let n, n1, n2 be Nat; for a being non trivial Nat st Py (a,n1), Py (a,n2) are_congruent_mod Px (a,n) & n > 0 & not n1,n2 are_congruent_mod 2 * n holds
n1, - n2 are_congruent_mod 2 * n
let a be non trivial Nat; ( Py (a,n1), Py (a,n2) are_congruent_mod Px (a,n) & n > 0 & not n1,n2 are_congruent_mod 2 * n implies n1, - n2 are_congruent_mod 2 * n )
assume that
A1:
Py (a,n1), Py (a,n2) are_congruent_mod Px (a,n)
and
A2:
n > 0
; ( n1,n2 are_congruent_mod 2 * n or n1, - n2 are_congruent_mod 2 * n )
reconsider m1 = n1 mod (4 * n), d1 = n1 div (4 * n) as Element of NAT ;
A3:
n1 - (d1 * (4 * n)) = m1
by A2, INT_1:def 10;
then
(sgn (((4 * d1) * n) + m1)) * (Py (a,|.(((4 * d1) * n) + m1).|)),(sgn m1) * (Py (a,|.m1.|)) are_congruent_mod Px (a,|.n.|)
by Th31;
then
Py (a,n1),(sgn m1) * (Py (a,|.m1.|)) are_congruent_mod Px (a,|.n.|)
by A3, Th23;
then
Py (a,n1), Py (a,m1) are_congruent_mod Px (a,n)
by Th23;
then A4:
Py (a,m1), Py (a,n1) are_congruent_mod Px (a,n)
by INT_1:14;
reconsider m2 = n2 mod (4 * n), d2 = n2 div (4 * n) as Element of NAT ;
A5:
( m1 <= 4 * n & m2 <= 4 * n )
by A2, INT_1:58;
A6:
( |.(Py (a,m1)).| = Py (a,m1) & |.(Py (a,m2)).| = Py (a,m2) )
;
A7:
n2 - (d2 * (4 * n)) = m2
by A2, INT_1:def 10;
then
(sgn (((4 * d2) * n) + m2)) * (Py (a,|.(((4 * d2) * n) + m2).|)),(sgn m2) * (Py (a,|.m2.|)) are_congruent_mod Px (a,|.n.|)
by Th31;
then
Py (a,n2),(sgn m2) * (Py (a,|.m2.|)) are_congruent_mod Px (a,|.n.|)
by A7, Th23;
then
Py (a,n2), Py (a,m2) are_congruent_mod Px (a,n)
by Th23;
then
Py (a,n1), Py (a,m2) are_congruent_mod Px (a,n)
by A1, INT_1:15;
then
Py (a,m1), Py (a,m2) are_congruent_mod Px (a,n)
by A4, INT_1:15;
then A8:
( m1,m2 are_congruent_mod 2 * n or m1, - m2 are_congruent_mod 2 * n )
by A6, A5, Th35;
n1 - m1 = (2 * d1) * (2 * n)
by A3;
then
n1,m1 are_congruent_mod 2 * n
by INT_1:def 5;
then A9:
( n1,m2 are_congruent_mod 2 * n or n1, - m2 are_congruent_mod 2 * n )
by A8, INT_1:15;
( m2 - n2 = (- (2 * d2)) * (2 * n) & (- m2) - (- n2) = (2 * d2) * (2 * n) )
by A7;
then
( m2,n2 are_congruent_mod 2 * n & - m2, - n2 are_congruent_mod 2 * n )
by INT_1:def 5;
hence
( n1,n2 are_congruent_mod 2 * n or n1, - n2 are_congruent_mod 2 * n )
by A9, INT_1:15; verum