND output is OR'd with C. The OR output is NOT'd to give X.'
A AND B.(A AND B) OR C.X = NOT ((A AND B) OR C).Brackets matter. Without them, NOT (A AND B) OR C is a different expression with a different truth table.
Cambridge tip. When unsure, work LEFT-TO-RIGHT (input to output) following each gate. Use brackets liberally to enforce the order of operations.
See the full worked example for logic expressions and circuits →
List every input combo; compute the output row by row.
Step 1. Count inputs. n inputs = 2^n rows.
Step 2. List inputs in BINARY ORDER. For three inputs (A, B, C):
| A | B | C |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
| 1 | 1 | 1 |
Step 3. Add INTERMEDIATE columns for each sub-expression. For X = (A AND B) OR C, add a column A AND B first, then X.
Step 4. Fill row by row. Compute each cell from the inputs.
Step 5. Check the output column has the right pattern.
Why intermediate columns. They earn METHOD MARKS even if the final answer has errors. Always show working.
Cambridge tip. Don't skip rows; don't reorder inputs. Mark schemes count rows in standard order. Inserting them out of sequence loses marks.
See the full worked example for logic expressions and circuits →
Verbatim phrases and definitions Cambridge mark schemes credit.
Logic expressions and circuits appear on most Paper 2s. Most-tested questions: draw a circuit from an expression (8 marks), derive an expression from a circuit (6 marks), complete a truth table for a 3-input expression (8 marks). Examiner reports flag missed brackets and reversed signal flow as the most common errors.
Sources: Cambridge IGCSE Computer Science 0478 syllabus (2026-2028); 0478 Examiner Reports 2022-2024; 0478/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-09.
Step-by-step solutions to past-paper-style questions on logic expressions and circuits, written exactly the way a tutor would explain them at the board.
Question
Complete the truth table for the logic expression X = A AND (NOT B). (4 marks)
Step-by-step solution
Step 1
List the rows (1 mark). Two inputs means 2² = 4 rows. Count up in binary: 00, 01, 10, 11.
Step 2
Work out NOT B (1 mark). NOT B is the opposite of B. So for B = 0,1,0,1 the column NOT B = 1,0,1,0.
Step 3
Apply the AND (2 marks). X = A AND (NOT B) is 1 only when A = 1 AND NOT B = 1. That is the single row A = 1, B = 0.
Answer
| A | B | NOT B | X |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
Examiner tip
Include the NOT B working column — it earns a method mark even if the final X column slips.
Question
A circuit has inputs A, B, C. A and B feed an AND gate; the result is OR'd with C; that result is NOT'd to give the output X. Write the logic expression for X. (3 marks)
Step-by-step solution
Step 1
Innermost (1 mark). The first gate gives A AND B.
Step 2
OR with C (1 mark). Next gate gives (A AND B) OR C.
Step 3
Final NOT (1 mark). The last gate inverts: X = NOT ((A AND B) OR C).
Answer
X = NOT ((A AND B) OR C). Brackets matter — they enforce the order in which the gates were connected.
Question
Describe, gate by gate, the logic circuit for the expression X = (A AND B) OR (NOT C). (4 marks)
Step-by-step solution
Step 1
Identify the gates needed (1 mark). Three gates: an AND (for A AND B), a NOT (for NOT C), and an OR to combine them.
Step 2
AND gate (1 mark). Inputs A and B enter an AND gate. Call its output P, so P = A AND B.
Step 3
NOT gate (1 mark). Input C enters a NOT gate. Call its output Q, so Q = NOT C.
Step 4
OR gate (1 mark). P and Q feed into an OR gate, giving the output X = P OR Q.
Answer
Circuit (left → right): A, B → AND → P; C → NOT → Q; P, Q → OR → X. Three gates, two intermediate signals (P, Q), one output X.
Question
A greenhouse fan turns ON (X = 1) when the temperature is high (T = 1) AND the window is closed (W = 1), OR when an override switch is pressed (S = 1). Write a logic expression for X. (4 marks)
Step-by-step solution
Step 1
Spot the inputs (1 mark). Three binary inputs: T (temperature high), W (window closed), S (override pressed).
Step 2
The 'high AND closed' clause (1 mark). 'High temperature AND window closed' becomes T AND W.
Step 3
The override clause (1 mark). The fan ALSO runs whenever S = 1, so the override is OR'd onto the first clause.
Step 4
Combine with brackets (1 mark). Bracket the AND clause so it is evaluated before the OR: X = (T AND W) OR S.
Answer
X = (T AND W) OR S. The bracket keeps T AND W together so the override S is correctly OR'd onto the whole condition.
Question
Draw a complete truth table for X = (A AND B) OR C. (6 marks)
Step-by-step solution
Step 1
8 rows (1 mark). Three inputs = 2³ = 8 rows. List in binary: 000, 001, 010, 011, 100, 101, 110, 111.
Step 2
Compute A AND B (2 marks). This is 1 only on the last two rows (A = 1, B = 1). Column: 0, 0, 0, 0, 0, 0, 1, 1.
Step 3
OR with C (3 marks). X = (A AND B) OR C is 1 whenever C = 1 OR the AND column is 1. Column: 0, 1, 0, 1, 0, 1, 1, 1.
Answer
| A | B | C | A AND B | X |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 |
Examiner tip
Mark scheme expects the 'A AND B' intermediate column for working — it earns method marks even if the final column has errors.
Question
Complete the truth table for X = (A XOR B) NAND C. (6 marks)
Step-by-step solution
Step 1
8 rows (1 mark). Three inputs A, B, C → 2³ = 8 rows in binary order 000 … 111.
Step 2
A XOR B (2 marks). XOR is 1 when the two inputs DIFFER. Column: 0, 0, 1, 1, 1, 1, 0, 0.
