Engineering Where Did the Minterm AB'C Come From in This Boolean Function?

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The discussion centers on the confusion surrounding the derivation of the minterm AB'C from a given Boolean function F(A,B,C,D) = ABC + (A + B + C) + AB. The original function was simplified, leading to the conclusion that AB is the only term involving a logical AND operation, while (A+B+C) is identified as a maxterm. The participants explore the possibility of using Karnaugh maps (k-maps) to clarify the minterms and their arrangements, suggesting that mapping the function could reveal the correct minterms. There is uncertainty about how to determine the values of the k-map entries and whether all entries are necessary for the analysis. Ultimately, the conversation emphasizes the utility of k-maps in resolving ambiguities in Boolean function simplifications.
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Homework Statement
Determine the minterms of F
Relevant Equations
Definition of minterm: A standard product term, also called a minterm, is a product term that includes each variable of the problem, either uncomplemented or
complemented.
I simplified the given function into a single minterm and a single maxterm
F(A,B,C,D) = ABC + (A + B + C) + AB
F(A,B,C,D) = AB(C+1) + (A+B+C)
F(A,B,C,D) = AB(1) + (A+B+C)
F(A,B,C,D) = AB + (A+B+C)
The only terms that involve a logical AND operation are AB as (A+B+C) is a maxterm of the function.
I am not sure if simplifying the function is necessary here and if I'm not supposed to for whatever reason I would think that either (AB) or (ABC) from the original function F(A,B,C,D) would be correct. Maybe ABC would be the best option because it includes all variables of the function. Regardless, none of the available solutions match what I believe the solution is and upon looking at the posted solution in the book it says AB'C is the solution. I'm totally lost as to where this minterm actually came from as it doesn' even appear in the original function. Is there some sort of algebraic manipulation going on here? What am I not seeing?
 

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Are you allowed to use k-maps?
 
Joshy said:
Are you allowed to use k-maps?
absolutely! should I plot this function on a k-map?
 
It's been a while since I've done this, but I think for example if

$$F = A B C' +C$$
then
$$\begin{array}
{c|c|c|c|}
& B'C' & B'C & BC & BC' \\
\hline A' & 0 & 1 & 1 & 0 \\
\hline A & 0 & 1 & 1 & 1 \\
\hline \end{array}$$

Then wouldn't your minterms be those 5 items A'B'C, A'BC, AB'C, ABC, and ABC'?

You can try it for your problem (above is fake one I did on the fly) and see where those 1's land at, then you can choose the correct answer.
 
I believe for:
$$F = [A+B+C] + [AB] + [ABC]$$
would map accordingly
$$\begin{array}
{c|c|c|c|}
& B'C' & B'C & BC & BC' \\
\hline A' & 0 & & & \\
\hline A & & & 1 & 1 \\
\hline \end{array}$$
Where each 1 and the 0 is grouped only by itself to produce the three terms given in the function. I think I'm confused about how to determine what the values of the other entries are. Do I need them?
 
If, for example:
$$F=B$$
then wouldn't the k-map look something like this?
$$\begin{array}
{c|c|c|c|}
& B'C' & B'C & BC & BC' \\
\hline A' & 0 & 0 & 1 & 1 \\
\hline A & 0 & 0 & 1 & 1 \\
\hline \end{array}$$

If:
$$F=AC$$
maybe?
$$\begin{array}
{c|c|c|c|}
& B'C' & B'C & BC & BC' \\
\hline A' & 0 & 0 & 0 & 0 \\
\hline A & 0 & 1 & 1 & 0 \\
\hline \end{array}$$

How about?
$$F=ABC$$
maybe?
$$\begin{array}
{c|c|c|c|}
& B'C' & B'C & BC & BC' \\
\hline A' & 0 & 0 & 0 & 0 \\
\hline A & 0 & 0 & 1 & 0 \\
\hline \end{array}$$

Can you see that when it's the one letter by itself there is an entire group of four. When you eliminate a letter it becomes a smaller group of just two; lastly the very specific term ##ABC## it's just standalone only one on the k-map.

I think you can use the same approach with your equation. You have an ##A## by itself; you also have a ##B## or ##C## by itself. I'm expecting that there should be a lot of 1's on your k-map. Then the other terms they are just a group of one or two.

Here's another example where I'll combine two of them with OR statements.
$$F = AB' + AC$$
$$\begin{array}
{c|c|c|c|}
& B'C' & B'C & BC & BC' \\
\hline A' & 0 & 0 & 0 & 0 \\
\hline A & 1 & 1 & 1 & 0 \\
\hline \end{array}$$
You'll notice it's the same as ##F=AB'C'+AB'C+ABC##.

Hopefully this isn't too far out of the way, but I always enjoyed k-maps whenever I got stuck with the algebraic approach. You'll notice in textbooks or maybe online references there's some circling around like word searches... kind of fun ;).

I didn't look through this example thoroughly, but couldn't think of a way to circle things in latex. So I just ran this through a search engine to find a circling example. I like that they show the four element one too although yours for this example is just three elements. https://www.geeksforgeeks.org/introduction-of-k-map-karnaugh-map/

edit:

I ended up scanning something just another example, but to show the circling.

kmap_example.jpg

Once you fill out the map, then you'll see all of your minterms no problem.
 
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