let o be infinite Ordinal; not GrLexOrder o is well-ordering
set R = GrLexOrder o;
set r = RelStr(# (Bags o),(GrLexOrder o) #);
set ir = the InternalRel of RelStr(# (Bags o),(GrLexOrder o) #);
set cr = the carrier of RelStr(# (Bags o),(GrLexOrder o) #);
assume
GrLexOrder o is well-ordering
; contradiction
then A1:
GrLexOrder o is well_founded
by WELLORD1:def 4;
the carrier of RelStr(# (Bags o),(GrLexOrder o) #) = field the InternalRel of RelStr(# (Bags o),(GrLexOrder o) #)
by ORDERS_1:100;
then
the InternalRel of RelStr(# (Bags o),(GrLexOrder o) #) is_well_founded_in the carrier of RelStr(# (Bags o),(GrLexOrder o) #)
by A1, WELLORD1:5;
then A2:
RelStr(# (Bags o),(GrLexOrder o) #) is well_founded
by WELLFND1:def 2;
defpred S1[ Nat, set ] means $2 = (o --> 0 ) +* $1,1;
consider f being Function of NAT ,the carrier of RelStr(# (Bags o),(GrLexOrder o) #) such that
A10:
for n being Element of NAT holds S1[n,f . n]
from FUNCT_2:sch 3(A3);
reconsider f = f as sequence of RelStr(# (Bags o),(GrLexOrder o) #) ;
f is descending
proof
let n be
Nat;
WELLFND1:def 7 ( not f . (n + 1) = f . n & [(f . (n + 1)),(f . n)] in the InternalRel of RelStr(# (Bags o),(GrLexOrder o) #) )
reconsider n0 =
n as
Element of
NAT by ORDINAL1:def 13;
set fn1 =
f . (n0 + 1);
set fn =
f . n0;
A11:
f . (n0 + 1) = (o --> 0 ) +* (n + 1),1
by A10;
A12:
f . n0 = (o --> 0 ) +* n,1
by A10;
reconsider fn1 =
f . (n0 + 1) as
bag of
o ;
reconsider fn =
f . n0 as
bag of
o ;
A13:
n0 in omega
;
A14:
omega c= o
by CARD_3:102;
n <> n + 1
;
then A15:
fn1 . n =
(o --> 0 ) . n
by A11, FUNCT_7:34
.=
0
by A13, A14, FUNCOP_1:13
;
A16:
dom (o --> 0 ) = o
by FUNCOP_1:19;
then A17:
fn . n = 1
by A12, A13, A14, FUNCT_7:33;
then A20:
fn1 < fn
by A15, A17, PRE_POLY:def 9;
thus
f . (n + 1) <> f . n
by A12, A13, A14, A15, A16, FUNCT_7:33;
[(f . (n + 1)),(f . n)] in the InternalRel of RelStr(# (Bags o),(GrLexOrder o) #)
fn1 <=' fn
by A20, PRE_POLY:def 10;
then A21:
[(f . (n + 1)),(f . n)] in LexOrder o
by PRE_POLY:def 14;
consider tn being
FinSequence of
NAT such that A22:
TotDegree fn = Sum tn
and A23:
tn = fn * (SgmX (RelIncl o),(support fn))
by Def4;
consider tn1 being
FinSequence of
NAT such that A24:
TotDegree fn1 = Sum tn1
and A25:
tn1 = fn1 * (SgmX (RelIncl o),(support fn1))
by Def4;
A26:
n + 1
in omega
;
omega c= o
by CARD_3:102;
then reconsider nn =
n,
n1n =
n + 1 as
Element of
o by A13, A26;
A27:
field (RelIncl o) = o
by WELLORD2:def 1;
RelIncl o is
well-ordering
by WELLORD2:7;
then A28:
RelIncl o linearly_orders o
by A27, ORDERS_1:107, ORDERS_1:133;
then
support fn1 = {n1n}
by TARSKI:2;
then A30:
SgmX (RelIncl o),
(support fn1) = <*n1n*>
by A28, Th12, ORDERS_1:134;
A31:
dom fn = o
by A12, A16, FUNCT_7:32;
A32:
dom fn1 = o
by A11, A16, FUNCT_7:32;
then
support fn = {nn}
by TARSKI:2;
then
SgmX (RelIncl o),
(support fn) = <*nn*>
by A28, Th12, ORDERS_1:134;
then A34:
tn =
<*(fn . n)*>
by A23, A31, FINSEQ_2:38
.=
<*1*>
by A12, A13, A14, A16, FUNCT_7:33
.=
<*(fn1 . n1n)*>
by A11, A16, FUNCT_7:33
.=
tn1
by A25, A30, A32, FINSEQ_2:38
;
for
a,
b,
c being
bag of
o st
[a,b] in LexOrder o holds
[(a + c),(b + c)] in LexOrder o
by Def7;
hence
[(f . (n + 1)),(f . n)] in the
InternalRel of
RelStr(#
(Bags o),
(GrLexOrder o) #)
by A21, A22, A24, A34, Def9;
verum
end;
hence
contradiction
by A2, WELLFND1:15; verum