Solution of one system of Boolean equations in 08.2016 format

Original system
((x1=>y1) =>z1) ^ ((x2 v y2 v z2) =>((x1≡y1)≡z1)) =1
((x2=>y2) =>z2) ^ ((x3 v y3 v z3) =>((x2≡y2)≡z2)) =1
((x3=>y3 )=>z3) ^ ((x4 v y4 v z4) =>((x3≡y3)≡z3)) =1
((x4=>y4) =>z4) ^ ((x5 v y5 v z5) =>((x4≡y4)≡z4)) =1
((x5=>y5) =>z5) ^ ((x6 v y6 v z6) =>((x5≡y5)≡z5)) =1
((x6=>y6) =>z6) ^ ((x7 v y7 v z7) =>((x6≡y6)≡z6)) =1
(x7≡y7)≡z7=1
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Consider the transmission on the line "0"
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((xj=yj)=zj) ======> !xj^!y^!z
True triples
((xj=yj)=zj) ===*3==> !xj*!yj*!zj
001                                 ==>110 
010                                 ==>101 
100                                 ==>011
111                                 ==>000
We obtain exactly ((xj = yj) = zj) four false triples
110 
101 
100 
111 
Thus, on the line "0" we transmit false triples
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Consider the transmission on the line "1"
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In the lower right node has total 24 triples. Notice that the left upper node gives diagonally 12 true bit triples, and the lower left node gives 12 false ones. The lower left node generates 12 triples that are not supposed to be transmitted (they are false). 12 bit triples from the upper left node give true triples and they are transmitted, 12 from the lower left node give 0 (stop on line 1)

Three "3" is the power of intersection of the truth regions (xj => yj) => zj and (xj≡yj) ≡zj
this is {001,100,111} = {001,011,100,101,111} ∩ {010,001,100,111}
Convert  implication     (x2 v y2 v z2) =>((x1≡y1)≡z1)  into  ((x1≡y1)≡z1) v  !x2^!y2^!z2
and so on

    All underlined passed Polyakov's control