Simplifying the Boolean expression $A'BC' + AB'C + ABC' + ABC$

Question

I have this Boolean to simplify.

$\bar AB\bar C + A\bar BC + AB\bar C + ABC$

I checked the final answer here. It gives the final simplification is $B\bar C + AC$

but I am confused here :

$\bar AB\bar C + A\bar BC + AB\bar C + ABC$

$\bar AB\bar C + A\bar BC + AB(\bar C+C)$

$\bar AB\bar C + A\bar BC + AB$

$\bar AB\bar C + A(\bar BC + B)$

$\bar AB\bar C + A(C + B)$

$\bar AB\bar C + AC + AB$

$B(\bar A\bar C + A) + AC$ (1st and 3rd considered)

$B(\bar C + A) + AC$

$B\bar C + AB + AC$

Here, How can I go further to simplify $B\bar C + AB + AC$ to $B\bar C + AC$

How to remove $AB$ ?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Aug 09 '19 at 14:24
Don't really see the need for MathJax here. In fact, I personally prefer monospaced font for this @JoséCarlosSantos — Simply Beautiful Art, Aug 09 '19 at 14:25

score 6 · Answer 1 · answered Aug 09 '19 at 14:25

6

I suggest you swap $AB'C$ and $ABC'$ before simplifying:

$$\begin{align}&A'BC' + ABC' + AB'C + ABC \\=& (A'+A)BC' + (B'+B)AC \\=& BC' + AC. \end{align}$$

answered Aug 09 '19 at 14:25

Magma

6,270

Yeah! I also did this. but the above alternate is the thing – test team Aug 09 '19 at 15:03

Simply Beautiful Art · Accepted Answer · 2019-08-09T15:21:39.230

3

Since we have AB = ABC' + ABC, it follows that

BC' + AB + AC
= BC' + ABC' + ABC + AC
= BC' + AC

The Karnaugh map is given by

from which it is visually clear that AB is covered by the other two, which lends itself to showing you can split AB into two parts and combine them with AC and BC'.

edited Aug 09 '19 at 15:21

answered Aug 09 '19 at 14:48

Simply Beautiful Art

74,685

AB = ABC' + ABC is ok. But How this BC' + ABC' + ABC + AC simplify to BC' + AC ? – test team Aug 09 '19 at 15:00
BC' + ABC' = ... and ABC + AC = ... – Simply Beautiful Art Aug 09 '19 at 15:01
Ah! Ok. Nice. But this is simplification. Then how can I think to expand (AB = ABC' + ABC). In simplification we think go further to simplify. – test team Aug 09 '19 at 15:05
There are algorithms for performing these kinds of reductions. I personally find Karnaugh maps very useful for seeing the end solution and how things combine. I can extend my answer if you want. – Simply Beautiful Art Aug 09 '19 at 15:09
Thank you. The K-Map explanation is nice. But how can we think in examination? We only think how to simplify go further and further, then we're not to expand. – test team Aug 09 '19 at 15:29
If you draw out the K-map, then you know which terms you can group to remove them. – Simply Beautiful Art Aug 09 '19 at 15:32
It is ok. If they asked to do by Boolean theorems, then ? – test team Aug 09 '19 at 15:45
The point is you can visualize what to do with the theorems from the K-map. – Simply Beautiful Art Aug 09 '19 at 15:46
Ah! ok. So K-map also to be considered in Boolean theorems simplification, no? – test team Aug 09 '19 at 15:48
It is simply a tool that let's you see what you should do more visually, so I don't see why not. – Simply Beautiful Art Aug 09 '19 at 16:14

poetasis · Answer 3 · 2019-08-09T16:36:33.660

0

I learned Veitch diagrams back in the early 70's and I find them easier to understand than the Karnaugh maps that are popular today. Here is an example using your problem. You fill in the squares and then see what they add up to. Note, the light gray area is $AC$ and the slightly darker gray is $BC'$. Note also how $B'$ spans part of $A$ and the rest of $B$ is under $A'$ so $AB$ is eliminated.

edited Aug 09 '19 at 16:36

answered Aug 09 '19 at 16:30

poetasis

6,338

New to Veitch diagram. And I'm unable to understand "Note also how B′ spans part of A and the rest of B is under A′ so AB is eliminated." – test team Aug 09 '19 at 17:05
I've seen Karnaugh maps that look $exactly$ like Veitch diagrams but the way they try to teach is more like the answer from Simply Beautiful Art and not intuitive to me. I hope you can see in my diagram that each square is filled with something-$A$, something-$B$, and something-$C$ that $maps$ to one of the terms of the original Boolean expression. Then we find that different terms are adjoined in one or more rectangles that represent new terms simpler than the originals. You can do the same with straight Boolean Algebra such as $(A'+B')'=AB$ but the visual aspect is easier to $see$. – poetasis Aug 09 '19 at 17:22
Eh, personally this is less intuitive, since the BC' is made up of two disconnected sections, for example. The Karnaugh map avoids this by ensuring adjacent squares differ by 1 variable, so you never have to do this (with the exception of wrapping from one side of the K-map to the other). – Simply Beautiful Art Aug 09 '19 at 21:10
I guess I'm just used to Veitch diagrams because I learned them so well when I was so much younger. In any case, I trust the Karnaugh maps were helpful. – poetasis Aug 10 '19 at 09:50

Simplifying the Boolean expression $A'BC' + AB'C + ABC' + ABC$

3 Answers3