AQA A-Level Computer Science Paper 2, June 2025: Question 13

4 marks · Medium difficulty · Calculation

Simplify the Boolean expression overline(A.B + NOT(A).C + B.C + NOT(B).A) using the rules of Boolean algebra, showing all working.

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Question

Question text

13 Using the rules of Boolean algebra, simplify the following Boolean expression.

�A��⋅��B��+��A��⋅�C��+��B��⋅��C��+��B��⋅��A�

You must show your working.

[4 marks]

Working

Answer

Mark scheme

Show the mark scheme

Total

Qu Pt Marking guidance

marks

13 Marks are AO2 (apply) 4

Marking guidance for examiners

• Award marks for working out until an incorrect step has been made.

• Ignore missing steps from the example solutions, as long as the jumps between

steps are logically correct.

• If, in any one step, a candidate is simplifying different parts of an expression

simultaneously and one part is simplified incorrectly, award all relevant marks

for this multiple stage but don’t award any further marks for subsequent stages.

For example, if the expression P.P.(P+Q) + P.P.1 was changed to P.(P+Q) +

P.0, the candidate would get one mark for correctly simplifying the first part to

P.(P+Q) even though the simplification of P.P.1 to P.0 is incorrect. However, no

further marks could be awarded for later simplification steps.

1 mark for final answer: ̅A̅̅ +̅̅̅ C̅ R. A̅ ⋅ C̅

Max 3 for working. Award up to three marks for applying any of these techniques

(can award multiple marks for applying the same technique multiple times):

• a successful application of De Morgan’s laws (and any associated cancellation

of NOTs)

• applying an identity other than cancelling NOTs that produces a simpler

expression (Note: A simpler expression is one that is logically equivalent to the

original expression but uses fewer logical operators)

• successfully expanding brackets // factorising.

Max 3 overall if any incorrect working.

Example Solution 1

̅A̅̅ ⋅̅ ̅B̅̅ +̅̅̅ A̅̅̅ ̅⋅̅ C̅̅ ̅+̅̅ ̅B̅̅ ⋅̅ ̅C̅̅+̅̅̅B̅̅̅ ⋅̅̅A̅

̅A̅̅ ⋅̅ ̅(̅B̅̅ ̅+̅̅ ̅B̅̅)̅̅ +̅̅̅ A̅̅̅̅ ⋅̅ ̅C̅̅ +̅̅̅B̅̅ ̅⋅̅ C̅ Factorising

̅A̅̅ ⋅̅̅1̅̅ +̅̅̅ A̅̅̅̅⋅̅ C̅̅̅+̅̅̅B̅̅ ̅⋅̅ C̅ By 𝑋 + 𝑋̅ = 1

̅A̅̅ +̅̅̅ A̅̅̅̅⋅̅ C̅̅̅+̅̅̅B̅̅ ̅⋅̅ C̅ By 𝑋 ∙ 1 = 𝑋

̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅̅ ̅ ̅̅̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ Application of De Morgan

A + A + C + B ⋅ C

̅̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅̅ ̅̅ ̅̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ Application of De Morgan

A ⋅ (A + C) + B ⋅ C

̅̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅̅ ̅̅ ̅̅ ̅̅ ̅̅̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ Expanding brackets

A ⋅ A + A ⋅ C + B ⋅ C

̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ By 𝑋 ∙ 𝑋̅ = 0

0 + A ⋅ C + B ⋅ C

̅̅̅ ̅̅ ̅̅ ̅̅ ̅̅̅̅̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ ̅ By 𝑋 + 0 = 𝑋

