Logic Laws
Boolean algebra provides laws (identities) that allow us to simplify logical expressions — reducing the number of gates in a circuit. Simplification means fewer components, lower cost, less power, and faster processing. This topic is all about applying rules to transform complex expressions into simpler equivalent ones.
Learning Objectives
- 12.3.3.3 Simplify logical expressions using the laws of logic
Notation Guide
| Operation | Symbols Used |
|---|---|
| AND | A · B or A ∧ B or AB |
| OR | A + B or A ∨ B |
| NOT | A̅ or ¬A or NOT A |
In this lesson, we use: · for AND, + for OR, and A̅ (overbar) for NOT.
Boolean Laws Reference Table
| Law | AND Form | OR Form |
|---|---|---|
| Identity | A · 1 = A | A + 0 = A |
| Null (Annulment) | A · 0 = 0 | A + 1 = 1 |
| Idempotent | A · A = A | A + A = A |
| Complement | A · A̅ = 0 | A + A̅ = 1 |
| Double Negation | NOT(NOT A) = A (A̿ = A) | |
| Commutative | A · B = B · A | A + B = B + A |
| Associative | (A · B) · C = A · (B · C) | (A + B) + C = A + (B + C) |
| Distributive | A · (B + C) = A·B + A·C | A + (B · C) = (A+B) · (A+C) |
| Absorption | A · (A + B) = A | A + (A · B) = A |
| De Morgan's | (A · B)̅ = A̅ + B̅ | (A + B)̅ = A̅ · B̅ |
De Morgan's Laws — Deep Dive
De Morgan's laws are the most important simplification rules. They show how to "break the bar" (distribute NOT across an expression):
1 First Law: NOT (A AND B) = (NOT A) OR (NOT B)
(A · B)̅ = A̅ + B̅
"Break the bar, change the sign"
→ The bar breaks over each variable
→ AND changes to OR
Proof by truth table:
A B | A·B | (A·B)̅ | A̅ | B̅ | A̅+B̅
0 0 | 0 | 1 | 1 | 1 | 1 ✓
0 1 | 0 | 1 | 1 | 0 | 1 ✓
1 0 | 0 | 1 | 0 | 1 | 1 ✓
1 1 | 1 | 0 | 0 | 0 | 0 ✓2 Second Law: NOT (A OR B) = (NOT A) AND (NOT B)
(A + B)̅ = A̅ · B̅
"Break the bar, change the sign"
→ The bar breaks over each variable
→ OR changes to ANDStep-by-Step Simplification
✓ Example 1: Simplify A · B + A · B̅
A · B + A · B̅
= A · (B + B̅) ← Factor out A (Distributive law)
= A · 1 ← Complement law: B + B̅ = 1
= A ← Identity law: A · 1 = A
Result: A · B + A · B̅ = A✓ Example 2: Simplify A + A · B
A + A · B
= A · 1 + A · B ← Identity law: A = A · 1
= A · (1 + B) ← Factor out A (Distributive)
= A · 1 ← Null law: 1 + B = 1
= A ← Identity law
OR simply apply Absorption law directly:
A + A · B = A ← Absorption law
Result: A + A · B = A✓ Example 3: Simplify (A + B) · (A + B̅)
(A + B) · (A + B̅)
= A·A + A·B̅ + B·A + B·B̅ ← Expand (FOIL / Distributive)
= A + A·B̅ + A·B + 0 ← Idempotent: A·A=A; Complement: B·B̅=0
= A + A·B̅ + A·B ← Remove 0
= A + A·(B̅ + B) ← Factor out A
= A + A·1 ← Complement: B̅ + B = 1
= A + A ← Identity: A·1 = A
= A ← Idempotent: A + A = A
Result: (A + B) · (A + B̅) = A✓ Example 4: Apply De Morgan's to NOT(A · B + C)
(A · B + C)̅
= (A · B)̅ · C̅ ← De Morgan's: break bar, OR→AND
= (A̅ + B̅) · C̅ ← De Morgan's again on (A · B)̅
Result: NOT(A·B + C) = (A̅ + B̅) · C̅Simplification Patterns to Memorize
| Pattern | Simplifies To | Law Used |
|---|---|---|
| A · B + A · B̅ | A | Complement + Identity |
| A + A · B | A | Absorption |
| A · (A + B) | A | Absorption |
| (A + B) · (A + B̅) | A | Distributive + Complement |
| A + A̅ · B | A + B | Distributive + Complement |
| A · A | A | Idempotent |
| A + A | A | Idempotent |
Pitfalls & Common Errors
Applying De Morgan's Incorrectly
Remember: break the bar AND change the sign. Students often break the bar but
forget to change AND↔OR. (A·B)̅ = A̅ + B̅ — NOT A̅ · B̅.
Not Showing Working
Always label each step with the law you're using. Examiners give marks for showing which law was applied, not just the answer.
Confusing + with Addition
In Boolean algebra, + means OR, not arithmetic addition. 1 + 1 = 1 (not 2). Think logically, not mathematically.
Pro-Tips for Exams
Simplification Strategy
- Step 1: Look for complement pairs (A + A̅ = 1, A · A̅ = 0)
- Step 2: Factor out common terms (Distributive law)
- Step 3: Apply absorption if you see A + A·B or A·(A+B)
- Step 4: Use De Morgan's to push NOT inward
- Step 5: Verify with a truth table — original and simplified must match
- Always name the law you apply at each step
Graded Tasks
Write out all Boolean laws from memory (Identity, Null, Idempotent, Complement, Commutative, Associative, Distributive, Absorption, De Morgan's).
Simplify: (a) A · B + A · C (b) A̅ · B + A̅ · B̅ (c) (A + B) · (A̅ + B)
Use De Morgan's Law to simplify: NOT(A AND B AND C). Show each step.
Simplify X = A · B · C + A̅ · B · C + A · B̅ · C + A · B · C̅. Show all steps and verify with a truth table.
Given the expression X = (A + B) · (A̅ + C) · (B + C), simplify it as far as possible. Then draw both the original and simplified logic circuits. How many fewer gates does the simplified version need?