boolean algebra simplification?

Question

what is the simplification of (AB + C'D + A'B + CD) ?

Answer 1

First method, using Laws of Boolean algebra:

Factor out the common terms:

[ B (A + A') + D (C' + C) ]

Inverse Laws: X + X' = 1;

So you'll get:

B(1) + D(1)

Identity Laws: X(1) = X;

So you final answer is simply [ B + D ].

Alternatively, you may use Karnaugh maps to simplify the equation so you won't have to memorize all the laws.

Hope it helps.

Answer 2

(AB + C'D + A'B + CD) = (AB + A'B + CD + C'D)
rearranging the order of the terms does not change the answer.
AB+A'B = B also CD+C'D = D
A can be on or off, but B must be on
C can be on or off, but D must be on

Therefore:
(AB + C'D + A'B + CD) = (AB + A'B + CD + C'D) =
(AB + A'B) + (CD + C'D) = B + D

Answer 3

I found a site that may help you simplify your question.

Go to:

http://www.cs.usask.ca/resources/tutorials/csconcepts/2000_8/html/boolean_algebra.htm

Guido