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March 9th, 2017, 12:09 PM   #1
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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
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March 9th, 2017, 12:21 PM   #2
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$\bar{A}\bar{B}CD + A + B = (A+B)CD+A+B ~~~~ \text{(DeMorgan's Law)}$

$(A+B)CD+(A+B) = (A+B)(CD+U) = (A+B)(U) = A+B$
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March 9th, 2017, 04:49 PM   #3
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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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