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I have the two expressions that I have already simplified as much as possible:

  1. C̅D̅ ∨ A̅B̅D ∨ A̅BC̅
  2. AB ∨ AC ∨ B̅C̅

However, I know that you can simplify both expressions even further per KV diagram:

  1. C̅D̅ ∨ A̅B̅D ∨ A̅C̅
  2. A ∨ B̅C̅

However, I don't know how to transform the boolean expressions from above to arrive at the solution of the KV diagrams.

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  • \$\begingroup\$ What is the origin of the supposed simplifications? Are they supposed to be minimal? \$\endgroup\$
    – greybeard
    May 30, 2023 at 11:07
  • \$\begingroup\$ Yes, they are supposed to be minimal \$\endgroup\$
    – Hector234
    May 30, 2023 at 11:36
  • \$\begingroup\$ (I couldn't guide you any better than the next online boolean algebra tool. For the second expression, it's simple enough for me to spot. For the first one, an insight may be waiting for you.) \$\endgroup\$
    – greybeard
    May 30, 2023 at 11:43
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    \$\begingroup\$ I think you should reverse the first formula back to show all terms then reduce. \$\endgroup\$
    – Andy aka
    May 30, 2023 at 12:18
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    \$\begingroup\$ As Andy says, expand the terms in (1) and resimplify or use a k-map. One term can be simpler. \$\endgroup\$ May 30, 2023 at 15:25

2 Answers 2

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To simplify further, you need to make use of the various laws of boolean algebra. Along with the obvious operations (e.g. A.A' = 0), you have other tools such as De Morgans theory:

A.B = (A'+B')'
A+B = (A'.B')'

You also have factorisation:

A.B + A.C = A.(B + C)

Along with a useful trick of adding in extra 0 and 1 terms (needed for you second example). This allows you to add in missing terms to be able to simplify others away.

If you are playing with inputs A and B, you can add in extra operations such as:

B  =>  1.B  =>  (A + 1).B  =>  A.B + B

Notice that the two statements are the same - OR any input with 1 and that input has no impact on the output. Why is that expansion useful? Consider if you end up with this:

A.B' + B

Instinctively you can tell that A.B' and B are mutually exclusive, but removing them formally requires the trick of adding in the extra terms to factorise.

A.B' + (A + 1).B
A.B' + A.B + B
...

You can then go from there to simplify. I've intentionally left the simplification of that to the reader.

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If you're familiar with K-Maps, simplifying Boolean expressions is much easier.

Let's take a look at the K-Map for the first expression, (C̅D̅ ∨ A̅B̅D ∨ A̅BC̅):

K-Map

One solution is taking the SOP form of the expression by grouping the 1's together:

K-Map

After that, we get C̅D̅ ∨ A̅B̅D ∨ A̅BC̅ = C̅D̅ ∨ A̅C̅ ∨ A̅B̅D.

Here's a great article about K-Maps and how to use them - hopefully after reading my answer and the article, it should be clear for you how to check if the second expression is minimized.

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