let R be domRing; for p, q being Polynomial of R st ex S being Subset of R st
( card S = (max ((deg p),(deg q))) + 1 & ( for a being Element of R st a in S holds
eval (p,a) = eval (q,a) ) ) holds
p = q
let p, q be Polynomial of R; ( ex S being Subset of R st
( card S = (max ((deg p),(deg q))) + 1 & ( for a being Element of R st a in S holds
eval (p,a) = eval (q,a) ) ) implies p = q )
assume
ex S being Subset of R st
( card S = (max ((deg p),(deg q))) + 1 & ( for a being Element of R st a in S holds
eval (p,a) = eval (q,a) ) )
; p = q
then consider S being Subset of R such that
A0:
( card S = (max ((deg p),(deg q))) + 1 & ( for a being Element of R st a in S holds
eval (p,a) = eval (q,a) ) )
;
now not p <> qassume HH:
p <> q
;
contradiction
(max ((deg p),(deg q))) + 1 is
Element of
NAT
then reconsider S =
S as
finite Subset of the
carrier of
R by A0;
per cases
( p is zero or not p is zero )
;
suppose
not
p is
zero
;
contradictionthen reconsider n =
deg p as
Element of
NAT by T8;
H2:
n = (len p) - 1
by HURWITZ:def 2;
then H2a:
len p = n + 1
;
per cases
( len q < len p or len p < len q or len p = len q )
by XXREAL_0:1;
suppose D:
len q < len p
;
contradictionthen
(len q) + 1
<= len p
by INT_1:7;
then
((len q) + 1) - 1
<= (len p) - 1
by XREAL_1:9;
then
q . n = 0. R
by H2, ALGSEQ_1:8;
then H3:
q . n <> p . n
by H2a, ALGSEQ_1:10;
(
deg q = (len q) - 1 &
deg p = (len p) - 1 )
by HURWITZ:def 2;
then H4:
max (
(deg p),
(deg q))
= n
by D, XREAL_1:9, XXREAL_0:def 10;
defpred S1[
Nat]
means p . $1
<> q . $1;
A2:
for
k being
Nat st
S1[
k] holds
k <= n
proof
let k be
Nat;
( S1[k] implies k <= n )
assume B0:
S1[
k]
;
k <= n
now for i being Nat st i > n holds
p . i = q . iend;
hence
k <= n
by B0;
verum
end; A3:
ex
k being
Nat st
S1[
k]
by H3;
consider m being
Nat such that A4:
(
S1[
m] & ( for
i being
Nat st
S1[
i] holds
i <= m ) )
from NAT_1:sch 6(A2, A3);
A5:
(
p . m <> q . m &
m <= n )
by A2, A4;
then
p - q <> 0_. R
by FUNCOP_1:7, ORDINAL1:def 12;
then reconsider r =
p - q as non
zero Polynomial of
R by UPROOTS:def 5;
then C2:
n + 1
<= card (Roots r)
by A0, H4, NAT_1:43, TARSKI:def 3;
len r = m + 1
then
(len r) - 1
= m
;
then C3:
deg r = m
by HURWITZ:def 2;
card (Roots r) <= deg r
by degpoly;
then
n + 1
<= m
by C2, C3, XXREAL_0:2;
then C4:
n + 1
<= n
by A5, XXREAL_0:2;
n <= n + 1
by NAT_1:11;
then
n + 1
= n
by C4, XXREAL_0:1;
hence
contradiction
;
verum end; suppose D:
len p < len q
;
contradictionthen
(len p) + 1
<= len q
by INT_1:7;
then D1:
((len p) + 1) - 1
<= (len q) - 1
by XREAL_1:9;
(len p) - 1
< (len q) - 1
