# Product of Sums to Sum of Products

I apologize if this is a dumb question, but I'm having some trouble seeing how we can go from the Boolean equation

P = (P1 + P2)(P3 + P4)(P1 + P3)(P5 + P6)(P2 + P5)(P4 + P6)

to

P = P1*P4*P5 + P1*P3*P5*P6 + P2*P3*P4*P5 + P2*P3*P5*P6 +
P1*P2*P4*P6 + P1*P2*P3*P6 + P2*P3*P4*P6 + P2*P3*P6


I keep getting P = P1 P4 P5 + P2 P3 P6 + P2 P3 P4 P5 + P1 P2 P4 P6 + P1 P3 P5 P6 as my minimization.

• What exactly are you asking about? How to get symbol expression or how to find boolean value once you have it ? Jan 26, 2013 at 3:28
• Remember that in a Boolean algebra, P1*P1=P1, P2*P2=P2, etc. I haven't checked out every term but I think this is just a case of distributing and then simplifying. Jan 26, 2013 at 3:29
• Why don't you post this as answer, @BrettFrankel Jan 26, 2013 at 4:02
• I keep getting P = P1 P4 P5 + P2 P3 P6 + P2 P3 P4 P5 + P1 P2 P4 P6 + P1 P3 P5 P6 as my minimization of this. Jan 26, 2013 at 4:16
• If there is no other conditions but the fact that Pi is bool variable, both your expressions are wrong Jan 26, 2013 at 4:27

## 2 Answers

The second answer is just a simplification of the first answer using $Q+1=1$

$P =P_1P_4P_5 + P_1P_3P_5P_6 + P_2P_3P_4P_5 + P_2P_3P_5P_6 + P_1P_2P_4P_6 + P_1P_2P_3P_6 + P_2P_3P_4P_6 + P_2P_3P_6$

$P =P_1P_4P_5 + P_1P_3P_5P_6 + P_2P_3P_4P_5 + P_1P_2P_4P_6 + (P_2P_3P_6)(P_5+P_1+P_4+1)$

$P =P_1P_4P_5 + P_1P_3P_5P_6 + P_2P_3P_4P_5 + P_1P_2P_4P_6 + P_2P_3P_6$

• So both answers are right. Nice observation. +1 Jan 26, 2013 at 9:28

Sorry for the bad format.

P = (P1+P2)(P3+P4)(P1+P3)(P5+P6)(P2+P5)(P4+P6)
P = (P1+P2)(P1+P3)(P4+P3)(P4+P6)(P5+P6)(P5+P2)         (rearranging terms)
P = (P1+P2*P3)(P4+P3*P6)(P5+P6*P2)                     (X+A)(X+B)=X+AB
P = (P1+P2*P3)(P4+P3*P6)P5+(P1+P2*P3)(P4+P3*P6)P6*P2   (distributive)
P = (P1+P2*P3)P4*P5+(P1+P2*P3)P3*P6*P5+(P1+P2*P3)P4*P6*P2+(P1+P2*P3)P3*P6*P2  (X*X=X)
P =  P1*P4*P5 + P2*P3*P4*P5 + P1*P3*P6*P5 + P2*P3*P6*P5 + P1*P4*P6*P2 + P2*P3*P4*P6 + P1*P3*P6*P2 + P2*P3*P6