A ⋅ C + B ⋅ C

̅A̅̅ +̅̅̅ C̅̅̅+̅̅̅B̅̅ ⋅̅̅ C̅ Application of De Morgan

̅A̅̅ +̅̅̅ C̅ By 𝑋 + 𝑋 ∙ 𝑌 = 𝑋

– A-LEVEL COMPUTER SCIENCE – –

Example Solution 2 33

̅A̅̅ ⋅̅ ̅B̅̅ +̅̅̅ A̅̅̅ ̅⋅̅ C̅̅ ̅+̅̅ ̅B̅̅ ⋅̅ ̅C̅̅+̅̅̅B̅̅̅ ⋅̅̅A̅

̅A̅̅ ⋅̅ ̅(̅B̅̅ ̅+̅̅ ̅B̅̅)̅̅ +̅̅̅ A̅̅̅̅ ⋅̅ ̅C̅̅ +̅̅̅B̅̅ ̅⋅̅ C̅ Factorising

̅A̅̅ ⋅̅̅1̅̅ +̅̅̅ A̅̅̅̅⋅̅ C̅̅̅+̅̅̅B̅̅ ̅⋅̅ C̅ By 𝑋 + 𝑋̅ = 1

̅A̅̅ +̅̅̅ A̅̅̅̅⋅̅ C̅̅̅+̅̅̅B̅̅ ̅⋅̅ C̅ By 𝑋 ∙ 1 = 𝑋

̅A̅̅ +̅̅̅ C̅̅ ̅+̅̅̅B̅̅ ⋅̅̅ C̅ By 𝑋 + (𝑋̅ ∙ 𝑌) = 𝑋 + 𝑌

̅A̅̅ +̅̅̅ C̅ By 𝑋 + 𝑋 ∙ 𝑌 = 𝑋

Example Solution 3

̅A̅̅ ⋅̅ ̅B̅̅ +̅̅̅ A̅̅̅ ̅⋅̅ C̅̅ ̅+̅̅ ̅B̅̅ ⋅̅ ̅C̅̅+̅̅̅B̅̅̅ ⋅̅̅A̅

̅A̅̅ ⋅̅ ̅(̅B̅̅ ̅+̅̅ ̅B̅̅)̅̅ +̅̅̅ A̅̅̅̅ ⋅̅ ̅C̅̅ +̅̅̅B̅̅ ̅⋅̅ C̅ Factorising

̅A̅̅ ⋅̅̅1̅̅ +̅̅̅ A̅̅̅̅⋅̅ C̅̅̅+̅̅̅B̅̅ ̅⋅̅ C̅ By 𝑋 + 𝑋̅ = 1

̅A̅̅ +̅̅̅ A̅̅̅̅⋅̅ C̅̅̅+̅̅̅B̅̅ ̅⋅̅ C̅ By 𝑋 ∙ 1 = 𝑋

̅A̅̅ +̅̅̅ C̅̅ ⋅̅̅(̅A̅̅̅̅+̅̅ ̅B̅̅) Factorising

A̅ ⋅ ̅C̅̅⋅̅ ̅(̅A̅̅̅ ̅+̅̅ ̅B̅̅) Application of De Morgan

A̅ ⋅ (C̅ + ̅A̅̅̅ +̅̅̅B̅) Application of De Morgan

A̅ ⋅ (C̅ + A ⋅ B̅) Application of De Morgan

A̅ ⋅ C̅ + A̅ ⋅ A ⋅ B̅ Expanding brackets

A̅ ⋅ C̅ + 0 ⋅ B̅ By 𝑋 ∙ 𝑋̅ = 0

A̅ ⋅ C̅ + 0 By 𝑋 ∙ 0 = 0

A̅ ⋅ C̅ By 𝑋 + 0 = 𝑋

̅A̅̅ +̅̅̅ C̅ Application of De Morgan

Partial Example Solution

This is the start of a solution when De Morgan’s is applied first to each of the OR

operators. Many further steps are required to arrive at a fully correct solution but

this partial solution is worth 3 marks.

