Boolean Algebra


In this blog post we are investigating different formulas than can be used to simplify a Boolean expression.

Double Negation

¬ ¬A = A

Complement Laws

A ∧ ¬A = 0
A ∨ ¬A = 1

Idempotent Laws

A ∧ A = A
A ∨ A = A

Identity Laws

A ∧ 1 = A
A ∧ 0 = 0
A ∨ 1 = 1
A ∨ 0 = A

Associative Laws

(A ∧ B) ∧ C = A ∧ (B ∧ C)
(A ∨ B) ∨ C = A ∨ (B ∨ C)

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)

Absorptive Laws

A ∧ (A ∨ B) = A
A ∨ (A ∧ B) = A

De Morgan’s Rules

¬(A ∨ B) = ¬A ∧ ¬B
¬(A ∧ B) = ¬A ∨ ¬B

Boolean Algebra Practice


Use the formulas listed above to simplify the following Boolean expressions:

#1#2#3#4#5#6
Boolean Expression
A ∨ ¬(A ∧ B)

Simplified Boolean Expresssion:



View Solution
Boolean Expression
(A ∧ B) ∨ (A ∧ C)

Simplified Boolean Expresssion:



View Solution
Boolean Expression
(A ∧ B) ∨ A ∧ (B ∨ C)

Simplified Boolean Expresssion:



View Solution
Boolean Expression
¬A ∨ C ∨ (A ∧ B)

Simplified Boolean Expresssion:



View Solution
Boolean Expression
¬(¬A ∧ (B ∧ C))

Simplified Boolean Expresssion:



View Solution
Boolean Expression
¬(A ∧ ¬B) ∨ (¬A ∧ B)

Simplified Boolean Expresssion:



View Solution

Did you like this challenge?

Click on a star to rate it!

Average rating 4.7 / 5. Vote count: 14

No votes so far! Be the first to rate this post.

As you found this challenge interesting...

Follow us on social media!