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Free A-Level Computer Science guide
The aim of simplification is to preserve every output while reducing the expression or circuit. Check a simplified result against the original truth conditions.
Content owner: Michael Print · Written for A-Level learners · Checked against official specifications
Boolean expressions combine true/false values. AND needs both inputs true, OR needs at least one true, and NOT reverses its input. Use parentheses to specify the intended grouping; arithmetic intuition alone is insufficient.
A Karnaugh map rearranges truth-table rows so adjacent cells differ in one input bit. Group one-valued cells in rectangular powers of two, including valid wrap-around groups, then retain only the variables constant within each group. OR the resulting terms.
OCR H446 · 1.4.3(a–d). Textual gate connections and tables accompany original two-, three- and four-variable reasoning.
Useful foundations
By the end, you should be able to
OCR Appendix 5d uses ∧ for AND, ∨ for inclusive OR and ¬ for NOT. An underlined OR means exclusive OR; XOR and ⊕ are accepted alternatives.
Equivalence uses ≡, with ↔ accepted. Equivalent expressions match on every supported input, not just one example. A dot can mean AND, plus OR and an overbar NOT.
¬(A ∨ B) ≡ (¬A ∧ ¬B) means NOT(A OR B) is equivalent to NOT A AND NOT B. Both sides are true only when A and B are false. Boolean plus means OR, rather than arithmetic addition.
An alarm sounds when armed and either a door is open or motion is detected. Define A=armed, O=door open and M=motion detected. The expression is F = A AND (O OR M).
Define what each true input means. Reversing open/closed meaning changes the condition. Four sensor combinations for each armed state give eight truth-table rows. OR is inclusive: both sensors can trigger together.
Combine sensors
Feed O and M into an OR gate. Its output is true when either or both sensors trigger.
Require arming
Feed the OR output and A into an AND gate to produce F.
Check the safeguard
If A=0, F=0 regardless of either sensor.
| A | O | M | F |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Commutation swaps operands: A AND B = B AND A, and likewise OR. Association regroups one operator: (A OR B) OR C = A OR (B OR C). It cannot arbitrarily regroup mixed operators. Double negation gives NOT NOT A = A.
Distribution: A AND (B OR C) = (A AND B) OR (A AND C). Its other form is A OR (B AND C) = (A OR B) AND (A OR C).
De Morgan: NOT(A AND B) = (NOT A) OR (NOT B), and NOT(A OR B) = (NOT A) AND (NOT B). Complement every operand and swap the joining operator.
Start
(A AND B) OR (A AND NOT B).
Distribution
Factor A: A AND (B OR NOT B).
Complement
B OR NOT B is true, giving A AND true.
Identity
The result is A. Substitute both values of A to check the behaviour for either B.
Two variables need four cells, three need eight and four need sixteen. A two-bit axis uses Gray order 00,01,11,10; ordinary binary order would create some neighbours differing in two bits.
Use rows A=0,1 and columns BC=00,01,11,10. For one-valued inputs 001,011,101,111, both rows contain ones in columns 01 and 11.
That four-cell group keeps C=1 while A and B vary, giving F=C. The map rearranges the same eight truth-table rows; it does not create new truth conditions.
Cover every one with large valid groups. Each group is a rectangle of 1,2,4,8 or 16 one-valued cells. Diagonal contact is not adjacency, and absorbing a zero would change a required false output.
Gray-ordered opposite edges are adjacent. Groups can wrap left/right, top/bottom or both. Overlap is allowed when it helps make larger groups.
Remove variables that change inside a group. Constant zero gives a complemented literal; constant one gives an uncomplemented literal. Larger groups usually need fewer literals. These examples have no don’t-care cells; those require an explicit statement.
Worked example
(NOT B) AND (NOT D).A AND C. Row 10, column 10 belongs to both groups; overlap is valid.F = ((NOT B) AND (NOT D)) OR (A AND C).| AB / CD | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | 1 | 0 | 0 | 1 |
| 01 | 0 | 0 | 0 | 0 |
| 11 | 0 | 0 | 1 | 1 |
| 10 | 1 | 0 | 1 | 1 |
Worked example
NOT NOT ((P AND Q) AND R). Double negation removes the paired NOT operations, giving (P AND Q) AND R.P AND (Q AND R). Commutation then swaps the two outer operands to (Q AND R) AND P.Worked example
(NOT A AND B) OR (A AND NOT B). There is no valid two-cell group using only the ones. The XOR gate can express this behaviour compactly, but diagonal grouping would incorrectly remove necessary conditions.7 original questions total 21 marks. Attempt each before opening the independently written indicative marking guidance.
Simplify (P AND Q) OR (P AND NOT Q), naming the useful laws. [3 marks]
Indicative answer
P AND (Q OR NOT Q) using distribution (1). Q OR NOT Q is true by complement (1); P AND true is P by identity (1).Apply De Morgan's laws to NOT(A OR (B AND C)). [3 marks]
Indicative answer
NOT A AND NOT(B AND C) (1). The inner negation becomes NOT B OR NOT C (1). Final expression: NOT A AND (NOT B OR NOT C) with those parentheses (1).On the four-variable map below, state the term for the four corners and explain which variables disappear. [3 marks]
Rows are labelled AB and columns CD. Both axes use Gray order 00,01,11,10; opposite edges are logically adjacent. There are no don't-care cells.
| AB / CD | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | 1 | 0 | 0 | 1 |
| 01 | 0 | 0 | 0 | 0 |
| 11 | 0 | 0 | 1 | 1 |
| 10 | 1 | 0 | 1 | 1 |
Indicative answer
NOT B AND NOT D (1). A changes between the selected rows (1), and C changes between the selected columns (1), so neither remains in the term.A learner groups three adjacent ones and one zero into a four-cell rectangle. Explain the error and give two other grouping rules. [3 marks]
Indicative answer
Model a fan that runs when hot and either manual override or automatic mode is enabled. Define three variables and give an expression. [3 marks]
Indicative answer
F=H AND (M OR A) (1 for the inner inclusive OR, 1 for requiring H with the whole grouped result).Apply double negation to NOT NOT ((X OR Y) OR Z), then association, then commutation of the two outer operands. Give the expression after each step. [3 marks]
Indicative answer
(X OR Y) OR Z (1). Association: X OR (Y OR Z) (1). Commutation: (Y OR Z) OR X (1). Equivalent correctly justified same-operator rearrangements are acceptable.Translate ¬(P ∧ Q) ≡ (¬P ∨ ¬Q) into words, and state what ≡ asserts. [3 marks]
Indicative answer
This guide addresses OCR H446 1.4.3(a–d). Textual gate connections and tables accompany original two-, three- and four-variable reasoning.. Check your examination year and the complete specification for the assessment scope.
These are independently written explanations and practice questions. CompSciTutoring.co.uk is not affiliated with or endorsed by an examination board. The marking guidance is indicative; always check the syllabus for your examination year.
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