reserve n, k, r, m, i, j for Nat;

theorem Th12:
  for k, m, m1, n1 being Element of NAT st k divides m & k divides
  n holds k divides m * m1 + n * n1
proof
  let k, m, m1, n1 be Element of NAT;
  assume k divides m & k divides n;
  then k divides m * m1 & k divides n * n1 by NAT_D:9;
  hence thesis by NAT_D:8;
end;
