OCR J277 · Component 2 · Topic 2.4.1b

Boolean Logic
Truth Tables

Building, completing and reading truth tables for multi-gate logic circuits

CSZone OCR GCSE Computer Science J277

Learning Objectives

By the end of this video you will be able to...

Set up a truth table correctly for any single or multi-gate circuit — listing every possible input combination in order, with one column per input, one column per intermediate signal, and one column for the final output

Complete a two-input truth table (4 rows) for circuits combining two logic gates — applying each gate's rule in turn and filling an intermediate column before calculating the final output

Construct a three-input truth table (8 rows) for circuits with inputs A, B and C — listing combinations 000 through 111 in binary order and working through each gate stage by stage

Write a Boolean expression for the output of any circuit by reading the gates left to right — and reverse the process by converting a Boolean expression back into a circuit diagram

Identify the output of any input row in a multi-gate circuit without re-drawing the table — by tracing through each gate one at a time and applying its rule to the values you already know

⚡ Every multi-mark Boolean logic question in the OCR exam asks you to complete a truth table for a circuit. Master the systematic column-by-column method and these questions become routine.

Method

How to tackle any truth table question

THREE-STEP METHOD

1Set up columns — one per input, one per intermediate signal, one for the final output Q

2List all input combinations — in binary counting order (00→11 for 2 inputs; 000→111 for 3 inputs)

3Fill left to right — complete each gate's output column before moving to the next gate. Never skip ahead

NUMBER OF ROWS

Rows = 2ⁿ where n = number of inputs.
1 input (NOT) → 2 rows
2 inputs (AND/OR/etc.) → 4 rows
3 inputs → 8 rows
The exam will always give you enough rows — count them first to check you have the right number.

THE THREE CORE GATES — QUICK REMINDER

Gate	Rule	Output 1 when
NOT	Invert input	A = 0
AND	All inputs = 1	A=1 AND B=1
OR	Any input = 1	A=1 OR B=1

Note: The OCR spec names AND, OR and NOT as the three core gates. NAND, NOR and XOR (from the previous video) may also appear — apply the same column-by-column method to them.

WHAT IS AN INTERMEDIATE COLUMN?

When a circuit has two gates in series, the output of the first gate feeds into the second. That output is an intermediate signal — usually labelled P, X, or similar. It gets its own column in the table. Without it, you cannot work out the final output correctly.

Two-Gate Circuit

Circuit 1 — NOT into AND

CIRCUIT

Input A feeds through a NOT gate → output P.
P and input B feed into an AND gate → final output Q.
Boolean expression: Q = A · B

STEP 1 — FIND P (NOT A)

P = NOT A. When A=0, P=1. When A=1, P=0. Fill the P column before touching Q.

STEP 2 — FIND Q (P AND B)

Q = P AND B. Use P from column 3 and B from column 2. Output 1 only when both P and B are 1.

COMPLETE TRUTH TABLE

A	B	P = A	Q = P · B
0	0	1	0
0	1	1	1
1	0	0	0
1	1	0	0

Yellow = intermediate column (marker checks this separately). Purple = final output (marks here too).

OBSERVATION

Q = 1 only in one case: A=0, B=1. In words: "B is true and A is false." The NOT gate on A means A must be off before the AND gate will pass B through. This pattern is sometimes called "B but not A."

Two-Gate Circuit

Circuit 2 — NOT into OR

CIRCUIT

Input A feeds through a NOT gate → output P.
P and input B feed into an OR gate → final output Q.
Boolean expression: Q = A + B

STEP 1 — FIND P (NOT A)

P = NOT A. When A=0 → P=1. When A=1 → P=0. Same as before — fill this column first.

STEP 2 — FIND Q (P OR B)

Q = P OR B. Output 1 when P=1 OR B=1. Only 0 when both P and B are 0.

COMPLETE TRUTH TABLE

A	B	P = A	Q = P + B
0	0	1	1
0	1	1	1
1	0	0	0
1	1	0	1

Compare Circuit 1 (NOT→AND) with Circuit 2 (NOT→OR): same P column but very different Q columns — choosing AND vs OR dramatically changes the behaviour.

CIRCUIT 1 vs CIRCUIT 2 — SIDE BY SIDE

A	B	NOT→AND	NOT→OR
0	0	0	1
0	1	1	1
1	0	0	0
1	1	0	1

Three Inputs

Three-input circuits — 8 rows

WHY 8 ROWS?

