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

    Convert system to equivalent
   
    (x1 =>¬y1) => (x2 ≡ ¬y2) =1  ≡   (¬x1 v ¬y1) => (x2⊕y2) =1
    (¬x1 v ¬y1) => (x2⊕y2) =1 ≡  ¬(x1^y1) => (x2⊕y2) =1
   ¬(x1^y1) => (x2⊕y2) =1 ≡  (x1^y1) v (x2⊕y2)=1

    Conversion is done
    New system looks like

    (x1^y1) v (x2⊕y2)=1
    (x2^y2) v (x3⊕y3)=1
    (x3^y3) v (x4⊕y4)=1
    (x4^y4) v (x5⊕y5)=1
    (x5^y5) v (x6⊕y6)=1

    The core block to replicate per 08/2016 technique
    

  Now fork 08/2016 diagram