Chains of tie nodes of a multi-part graph && reverse pass to center per Helen Mironchick's "The Mapping method with a reverse calculation" - Computer science at school #10 2019

Original system

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

Solution below is based on Helen Mironchick article in meantime pending publication. Intellectual property for idea of "Reverse pass to center" completely belongs Helen Mironchick. We just place into play her technique awaiting publication.

For X-LINE

#2 is (x1=>x2)
#3 is (x1=>x2)=>x3
#4 is ((x1=>x2)=>x3)=x4
#5 is (((x1=>x2)=>x3)=x4)=>x5

For Y-LINE and Z-LINE replace x by y and x by z correspondingly

*******************
Chaining order
*******************
1) We get x3 via reverse pass to center
2) Getting y3 via simple implication
3) Getting y4  via reverse pass to center
    Here we have y1=>y2=>y3 (#3) Y-LINE.
    Perform move from #5 to #4 in the opposite direction.
    Now calculate y4 as usual.
4) Getting z4 via simple implication
5) Completing Z-LINE having z4  (on line-0 213,on line-1-333)

Another sample

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

*******************
Chaining order
*******************
1) We get x3 via reverse pass to center
2) Getting y3 via direct XOR
3) Getting y4  via reverse pass to center
    Here we calculate y1=>y2=>y3 (#3) Y-LINE.
    Perform move from #5 to #4 in the opposite direction.
    Now calculate y4 as usual.
4) Getting z4 via direct XOR
5) Completing Z-LINE having z4  (on line-0 112 , on line-1-200)