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