Solution of problems kind of P154,P175 from ege23.doc via Classic Mapping Method

Original system  kind of P154

(((x1≡y1)≡z1)=>((x2≡y2)≡z2))^(x1=>x2)^(y1=>y2)^(z1=>z2)=1
(((x2≡y2)≡z2)=>((x3≡y3)≡z3))^(x2=>x3)^(y2=>y3)^(z2=>z3)=1
(((x3≡y3)≡z3)=>((x4≡y4)≡z4))^(x3=>x4)^(y3=>y4)^(z3=>z4)=1
(((x4≡y4)≡z4)=>((x5≡y5)≡z5))^(x4=>x5)^(y4=>y5)^(z4=>z5)=1
(((x5≡y5)≡z5)=>((x6≡y6)≡z6))^(x5=>x6)^(y5=>y6)^(z5=>z6)=1
(((x6≡y6)≡z6)=>((x7≡y7)≡z7))^(x6=>x7)^(y6=>y7)^(z6=>z7)=1

Solution via Mapping Method

     Passing Polyakov's Control separate screen

Original system

Video https://www.youtube.com/watch?v=R7UqPN3-l4k  contains a special consideration been done by Informatic BU. BU's logic is obviously driven by bit chains .
Consider  the same task for 3 variables and avoid wasting time doing bit chains logic,
we would apply Classic version of Mapping Method (2013)
See http://kpolyakov.spb.ru/download/mea-2013-10.pdf

I guess in particular case both truth tables && arrows diagrams are quite trivial
and mapping method works with no issues.

(¬x1 v ¬y1 v ¬z1)^((x1 v y1 v z1) =>(x2 v y2 v z2)) = 1
(¬x2 v ¬y2 v ¬z2)^((x2 v y2 v z2) =>(x3 v y3 v z3)) = 1
(¬x3 v ¬y3 v ¬z3)^((x3 v y3 v z3) =>(x4 v y4 v z4)) = 1
(¬x4 v ¬y4 v ¬z4)^((x4 v y4 v z4) =>(x5 v y5 v z5)) = 1
(¬x5 v ¬y5 v ¬z5)^((x5 v y5 v z5) =>(x6 v y6 v z6)) = 1
(¬x6 v ¬y6 v ¬z6)^((x 6 v y6 v z6) =>(x7 v y7 v z7)) = 1
(¬x7 v ¬y7 v ¬z7)=1