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Lesson 0011: Simplifying Boolean Expressions

Boolean algebra follows some of the same patterns as regular algebra. The AND and OR operations can be changed according to the commutative property, the associative property, and the distribuitive property.

Commutative Property
AB=BA
A+B=B+A.

Associative Property
(A+B)+C = A+(B+C)

Distribuitive Property
A(B+C) = AB+AC

De Morgans Laws

There is a very important rule that applies to the NOR and NAND functions. It is called De Morgans Law. It states that:

~(A*B) is the same as ~A+~B and ~(A+B) is the same as ~A~B

Other Laws
Some other laws state that:
A+0 = A
(because it means A OR 0, which the 0 cancels out, leaving A)
A*0 = 0
(because 0 AND-ed with anything is false.)
A+1 = 1
(because 1 OR anything is true.)
A*1 = A
(because 1 AND any constant is the same as the constant. The 1 cancels out.)
A+~A = 1.
(because either A or ~A is true, and a true OR a false is true, or 1.)
A*~A = 0
(because either A or ~A is false and a false AND a true is false, or 0.)
A+A = A
(because A OR itself is the same as A.)
A*A = A
(because if A is true, the result will be true and if A is false the result will be false.)
~(~A) = A
(because A negated is ~A, and this result negated will be the same as the original.)


Check your progress

  1. Simplify the following expression using De Morgans Law and the distribuitive property.

    ~(A+B)+(~AB)

      ~B + ~A
      ~BA
      ~(BA)
      ~B+A
  2. Which of the following equations is always true?
      A+B
      A*~A
      ~A+A
      AA
      none of the above

  3. Which of the following expressions is always false?
      A+0
      AA
      A+~A
      A+1
      none of the above

  4. What properties are used in solving the following expression?
    ~(AB)
      Distribuitive
      De Morgans
      De Morgans and the distribuitive
      De Morgans or the distribuitive

  5. Simplify the following expression: (list the inputs in alphabetical order using capital letters)
    ~AB+BA+A+B



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