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OCR H446 1.4.3 Boolean laws and simplification
Part 4 of 11 · H446 1.4.3 · Boolean algebra
Simplification earns marks for the stages shown, not only the final line, and OCR H446 1.4.3 names rule families that students must apply by name. This worksheet works through commutation, association, distribution, absorption and De Morgan's laws with every intermediate transformation visible.
Students will:
- name the law that justifies a given transformation
- apply De Morgan's laws so the operator switches as well as the terms
- distribute a repeated term without losing the operator that joins the brackets
- simplify an expression stage by stage, naming the law used at each step
- explain the practical hardware benefits of using fewer gates
Inside: 6 explanation cells, 2 multiple-choice questions, 2 fill-in-the-blanks cells and 2 written answers. 18 marks, about 45 to 55 minutes.
Series: H446 1.4.3 · Boolean algebra, part 4 of 11.
Shared by Coding PathwayVerified teacher
- 12 cells
- About 45 minutes
- CC BY-SA 4.0
- Shared 31 Aug 2026
- Updated 3 Sept 2026
Preview
The whole resource, exactly as a class sees it. Answers and marking are held back.
Boolean laws and simplification
A simpler Boolean expression can use fewer gates, reducing circuit size, cost, power use and propagation delay. OCR names five rule families that you must apply accurately.
You should already know: OCR operator notation and brackets.
The required laws
| Law | Pattern | What it permits |
|---|---|---|
| Commutation | A ∧ B ≡ B ∧ A; A ∨ B ≡ B ∨ A | reorder terms |
| Association | (A ∧ B) ∧ C ≡ A ∧ (B ∧ C) | regroup repeated operators |
| Distribution | (A ∨ B) ∧ (A ∨ C) ≡ A ∨ (B ∧ C) | factor or expand |
| Double negation | ¬(¬A) ≡ A | remove two negations |
| De Morgan | ¬(A ∧ B) ≡ ¬A ∨ ¬B; ¬(A ∨ B) ≡ ¬A ∧ ¬B | move a negation through brackets and switch the operator |
Association changes brackets but not operator order. Commutation changes order but not the operators. De Morgan changes both inputs and the operator.
Worked model: distribution without losing the operator
Simplify (A ∨ B) ∧ (A ∨ C).
The repeated term is A. Keep one A outside. The remaining terms B and C are joined by the operator that sits between the two brackets, AND:
(A ∨ B) ∧ (A ∨ C) ≡ A ∨ (B ∧ C)
B ∨ C is not a valid distribution result here. Write the law pattern before substituting the letters.
Why is ¬(A ∨ B) not equivalent to ¬A ∨ ¬B?
- ADe Morgan requires the operator to switch from OR to AND
- BNOT cannot be used with brackets
- CA and B must be reordered
- DDouble negation must be applied first
Which law justifies changing (P ∨ Q) ∨ R into P ∨ (Q ∨ R)?
- ACommutation
- BAssociation
- CDistribution
- DDouble negation
Simplify ¬(¬A ∨ B) using named Boolean laws. Show each stage and name the law used.
Apply De Morgan to the outer negation, then double negation.
Students type their answer here.
Explain two practical benefits of simplifying a Boolean expression used to build a hardware circuit.
Link each benefit to using fewer gates or connections.
Students type their answer here.
Examination lens
When a question names a law, use that law rather than jumping to an unexplained answer. OCR can award a second mark for the correct intermediate transformation. Keep ∧ and ∨ visually distinct.
Closed-book checkpoint
These statements retrieve the decision process, not the earlier worked expressions.
Review your responses
Check every response against its command word and the supplied constraints. Strengthen unsupported answers with accurate method, mechanism, state or contextual consequence before submitting.