How does this Boolean simplification work?

Question

The following expression:

$$ABC + BCD + A'D$$

simplifies to:

$$ABC + A'D$$

I just cannot understand how this comes about.

score 1 · Accepted Answer · answered Mar 26 '17 at 08:48

1

Treat $BC=X$ as a single variable. Then: $$ AX + XD + A'D = AX + (A+A')XD +A'D = AX + AXD + A'XD + A'D \\ = AX(1+D) + A'D(1+X) = AX + A'D $$

Please tell if you did not understand anything.

answered Mar 26 '17 at 08:48

Sarvesh Ravichandran Iyer

76,105

You are welcome. – Sarvesh Ravichandran Iyer Mar 26 '17 at 09:08

Jared Cannon · Answer 2 · 2017-03-26T09:05:13.300

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Using a Karnaugh Map will make this a lot easier to understand. If you input your original statement into a 4x4 map, you end up with the following expression:

$$A'C'D + A'CD + ABC$$

simplifies to:

$$A'D + ABC$$

Here is the Wikipedia article on Karnaugh Maps: https://en.wikipedia.org/wiki/Karnaugh_map.

Edit: The Karnaugh Map actually outputs the fully-simplified expression ABC + A'D, I just wanted to show where in the process the BCD term disappeared.

edited Mar 26 '17 at 09:05

answered Mar 26 '17 at 08:53

Jared Cannon

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Sorry, I should've mentioned it in the post, but I wanted to do this via Boolean Algebra. Thank you for taking the time to answer though. – chomprrr Mar 26 '17 at 09:06
Not a problem, I hope it at least provided some insight! – Jared Cannon Mar 26 '17 at 09:10

score 1 · Answer 3 · answered Mar 26 '17 at 22:25

1

$ABC + BCD + A'D = ABC + ABCD + A'BCD + A'D = ABC + A'D$

answered Mar 26 '17 at 22:25

Bram28

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How does this Boolean simplification work?

3 Answers3