Can Boolean Algebra Simplify ab'c + a'b + bc' + abc to B + AC?

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Discussion Overview

The discussion revolves around the simplification of the Boolean expression ab'c + a'b + bc' + abc, specifically exploring whether it can be simplified using Boolean algebra to the form B + AC, as opposed to using Karnaugh maps (K-maps).

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Some participants express that the simplification to B + AC can be achieved using K-maps, suggesting that this method is straightforward.
  • Others inquire about the possibility of achieving the same simplification through Boolean algebra, indicating curiosity about algebraic methods.
  • One participant suggests that De Morgan's theorem could be a useful first step in the algebraic reduction process.
  • Another participant asserts that any simplification possible with K-maps can also be done algebraically, though it may be more complex.
  • There is a mention of a specific reduction step involving the expression ac + a'b + bc', with a suggestion to reduce it further using De Morgan's theorem.
  • Some participants confirm their familiarity with K-maps while seeking clarification on algebraic methods.

Areas of Agreement / Disagreement

Participants generally agree that K-maps provide a clear method for simplification, but there is no consensus on the effectiveness or ease of algebraic methods. The discussion remains unresolved regarding the specific algebraic steps to achieve the simplification.

Contextual Notes

The discussion highlights the potential complexity of algebraic simplification compared to K-map methods, but does not resolve the specific algebraic steps or assumptions involved.

kukumaluboy
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ab'c + a'b + bc' + abc

= ac + a'b + bc' (How to further reduce this?)

Kmap gives B + AC
 
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kukumaluboy said:
ab'c + a'b + bc' + abc

= ac + a'b + bc' (How to further reduce this?)

Kmap gives B + AC
Do you know about karnaugh maps? The reduction to B + AC is trivial and obvious if you do.
 
As in is there a boolean algebra way? Yea i know it can be done in kmap. Just curious
 
kukumaluboy said:
As in is there a boolean algebra way?
You can reduce ac + a'b + bc' further using DeMorgan's theorem as a first step.
 
kukumaluboy said:
As in is there a boolean algebra way? Yea i know it can be done in kmap. Just curious
Sorry, yes, your subject line does say algebraic simplification. As milesyoung said, use deMorgan's theorems.
 
One thing I should add: if you can reduce the complexity of a Boolean statement using a K-map then you can ALWAYS do the same thing algebraically. It would not make any sense for it to be otherwise. It may not be as easy as w/ a K-map but it has got to be doable.
 
kukumaluboy said:
As in is there a boolean algebra way? Yea i know it can be done in kmap. Just curious
First question: do you know De Morgan's theorems?

If you can reduce your logic expression to ac + b.¬(ac) [/size] I can give you the next step after that, if needed.
 

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