let u be integer-valued FinSequence; for m being CR_Sequence st dom u = dom m holds
{ z where z is positive Nat : z solves_CRT u,m } is infinite
let m be CR_Sequence; ( dom u = dom m implies { z where z is positive Nat : z solves_CRT u,m } is infinite )
assume A1:
dom u = dom m
; { z where z is positive Nat : z solves_CRT u,m } is infinite
set X = { x where x is positive Nat : x solves_CRT u,m } ;
CRT (u,m) solves_CRT u,m
by A1, Def4;
then
(CRT (u,m)) + (1 * (Product m)) solves_CRT u,m
by A1, Th30;
then A2:
(CRT (u,m)) + (1 * (Product m)) in { x where x is positive Nat : x solves_CRT u,m }
;
A3:
{ x where x is positive Nat : x solves_CRT u,m } is natural-membered
for a being Nat st a in { x where x is positive Nat : x solves_CRT u,m } holds
ex b being Nat st
( b > a & b in { x where x is positive Nat : x solves_CRT u,m } )
proof
let a be
Nat;
( a in { x where x is positive Nat : x solves_CRT u,m } implies ex b being Nat st
( b > a & b in { x where x is positive Nat : x solves_CRT u,m } ) )
assume
a in { x where x is positive Nat : x solves_CRT u,m }
;
ex b being Nat st
( b > a & b in { x where x is positive Nat : x solves_CRT u,m } )
then consider x being
positive Nat such that A4:
a = x
and A5:
x solves_CRT u,
m
;
take b =
x + (1 * (Product m));
( b > a & b in { x where x is positive Nat : x solves_CRT u,m } )
x + (Product m) > x + 0
by XREAL_1:8;
hence
b > a
by A4;
b in { x where x is positive Nat : x solves_CRT u,m }
b solves_CRT u,
m
by A1, A5, Th30;
hence
b in { x where x is positive Nat : x solves_CRT u,m }
;
verum
end;
hence
{ z where z is positive Nat : z solves_CRT u,m } is infinite
by A2, A3, NUMBER04:1; verum