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

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

WJEC Eduqas Eduqas 2020-style practice: Simplify the Boolean expression A · B + A · B using the laws of Boolean algebra, naming each law used, and verify your answer with a truth table.

Simplification (up to 3 marks): factor out A using the distributive law to get A · (B + B). By the complement law B + B = 1, so the expression becomes A · 1. By the identity law A · 1 = A. The simplified expression is A. Verification (up to 2 marks): a truth table for A · B + A · B gives output 0,0,1,1 for A,B = 00,01,10,11, which equals the column for A. Markers reward the named distributive, complement and identity laws, and a correct truth table showing the output matches A.

WJEC Eduqas Eduqas 2022-style practice: State De Morgan's two laws, and apply the appropriate one to simplify A + B · A.

De Morgan's laws (up to 2 marks): A + B = A · B, and A · B = A + B. In words, break the bar over the bracket and swap the operator. Application (up to 2 marks): A + B · A = (A · B) · A = A · A · B. Since A · A = 0 (complement law), the whole expression is 0. Markers reward the correct statement of both laws and the simplification to 0 using De Morgan's plus the complement law.

EnglandComputer ScienceSyllabus dot point

How are logic gates, truth tables and Boolean algebra used to design and simplify digital logic?

Logical operations: the logic gates (AND, OR, NOT, NAND, NOR, XOR) and their truth tables, building and reading truth tables for expressions, the laws of Boolean algebra and De Morgan's laws, and simplifying expressions with Karnaugh maps.

An Eduqas Component 1 answer on logical operations: the six logic gates and their truth tables, building truth tables for expressions, simplifying with the laws of Boolean algebra and De Morgan's laws, and using Karnaugh maps to minimise a Boolean expression.

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

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

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Quick answer

A logic gate implements a Boolean operation on $0$ / $1$ inputs. AND ( $A \cdot B$ ) outputs $1$ only when both inputs are $1$ ; OR ( $A + B$ ) outputs $1$ when either is $1$ ; NOT ( $\overline{A}$ ) inverts; NAND and NOR are AND and OR followed by NOT; XOR ( $A \oplus B$ ) outputs $1$ when the inputs differ. A truth table lists the output for every combination of inputs and is how you verify or define an expression. The laws of Boolean algebra (identity, complement, idempotent, absorption, distributive) let you simplify expressions algebraically, and De Morgan's laws ( $\overline{A + B} = \overline{A} \cdot \overline{B}$ and $\overline{A \cdot B} = \overline{A} + \overline{B}$ ) handle a NOT over a whole bracket. A Karnaugh map is a grid that arranges the truth table so adjacent $1$ s can be grouped to read off a minimal expression. Minimising logic means fewer gates, which is cheaper and faster in hardware.

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What this dot point is asking

Eduqas wants you to know the logic gates and their truth tables, build and read truth tables for expressions, simplify Boolean expressions with the named laws and De Morgan's laws, and minimise expressions with Karnaugh maps. This is a calculation-style topic assessed without a calculator, and shown working matters.

The answer

The gates and their truth tables

Building truth tables and the laws of Boolean algebra

De Morgan's laws and Karnaugh maps

Simplify with the laws and check with a Karnaugh map

Simplify $A \cdot B + A \cdot \overline{B} + \overline{A} \cdot B$ .

step 1: Group the first two terms

$A \cdot B + A \cdot \overline{B} = A \cdot (B + \overline{B})$ by the distributive law, and $B + \overline{B} = 1$ by complement, so this is $A$ .

step 2: Rewrite the expression

The expression is now $A + \overline{A} \cdot B$ .

step 3: Apply the absorption-style rule

$A + \overline{A} \cdot B = A + B$ (a standard identity: $A + \overline{A} \cdot B = A + B$ ). The simplified expression is $A + B$ .

step 4: Confirm with a Karnaugh map

Filling a 2-variable K-map, the cells for $AB = 01, 10, 11$ are $1$ and only $00$ is $0$ ; the largest groups cover the $A$ row and the $B$ column, reading off $A + B$ , which matches.

Examples in context

Logic gates are the physical building blocks of every processor: the ALU is made of gate networks, and the adder (covered under architecture) is built from XOR and AND gates. Boolean simplification matters because fewer gates mean a cheaper, faster, lower-power circuit, the same expression can often be built with far fewer components after minimisation. De Morgan's laws let designers build any function from a single gate type (all NAND or all NOR), which simplifies manufacturing. This dot point connects to number representation and to the CPU architecture topics in Component 2.

Try this

Q1. Give the truth-table output of an XOR gate for inputs $A,B = 11$ . [1 mark]

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

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

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

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

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

Exam-style practice questions

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

Eduqas 20205 marksSimplify the Boolean expression

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

using the laws of Boolean algebra, naming each law used, and verify your answer with a truth table.

Show worked answer →

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

Verification (up to 2 marks): a truth table for $A \cdot B + A \cdot \overline{B}$ gives output $0,0,1,1$ for $A,B = 00,01,10,11$ , which equals the column for $A$ .

Markers reward the named distributive, complement and identity laws, and a correct truth table showing the output matches $A$ .

Eduqas 20224 marksState De Morgan's two laws, and apply the appropriate one to simplify

\overline{A + B} \cdot A

Show worked answer →

De Morgan's laws (up to 2 marks): $\overline{A + B} = \overline{A} \cdot \overline{B}$ , and $\overline{A \cdot B} = \overline{A} + \overline{B}$ . In words, break the bar over the bracket and swap the operator.

Application (up to 2 marks): $\overline{A + B} \cdot A = (\overline{A} \cdot \overline{B}) \cdot A = \overline{A} \cdot A \cdot \overline{B}$ . Since $\overline{A} \cdot A = 0$ (complement law), the whole expression is $0$ .

Markers reward the correct statement of both laws and the simplification to $0$ using De Morgan's plus the complement law.

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Sources & how we know this

WJEC Eduqas GCE AS/A Level Computer Science specification (from 2015) — Eduqas (2015)