Boolean Algebra Simplification Help

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Frankie715
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Homework Statement


A'B'C' + ABC' + A'BC + A'BC'.

The ' denotes a bar over the previous letter.


Homework Equations



Simplification Rules

The Attempt at a Solution



C' (A'B' + AB) + A'B(C+C')

C' (A'B' + AB) + A'B ... or like so: A'B'C' + ABC' + A'B

Is it possible to simplify the left expression any more? Does A'B' + AB = 1? When I did the K-map I got answers of

A'C' + A'B + BC' (Although I am not entirely sure I did the K-map right because it came out like this.
Code:
AB (00)   (01)    (11)    (10)
c(0) 1...1...1...
 (1)  ...1.....

That is the K-map of the original equation. When getting the simplified answer using the K-map am I supposed to circle EACH group of two consecutive 1's? I will arrive at the answer I got when simplifying above (A'B'C' + ABC' + A'B) if I select the top left most 1's in a group, then a group of the middle 1 and bottom 1, and then the right most 1 all by itself. If I select them in a different way (left most top 2, then middle and bottom, then middle and top right) I will get the answer of (A'C' + A'B + BC').

I am not sure which one is right and it is driving me crazy!

I would appreciate any help guys! Thanks a LOT!

The answer I arrived to also gives me the correct answer for the truth table, but it looks like it could be simplified to the above form. How do I go about doing that?

Thanks for any help!
 
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on Phys.org
I THINK I figured it out, but I am not sure. I will pick up where I left off and leave it to you guys to tell me if I did it correctly.

A'B'C' + ABC' + A'B=

A'(B+B'C') + ABC' =

A'(B+C') + ABC' =

A'B + A'C' + ABC' =

B(A' + AC') + A'C' =

B(A' + C') + A'C' =

A'B + BC' + A'C'

This final answer is what I arrived to when I did the k-map. I am just not 100% sure if my methods for reaching the final answer is correct.

Thanks for any help guys!