Boolean Algebra simplification (3 inputs)

Question

I may need your help with this quite simple thing: This -> abc' + ab'c + a'bc + abc can (I guess) be simplified to this -> ab+ac+bc.

But how in the world is this done with Boolean algebra?

I already reduced it to -> abc'+ab'c+bc by using the absorption rule on the last two terms [a'bc + abc]. But how can I reduce the rest of it to get the end result?

Answer 1

Before simplifying the expression I'll show you one nice trick that will be used later. For any two logical variables A and B the following holds:

A + AB = A(B + 1) = A

With this in mind let's simplify your expression:

abc' + ab'c + a'bc + abc = ac(b + b') + abc' + a'bc = ac + abc' + a'bc

We can expand ac in the following way using that 'trick' I mentioned before:

ac = ac + abc = ac(b + 1) = ac

Using this we get:

ac + abc' + a'bc = 
ac + abc + abc' + a'bc =
ac + ab(c + c') + a'bc = 
ac + ab + a'bc =
ac + ab + abc + a'bc =
ab + bc(a + a') + ac =
ab + ac + bc

Leading to the final expression you wanted to get in the first place.