Complicated system of boolean equations kind of P-40 (ege23.pdf) && Mapping method

Use case bellow has obvious typos as missing right hand side values of equations. Just assume  that right hand side of equations bellow has 0 (zerro) values and attempt to create approach to generate matrix with given values.
In case of values 1-th it is obvious that both diagrams will be updated corespondently  and will generate different matrix. Same schema would work
for values sequence 1,0,1,0,1,0,1,0 ..... with another set of fork diagrams.

((x1^y1 = x3^y3)) =>  (x2^y2) = 1
((x2^y2 = x4^y4)) => ¬(x3^y3) = 1
((x3^y3 = x5^y5)) =>  (x4^y4) = 1
((x4^y4 = x6^y6)) => ¬(x5^y5) = 1
((x5^y5 = x7^y7)) =>  (x6^y6) = 1
((x6^y6 = x8^y8)) => ¬(x7^y7) = 1

zj=zj^yj

Just repeat.  If they are 1-th  (on right hand side)  or changing 1 with 0 on regular basis, same schema would work as well with corespondently  updated  fork diagrams  per right hand side value of first and second equations.
I mean there should be just two  diagrams generating matrix.
Each outgoing value of zj having value 0 should mutiply cell's value
content by 3. Here situation looks like P-40 outgoing values 0 due to in P-40
has zj=xj^yj . Setup  Helen's Mironchick may be viewed in ege23.pdf  close to top of document
  

  

Consider "0" right hand size case

((x1^y1 = x3^y3)) =>  (x2^y2) = 0
((x2^y2 = x4^y4)) => ¬(x3^y3) = 0
((x3^y3 = x5^y5)) =>  (x4^y4) = 0
((x4^y4 = x6^y6)) => ¬(x5^y5) = 0
((x5^y5 = x7^y7)) =>  (x6^y6) = 0
((x6^y6 = x8^y8)) => ¬(x7^y7) = 0

zj=xj^yj

  

((x1vy1 = x3vy3)) =>  (x2vy2) = 0
((x2vy2 = x4vy4)) => ¬(x3vy3) = 0
((x3vy3 = x5vy5)) =>  (x4vy4) = 0
((x4vy4 = x6vy6)) => ¬(x5vy5) = 0
((x5vy5 = x7vy7)) =>  (x6vy6) = 0
((x6vy6 = x8vy8)) => ¬(x7vy7) = 0
((x7vy7 = x9vy9)) => ¬(x8vy8) = 0

zj=xjvyj