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 $AB$ ?

Answer 1

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

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

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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'.

Answer 2

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

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

Answer 3

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.