Solution Another 23rd problem in 08/2016 format from VK's Informatics_100 News Wire as of 12/09/2019

    https://vk.com/informatics_100?z=photo-40390768_457275192%2Fwall-40390768_196117
  
   (x1^y1)=>(¬x2 v ¬y2) ≡  ¬(x1^y1) v ¬(x2^y2) = ¬((x1^y1)^(x2^y2))
   ¬((x1^y1)^(x2^y2)) =1 ≡ (x1^y1)^(x2^y2)=0

   

   In particular case 08/2016 schema appears to be much easier
   then analysis of truth tables and arrows (pair) diagram

   

  Original system is equivalent.

   (x1^y1)^(x2^y2) =0
   (x2^y2)^(x3^y3) =0
   (x3^y3)^(x4^y4) =0
   (x4^y4)^(x5^y5) =0
   x2^y4=0
   Fork two 08/2016 diagrams ( free your mind )