Boolean Logic · 5 question types
Past paper frequency (2018 to 2024)
This topic accounts for approximately 6% of your exam marks.
Writing Boolean expressions from logic diagrams and simplifying using laws appear regularly.
A typical exam question gives you a Boolean expression and asks you to simplify it, stating the law used at each step. The marks usually go on showing the working rather than just the final answer.
General strategy. (1) Identify any constants (
0or1) inside the expression and clear them with the identity or annihilator laws. (2) Look for complement pairs (A · ¬AorA + ¬A) and replace them. (3) Apply absorption if a term is fully contained in another. (4) Use De Morgan's to push NOTs onto individual variables when that helps. (5) Repeat until no law applies.
Example — simplify A · 1 + A · ¬A + B. State the law used at each step.
| Step | Working | Law |
|---|---|---|
| Start | A · 1 + A · ¬A + B | — |
Replace A · 1 with A | A + A · ¬A + B | Identity |
Replace A · ¬A with 0 | A + 0 + B | Complement |
Replace A + 0 with A | A + B | Identity |
Final answer: A · 1 + A · ¬A + B = A + B.
Example — simplify A + (A · B). State the law used.
| Step | Working | Law |
|---|---|---|
| Start | A + (A · B) | — |
| Apply absorption directly | A | Absorption |
Final answer: A + A · B = A.
Example — simplify ¬(A · B) + B. State the laws used.
| Step | Working | Law |
|---|---|---|
| Start | ¬(A · B) + B | — |
Apply De Morgan's to ¬(A · B) | (¬A + ¬B) + B | De Morgan |
| Drop the outer brackets (associative) | ¬A + ¬B + B | Associative |
Replace ¬B + B with 1 | ¬A + 1 | Complement |
Replace ¬A + 1 with 1 | 1 | Annihilator |
Final answer: ¬(A · B) + B = 1 (always true).
A small tip: when an exam question asks you to "show that two expressions are equivalent", a clean approach is to simplify each one separately and show they reduce to the same result. If you cannot find a simplification, you can also build truth tables for both expressions and show the columns agree on every row.