With 3 inputs (A, B, C), each can be 0 or 1. That gives 2 × 2 × 2 = 8 combinations. You list them in binary counting order — starting at 000 and ending at 111. Think of the A column as the hundreds column of a binary number, B as tens, C as units.

PATTERN TO REMEMBER

C column (rightmost input): alternates every row → 0,1,0,1,0,1,0,1
B column: alternates every 2 rows → 0,0,1,1,0,0,1,1
A column: alternates every 4 rows → 0,0,0,0,1,1,1,1

EXAM TECHNIQUE

Fill the input columns first using the pattern above — before touching any gate output. This guarantees you never miss a row or repeat a combination. The examiner checks all 8 rows, so a missing row costs marks.

TEMPLATE — 3 INPUTS, 2 GATES

A	B	C	P = ?	Q = ?
0	0	0	—	—
0	0	1	—	—
0	1	0	—	—
0	1	1	—	—
1	0	0	—	—
1	0	1	—	—
1	1	0	—	—
1	1	1	—	—

The input columns follow the pattern exactly: A changes every 4 rows, B every 2, C every row. Memorise this pattern and you'll never have to think about it — just write it out automatically.

Three-Input Circuit

Circuit 3 — AND into OR (three inputs)

CIRCUIT

Inputs A and B feed into an AND gate → P.
P and input C feed into an OR gate → Q.
Boolean: Q = (A · B) + C

HOW TO FILL

P column first: P = A AND B → only 1 when both A and B are 1 (rows 7 and 8).

Q column second: Q = P OR C → 1 when P=1 OR C=1.

COMPLETED TRUTH TABLE

A	B	C	P=A·B	Q=P+C
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

P is 1 in only 2 of 8 rows (when A=1 AND B=1). C alone can make Q=1 in another 4 rows. The OR gate makes Q=1 in 5 of 8 rows total — OR is always generous.

Three-Input Circuit

Circuit 4 — OR and NOT into AND

CIRCUIT — TWO GATE OUTPUTS COMBINE

A and B feed into an OR gate → P.
C feeds through a NOT gate → X.
P and X feed into an AND gate → Q.
Boolean: Q = (A + B) · C

TWO INTERMEDIATE COLUMNS

P = A OR B — 1 when A=1 or B=1 (0 only when A=0 and B=0)

X = NOT C — simply inverts C. When C=0, X=1. When C=1, X=0.

COMPLETED TRUTH TABLE

A	B	C	P=A+B	X=C	Q=P·X
0	0	0	0	1	0
0	0	1	0	0	0
0	1	0	1	1	1
0	1	1	1	0	0
1	0	0	1	1	1
1	0	1	1	0	0
1	1	0	1	1	1
1	1	1	1	0	0

Two intermediate columns — P and X — are both needed. The AND gate at the end is strict: P must be 1 AND X must be 1. Notice Q=1 only when (A or B is 1) AND C=0. The NOT gate on C means C must be off for the circuit to pass a signal through.

Boolean Expressions

Reading and writing Boolean expressions

EXPRESSION → CIRCUIT

Given: Q = A + B

1. Identify the outermost operation: + (OR gate) → final gate is OR
2. Left input to OR: A → that's A through a NOT gate → intermediate P
3. Right input to OR: B → direct wire
Result: A → NOT → P, then P and B → OR → Q

READING ORDER — INSIDE → OUT

Brackets or overbars are computed first (innermost gates). The outermost operator is always the last gate — closest to the output Q. Work inside-out when reading an expression, left-to-right when drawing the circuit.

NOTATION SUMMARY

Symbol	Means	Gate
A	NOT A	NOT
A · B	A AND B	AND
A + B	A OR B	OR

CIRCUIT → EXPRESSION

Given a circuit: A and B → AND → P, then P and C → OR → Q.

1. Write the output of the first gate: P = A · B
2. Substitute P into the next gate: Q = P + C
3. Replace P: Q = (A · B) + C

BRACKETS = INTERMEDIATE GATE

Whenever you substitute one gate's output into another, wrap it in brackets. (A · B) + C tells you: compute A AND B first (intermediate P), then OR that result with C. The brackets correspond exactly to the intermediate signal column in your truth table.

