 If someone says (as Brouwer does) that for \((x) f_1 x = f_2 x\), there is, as well as yes and no, also the case of undecidability, this implies that ‘\((x)\)…’ is meant extensionally and we may talk of the case in which all \(x\) happen to have a property. In truth, however, it’s impossible to talk of such a case at all and the ‘\((x)\)…’ in arithmetic cannot be taken extensionally. (PR §174)