Mapping Method and solution of system of boolean equations as of 12/06/2019

Original system

(x1⊕y1) => (x2⊕y2) =1
(x1⊕y1) => (x2⊕y2) =1
(x2⊕y2) => (x3⊕y3) =1
(x3⊕y3) => (x4⊕y4) =1
(x1⊕x4) => (y1⊕y4) =1

Solve first system 1

(x1⊕y1) => (x2⊕y2) =1
(x1⊕y1) => (x2⊕y2) =1
(x2⊕y2) => (x3⊕y3) =1
(x3⊕y3) => (x4⊕y4) =1

Solve second system 2

(x1⊕y1) => (x2⊕y2) =1
(x1⊕y1) => (x2⊕y2) =1
(x2⊕y2) => (x3⊕y3) =1
(x3⊕y3) => (x4⊕y4) =1
(x1⊕x4) => (y1⊕y4) =0

and deduct from number of solutions
system 1 number of solutions system 2

    Passing Polyakov's control