>> as the ietf is legally f(isoc), > The IETF is not legally ISOC. you are correct. legally, it is a function of isoc. sorry about the notational shortening. randy
>> as the ietf is legally f(isoc), > The IETF is not legally ISOC. you are correct. legally, it is a function of isoc. sorry about the notational shortening. randy