theorem LCM1:
  for a,b be Integer holds
    a divides b iff a lcm b = |.b.|
  proof
    let a,b be Integer;
    thus a divides b implies a lcm b= |.b.|
    proof
      assume a divides b; then
      |.b.| = |.a.| lcm |.b.| by INT_2:16, NEWTON:44;
      hence thesis by Def2;
    end;
    assume a lcm b = |.b.|; then
    |.a.| lcm |.b.| = |.b.| by Def2;
    hence thesis by NEWTON:44,INT_2:16;
  end;
