Chains of tie nodes of a multi-part graph && reverse pass to center VS Bit's Chains Method proposed by Informatick BU in his streams 2019

Once again I strongly believe that blindly drive under bit's chains pressure 
is not a smart way of 23-rd problem development of USE's in Informatics.
Almost any case been treated with bit's chains might be converted to 08/2016
system's design. Otherwise, Mapping Method utilising pairs for establishing
connection between equations of system would solve the problem with no issues at all.

Solution provided by Informatick BU on snapshot below
Stream 51 2019 (2 hr 32 min )
https://www.youtube.com/watch?v=bzaBV7GkW74

System   itself

Here we intend to convert system above to equivalent  with minimal knowledge of Boolean Algebra

(x10=>x9)=>(x8=>x7)=1
(x8=>x7)=>(x6=>x5)=1
(x6=>x5)=>(x4=>x3)=1
(x4=>x3)=>(x2=>x1)=1

Generation simplest 08/2016 style chart  for converted system

Now consider
https://www.youtube.com/watch?time_continue=2844&v=0Aq8fell3gE&feature=emb_logo
( 51 min )

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

   (2)
   (x1=>x2)^(x2=>x3)^(x3=>x4)^(x4=>x5)=1
   (y1=>y2)^(y2=>y3)^(y3=>y4)^(y4=>y5)=1
   (z1=>z2)^(z2=>z3)^(z3=>z4)^(z4=>z5)=1
   x2 v y3 v z4=1