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?
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.