Another (like P-40) Complicated system of boolean equations && 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.

    Assume right hand values are equal  0 (all), otherwise regenerate fork diagrams correspodently

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

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 is exactly like  in P-40 :-  zj=xj^yj . Setup with no
changing forking diagrams belongs to Helen Mironchick and may be viewed in ege23.pdf  close to top of document

   Solution   zj = xj^yj

Case "1"
  
(z1~z2) =>z3 = 1
(z3=>z2) => z4 = 1
(z3~z4) => z5 = 1
(z5=>z4)=>z6 = 1
  

  

Case "0"
  
((x1 ^ y1) ~ (x2 ^ y2)) => (x3 ^ y3) = 0
((x2 ^ y2) v ¬(x3 ^ y3)) => (x4 ^ y4)  =  0
((x3 ^ y3) ~ (x4 ^ y4)) => (x5  ^ y5) = 0
((x4 ^ y4) v ¬(x5^ y5)) => (x6 ^ y6)  = 0

zj=xj^yj    Solution

(z1~z2) =>z3 = 0
(z3=>z2) => z4 = 0
(z3~z4) => z5 = 0
(z5=>z4)=>z6 = 0