let n, n1, n2 be Nat; for x, y being Integer
for a being non trivial Nat st n1 < n2 & n2 <= 2 * n & |.x.| = Py (a,n1) & |.y.| = Py (a,n2) & x,y are_congruent_mod Px (a,n) holds
n1 = (2 * n) - n2
let x, y be Integer; for a being non trivial Nat st n1 < n2 & n2 <= 2 * n & |.x.| = Py (a,n1) & |.y.| = Py (a,n2) & x,y are_congruent_mod Px (a,n) holds
n1 = (2 * n) - n2
let a be non trivial Nat; ( n1 < n2 & n2 <= 2 * n & |.x.| = Py (a,n1) & |.y.| = Py (a,n2) & x,y are_congruent_mod Px (a,n) implies n1 = (2 * n) - n2 )
assume A1:
( n1 < n2 & n2 <= 2 * n & |.x.| = Py (a,n1) & |.y.| = Py (a,n2) & x,y are_congruent_mod Px (a,n) & n1 <> (2 * n) - n2 )
; contradiction
per cases
( n2 <= n or n2 > n )
;
suppose A2:
n2 > n
;
contradictionthen reconsider m =
n2 - n as
Nat by NAT_1:21;
(n + n) - n2 >= n2 - n2
by A1, XREAL_1:9;
then reconsider nm =
n - m as
Element of
NAT by INT_1:3;
A3:
(
(sgn n2) * (Py (a,|.n2.|)) = Py (
a,
n2) &
(sgn nm) * (Py (a,|.nm.|)) = Py (
a,
nm) )
by Th23;
n2 = n + m
;
then
(sgn n2) * (Py (a,|.n2.|)),
(sgn nm) * (Py (a,|.nm.|)) are_congruent_mod Px (
a,
|.n.|)
by Th32;
then
(sgn y) * (Py (a,n2)),
(sgn y) * (Py (a,nm)) are_congruent_mod Px (
a,
n)
by A3, INT_4:11;
then
y,
(sgn y) * (Py (a,nm)) are_congruent_mod Px (
a,
n)
by A1, ABSVALUE:17;
then A4:
x,
(sgn y) * (Py (a,nm)) are_congruent_mod Px (
a,
n)
by A1, INT_1:15;
then A5:
(sgn y) * (Py (a,nm)),
x are_congruent_mod Px (
a,
n)
by INT_1:14;
nm = (n + n) - n2
;
then A6:
nm < (n + n) - n
by A2, XREAL_1:15;
A7:
(
n1 < nm or
nm < n1 )
by A1, XXREAL_0:1;
y <> 0
by A1;
then
(
y > 0 or
y < 0 )
;
then
(
sgn y = 1 or
sgn y = - 1 )
by ABSVALUE:def 2;
then
(
(sgn y) * (Py (a,nm)) = Py (
a,
nm) or
(sgn y) * (Py (a,nm)) = - (Py (a,nm)) )
;
then A8:
|.((sgn y) * (Py (a,nm))).| = Py (
a,
nm)
;
per cases
( n1 <= n or n1 > n )
;
suppose
n1 <= n
;
contradictionthen A9:
(
Py (
a,
n1)
= Py (
a,
nm) or
Py (
a,
n1)
= - (Py (a,nm)) )
by A1, A4, INT_1:14, A8, A6, A7, Th33;
Py (
a,
n1)
<> - (Py (a,nm))
proof
assume A10:
Py (
a,
n1)
= - (Py (a,nm))
;
contradiction
(
n1 = 0 &
nm = 0 )
by A10;
hence
contradiction
by A1;
verum
end; hence
contradiction
by A9, A1, Th15;
verum end; suppose A11:
n1 > n
;
contradictionthen reconsider k =
n1 - n as
Nat by NAT_1:21;
n + n >= n1
by A1, XXREAL_0:2;
then
(n + n) - n1 >= n1 - n1
by XREAL_1:9;
then reconsider nk =
n - k as
Element of
NAT by INT_1:3;
A12:
(
(sgn n1) * (Py (a,|.n1.|)) = Py (
a,
n1) &
(sgn nk) * (Py (a,|.nk.|)) = Py (
a,
nk) )
by Th23;
n1 = n + k
;
then
(sgn n1) * (Py (a,|.n1.|)),
(sgn nk) * (Py (a,|.nk.|)) are_congruent_mod Px (
a,
|.n.|)
by Th32;
then
(sgn x) * (Py (a,n1)),
(sgn x) * (Py (a,nk)) are_congruent_mod Px (
a,
n)
by A12, INT_4:11;
then
x,
(sgn x) * (Py (a,nk)) are_congruent_mod Px (
a,
n)
by A1, ABSVALUE:17;
then A13:
(sgn y) * (Py (a,nm)),
(sgn x) * (Py (a,nk)) are_congruent_mod Px (
a,
n)
by A5, INT_1:15;
nk = (n + n) - n1
;
then A14:
nk < (n + n) - n
by A11, XREAL_1:15;
A15:
nk <> nm
by A1;
then A16:
(
nk < nm or
nm < nk )
by XXREAL_0:1;
x <> 0
by A1, A11;
then
(
x > 0 or
x < 0 )
;
then
(
sgn x = 1 or
sgn x = - 1 )
by ABSVALUE:def 2;
then
(
(sgn x) * (Py (a,nk)) = Py (
a,
nk) or
(sgn x) * (Py (a,nk)) = - (Py (a,nk)) )
;
then A17:
|.((sgn x) * (Py (a,nk))).| = Py (
a,
nk)
;
A18:
(
Py (
a,
nk)
= Py (
a,
nm) or
Py (
a,
nk)
= - (Py (a,nm)) )
by Th33, A16, A14, A6, A13, INT_1:14, A8, A17;
Py (
a,
nk)
<> - (Py (a,nm))
proof
assume A19:
Py (
a,
nk)
= - (Py (a,nm))
;
contradiction
(
nk = 0 &
nm = 0 )
by A19;
hence
contradiction
by A1;
verum
end; hence
contradiction
by A18, A15, Th15;
verum end; end; end; end;