̅A̅̅ ⋅̅ ̅B̅̅ +̅̅̅ A̅̅̅ ̅⋅̅ C̅̅ ̅+̅̅ ̅B̅̅ ⋅̅ ̅C̅̅+̅̅̅B̅̅̅ ⋅̅̅A̅

̅A̅̅ ⋅̅̅B̅ ⋅ ̅A̅̅̅ ⋅̅̅ C̅ ⋅ ̅B̅̅ ⋅̅̅ C̅ ⋅ ̅B̅̅̅ ⋅̅̅A̅ Application of De Morgan

(A̅ + B̅) ⋅ (A + C̅) ⋅ (B̅ + C̅) ⋅ (B + A̅) 4 x Application of De Morgan

Incorrect Example

This is an incorrect application of De Morgan’s or it has been correctly applied

multiple times and the order of precedence has been lost. No marks can be

awarded if this is the first step that a student has shown.

A̅ + B̅ ⋅ A + C̅ ⋅ B̅ + C̅ ⋅ B + A̅

How to answer it

AQA A-Level Computer Science • Paper 2 • Boolean Algebra

Simplification of Compound Boolean Expressions

What this question tests

This question assesses your ability to manipulate and simplify complex Boolean logic expressions using standard algebraic laws and identities (AO2 - Application). Key skills include:

Applying factorisation to isolate common terms like A · (B + B ).
Using inverse identities (X + X = 1) and identity laws (X · 1 = X).
Recognising simplification shortcuts such as absorption rules (X + X · Y = X) and redundancy laws (X + X · Y = X + Y).
Correctly managing overarching negation bars and applying De Morgan's Laws: X + Y = X · Y .

Question 13 (4 Marks)

Simplify the expression: A · B + A · C + B · C + B · A

📐 Step-by-Step Simplification (Recommended Method)

The cleanest approach simplifies inside the main overline first, leaving the overarching NOT until the very end.

Step	Expression	Law / Rule Applied	Reasoning
Start	A · B + A · C + B · C + B · A	Given Expression	Note the single large NOT bar over the entire expression.
Step 1	A · (B + B ) + A · C + B · C	Factorisation	Group terms containing A : A · B + B · A = A · (B + B ) .
Step 2	A · 1 + A · C + B · C	Inverse Law: X + X = 1	Any variable ORed with its complement evaluates to True (1).
Step 3	A + A · C + B · C	Identity Law: X · 1 = X	ANDing any variable with 1 leaves it unchanged.
Step 4	A + C + B · C	Redundancy Law: X + X · Y = X + Y	Alternatively proved via De Morgan's / distribution: A + A · C = (A + A )(A + C) = 1 · (A + C) .
Step 5	A + C	Absorption Law: C + B · C = C	Because C + B · C = C · (1 + B) = C · 1 = C .
Step 6	A + C or A · C	De Morgan's Law	Both forms are fully simplified and accepted.

✅ Acceptable Final Answers (1 Mark)

A + C
A · C

Mark Scheme Rule: Exactly 1 mark is awarded for reaching one of these two final simplified expressions, provided prior working contains no invalid logic jumps.

🧠 How the Working Marks are Awarded (Max 3)

You can earn up to 3 working marks for correctly applying any valid Boolean rules (even multiple times for the same rule):

Successful factorisation or expansion of brackets.
Applying an identity (other than cancelling double NOTs) that creates a strictly simpler expression.
Correct application of De Morgan's laws (and cancelling resulting NOTs).

Note: Max 3/4 if any incorrect working appears before reaching the final answer.

💡 Key Identities Used in This Question

Inverse: B + B = 1
Identity: A · 1 = A and A · 0 = 0
Absorption: C + C · B = C
Redundancy: A + A · C = A + C
De Morgan's: X + Y = X · Y and X · Y = X + Y

❌ Common Traps & Examiner Warnings

Splitting the overarching bar incorrectly:
Writing A + B · A + C ... directly without parentheses destroys operator precedence. The examiner notes this scores 0 marks immediately.
Over-complicating De Morgan's first:
Splitting the outer bar across all terms right at Step 1 gives A · B · A · C · B · C · B · A . While technically valid, it requires many more steps and creates high risk of algebraic slips.
Missing the factorisation:
Not spotting that A · B and B · A both share the factor A is the most common reason students get stuck.

Topics

4.6 Fundamentals of computer systems · 4.6.5 Boolean algebra

Question and mark scheme from the AQA A-Level Computer Science examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.