Lessons been taught from Helen Mironchick regarding technique of solving the systems of equations in boolean variables as of 08.2016

Solution of system 

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

Down here we would have the only one transition variable between equations of the system.
Basic diagram

 
Solving first four equations

   Next consider the System
   (y5+x5) =1
   (x5+x6) =1
   (x6+y6) =1
  

  

  With node x5=0 system would have  55*2 = 110
  solutions. With node x5=1 system would have  34*6 = 204
  solutions. Total number of solutions would be 314

 
System usually treated via bitmasks tables
concatenation. In particular case could hardly be
solved as it happens with fourth equation kind of
x6^y6^z6=0

(x1=>x2)^(x2=>x3)^(x3=>x4)^(x4=>x5)^(x5=>x6)=1
(y1=>y2)^(y2=>y3)^(y3=>y4)^(y4=>y5)^(y5=>y6)=1
(z1=>z2)^(z2=>z3)^(z3=>z4)^(z4=>z5)^(z5=>z6)=1
(x6=>y6)=>z6=1

works as follows

 
   

 Refrences

 1. ВК1.PDF is attached to page
 https://vk.com/club180658320?w=wall-180658320_68%2Fall