Style 08.2016 on the go (Solving systems of Boolen Equations for USE Informatics)

Original system

(x1=>y1)^(x1 v x2)^¬(x1^x2)=1
(x2=>y2)^(x2 v x3)^¬(x2^x3)=1
(x3=>y3)^(x3 v x4)^¬(x3^x4)=1
(x4=>y4)^(x4 v x5)^¬(x4^x5)=1
(x5=>y5)^(x5 v x6)^¬(x5^x6)=1
(x6=>y6)^(x6 v x7)^¬(x6^x7)=1
(x7=>y7)=1

Notice first ( due to De Morgan rules ) :-

(x1 v x2)^¬(x1^x2)=(x1 v x2)^(¬x1 v ¬x2)=
=x1^¬x2 v x2^¬x1 = x1⊕x2

Convert to equivalent

(x1=>y1)^(x1⊕x2)=1
(x2=>y2)^(x2⊕x3)=1
(x3=>y3)^(x3⊕x4)=1
(x4=>y4)^(x4⊕x5)=1
(x5=>y5)^(x5⊕x6)=1
(x6=>y6)^(x6⊕x7)=1
(x7=>y7)=1

Fork 08.2016 chart to solve the system