Math Problem Statement

ab + c’ + c(a’ + b’)

Solution

We can simplify the Boolean expression step by step:

Given: ab+c+c(a+b)ab + c' + c(a' + b')

Step 1: Expand the third term

Use distribution to expand c(a+b)c(a' + b'): ab+c+ca+cbab + c' + ca' + cb'

Step 2: Group like terms

ab+c+ca+cbab + c' + ca' + cb'

Step 3: Simplify using Boolean algebra rules

  • Identity Rule: x+x=1x + x' = 1 In the current form, we can't simplify further unless we apply specific scenarios. Therefore, this expression can't be reduced beyond this point using fundamental Boolean algebra rules.

The simplified expression is: ab+c+ca+cbab + c' + ca' + cb'

Would you like further details or have any questions? Here are some related questions to practice:

  1. How would you simplify (a+b)(a+c)(a + b)(a' + c)?
  2. What is De Morgan's Law in Boolean algebra?
  3. How do you factor the expression ab+ac+bcab + ac + bc?
  4. What is the distributive property in Boolean algebra?
  5. How do you use Karnaugh maps to simplify Boolean expressions?

Tip: When simplifying Boolean expressions, always look for terms that can be combined using fundamental laws such as the distributive, absorption, or De Morgan’s laws.

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Math Problem Analysis

Mathematical Concepts

Boolean Algebra
Logic Simplification

Formulas

x + x' = 1 (Identity Rule)

Theorems

Distribution in Boolean Algebra
De Morgan’s Law

Suitable Grade Level

High School (Grades 9-12)