One 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.

((x1 v y1) ~ (x2 v y2)) => (x3  v y3) = 1
((x2 v y2) v ¬(x3 v y3)) => (x4 v y4)  = 1
((x3 v y3) ~ (x4 v y4)) => (x5  v y5) = 1
((x4 v y4) v ¬(x5 v y5)) => (x6 v y6)  = 1
((x5 v y5) ~ (x6 v y6)) => (x7  v y7) = 1
((x6 v y6) v ¬(x7 v y7)) => (x8 v y8)  = 1
((x7 v y7) ~ (x8 v y8)) => (x9  v y9) = 1

zj = xj v yj (j=1,2,3,.....,9)


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 1 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 . Those setup belongs Helen Mironchick and may be viewed in ege23.pdf  close to top of document

(z1 ~ z2)  => z3 =1
(z2 v ¬z3) => z4 =1
(z3 ~ z4 ) => z5 =1
(z4 v ¬z5) => z6 =1
(z5 ~ z6)  => z7 =1
(z6 v ¬z7) => z8 =1
(z7 ~ z8) => z9  =1

((x1 v y1) ~ (x2 v y2)) => (x3 v y3) = 1   (z1z2 z2z3 -diagram1)
((x2 v y2) v ¬(x3 v y3)) => (x4 v y4) = 1 (z2z3 z3z4 -diagram2)
((x3 v y3) ~ (x4 v y4)) => (x5 v y5) = 1   diagram1
((x4 v y4) v ¬(x5 v y5)) => (x6 v y6) = 1 diagram2
((x5 v y5) ~ (x6 v y6)) => (x7 v y7) = 1   diagram1
((x6 v y6) v ¬(x7 v y7)) => (x8 v y8) = 1 diagram2
((x7 v y7) ~ (x8 v y8)) => (x9 v y9) = 1  diagram1

   Loading original system matches Polyakov's calculator result
 

Another sample

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

zj=xj v yj

(z1 ~ z2) => z3 = 1
(z2 ~ z3) => ¬z4 = 1
(z3 ~ z4) => z5 = 1
(z4 ~ z5) => ¬z6 = 1
(z5 ~ z6) => z7 = 1
(z6 ~ z7) =>¬z8 = 1