Solution of problem type 23 from VK's News Wire 10.03.18 via technique developed in mea-2016-8.pdf

    Bellow we intend to benefit from idea of H.A, Mironchick  building a graph
    for system of boolean equations using just one variable for establishing
    connection  between equations in system on snapshot  above.

   Perform substitute :-

     x1 ~ x3 = z1; x2 ~ x4 = z2; x5 ~ x7 = z3; x6 ~ x8 = z4; x9 ~ x10 = z5

    Convert original system into

    (z1 v z2)^(z1 => z2) = 0
    (z2 v z3)^(z2 => z3) = 0 
    (z3 v z4)^(z3 => z4) = 0
    (z4 v z5)^(z4 => z5) = 0
    (z1 => z2) => (z5 => z4) =1

    Build graph to solve system of first four equations
   

   (z1 => z2) =>(z5=> z4) = ( ¬z1 v z2) => (¬z5 v z4) =
   ¬(¬z1 v z2) v ( ¬z5 v z4) = z1^(¬z2)  v ¬z5 v z4  = 1

Passing Polyakov's control

  

  In other words we get 2^5 + 2^5 = 2^6 = 64 ( solutions for {x})

 Straight forward mapping would look like

  ((x1~x3)v(x2~x4))^((x1~x3)=>(x2~x4)) = 0
  ((x2~x4)v(x5~x7))^((x2~x4)=>(x5~x7)) = 0
  ((x5~x7)v(x6~x8))^((x5~x7)=>(x6~x8)) = 0
  ((x6~x8)v(x9~x10))^((x6~x8)=>(x9~x10)) = 0
  ((x1~x3)=>(x2~x4))=>((x9~x10)=>(x6~x8)) = 1

   Build diagrama based on first equation which would work
   for first four equations. Notice that transition pairs are x2x4 , x5x7 , x6x8  

   In fact adding 
         ((x1~x3)=>(x2~x4))=>((x9~x10)=>(x6~x8)) = 1 
   would add  ( 0 => 0) => (0 =>0)  =1 on lines "01" and "10"
   e.g. True => True = 1 . So adding  this 5-th equation won't decrease number      of solutions of original sytem

   Thus answer is 64