let a1, a2 be Integer; for n1, n2 being Nat st n1,n2 are_coprime & n1 > 0 & n2 > 0 holds
{ x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } is infinite
let n1, n2 be Nat; ( n1,n2 are_coprime & n1 > 0 & n2 > 0 implies { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } is infinite )
assume that
A1:
n1,n2 are_coprime
and
A2:
( n1 > 0 & n2 > 0 )
; { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } is infinite
set X = { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } ;
CRT (a1,n1,a2,n2) solves_CRT a1,n1,a2,n2
by A1, A2, Def2;
then
(CRT (a1,n1,a2,n2)) + ((1 * n1) * n2) solves_CRT a1,n1,a2,n2
by Th26;
then A3:
(CRT (a1,n1,a2,n2)) + (n1 * n2) in { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 }
by A2;
A4:
{ x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } is natural-membered
for a being Nat st a in { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } holds
ex b being Nat st
( b > a & b in { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } )
proof
let a be
Nat;
( a in { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } implies ex b being Nat st
( b > a & b in { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } ) )
assume
a in { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 }
;
ex b being Nat st
( b > a & b in { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } )
then consider x being
positive Nat such that A5:
a = x
and A6:
x solves_CRT a1,
n1,
a2,
n2
;
take b =
x + ((1 * n1) * n2);
( b > a & b in { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } )
x + (n1 * n2) > x + 0
by A2, XREAL_1:8;
hence
b > a
by A5;
b in { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 }
b solves_CRT a1,
n1,
a2,
n2
by A6, Th26;
hence
b in { x where x is positive Nat : x solves_CRT a1,n1,a2,n2 }
;
verum
end;
hence
{ x where x is positive Nat : x solves_CRT a1,n1,a2,n2 } is infinite
by A3, A4, NUMBER04:1; verum