Solution of one system of boolean equations via reverse pass per Helen Mironchick

Down here we rely on technique proposed in
https://mapping-metod.blogspot.com/2019/03/blog-post.html

Original system

((x1^x2=>x3)^x4=>x5)^x6=>x7=1
((y1^y2=>y3)^y4=>y5)^y6=>y7=1
x1=>y7=1

#1 - x1
#2 - x1^x2
#3 - (x1^x2=>x3)
#4 - (x1^x2=>x3)^x4
#5 - (x1^x2=>x3)^x4=>x5)
#6 - (x1^x2=>x3)^x4=>x5)^x6
#7 - ((x1^x2=>x3)^x4=>x5)^x6=>x7=1

  
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