2

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?

Mike Nakis
  • 56,297
  • 11
  • 110
  • 142
AudioGuy
  • 413
  • 5
  • 18
  • If you just google "boolean evaluator" or something you will find web sites that will simplify your expression ***and*** list the rules they used to simplify it. "Wolfram Alpha" is known to be a good one. – Mike Nakis Oct 11 '15 at 18:46
  • Try using de-morgans law as well – Ed Heal Oct 11 '15 at 18:48
  • thanks, I already tried wolfram alpha and another one (in german). they all come to the same result. but there is no real explanation for the result. For example, they say: "a~bc + ab" results in "ac + ab". But they do not say WHY :-) – AudioGuy Oct 11 '15 at 19:02

1 Answers1

2

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.

dbajgoric
  • 1,447
  • 11
  • 17