reserve A for set, x,y,z for object,
  k for Element of NAT;
reserve n for Nat,
  x for object;
reserve V, C for set;

theorem
  for X be set, b being bag of X holds b+EmptyBag X = b
proof
  let X be set, b be bag of X;
  now
    let x be object;
    assume
    x in X;
    thus (b+EmptyBag X).x = b.x+((EmptyBag X).x qua Nat) by Def5
      .= b.x;
  end;
  hence thesis;
end;
