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

[kukumaluboy]

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

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

Kmap gives B + AC

 

[phinds]

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.

 

[kukumaluboy]

As in is there a boolean algebra way? Yea i know it can be done in kmap. Just curious

 

[milesyoung]

As in is there a boolean algebra way?

You can reduce ac + a'b + bc' further using DeMorgan's theorem as a first step.

 

[phinds]

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.

 

[phinds]

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.

 

[NascentOxygen]

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) I can give you the next step after that, if needed.