PRACTICE — MATCH THE EXPRESSION

A · B→ NOT A, then AND with B

(A + B) · C→ OR first, then AND with C

(A · B) + C→ AND + NOT C, then OR

Worked Example

Full worked example — three gates, two inputs

CIRCUIT

A → NOT → P
P and B → AND → X
A and B → OR → Y
X and Y → OR → Q

Boolean: Q = (A · B) + (A + B)

FILL ORDER — 4 EXTRA COLUMNS

Step 1: P = NOT A

Step 2: X = P AND B

Step 3: Y = A OR B

Step 4: Q = X OR Y

INTERESTING RESULT

Look at the Q column — it equals the OR gate output in every row. That's because (A+B) already captures all the cases where Q=1. The first branch (A·B) is a subset of (A+B), so the final OR adds nothing new. A circuit can be simplified.

COMPLETED TRUTH TABLE

A	B	P=A	X=P·B	Y=A+B	Q=X+Y
0	0	1	0	0	0
0	1	1	1	1	1
1	0	0	0	1	1
1	1	0	0	1	1

Q column is identical to Y column — the three-gate circuit simplifies to a single OR gate. In an exam, mark each intermediate column carefully — you get marks for each correct column.

WHAT TO WRITE IN THE EXAM

Always show intermediate columns even if you spot a shortcut — the examiner awards marks per column. Getting X wrong but Q right (because you noticed the simplification) loses X marks while gaining Q marks. Show all working.

Exam Practice

Truth tables — exam questions

Question 1 — 1 mark

A circuit has input A feeding through a NOT gate to produce output Q. State the output Q when A = 1.

Answer — Q1

Q = 0. A NOT gate inverts its input — when A=1 the output is 0. (1 mark)

Question 2 — 3 marks

A logic circuit has inputs A and B. A and B feed into an OR gate to produce P. A feeds through a NOT gate to produce X. P and X feed into an AND gate to produce Q.

(a) Write a Boolean expression for Q. [1]
(b) Complete the truth table: [2]

A	B	P	X	Q
0	0	?	?	?
0	1	?	?	?
1	0	?	?	?
1	1	?	?	?

Answer — Q2

(a) Q = (A + B) · A [1]
(b)

A	B	P	X	Q
0	0	0	1	0
0	1	1	1	1
1	0	1	0	0
1	1	1	0	0

[1] P and X columns correct · [1] Q column correct

Question 3 — 4 marks

A logic circuit has three inputs A, B and C. Inputs A and B feed into an AND gate to produce P. Input C feeds through a NOT gate to produce X. P and X then feed into an OR gate to produce the final output Q.

(a) Write a Boolean expression for Q in terms of A, B and C. [1]
(b) Complete the truth table for this circuit. [3]

A	B	C	P	X	Q
0	0	0	?	?	?
0	0	1	?	?	?
0	1	0	?	?	?
0	1	1	?	?	?
1	0	0	?	?	?
1	0	1	?	?	?
1	1	0	?	?	?
1	1	1	?	?	?

Exam Answers

Question 3 — mark scheme

MARK SCHEME

(a) Q = (A · B) + C [1]

(b) [3 marks — see table]

A	B	C	P=A·B	X=C	Q=P+X
0	0	0	0	1	1
0	0	1	0	0	0
0	1	0	0	1	1
0	1	1	0	0	0
1	0	0	0	1	1
1	0	1	0	0	0
1	1	0	1	1	1
1	1	1	1	0	1

[1] P column correct · [1] X column correct · [1] Q column correct

KEY OBSERVATION

Q = 0 only in rows where C=1 AND A and B are not both 1. The NOT on C means C=1 kills the X path; the AND on A·B means you need both to make P=1. Check the last row: P=1 AND X=0 but Q=P+X=1+0=1 — P alone is enough for OR.

COLUMN-BY-COLUMN APPROACH — ALWAYS

✓ Add a column for every intermediate signal (P, X, etc.) — even if the exam question doesn't show them. Examiners accept and reward this

✓ Fill P before X before Q — never work diagonally row by row

✓ A correct P column earns a mark even if Q is wrong — column marks are independent

QUICK REFERENCE — GATE RULES

Gate	Output 0 when	Output 1 when
NOT	A = 1	A = 0
AND	any input = 0	all inputs = 1
OR	all inputs = 0	any input = 1

⚡ Exam tip: if a question asks "state the output Q when A=1, B=0, C=1" — you don't need to build the whole table. Just trace through the circuit: P = A·B = 1·0 = 0. X = NOT C = NOT 1 = 0. Q = P+X = 0+0 = 0. One value per gate, left to right.

