Douglas Otis wrote:
> 
> On Apr 12, 2009, at 9:21 PM, Hector Santos wrote:
> 
>> In this case, the ADSP function might include part of the DKIM-BASE 
>> checking and for ADSP that means d=/i= checking. The ADSP docs does 
>> not talk about this approach because it is presumed this DKIM-BASE 
>> work was already done.
>>
>> IMV, the ADSP docs should have semantics about what i= means - 
>> technically:
>> From.domain == d.domain == i.domain (or subdomain of d=)
> 
> Disagree.  Constrains on the i=value are well defined by RFC 4871.
> 
> A domain within the i= value MUST be at or below the signing domain, and 
> when the "s" flag is set within the key record's t= value, a domain 
> within the i= value MUST NOT include any sub-domain for the signature to 
> be valid.

Agree. So which part of my message are disagreeing which.

I'm settle on what DKIM-BASE and ADSP is leaning towards. I've waited 
a long time to reach this point and although I am not at all happy 
with how diminished the POLICY component has become, I believe the 
most important policy (EXCLUSIVE declarations) is possible.  So 
personally, I don't see what all the haggling is about about i=.

> By having ADSP's reference the From email-address domain constrain d= 
> values, signing domains are provided full control over the i= value.  

Sure, but it still must comply as 4871 has it defined.  I would of 
thought making it open ended would of satisfied the 3rd party market, 
but even that was screwed up.  So it is what it is d=/i= have a 
specific compliance to it.

> There is no reason for recipients to enforce i= value use beyond what is 
> currently defined by RFC 4871.

Oh I see what you are thinking.  I was thinking of terms of stupid 
software with no AI to it, no subjective meaning to it, no accessors 
involved and that is basically applying a deterministic algorithm as 
defined by DKIM-BASE for d= and i=.

I'm sorry, if we are not talking about the same thing, I'm talking 
about non-subjective interpretations and simply following a 
deterministic protocol. 1 + 1 = 2, not 3.

-- 
Sincerely

Hector Santos
http://www.santronics.com


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