 Hence, since a portion of S is borne by the countervailing representation, the weaker representation can (and must) remain uninhibited by just that amount.  If we presuppose the total opposition of representations, then we can extend the above determination from two to any number of representations, so that a would be opposed by \(b, c, d, \ldots n\). S would then not be smaller than \((b + c + d + \ldots + n)\), but also not larger. 