thus
( p <> 0_ (n,L) implies ex d being Nat st
( ex s being bag of n st
( s in Support p & d = degree s ) & ( for s1 being bag of n st s1 in Support p holds
d >= degree s1 ) ) )
( not p <> 0_ (n,L) implies ex b1 being Nat st b1 = 0 )proof
assume
p <> 0_ (
n,
L)
;
ex d being Nat st
( ex s being bag of n st
( s in Support p & d = degree s ) & ( for s1 being bag of n st s1 in Support p holds
d >= degree s1 ) )
then
p <> (dom p) --> (0. L)
;
then consider b being
object such that A1:
(
b in dom p &
p . b <> 0. L )
by FUNCOP_1:11;
reconsider b =
b as
bag of
n by A1;
defpred S1[
object ,
object ]
means for
s being
bag of
n st
s = $1 holds
$2
= degree s;
A2:
for
e being
object st
e in Support p holds
ex
u being
object st
S1[
e,
u]
consider D being
Function such that A3:
(
dom D = Support p & ( for
e being
object st
e in Support p holds
S1[
e,
D . e] ) )
from CLASSES1:sch 1(A2);
b in Support p
by A1, POLYNOM1:def 3;
then
dom D <> {}
by A3, XBOOLE_0:def 1;
then A4:
rng D <> {}
by RELAT_1:42;
then reconsider R =
rng D as non
empty finite natural-membered set by A4, A3, FINSET_1:8, MEMBERED:def 6;
max R in R
by XXREAL_2:def 8;
then consider s being
object such that A6:
(
s in dom D &
max R = D . s )
by FUNCT_1:def 3;
reconsider s =
s as
bag of
n by A6, A3;
reconsider m =
max R as
Nat by A6;
take
m
;
( ex s being bag of n st
( s in Support p & m = degree s ) & ( for s1 being bag of n st s1 in Support p holds
m >= degree s1 ) )
(
s in Support p &
m = degree s )
by A6, A3;
hence
ex
s being
bag of
n st
(
s in Support p &
m = degree s )
;
for s1 being bag of n st s1 in Support p holds
m >= degree s1
let s1 be
bag of
n;
( s1 in Support p implies m >= degree s1 )
assume
s1 in Support p
;
m >= degree s1
then
(
D . s1 in R &
D . s1 = degree s1 )
by FUNCT_1:def 3, A3;
hence
m >= degree s1
by XXREAL_2:def 8;
verum
end;
thus
( not p <> 0_ (n,L) implies ex b1 being Nat st b1 = 0 )
; verum