Common Mistakes

Common mistakes — avoid these in the exam

MISTAKE 1 — Wrong number of rows

Students with two inputs write only 2 rows (treating it like a NOT gate), or with three inputs write only 4 rows (as if there were 2 inputs). The number of rows is 2ⁿ where n is the number of inputs. Two inputs = 4 rows. Three inputs = 8 rows. Counting rows first takes five seconds and prevents this error entirely

✓ Count inputs → 2ⁿ → draw all rows before filling any values

MISTAKE 2 — Wrong input combinations or wrong order

Students list inputs randomly (e.g. 00, 11, 01, 10) or repeat combinations, missing some rows. The correct order is binary counting: for two inputs 00→01→10→11; for three inputs 000→001→010→011→100→101→110→111. The column pattern (each input alternates at 1, 2, 4 row intervals) is the safest way to fill these

✓ Use the alternating pattern: rightmost input alternates every row, next every 2 rows, next every 4 rows

MISTAKE 3 — Skipping intermediate columns

Students try to jump straight from A and B to Q without working out P first. With two gates this sometimes works by luck, but with three gates it almost always leads to errors. More importantly, intermediate columns are worth separate marks — a correct P column earns a mark even if Q is wrong. Never skip them

✓ Add a column for every signal that is not an input and not the final output. Fill them in order, left to right

MISTAKE 4 — Applying the wrong gate rule

Students confuse AND and OR — writing 1 for AND when only one input is 1, or writing 0 for OR when at least one input is 1. AND is strict: ALL inputs must be 1. OR is generous: ANY input being 1 is enough. A single wrong gate rule corrupts an entire column. Check the gate symbol before filling each column

✓ Before filling a column, state the gate rule out loud: "AND — all must be 1." Then apply it row by row

Summary

Key points — 2.4.1b

A truth table lists every possible input combination and the corresponding output. Rows = 2ⁿ where n = number of inputs. Two inputs → 4 rows. Three inputs → 8 rows. Input combinations are listed in binary counting order: 00→01→10→11 for two inputs

For multi-gate circuits: always add a column for every intermediate signal. Fill columns left to right — complete the first gate's column before moving to the next. Skipping ahead causes errors and loses marks, since each column is marked independently

A Boolean expression describes the circuit in algebra. NOT A = A. A AND B = A·B. A OR B = A+B. Brackets show which gate is computed first. (A·B)+C means: compute A AND B first (intermediate P), then OR that result with C

To convert expression → circuit: identify the outermost operation (last gate), work inward for each operand. To convert circuit → expression: name each gate's output, substitute step by step, wrap in brackets where needed

To find the output for one specific input row without the full table: trace through the circuit gate by gate — compute each signal in order. P = ? then X = ? then Q = ? No need to build all rows if only one is asked for

⚡ Next topic: 2.5.1 — High and Low Level Languages and the purpose of translators.

2.4.1b Complete

Truth Tables
Multi-gate circuits · Three inputs · Boolean expressions

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2.5.1 — High and Low Level Languages

A	B	C	P = ?	Q = ?
0	0	0	—	—
0	0	1	—	—
0	1	0	—	—
0	1	1	—	—
1	0	0	—	—
1	0	1	—	—
1	1	0	—	—
1	1	1	—	—

A	B	C	P = ?	Q = ?
0	0	0	—	—
0	0	1	—	—
0	1	0	—	—
0	1	1	—	—
1	0	0	—	—
1	0	1	—	—
1	1	0	—	—
1	1	1	—	—

Boolean LogicTruth Tables

By the end of this video you will be able to...

How to tackle any truth table question

Circuit 1 — NOT into AND

Circuit 2 — NOT into OR

Three-input circuits — 8 rows

Circuit 3 — AND into OR (three inputs)

Circuit 4 — OR and NOT into AND

Reading and writing Boolean expressions

Full worked example — three gates, two inputs

Truth tables — exam questions

Question 3 — mark scheme

Common mistakes — avoid these in the exam

Key points — 2.4.1b

Truth TablesMulti-gate circuits · Three inputs · Boolean expressions

Boolean Logic
Truth Tables

Truth Tables
Multi-gate circuits · Three inputs · Boolean expressions

A	B	C	P = ?	Q = ?
0	0	0	—	—
0	0	1	—	—
0	1	0	—	—
0	1	1	—	—
1	0	0	—	—
1	0	1	—	—
1	1	0	—	—
1	1	1	—	—