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// Logic Engine v3.0

Boolean Algebra Simplifier

Supports LaTeX  ·  standard symbols  ·  keyword syntax  ·  step-by-step solutions

Expression Input
Boolean Expression
Parsed as
Syntax Guide

All formats can be mixed freely.

LaTeX / Typeset
\overline{X}NOT X
\cdotAND
X'NOT X (prime)
\overline{AB}NOT(A·B)
Symbols
+OR
· * &AND
~ !NOT (prefix)
^XOR
Keywords
AND ORbasic gates
NOTnegation
NAND NORuniversal
XOR XNORexclusive

Try these examples:
A+B
\overline{A}B+A\overline{B}
(\overline{C}A+D0)A+\overline{1C}
A NAND (B XOR C')
A+AB+A\overline{B}

Simplified Result
— enter an expression above —
Variables
Truth rows
Output = 1
Steps
Detailed Analysis
Step-by-step simplification will appear here
Truth table will appear here
Canonical forms will appear here
Laws applied will appear here
Math

Simplify Boolean expressions for logic circuits and digital design. Enter expressions to simplify, making it easier to optimize logic gates and circuits in computer engineering.

 

Boolean Algebra Simplifier

Boolean Algebra Laws

1. Commutative Law

The Commutative Law states that the order of the operands does not affect the result of the operation.

  • For OR: A + B = B + A
  • For AND: A & B = B & A

2. Associative Law

The Associative Law states that the grouping of operands does not affect the result of the operation.

  • For OR: (A + B) + C = A + (B + C)
  • For AND: (A & B) & C = A & (B & C)

3. Distributive Law

The Distributive Law allows you to distribute one operation over another.

  • For OR over AND: A + (B & C) = (A + B) & (A + C)
  • For AND over OR: A & (B + C) = (A & B) + (A & C)

4. Identity Law

The Identity Law states that a variable combined with 1 (for OR) or 0 (for AND) does not change the value of the variable.

  • For OR: A + 0 = A
  • For AND: A & 1 = A

5. Null Law

The Null Law states that a variable combined with 0 (for OR) or 1 (for AND) results in a null value.

  • For OR: A + 1 = 1
  • For AND: A & 0 = 0

6. Complement Law

The Complement Law states that a variable combined with its complement (inverted value) results in 1 (for OR) or 0 (for AND).

  • For OR: A + \overline{A} = 1
  • For AND: A & \overline{A} = 0

7. Idempotent Law

The Idempotent Law states that a variable combined with itself does not change its value.

  • For OR: A + A = A
  • For AND: A & A = A

8. Absorption Law

The Absorption Law allows you to "absorb" one term into another.

  • For OR: A + (A & B) = A
  • For AND: A & (A + B) = A

9. De Morgan's Law

De Morgan's Law relates to the negation of AND and OR operations.

  • For negation of AND: \( \overline{A \cdot B} = \overline{A} + \overline{B} \)
  • For negation of OR: \( \overline{A + B} = \overline{A} \cdot \overline{B} \)

10. Double Negation Law

The Double Negation Law states that negating a negation brings back the original variable.

  • For negation: \( \overline{\overline{A}} = A \)

Boolean Algebra Examples

  1. Expression: \( A \cdot (A + B) \)
    Solution: Using the Absorption Law: \( A \cdot (A + B) = A \)
  2. Expression: \( A + \overline{A} \cdot B \)
    Solution: Using the Absorption and Complement Laws: \( A + \overline{A} \cdot B = A + B \)
  3. Expression: \( (A + B) \cdot (A + \overline{B}) \)
    Solution: Using the Distributive Law and Identity Law: \( (A + B) \cdot (A + \overline{B}) = A \)
  4. Expression: \( \overline{\overline{A}} + \overline{A \cdot B} \)
    Solution: Using Double Negation Law and De Morgan’s Theorem: \( \overline{\overline{A}} + \overline{A \cdot B} = A + \overline{A} + \overline{B} = 1 \)
  5. Expression: \( A \cdot \overline{A \cdot B} \)
    Solution: Using De Morgan’s Theorem and Absorption: \( A \cdot \overline{A \cdot B} = A \cdot (\overline{A} + \overline{B}) = A \cdot \overline{B} \)
  6. Expression: \( (A \cdot B) + (\overline{A} \cdot B) \)
    Solution: Using Distribution: \( (A \cdot B) + (\overline{A} \cdot B) = B \)
  7. Expression: \( A + (A \cdot B) + \overline{A} \)
    Solution: Using Absorption: \( A + (A \cdot B) + \overline{A} = A + \overline{A} = 1 \)
  8. Expression: \( \overline{A \cdot \overline{B}} + A \cdot B \)
    Solution: Using De Morgan's Theorem: \( \overline{A \cdot \overline{B}} + A \cdot B = (\overline{A} + B) + A \cdot B = B + \overline{A} \)
  9. Expression: \( A \cdot (A + \overline{B}) + B \)
    Solution: Using Distribution and Absorption: \( A \cdot (A + \overline{B}) + B = A + B \)
  10. Expression: \( (A + B) \cdot (A + \overline{B} + C) \)
    Solution: Using Distribution: \( (A + B) \cdot (A + \overline{B} + C) = A + B \cdot C \)

Published
2026-04-22 13:14:38
Updated
2026-05-16 09:28:07
Author
Taylor Bennett