Solution of several systems in Boolean variables from VK Informatics_100 admin's pool as of 05/06/2019

Down here goes a pretty standard systems a kind of often appear to be placed on group Informatics_100 News Wire. Just click on snapshot to obtain a perfect slide show of pictures been placed into current posting 

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

For instance convert first equation via de Morgan rules.

((x1~x2) v (x3~x4))^(¬((x1~x2)=>(x3~x4)))=1
((x1~x2) v (x3~x4))^(¬((x1⊕x2) v (x3~x4)))=1
((x1~x2) v (x3~x4))^( (x1~x2)^(x3⊕x4) )=1
(x1~x2)^(x3⊕x4) v (x3~x4)^(x1~x2)^(x3⊕x4) =1

Due to
   (x3~x4)^(x3⊕x4)=0
We get
(x1~x2)^(x3⊕x4)=1

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Continue those conversion the same way we finally get
***************************************************
(x1~x2)^(x3⊕x4)=1
(x2~x4)^(x5~x6)^(x7⊕x8) =1
(x1~x2)^(x7⊕x8)=1
(x1~x4)=>(x9~x10) =1

***************************************************
System which is ready to fork the diagram in style 08.2016
***************************************************
(1) x1~x2 =1
(2) x2~x4 =1
(3) x4⊕x3 =1
(4) x5~x6 =1
(5) x7⊕x8 =1
(6) x9~x10 =1

Build diagram per 08.2016 technique (Helen Mironchick)

   Passing Polyakov's Control

  Another sample from the same pool

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

  Fork diagram in 08.2016 style right away
  

    Passing Ployakov's Control