Boolean algebra simplification

 Boolean algebra simplification examples and solutions

 Example 1: Simplify the Boolean expression 

 

Solution 

 

 

 

Example 2:  Find the complement of the boolean expression given by

 

 

 Solution

a)

   Using De Morgan’s theorem

 

 

b)

 Let Y be the complement of 

  

 

 

    Applying De Morgan’s theorem

  

Example 3: Proof that the following algebra (boolean expression) is equal 

 

Solution 

 

 

Example 4: Simplify the following Boolean expressions

 

Solution 

  a)

 

 

b) 

 

 

Example 5: Proof of the Dual of Distributive Law

Which states that (x + y.z) = (x + y).(x + z)

Solution

 

Using the RHS