All subjects · Computer Science · WCP03 Advanced Principles of Computer Science

Boolean algebra: laws and simplification

Edexcel IAL Computer Science · WCP03 Advanced Principles of Computer Science

Boolean algebra simplifies logic expressions so circuits need fewer gates and conditions in code are easier to read. You need six laws and must be able to show that two expressions are equivalent.

Practise this in the app: instant marking and a fix for every mistake. Free, no email needed to start.

Practise this lesson

Notation

The specification asks you to use the symbols in its Appendix 2. The Unit 3 sample paper writes A.B for A AND B, A + B for A OR B, and NOT A with a bar over A; check the appendix in your copy of the specification. On this site the bar is shown as ¬A because overbars do not display reliably; in the exam, draw the bar. AND binds more tightly than OR, so A + B.C means A + (B.C).

Useful identities (the sample mark scheme names some of these laws in its steps):

The six laws

Each law has an AND form and an OR form. Name the law on every line of a simplification.

LawAND formOR form
CommutativeA.B = B.AA + B = B + A
Associative(A.B).C = A.(B.C)(A + B) + C = A + (B + C)
DistributiveA.(B + C) = A.B + A.CA + B.C = (A + B).(A + C)
AbsorptionA.(A + B) = AA + A.B = A
De Morgan's¬(A.B) = ¬A + ¬B¬(A + B) = ¬A.¬B
Double negation¬¬A = A

De Morgan's: break the bar and change the sign

Simplifying step by step

Write one law per line so the examiner can award each step.

Example 1: X.Y + Y.¬X

= Y.X + Y.¬X (commutative)

= Y.(X + ¬X) (distributive)

= Y.1 (X + ¬X = 1)

= Y

Example 2: ¬(¬A + B)

= ¬¬A.¬B (De Morgan's)

= A.¬B (double negation)

Example 3: (A + B).(A + ¬B)

= A + B.¬B (distributive, OR form)

= A + 0 = A

Checking with a truth table

Two expressions are equivalent if their truth tables match on every row. If you are unsure of an algebraic step, check it with a small truth table. This is also how every answer in this lesson was checked.

Worked example

Simplify A.B + A.¬B + ¬A.B.

  1. A.B + A.¬B = A.(B + ¬B) = A.1 = A (distributive, complement)
  2. So the expression is A + ¬A.B.
  3. A + ¬A.B = (A + ¬A).(A + B) (distributive, OR form) = 1.(A + B) = A + B

Answer: A + B

Key idea: Laws: commutative, associative, distributive, absorption (A + A.B = A), De Morgan's (¬(A.B) = ¬A + ¬B; ¬(A + B) = ¬A.¬B), double negation. Plus A + ¬A = 1, A.¬A = 0.

Check yourself

  1. Which law shows that A + A.B = A?
    • De Morgan's
    • Commutative
    • Double negation
    • Absorption
    Show answer

    Absorption. A + A.B = A.(1 + B) = A.

  2. Which law gives A + 1 = 1 and A.0 = 0?
    • Identity
    • Annulment
    • Absorption
    • Double negation
    Show answer

    Annulment. Annulment: OR with 1 always gives 1; AND with 0 always gives 0. Identity is A + 0 = A and A.1 = A.

  3. Which expression is equivalent to ¬(A.B)?
    • ¬A.¬B
    • A + B
    • ¬A.B
    • ¬A + ¬B
    Show answer

    ¬A + ¬B. De Morgan's: break the bar, change AND to OR.

  4. Which expression is equivalent to ¬(A + ¬B)?
    • ¬A.B
    • ¬A + B
    • A.¬B
    • ¬A.¬B
    Show answer

    ¬A.B. ¬(A + ¬B) = ¬A.¬¬B = ¬A.B.

  5. Simplify A.(A + B).
    • B
    • A.B
    • A
    • A + B
    Show answer

    A. Absorption (AND form): A.(A + B) = A.

  6. Simplify (A + B).(A + C).
    • A + B.C
    • A.B + C
    • A + B + C
    • A.B.C
    Show answer

    A + B.C. Distributive (OR form): A + B.C = (A + B).(A + C).

  7. Simplify ¬(¬A.¬B).
    • A.B
    • ¬A + ¬B
    • A + B
    • ¬(A + B)
    Show answer

    A + B. ¬¬A + ¬¬B = A + B.

  8. Simplify A.B + ¬A.B + A.¬B.
    • A.B
    • B
    • 1
    • A + B
    Show answer

    A + B. A.B + ¬A.B = B, so B + A.¬B = (B + A).(B + ¬B) = A + B.

Practise with instant feedback

Sources

Last updated 8 November 2026. Spotted a mistake? Tell us.