Boolean Algebra Simplification Help

[Frankie715]

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.

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!

 

[Frankie715]

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!

 

[NascentOxygen]

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

What you have in A'B is an INEQUALITY function--it evaluates TRUE iff A≠B
What you see in blue is an EQUALITY function--it is TRUE iff A=B

 

[NascentOxygen]

There is no need to guess or hope. You can always draw up truth tables to show whether your simplification is right or not.

 

[DeeNos]

I understand your frustration with trying to simplify Boolean algebra expressions. It can definitely be challenging at times. First, let me clarify that A'B' + AB does indeed simplify to 1. This is known as the "complement law" in Boolean algebra.

Now, for your specific expression, A'B'C' + ABC' + A'BC + A'BC, there are a few ways you can approach simplifying it. One method is using the K-map, as you have mentioned. When using the K-map, you are correct in circling groups of two or more consecutive 1's. However, it is important to note that these groups must be in a rectangular shape, with sides that are powers of 2 (e.g. 2, 4, 8, etc.). In your K-map, the group of 1's in the top left corner cannot be circled as it is not a rectangle.

Another method for simplifying Boolean algebra expressions is using the laws and rules of Boolean algebra. In this case, you can use the "distributive law" to rewrite the expression as A'B'(C' + C) + A'B(C + C') + A'BC. Since C + C' is equivalent to 1, this simplifies further to A'B' + A'B + A'BC, which can then be further simplified to A'B + A'BC.

I hope this helps clarify the process for simplifying Boolean algebra expressions. Remember to always check your work and make sure your final expression is equivalent to the original expression. Good luck with your homework!