Activity: Boolean Algebra & Digital Logic

Laws, simplification, De Morgan's theorem, and logic gate analysis with interactive tools

Grade XI • Computer Science ⏱️ ~35 min

Part 1 Boolean Laws & Postulates

Boolean Algebra, developed by George Boole, is used to simplify digital circuits. Just like regular algebra has rules, Boolean algebra has laws for AND, OR, and NOT operations.

Task 1: The Core Postulates

Every variable can only be 0 or 1. Master these fundamental behaviors:

Law Name	OR (Addition)	AND (Multiplication)
Annulment	A + 1 = 1	A · 0 = 0
Identity	A + 0 = A	A · 1 = A
Idempotent	A + A = A	A · A = A
Complement	A + A̅ = 1	A · A̅ = 0
Double Negation	A̿ = A (Double bar cancels out)

Practice: Simplify (A + 0) · (B · 0)

    Step 1: (A + 0) = A (Identity)
    Step 2: (B · 0) = 0 (Annulment)
    Step 3: A · 0 = 0 (Annulment)
    Result: 0

Task 2: Commutative, Associative, & Distributive Laws

Commutative: Order doesn't matter (A + B = B + A)
Associative: Grouping doesn't matter [A + (B + C) = (A + B) + C]
Distributive: A · (B + C) = A·B + A·C

Challenge: Which law is used in A + B·C = (A + B)(A + C)?

This is the second Distributive Law (OR over AND).

Task 3: Absorption Law

A term that contains its own component is "absorbed."

Rule: A + (A · B) = A

Practice: Simplify B · (B + A)

By Absorption Law (AND version), B · (B + A) = B.

Task 4: Law Identification Challenge

Name the law applied in each step:

1. X + (Y + Z) = (X + Y) + Z

Associative Law

2. X · X̅ = 0

Complement Law

3. X + 1 = 1

Annulment Law

Part 2 De Morgan's & Simplification

Task 5: De Morgan's Theorems

The mnemonic is "Break the bar, change the sign."

Theorem 1: A · B = A̅ + B̅
Theorem 2: A + B = A̅ · B̅

Practice: Rewrite X + Y + Z

X̅ · Y̅ · Z̅

Practice: Rewrite P̅ · Q

P̿ + Q̅ = P + Q̅

Task 6: Step-by-Step Simplification

Simplify: (A + B)(A + C)

Expand using Distributive Law: A·A + A·C + A·B + B·C
Apply Idempotent Law (A·A = A): A + A·C + A·B + B·C
Factor out A: A(1 + C) + A·B + B·C
Apply Annulment (1 + C = 1): A·1 + A·B + B·C
Identity (A·1 = A): A + A·B + B·C
Apply Absorption (A + AB = A): A + B·C

Result check: Is (A + B)(A + C) = A + BC?

Yes, this proves the Distributive Law A + BC = (A+B)(A+C).

Tool: Expression Truth Table Builder

Select a Boolean expression to see its complete behavior.

Part 3 Logic Gate Analysis

Task 7: Basic Gates (AND, OR, NOT)

The building blocks of logic.

Practice: If A=1, B=0, what is (A AND B) OR (NOT A)?

    (1 AND 0) = 0
    NOT 1 = 0
    0 OR 0 = 0

Tool: Interactive Logic Gate Explorer

Input A Input B

AND

Output: 0

Task 8: XOR vs XNOR

XOR (Exclusive OR) is true if inputs are DIFFERENT. XNOR is its complement (true if inputs are SAME).

A	B	XOR	XNOR
0	0	0	1
0	1	1	0
1	0	1	0
1	1	0	1

Puzzle: When is XOR = 1 and XNOR = 0?

When inputs are (0,1) or (1,0).

Task 9: Universal Gates

NAND and NOR are called Universal Gates because any other logic gate can be built using only them.

Think: How to make a NOT gate using NAND?

Connect both inputs of the NAND gate together to the same signal. A NAND (A, A) results in A̅.

Task 10: Gate Identification Puzzle

Identify the gate from its behavior:

Gate 1: Output is 0 ONLY if both inputs are 1.

NAND Gate

Gate 2: Output is 1 ONLY if both inputs are 0.

NOR Gate

Gate 3: Output is the same as the input but reversed.

NOT Gate (Inverter)

Task 11: Multi-gate Circuit Trace

Trace this circuit: (A, B) → NAND → (Result 1). Then (Result 1, C) → OR → Final Out.

Trace: A=1, B=1, C=0. What is Final Out?

    Step 1: (1 NAND 1) = 0
    Step 2: (0 OR 0) = 0
    Final Out: 0

Trace: A=0, B=1, C=0. What is Final Out?

    Step 1: (0 NAND 1) = 1
    Step 2: (1 OR 0) = 1
    Final Out: 1

Task 12: Application Summary

Logic gates are used to build adders, multiplexers, and memory units in a computer's CPU.

Mastering these 7 gates allows you to understand how a processor calculates anything!

Logic puzzles await!

Ready to trace circuits and simplify expressions under pressure?

Launch Assessment →