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March 9th, 2017, 01:09 PM  #1 
Member Joined: Jan 2014 Posts: 34 Thanks: 0  boolean Algebra Problems
Why does A'B' cancel out in this problem: A'B'CD+A+B To get: CD+A+B (image boolean algebra.jpg) and can some one check to see if this is correct (refer to imageboolAlg.jpg) Thanks 
March 9th, 2017, 01:21 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 898 Thanks: 483 
$\bar{A}\bar{B}CD + A + B = (A+B)CD+A+B ~~~~ \text{(DeMorgan's Law)}$ $(A+B)CD+(A+B) = (A+B)(CD+U) = (A+B)(U) = A+B$ 
March 9th, 2017, 05:49 PM  #3 
Newbie Joined: Jan 2016 From: Athens, OH Posts: 22 Thanks: 11 
I think Romsek must have misread your question. For any X and A, $$\bar{A}X+A=(\bar{A}+A)(X+A)=1(X+A)=X+A$$ Here I used that + distributes over product. Use the above twice to get your result. 

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