Boolean Algebra and De Morgan's Laws | OCR H446

Learning Objectives

By the end of this topic you will be able to:

Apply Boolean algebra laws to simplify expressions

State and apply De Morgan's Laws to transform expressions

Simplify Boolean expressions algebraically showing each step

Verify simplification using truth tables

Boolean Laws

Key Boolean Algebra Laws

Identity Laws

A + 0 = A

A · 1 = A

A + 1 = 1

A · 0 = 0

Idempotent & Complement

A + A = A

A · A = A

A + ¬A = 1

A · ¬A = 0

Commutative Laws

A + B = B + A
A · B = B · A

Distributive Laws

A·(B+C) = A·B + A·C
A+(B·C) = (A+B)·(A+C)

Absorption & Double Negation

More Laws

Absorption Laws

A + (A · B) = A

A · (A + B) = A

Double Negation & Associative

¬(¬A) = A

A+(B+C) = (A+B)+C

A·(B·C) = (A·B)·C

These laws allow Boolean expressions to be simplified — a simpler expression requires fewer logic gates, reducing the cost, size and power consumption of a circuit.

De Morgan's Laws

Law 1: ¬(A · B) = ¬A + ¬B

Law 2: ¬(A + B) = ¬A · ¬B

Rule of thumb: break the bar (NOT), change the operator (AND↔OR). Apply to each variable. Double negation cancels (¬¬A = A).

Application: NAND → OR

¬(A·B) = ¬A + ¬B
NAND gate = NOT-AND = NOT A OR NOT B

Application: NOR → AND

¬(A+B) = ¬A · ¬B
NOR gate = NOT-OR = NOT A AND NOT B

Simplification Example

Simplifying a Boolean Expression

Simplify: Q = A·B + A·¬B + ¬A·B

= A·(B + ¬B) + ¬A·B           [Distributive]

= A·1 + ¬A·B                   [A + ¬A = 1]

= A + ¬A·B                      [A·1 = A]

= (A + ¬A)·(A + B)            [Distributive]

= 1·(A + B) = A + B          [A + ¬A = 1]

From 3 AND gates + 2 OR gates → just 1 OR gate. This is the power of Boolean simplification in reducing circuit complexity.

Exam Practice

OCR H446 Style · 4 marks

Apply De Morgan's Laws to simplify the expression Q = ¬(¬A · ¬B). Show your working and state which law you used at each step.

[4 marks]

Apply De Morgan's Law 1: ¬(¬A · ¬B) = ¬(¬A) + ¬(¬B)

Apply Double Negation: ¬(¬A) = A and ¬(¬B) = B

Result: Q = A + B

This shows that NOT(NOT A AND NOT B) is equivalent to A OR B — this is the principle behind implementing an OR gate using NOR gates.

Common Mistakes

Don't Lose Marks

Applying De Morgan's Law without changing the operator — the law requires BOTH steps: break the NOT over the whole expression AND change AND to OR (or vice versa). ¬(A·B) ≠ ¬A · ¬B. You must also change · to +.

Applying De Morgan's Law to individual terms instead of the whole expression — the NOT bar must cover the entire expression being transformed. ¬A·¬B is already simplified; De Morgan applies to ¬(A·B). Don't apply it incorrectly to subexpressions.

Not showing step-by-step working in simplification — OCR mark schemes award marks for each correct step (identifying the law used, applying it correctly). Writing only the final answer with no working loses intermediate marks even if the final answer is correct.

Boolean Algebra& De Morgan's Laws