Monday 19 August 2019

Solution of systems of Boolean equations with three variables via format 08/2016

Система

(((x1≡y1)≡z1) v ((x2≡y2)≡z2)) ^ (x1 v y1 v z1)=1
(((x2≡y2)≡z2) v ((x3≡y3)≡z3)) ^ (x2 v y2 v z2)=1
(((x3≡y3)≡z3) v ((x4≡y4)≡z4)) ^ (x3 v y3 v z3)=1
(((x4≡y4)≡z4) v ((x5≡y5)≡z5)) ^ (x4 v y4 v z4)=1
(((x5≡y5)≡z5) v ((x6≡y6)≡z6)) ^ (x5 v y5 v z5)=1
(((x6≡y6)≡z6) v ((x7≡y7)≡z7)) ^ (x6 v y6 v z6)=1
(x7 v y7 v z7)=1

Решение в формате 08.2016


    Кортроль по Полякову


No comments:

Post a Comment

Featured Post

Solution of one USE Informatics system of Boolean equations in 08.2016 style

Original system Orinal system ¬(x1≡x2)v¬(x1≡x3)^(x2≡x3)=1 ¬(x3≡x4)v¬(x3≡x5)^(x4≡x5)=1 ¬(x5≡x6)v¬(x5≡x7)^(x6≡x7)=1 ¬(x7≡x8)v¬(x7≡x9...