let G be _Graph; :: thesis:
let C be Path of G; :: thesis:
assume that
A1: C is Cycle-like and
A2: C .length() > 3 ; :: thesis:
C .length() >= 3 + 1 by A2, NAT_1:13;
then 2 * (C .length()) >= 2 * 4 by XREAL_1:64;
then (2 * (C .length())) + 1 >= 8 + 1 by XREAL_1:7;
then A3: len C >= 9 by GLIB_001:112;
let x be Vertex of G; :: thesis:
assume x in C .vertices() ; :: thesis:
then consider n being odd Element of NAT such that
A4: n <= len C and
A5: C . n = x by GLIB_001:87;
A6: 0 + 1 <= n by ABIAN:12;

per cases ;

suppose A7: ( n = 1 or n = len C ) ; :: thesis:

reconsider k = (2 * 0) + 1 as odd Nat ;

A8: now :: thesis:

per cases by A7;

suppose n = 1 ; :: thesis:

hence x = C . k by A5; :: thesis:

end;

suppose n = len C ; :: thesis:

then x = C .last() by A5;
then x = C .first() by A1, GLIB_001:def 24;
hence x = C . k ; :: thesis:

end;

(len C) + (- 2) >= 9 + (- 2) by A3, XREAL_1:7;
then (len C) - (2 * 1) >= 0 by XXREAL_0:2;
then (len C) - 2 is odd Element of NAT by INT_1:3;
then reconsider j = (len C) - 2 as odd Nat ;
reconsider i = (2 * 1) + 1 as odd Nat ;
take i ; :: thesis:
take j ; :: thesis:
A9: (len C) + (- 2) >= 9 + (- 2) by A3, XREAL_1:7;
hence i + 2 < j by XXREAL_0:2; :: thesis:
A10: (len C) + 0 > (len C) + (- 2) by XREAL_1:8;
hence j <= len C ; :: thesis:
i < j by A9, XXREAL_0:2;
then A11: i < len C by A10, XXREAL_0:2;

hereby :: thesis:

assume A12: C . i = C . j ; :: thesis:
(i + 2) + (- 2) < j + 0 by A9, XXREAL_0:2;
hence contradiction by A10, A12, GLIB_001:def 28; :: thesis:

end;

A13: (len C) + 0 > 9 + (- 6) by A3, XREAL_1:8;
then reconsider Ci = C . i as Vertex of G by GLIB_001:7;
(len C) + 0 > 9 + (- 8) by A3, XREAL_1:8;
then C . (k + 1) Joins C . k,C . i,G by GLIB_001:def 3;
then A14: x,Ci are_adjacent by A8;
x <> Ci by A13, A8, GLIB_001:def 28;
hence C . i in G .AdjacentSet {x} by A14, Th51; :: thesis:
A15: j <> 1 by A9;
A16: j in NAT by ORDINAL1:def 12;
then reconsider Cj = C . j as Vertex of G by A10, GLIB_001:7;

A17: now :: thesis:

per cases by A7;

suppose n = 1 ; :: thesis:

then x = C .first() by A5;
then x = C .last() by A1, GLIB_001:def 24;
hence x = C . (j + 2) ; :: thesis:

end;

suppose n = len C ; :: thesis:

hence x = C . (j + 2) by A5; :: thesis:

end;

A18: now :: thesis:

assume x = Cj ; :: thesis:
then A19: j = 1 by A16, A10, A17, GLIB_001:def 28;
j + 2 = len C ;
hence contradiction by A3, A19; :: thesis:

end;

C . (j + 1) Joins C . j,C . (j + 2),G by A16, A10, GLIB_001:def 3;
then x,Cj are_adjacent by A17, Def3;
hence C . j in G .AdjacentSet {x} by A18, Th51; :: thesis:
let e be object ; :: thesis:
assume that
A20: e in C .edges() and
A21: e Joins C . i,C . j,G ; :: thesis:
consider k being even Element of NAT such that
A22: 1 <= k and
A23: k <= len C and
A24: C . k = e by A20, GLIB_001:99;
A25: ( ( (the_Source_of G) . e = C . i & (the_Target_of G) . e = C . j ) or ( (the_Source_of G) . e = C . j & (the_Target_of G) . e = C . i ) ) by A21;
k in dom C by A22, A23, FINSEQ_3:25;
then consider ku1 being odd Element of NAT such that
A26: ku1 = k - 1 and
A27: k - 1 in dom C and
A28: k + 1 in dom C and
A29: C . k Joins C . ku1,C . (k + 1),G by GLIB_001:9;
A30: ku1 <= len C by A26, A27, FINSEQ_3:25;
A31: k + 1 <= len C by A28, FINSEQ_3:25;

per cases by A24, A29, A25;

suppose A32: ( C . i = C . ku1 & C . j = C . (k + 1) ) ; :: thesis:

then i + 2 = (k - 1) + (1 + 1) by A26, A30, A11, Th25
.= j by A10, A31, A15, A32, Th25 ;
hence contradiction by A9; :: thesis:

end;