by D, XREAL_1:9;
then
deg p < (len q) - 1
by HURWITZ:def 2;
then
(
deg p < deg q &
0 <= n )
by HURWITZ:def 2;
then reconsider l =
deg q as
Element of
NAT by INT_1:3;
H2b:
l = (len q) - 1
by HURWITZ:def 2;
then H2c:
len q = l + 1
;
p . l = 0. R
by D1, H2b, ALGSEQ_1:8;
then H3:
q . l <> p . l
by H2c, ALGSEQ_1:10;
(
deg q = (len q) - 1 &
deg p = (len p) - 1 )
by HURWITZ:def 2;
then H4:
max (
(deg p),
(deg q))
= l
by D, XREAL_1:9, XXREAL_0:def 10;
defpred S1[
Nat]
means p . $1
<> q . $1;
A2:
for
k being
Nat st
S1[
k] holds
k <= l
proof
let k be
Nat;
( S1[k] implies k <= l )
assume B0:
S1[
k]
;
k <= l
now for i being Nat st i > l holds
q . i = p . iend;
hence
k <= l
by B0;
verum
end; A3:
ex
k being
Nat st
S1[
k]
by H3;
consider m being
Nat such that A4:
(
S1[
m] & ( for
i being
Nat st
S1[
i] holds
i <= m ) )
from NAT_1:sch 6(A2, A3);
A5:
(
p . m <> q . m &
m <= l )
by A2, A4;
then
p - q <> 0_. R
by FUNCOP_1:7, ORDINAL1:def 12;
then reconsider r =
p - q as non
zero Polynomial of
R by UPROOTS:def 5;
then C2:
l + 1
<= card (Roots r)
by A0, H4, NAT_1:43, TARSKI:def 3;
len r = m + 1
then
(len r) - 1
= m
;
then C3:
deg r = m
by HURWITZ:def 2;
card (Roots r) <= deg r
by degpoly;
then
l + 1
<= m
by C2, C3, XXREAL_0:2;
then C4:
l + 1
<= l
by A5, XXREAL_0:2;
l <= l + 1
by NAT_1:11;
then
l + 1
= l
by C4, XXREAL_0:1;
hence
contradiction
;
verum end; suppose D:
len p = len q
;
contradiction
n = (len p) - 1
by HURWITZ:def 2;
then H2:
len p = n + 1
;
H4:
(
deg q = (len q) - 1 &
deg p = (len p) - 1 )
by HURWITZ:def 2;
consider k being
Nat such that A1:
(
k < len p &
p . k <> q . k )
by HH, D, ALGSEQ_1:12;
defpred S1[
Nat]
means p . $1
<> q . $1;
A2:
for
k being
Nat st
S1[
k] holds
k <= n
proof
let k be
Nat;
( S1[k] implies k <= n )
assume B0:
S1[
k]
;
k <= n
now for i being Nat st i > n holds
p . i = q . iend;
hence
k <= n
by B0;
verum
end; A3:
ex
k being
Nat st
S1[
k]
by A1;
consider m being
Nat such that A4:
(
S1[
m] & ( for
i being
Nat st
S1[
i] holds
i <= m ) )
from NAT_1:sch 6(A2, A3);
A5:
(
p . m <> q . m &
m <= n )
by A2, A4;
then
p - q <> 0_. R
by FUNCOP_1:7, ORDINAL1:def 12;
then reconsider r =
p - q as non
zero Polynomial of
R by UPROOTS:def 5;
then C2:
n + 1
<= card (Roots r)
by H4, A0, D, NAT_1:43, TARSKI:def 3;
len r = m + 1
then
(len r) - 1
= m
;
then C3:
deg r = m
by HURWITZ:def 2;
card (Roots r) <= deg r
by degpoly;
then
n + 1
<= m
by C2, C3, XXREAL_0:2;
then C4:
n + 1
<= n
by A5, XXREAL_0:2;
n <= n + 1
by NAT_1:11;
then
n + 1
= n
by C4, XXREAL_0:1;
hence
contradiction
;
verum end; end; end; end; end;
hence
p = q
; verum