Apply De Morgan's law to A · B. [1 mark]

Simplify A + A · B and name the law. [2 marks]

OCR OCR 2019-style practice: Simplify the Boolean expression A · B + A · B using the laws of Boolean algebra, naming the laws used, and verify the result with a truth table.

Simplification (up to 3): factor out A using the distributive law: A · B + A · B = A · (B + B). By the complement law, B + B = 1. So A · 1 = A by the identity law. The simplified expression is A. Truth table verification (up to 2): | A | B | A · B | A · B | sum | | --- | --- | --- | --- | --- | | 0 | 0 | 0 | 0 | 0 | | 0 | 1 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 1 | | 1 | 1 | 1 | 0 | 1 | The sum column equals A, confirming the simplification. Markers reward distribution, the complement law B + B = 1, the identity law, and a correct truth table.

OCR OCR 2021-style practice: Apply De Morgan's laws to simplify A + B · A, and construct a truth table for the original expression.

Apply De Morgan (up to 3): A + B = A · B. So the expression becomes A · B · A = A · B (since A · A = A by idempotence). The simplified result is A · B, equivalent to A + B. Truth table (up to 2): | A | B | A + B | A + B | A | result | | --- | --- | --- | --- | --- | --- | | 0 | 0 | 0 | 1 | 1 | 1 | | 0 | 1 | 1 | 0 | 1 | 0 | | 1 | 0 | 1 | 0 | 0 | 0 | | 1 | 1 | 1 | 0 | 0 | 0 | The result is 1 only when both are 0, matching A · B. Markers reward the correct De Morgan expansion, the idempotent simplification, and a correct truth table.

EnglandComputer ScienceSyllabus dot point

How do logic gates implement Boolean expressions, and how do we simplify those expressions using the laws of Boolean algebra?

Logic gates (AND, OR, NOT, XOR, NAND, NOR) and their truth tables, constructing and interpreting truth tables, and simplifying Boolean expressions using the laws of Boolean algebra including De Morgan's laws, distribution, association and commutation, and Karnaugh maps.

An OCR H446 answer on Boolean algebra and logic: the logic gates (AND, OR, NOT, XOR, NAND, NOR) and their truth tables, constructing truth tables for an expression, and simplifying Boolean expressions using the laws of Boolean algebra, De Morgan's laws and Karnaugh maps.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Quick answer

Logic gates implement Boolean operations. AND ( $A \cdot B$ ) outputs 1 only if both inputs are 1; OR ( $A + B$ ) outputs 1 if either is 1; NOT ( $\overline{A}$ ) inverts; XOR ( $A \oplus B$ ) outputs 1 if the inputs differ; NAND and NOR are AND and OR with the output inverted. A truth table lists the output for every input combination ( $2^n$ rows for $n$ inputs). Boolean expressions are simplified with laws: identity ( $A \cdot 1 = A$ , $A + 0 = A$ ), complement ( $A + \overline{A} = 1$ , $A \cdot \overline{A} = 0$ ), idempotent ( $A \cdot A = A$ ), distributive ( $A(B+C) = AB + AC$ ), commutative and associative, and De Morgan's laws ( $\overline{A + B} = \overline{A} \cdot \overline{B}$ and $\overline{A \cdot B} = \overline{A} + \overline{B}$ ), which let you push a NOT through a bracket by swapping AND and OR. A Karnaugh map is a grid that groups adjacent 1s to read off a simplified expression.

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

OCR wants the six logic gates with their truth tables, the ability to build and read a truth table for an expression, and fluent simplification of Boolean expressions using the laws of Boolean algebra, De Morgan's laws and Karnaugh maps. This is calculation and manipulation: expect "simplify this expression" and "complete this truth table" questions.

The answer

The logic gates and truth tables

The laws of Boolean algebra

De Morgan's laws and Karnaugh maps

Key fact

De Morgan's laws handle a NOT over a whole bracket: $\overline{A + B} = \overline{A} \cdot \overline{B}$ and $\overline{A \cdot B} = \overline{A} + \overline{B}$ . In words, break the bar over the bracket and swap AND for OR (or OR for AND). They are essential for simplifying expressions and for building any circuit from only NAND or only NOR gates. A Karnaugh map (K-map) is a grid with one cell per input combination, arranged so adjacent cells differ in only one variable (Gray code order); you plot the 1s of the truth table, then group adjacent 1s in blocks of $1, 2, 4, 8$ (the largest possible, including wrap-around). Each group becomes one simplified term containing only the variables that stay constant across it, and the grouped terms are ORed together to give a minimal expression.