suppose A33: ( C . i = C . (k + 1) & C . j = C . ku1 ) ; :: thesis:

then i = k + 1 by A31, A11, Th25;
hence contradiction by A10, A26, A30, A15, A33, Th25; :: thesis:

end;

suppose A34: ( not n = 1 & not n = len C ) ; :: thesis:

then (2 * 0) + 1 < n by A6, XXREAL_0:1;
then 1 + 2 <= n by Th4;
then 3 + (- 2) <= n + (- 2) by XREAL_1:7;
then n - (2 * 1) is odd Element of NAT by INT_1:3;
then reconsider i = n - (2 * 1) as odd Nat ;
A35: i + 0 < i + 2 by XREAL_1:8;
then reconsider Ci = C . i as Vertex of G by A4, GLIB_001:7, XXREAL_0:2;
reconsider j = n + 2 as odd Nat ;
take i ; :: thesis:
take j ; :: thesis:
n + 0 < n + 2 by XREAL_1:8;
hence i + 2 < j ; :: thesis:
A36: n < len C by A4, A34, XXREAL_0:1;
hence A37: j <= len C by Th4; :: thesis:
then reconsider Cj = C . j as Vertex of G by GLIB_001:7;
A38: i in NAT by ORDINAL1:def 12;

A39: now :: thesis:

A40: n + 0 > n + (- 2) by XREAL_1:8;
assume Ci = x ; :: thesis:
hence contradiction by A4, A5, A34, A38, A40, GLIB_001:def 28; :: thesis:

end;

A41: now :: thesis:

assume that
A42: i = 1 and
A43: j = len C ; :: thesis:
j = i + 4 ;
hence contradiction by A3, A42, A43; :: thesis:

end;

hereby :: thesis:

A44: (i + 2) + (- 2) < j + 0 by XREAL_1:8;
assume C . i = C . j ; :: thesis:
hence contradiction by A38, A37, A41, A44, GLIB_001:def 28; :: thesis:

end;

A45: i < len C by A4, A35, XXREAL_0:2;
then C . (i + 1) Joins C . i,C . (i + 2),G by A38, GLIB_001:def 3;
then x,Ci are_adjacent by A5, Def3;
hence C . i in G .AdjacentSet {x} by A39, Th51; :: thesis:
1 + 2 <= j by A6, XREAL_1:7;
then A46: j <> 1 ;
A47: n + 2 <= ((len C) - 2) + 2 by A36, Th4;

A48: now :: thesis:

A49: n + 2 > n + 0 by XREAL_1:8;
assume Cj = x ; :: thesis:
hence contradiction by A5, A34, A47, A49, GLIB_001:def 28; :: thesis:

end;

C . (n + 1) Joins C . n,C . j,G by A36, GLIB_001:def 3;
then x,Cj are_adjacent by A5;
hence C . j in G .AdjacentSet {x} by A48, Th51; :: thesis:
let e be object ; :: thesis:
assume that
A50: e in C .edges() and
A51: e Joins C . i,C . j,G ; :: thesis:
consider k being even Element of NAT such that
A52: 1 <= k and
A53: k <= len C and
A54: C . k = e by A50, GLIB_001:99;
A55: ( ( (the_Source_of G) . e = C . i & (the_Target_of G) . e = C . j ) or ( (the_Source_of G) . e = C . j & (the_Target_of G) . e = C . i ) ) by A51;
1 + 1 <= k + 1 by A52, XREAL_1:7;
then A56: k + 1 <> 1 ;
A57: k - 1 < len C by A53, XREAL_1:146, XXREAL_0:2;
k in dom C by A52, A53, FINSEQ_3:25;
then consider ku1 being odd Element of NAT such that
A58: ku1 = k - 1 and
k - 1 in dom C and
A59: k + 1 in dom C and
A60: C . k Joins C . ku1,C . (k + 1),G by GLIB_001:9;
A61: k + 1 <= len C by A59, FINSEQ_3:25;

per cases by A54, A60, A55;

suppose A62: ( C . i = C . ku1 & C . j = C . (k + 1) ) ; :: thesis:

then i + 2 = (k - 1) + (1 + 1) by A45, A58, A57, Th25
.= j by A37, A61, A46, A56, A62, Th25 ;
hence contradiction ; :: thesis:

end;

suppose A63: ( C . i = C . (k + 1) & C . j = C . ku1 ) ; :: thesis:

per cases by A37, A45, A58, A57, A61, A63, Th25;

suppose ( i = k + 1 & j = k - 1 ) ; :: thesis:

hence contradiction ; :: thesis:

end;

suppose A64: ( i = k + 1 & k - 1 = 1 & j = len C ) ; :: thesis:

j = i + 4 ;
hence contradiction by A3, A64; :: thesis:

end;

suppose A65: ( i = 1 & k + 1 = len C & j = k - 1 ) ; :: thesis:

then k + 1 = 7 ;
hence contradiction by A3, A65; :: thesis:

end;

suppose ( i = 1 & k + 1 = len C & k - 1 = 1 & j = len C ) ; :: thesis:

hence contradiction ; :: thesis:

end;