Math Problem Statement

Simplify the Boolean expression: A + B·A' + B'·A

Solution

The expression shown in the image is:

A+BA+BAA + B\cdot \overline{A} + \overline{B} \cdot A

This looks like a Boolean algebra expression. Let's simplify it step by step.

  1. Apply the Distributive Law: Expand the terms.

    A+(BA)+(BA)A + (B \cdot \overline{A}) + (\overline{B} \cdot A)

  2. Apply the Absorption Law:

    Since AA=0A \cdot \overline{A} = 0 (because a variable AND its complement is always 0), simplify the second term:

    A+0+BAA + 0 + \overline{B} \cdot A

    This simplifies to:

    A+BAA + \overline{B} \cdot A

  3. Apply the Absorption Law again:

    A+AB=AA + A \cdot \overline{B} = A (because A+Aanything=AA + A\cdot anything = A).

Thus, the simplified form of the Boolean expression is:

AA

Would you like further clarification or additional details on any steps?

Here are 5 related questions to consider:

  1. What are the basic laws of Boolean algebra?
  2. How does De Morgan's Law apply to Boolean expressions?
  3. Can you explain the difference between the Distributive and Absorption Laws in Boolean algebra?
  4. What real-world applications can Boolean algebra have in digital logic design?
  5. How would you simplify more complex Boolean expressions involving XOR and NAND?

Tip: Mastering the Absorption Law can help you quickly reduce Boolean expressions and avoid unnecessary intermediate steps!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Boolean Algebra
Logic Gates
Simplification

Formulas

A + A·B = A (Absorption Law)
A·A' = 0 (Complement Law)

Theorems

Distributive Law
Absorption Law
Complement Law

Suitable Grade Level

Grades 10-12