Simplify an expression with the laws

Simplify $A \cdot B + A \cdot \overline{B} + \overline{A} \cdot B$ using Boolean algebra, naming each step.

step 1: Factor the first two terms

$A \cdot B + A \cdot \overline{B}$ shares $A$ . By the distributive law this is $A \cdot (B + \overline{B})$ .

step 2: Apply the complement and identity laws

$B + \overline{B} = 1$ (complement), so $A \cdot 1 = A$ (identity). The expression is now $A + \overline{A} \cdot B$ .

step 3: Apply the absorption-style rule

$A + \overline{A} \cdot B = A + B$ . (Because where $A$ is 0, the term needs $B$ ; where $A$ is 1 the whole is already 1, so it equals $A + B$ .)

step 4: State the result

The expression simplifies to $A + B$ . A circuit of three AND/OR terms collapses to a single OR gate, which is the point of simplification.

Examples in context

Boolean simplification reduces the number of gates in a circuit, saving cost, power and propagation delay, which is why chip designers minimise expressions. De Morgan's laws let an entire processor be built from NAND gates alone (a universal gate). Karnaugh maps are the standard hand method for minimising a function of up to four variables. OCR links this directly to logic circuits such as adders, where these simplified expressions are realised, and to thinking logically in Component 02.

Try this

Q1. State the output of an XOR gate when both inputs are 1. [1 mark]

Cue. 0 (XOR outputs 1 only when the inputs differ).

Q2. Apply De Morgan's law to $\overline{A \cdot B}$ . [1 mark]

Cue. $\overline{A} + \overline{B}$ .

Q3. Simplify $A + A \cdot B$ and name the law. [2 marks]

Cue. $A$ , by the absorption law (since $A + A \cdot B = A \cdot (1 + B) = A \cdot 1 = A$ ).

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20195 marksSimplify the Boolean expression

A \cdot B + A \cdot \overline{B}

using the laws of Boolean algebra, naming the laws used, and verify the result with a truth table.

Show worked answer →

Simplification (up to 3): factor out $A$ using the distributive law: $A \cdot B + A \cdot \overline{B} = A \cdot (B + \overline{B})$ . By the complement law, $B + \overline{B} = 1$ . So $A \cdot 1 = A$ by the identity law. The simplified expression is $A$ .

Truth table verification (up to 2):

$A$	$B$	$A \cdot B$	$A \cdot \overline{B}$	sum
0	0	0	0	0
0	1	0	0	0
1	0	0	1	1
1	1	1	0	1

The sum column equals $A$ , confirming the simplification. Markers reward distribution, the complement law $B + \overline{B} = 1$ , the identity law, and a correct truth table.

OCR 20215 marksApply De Morgan's laws to simplify

\overline{A + B} \cdot \overline{A}

, and construct a truth table for the original expression.

Show worked answer →

Apply De Morgan (up to 3): $\overline{A + B} = \overline{A} \cdot \overline{B}$ . So the expression becomes $\overline{A} \cdot \overline{B} \cdot \overline{A} = \overline{A} \cdot \overline{B}$ (since $\overline{A} \cdot \overline{A} = \overline{A}$ by idempotence). The simplified result is $\overline{A} \cdot \overline{B}$ , equivalent to $\overline{A + B}$ .

Truth table (up to 2):

$A$	$B$	$A + B$	$\overline{A + B}$	$\overline{A}$	result
0	0	0	1	1	1
0	1	1	0	1	0
1	0	1	0	0	0
1	1	1	0	0	0

The result is 1 only when both are 0, matching $\overline{A} \cdot \overline{B}$ . Markers reward the correct De Morgan expansion, the idempotent simplification, and a correct truth table.

Related dot points

Sources & how we know this

OCR AS and A Level Computer Science (H046, H446) specification — OCR (2015)