Solution one more system (another draft ) of Boolean equations kind of P40 with different outgoing multipliers for different Mapping Method table rows as of 18/07/19

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

Follow an idea proposed in P-40 solution (Helen Mironchick ) just with more or less complicated expressions for w1,w2,w3,w4,w5,w6. However, in particular case we would have different outgoing multipliers for bit 0 it would be 5 and for bit 1 it would be 3 vs previously considered ones.

Now make a substitutions ( kind of design had been suggested in P-40 from ege23.pdf )

   w1=(x1 v y1)^z1

   w2=(x2 v y2)^z2
   w3=(x3 v y3)^z3
   w4=(x4 v y4)^z4
   w5=(x5 v y5)^z5
   w6=(x6 v y6)^z6

   Outgoing multipliers are located  at right hand side of diagrams for  "identity" and "xor"
  

    Passing Polyakov's Control