let L be non empty reflexive antisymmetric RelStr ; for R being auxiliary(i) Relation of L
for C being strict_chain of R st ( for c being Element of L st c in C holds
( ex_sup_of SetBelow (R,C,c),L & c = sup (SetBelow (R,C,c)) ) ) holds
R satisfies_SIC_on C
let R be auxiliary(i) Relation of L; for C being strict_chain of R st ( for c being Element of L st c in C holds
( ex_sup_of SetBelow (R,C,c),L & c = sup (SetBelow (R,C,c)) ) ) holds
R satisfies_SIC_on C
let C be strict_chain of R; ( ( for c being Element of L st c in C holds
( ex_sup_of SetBelow (R,C,c),L & c = sup (SetBelow (R,C,c)) ) ) implies R satisfies_SIC_on C )
assume A1:
for c being Element of L st c in C holds
( ex_sup_of SetBelow (R,C,c),L & c = sup (SetBelow (R,C,c)) )
; R satisfies_SIC_on C
let x, z be Element of L; WAYBEL35:def 6 ( x in C & z in C & [x,z] in R & x <> z implies ex y being Element of L st
( y in C & [x,y] in R & [y,z] in R & x <> y ) )
assume that
A2:
x in C
and
A3:
z in C
and
A4:
[x,z] in R
and
A5:
x <> z
; ex y being Element of L st
( y in C & [x,y] in R & [y,z] in R & x <> y )
A6:
z = sup (SetBelow (R,C,z))
by A1, A3;
per cases
( for y being Element of L holds
( not y in SetBelow (R,C,z) or not x < y ) or ex y being Element of L st
( y in SetBelow (R,C,z) & x < y ) )
;
suppose A7:
for
y being
Element of
L holds
( not
y in SetBelow (
R,
C,
z) or not
x < y )
;
ex y being Element of L st
( y in C & [x,y] in R & [y,z] in R & x <> y )reconsider x =
x as
Element of
L ;
A8:
SetBelow (
R,
C,
z)
is_<=_than x
A13:
for
a being
Element of
L st
SetBelow (
R,
C,
z)
is_<=_than a holds
x <= a
by A2, A4, Th15;
ex_sup_of SetBelow (
R,
C,
z),
L
by A1, A3;
hence
ex
y being
Element of
L st
(
y in C &
[x,y] in R &
[y,z] in R &
x <> y )
by A13, A5, A6, A8, YELLOW_0:def 9;
verum end; suppose
ex
y being
Element of
L st
(
y in SetBelow (
R,
C,
z) &
x < y )
;
ex y being Element of L st
( y in C & [x,y] in R & [y,z] in R & x <> y )then consider y being
Element of
L such that A14:
y in SetBelow (
R,
C,
z)
and A15:
x < y
;
take
y
;
( y in C & [x,y] in R & [y,z] in R & x <> y )thus
y in C
by A14, Th15;
( [x,y] in R & [y,z] in R & x <> y )hence
[x,y] in R
by A2, A15, Th2;
( [y,z] in R & x <> y )thus
[y,z] in R
by A14, Th15;
x <> ythus
x <> y
by A15;
verum end; end;