let o be infinite Ordinal; :: thesis:
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 ; :: thesis:
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[ Element of NAT , set ] means $2 = (o --> 0 ) +* $1,1;

A3: now

let n be Element of NAT ; :: thesis:
set y = (o --> 0 ) +* n,1;
A4: dom (o --> 0 ) = o by FUNCOP_1:19;
reconsider y = (o --> 0 ) +* n,1 as ManySortedSet of ;
A5: ( n in omega & omega c= o ) by CARD_3:102;

now

let x be set ; :: thesis:

hereby :: thesis:

assume x in {n} ; :: thesis:
then x = n by TARSKI:def 1;
hence y . x <> 0 by A4, A5, FUNCT_7:33; :: thesis:

end;

assume that
A6: y . x <> 0 and
A7: not x in {n} ; :: thesis:
x <> n by A7, TARSKI:def 1;
then A8: y . x = (o --> 0 ) . x by FUNCT_7:34;

per cases ;

suppose x in dom (o --> 0 ) ; :: thesis:

hence contradiction by A6, A8, FUNCOP_1:13; :: thesis:

end;

suppose not x in dom (o --> 0 ) ; :: thesis:

hence contradiction by A6, A8, FUNCT_1:def 4; :: thesis:

end;

then support y = {n} by POLYNOM1:def 7;
then y is finite-support by POLYNOM1:def 8;
then reconsider y = y as Element of the carrier of RelStr(# (Bags o),(GrLexOrder o) #) by POLYNOM1:def 14;
take y = y; :: thesis:
thus S₁[n,y] ; :: thesis:

end;

consider f being Function of NAT ,the carrier of RelStr(# (Bags o),(GrLexOrder o) #) such that
A9: 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 Element of NAT ; :: according to WELLFND1:def 7 :: thesis:
set fn1 = f . (n + 1);
set fn = f . n;
A10: f . (n + 1) = (o --> 0 ) +* (n + 1),1 by A9;
A11: f . n = (o --> 0 ) +* n,1 by A9;
reconsider fn1 = f . (n + 1) as bag of by POLYNOM1:def 14;
reconsider fn = f . n as bag of by POLYNOM1:def 14;
A12: ( n in omega & omega c= o ) by CARD_3:102;
n <> n + 1 ;
then A13: fn1 . n = (o --> 0 ) . n by A10, FUNCT_7:34
.= 0 by A12, FUNCOP_1:13 ;
A14: dom (o --> 0 ) = o by FUNCOP_1:19;
then A15: fn . n = 1 by A11, A12, FUNCT_7:33;

now

let l be Ordinal; :: thesis:
assume A16: l in n ; :: thesis:
then A17: l <> n ;
n < n + 1 by NAT_1:13;
then n in { i where i is Element of NAT : i < n + 1 } ;
then n in n + 1 by AXIOMS:30;
then n c= n + 1 by ORDINAL1:def 2;
then l in n + 1 by A16;
then l <> n + 1 ;
hence fn1 . l = (o --> 0 ) . l by A10, FUNCT_7:34
.= fn . l by A11, A17, FUNCT_7:34 ;
:: thesis:

end;

then A18: fn1 < fn by A13, A15, POLYNOM1:def 11;
thus f . (n + 1) <> f . n by A11, A12, A13, A14, FUNCT_7:33; :: thesis:
fn1 <=' fn by A18, POLYNOM1:def 12;
then A19: [(f . (n + 1)),(f . n)] in LexOrder o by POLYNOM1:def 16;
consider tn being FinSequence of NAT such that
A20: TotDegree fn = Sum tn and
A21: tn = fn * (SgmX (RelIncl o),(support fn)) by Def4;
consider tn1 being FinSequence of NAT such that
A22: TotDegree fn1 = Sum tn1 and
A23: tn1 = fn1 * (SgmX (RelIncl o),(support fn1)) by Def4;
( n + 1 in omega & omega c= o ) by CARD_3:102;
then reconsider nn = n, n1n = n + 1 as Element of o by A12;
A24: field (RelIncl o) = o by WELLORD2:def 1;
RelIncl o is well-ordering by WELLORD2:7;
then A25: RelIncl o linearly_orders o by A24, ORDERS_1:107, ORDERS_1:133;

now

let x be set ; :: thesis:

hereby :: thesis:

assume A26: x in support fn1 ; :: thesis:

now

assume x <> n1n ; :: thesis:
then fn1 . x = (o --> 0 ) . x by A10, FUNCT_7:34;
then fn1 . x = 0 by A26, FUNCOP_1:13;
hence contradiction by A26, POLYNOM1:def 7; :: thesis:

end;

hence x in {n1n} by TARSKI:def 1; :: thesis:

end;

assume x in {n1n} ; :: thesis:
then x = n1n by TARSKI:def 1;
then fn1 . x = 1 by A10, A14, FUNCT_7:33;
hence x in support fn1 by POLYNOM1:def 7; :: thesis:

end;

then support fn1 = {n1n} by TARSKI:2;
then A27: SgmX (RelIncl o),(support fn1) = <*n1n*> by A25, Th12, ORDERS_1:134;
A28: ( dom fn = o & dom fn1 = o ) by A10, A11, A14, FUNCT_7:32;

now

let x be set ; :: thesis:

hereby :: thesis:

assume A29: x in support fn ; :: thesis:

now

assume x <> nn ; :: thesis:
then fn . x = (o --> 0 ) . x by A11, FUNCT_7:34;
then fn . x = 0 by A29, FUNCOP_1:13;
hence contradiction by A29, POLYNOM1:def 7; :: thesis:

end;

hence x in {nn} by TARSKI:def 1; :: thesis:

end;

assume x in {nn} ; :: thesis:
then x = nn by TARSKI:def 1;
then fn . x = 1 by A11, A14, FUNCT_7:33;
hence x in support fn by POLYNOM1:def 7; :: thesis:

end;

then support fn = {nn} by TARSKI:2;
then SgmX (RelIncl o),(support fn) = <*nn*> by A25, Th12, ORDERS_1:134;
then A30: tn = <*(fn . n)*> by A21, A28, Th3
.= <*1*> by A11, A12, A14, FUNCT_7:33
.= <*(fn1 . n1n)*> by A10, A14, FUNCT_7:33
.= tn1 by A23, A27, A28, Th3 ;
for a, b, c being bag of 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 A19, A20, A22, A30, Def9; :: thesis:

end;

hence contradiction by A2, WELLFND1:15; :: thesis: