Running cross-reference tables with systems of Boolean Equations of triple predicates

Original system

(x1 v y1)^z1 v ((x2 =>y2)=>z2)=1
(x2 v y2)^z2 v ((x3 =>y3)=>z3)=1
(x3 v y3)^z3 v ((x4 =>y4)=>z4)=1
(x4 v y4)^z4 v ((x5 =>y5)=>z5)=1

    In particular,  #149 from ege23.pdf might be solved same way pretty shortly vs classical solution via triples connectivity.
  You might also want to compare suggested approach with Demo provided by Informatik BU   for #149  https://www.youtube.com/watch?v=W1oGIwgpw8A  (1 hr  33 minutes )

     Running cross-reference tables based on  08.2016 Charts style VERSUS classic BU's calculation
   

      

   Polyakov's Control for system (1)

  

Another system

(x1 v y1)^z1=>((x2=>y2)=>z2)=1

(x2 v y2)^z2=>((x3=>y3)=>z3)=1

(x3 v y3)^z3=>((x4=>y4)=>z4)=1

(x4 v y4)^z4=>((x5=>y5)=>z5)=1

    Polyakov's Control