Boolean Algebra Simplification Help

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Discussion Overview

The discussion focuses on simplifying a Boolean algebra expression, specifically A'B'C' + ABC' + A'BC + A'BC'. Participants explore various methods for simplification, including algebraic manipulation and Karnaugh maps (K-maps), while questioning the validity of their approaches and results.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents an initial expression and attempts a simplification, questioning whether A'B' + AB equals 1.
  • The same participant expresses uncertainty about their K-map results and the method for grouping ones in the K-map.
  • Another participant proposes a series of algebraic transformations to simplify the expression, but expresses uncertainty about the correctness of their methods.
  • A third participant points out the distinction between inequality and equality functions in the context of the expression.
  • A later reply suggests using truth tables as a definitive method to verify the correctness of the simplifications.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the simplification methods or the correctness of the results. Multiple approaches and interpretations are presented, indicating ongoing uncertainty and debate.

Contextual Notes

Participants mention the use of K-maps and truth tables, but there are unresolved questions about the grouping of terms in the K-map and the validity of the simplifications made. The discussion reflects a reliance on personal methods and interpretations without definitive resolution.

Who May Find This Useful

Students and individuals interested in Boolean algebra simplification techniques, K-map applications, and verification methods in mathematical logic may find this discussion relevant.

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!
 
Last edited by a moderator:
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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!
 
Frankie715 said:
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
 
There is no need to guess or hope. You can always draw up truth tables to show whether your simplification is right or not.
 

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