Step 3
Form (A XOR B) AND C, then invert (NAND) (3 marks). NAND = NOT(AND). Take (A XOR B) AND C = 0,0,0,1,0,1,0,0 then invert every bit. X = 1,1,1,0,1,0,1,1.
Answer
| A | B | C | A XOR B | X |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 0 | 1 |
Examiner tip
NAND is the inverse of AND, so X = 0 only when BOTH (A XOR B) = 1 AND C = 1 — exactly two rows. Show the XOR working column for the method mark.
High-scoring sample answers for logic expressions and circuits on the Cambridge IGCSE 0478 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.
A door buzzer sounds (X = 1) when a button is not pressed (P = 1 means pressed). Write a logic expression for X. [2]
Model answer
X = NOT P
Why this scores
1 mark for recognising a NOT gate is required; 1 mark for the correct expression X = NOT P. Writing X = P (forgetting the inversion) scores 0 for the expression.
A car warning light comes ON (X = 1) when the engine is running (E = 1) AND either a door is open (D = 1) OR a seatbelt is unfastened (S = 1). Write a logic expression for X. [3]
Model answer
X = E AND (D OR S)
Why this scores
1 mark D OR S; 1 mark AND with E; 1 mark correct brackets. The bracket is essential — E AND D OR S evaluates AND before OR and gives the wrong logic.
A pump runs (X = 1) only when the tank is not full (F = 1 means full) AND the system is switched on (W = 1). Complete a truth table for X = (NOT F) AND W. [4]
Model answer
| F | W | NOT F | X |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 |
Why this scores
1 mark for the NOT F column; up to 3 marks for the X column (typically 1 mark per correct pair of rows). X is 1 only when the tank is not full AND the system is on — the single row F = 0, W = 1.
An air-conditioning unit turns ON (X = 1) for the expression X = (A AND B) OR (NOT C), where A = temperature high, B = room occupied, C = energy-saver mode. Describe the logic circuit gate by gate. [6]
Model answer
Three inputs A, B, C enter from the left.
P = A AND B).Q = NOT C).X = P OR Q).Signals flow left → right; the output X is on the right.
Why this scores
2 marks per correctly placed and connected gate (AND, NOT, OR), to a max of 6. Common errors: omitting the NOT gate on C, swapping AND/OR, or reversing signal flow. Inputs must be correctly fed (A and B to the AND gate; C to the NOT gate).
A factory machine raises an alarm (X = 1) when a guard is open (G = 1) OR when the temperature is too high (T = 1) AND the machine is running (R = 1). Write a logic expression for X and describe the matching circuit gate by gate. [6]
Model answer
Expression: X = G OR (T AND R)
Circuit (left → right):
P = T AND R).X = G OR P).Two gates, one intermediate signal P, one output X.
Why this scores
Up to 3 marks for the expression: T AND R (1), OR with G (1), correct brackets (1). Up to 3 marks for the circuit: AND gate fed by T and R (1+1) and OR gate fed by G and P giving X (1). Watch the precedence — G OR T AND R without brackets still evaluates AND first, but candidates should bracket for clarity and to match the circuit.
A security system sounds an alarm (X = 1) when the system is armed (A = 1) AND either a window sensor (B = 1) or a motion sensor (C = 1) is triggered, expressed as X = A AND (B OR C). Draw the complete truth table and state in which rows the alarm sounds. [8]
Model answer
| A | B | C | B OR C | X |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 |
The alarm sounds (X = 1) in the last three rows: A = 1 with at least one of B or C also 1 — i.e. (A,B,C) = (1,0,1), (1,1,0) and (1,1,1).
Why this scores
1 mark for the 8 rows in correct binary order; 2 marks for the B OR C working column; up to 4 marks for the X column (1 mark per correct group of rows); 1 mark for correctly stating where the alarm sounds. Key check: X is 0 on every row where A = 0, regardless of B and C — a frequent error is letting B OR C switch X on when the system is not armed.
Definitions to memorise and the exact keywords mark schemes credit for logic expressions and circuits answers — sharpened from recent examiner reports for the 2026 0478 sitting.
A formula combining inputs with logical operators (AND, OR, NOT, NAND, NOR, XOR) to compute an output. Equivalent to a Boolean algebra expression.
A diagram of logic gates connected to compute an output from one or more inputs.
Inputs are the variables (typically labelled A, B, C, ...). Output is the computed result (typically X or Y).
The output of an internal gate, used as input to a later gate. Often labelled P, Q, R for clarity.
Logic expressions are evaluated by precedence: NOT first, then AND, then OR. Brackets override.
The traps other students keep falling into on logic expressions and circuits questions — taken from recent Cambridge IGCSE 0478 examiner reports and mark schemes — and how to avoid them.
0478 Examiner Reports 2022-2024
Why it happens
Students reverse the convention.
How to avoid it
Convention: inputs on the LEFT, output on the RIGHT. Signals flow LEFT → RIGHT through gates. Mark schemes deduct for reversed flow.
Why it happens
Students compute left-to-right.
How to avoid it
Brackets enforce order. (A AND B) OR C ≠ A AND (B OR C). Different brackets give different truth tables.
Why it happens
Students forget to list every combination.
How to avoid it
Always 2^n rows for n inputs. 2 inputs = 4. 3 inputs = 8. List inputs in binary order: 000, 001, 010, ..., 111.
+, ., ' for OR, AND, NOT in Cambridge answersWhy it happens
Boolean algebra uses these symbols.
How to avoid it
Cambridge expects WORD operators: AND, OR, NOT (and NAND, NOR, XOR). Symbol notation may not be marked.
The things students keep getting wrong in this sub-topic, answered.