theorem
  m > 0 & n > 0 implies m lcm n > 0
proof
  assume that
A1: m>0 and
A2: n>0 and
A3: m lcm n <= 0;
A4: (m lcm n) divides m*n by Th47;
  m lcm n = 0 or m lcm n < 0 by A3;
  then ex r being Nat st m*n = 0*r by A4,NAT_D:def 3;
  hence contradiction by A1,A2;
end;
