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Subject: Re: [xdi] Inference and equivalence defintion/notation (and $has)


Hello,

sure, I agree that the +a/$has/+b statement "produces" the new  
composite subject +a+b (and viceversa); formally I think this can be  
expressed as

+a/$has/+b  <=>  +a+b

but, as Bill pointed out:

saying "Bill's email" is somehow different that saying "Bill has an  
email". Therefore I think that they are not aliases:

+a/$has/+b != +a+b

Does it make sense?

Kind Regards,
Giovanni

Def. Quota "Drummond Reed" <drummond.reed@xdi.org>:

> Giovanni,
>
> I'm missing something. The whole premise of $has statements is that they
> produce a new XDI subject which is the concatenation of the $has subject and
> the $has object.
>
> So why do you say +a/$has/+b != +a+b ?
>
> =Drummond
>
> On Sun, Jan 17, 2010 at 1:52 PM, Giovanni Bartolomeo <
> giovanni.bartolomeo@uniroma2.it> wrote:
>
>> Can we have && for AND maybe (C, C++, Java et al.)?
>>
>>
>>  You asked about the following
>>>>
>>>> :+a/$has/+b  =  (+a/+b)  =  +a+b
>>>>
>>>> Yes, I think so. Though this presents a hefty practicality concern that
>>>> =Bill+email now resolves to the graph of statements about the statement
>>>> (=Bill/+email), rather than a set of statements describing Bill's email.
>>>>
>>>
>> I also fear they represent different concepts.
>>
>> +a/$has/+b  =  (+a/+b)  maybe, but
>>
>>
>> +a/$has/+b !=  +a+b
>>
>> Kind Regards,
>> Giovanni
>>
>> Def. Quota "Drummond Reed" <drummond.reed@xdi.org>:
>>
>>
>>  Bill,
>>>
>>> Just a quick note to say that of course I agree let's use the terms that
>>> are
>>> widely established in the semantic community. I only typed /= last night
>>> because I was too tired to remember the notation (I really shouldn't stay
>>> up
>>> that late) and I wanted to follow up on yesterday's conversation quickly.
>>>
>>> I like all of your notation and suggest we adopt it. Doe everyone else
>>> agree? (Let me put it this way - anyone who does not agree, please post,
>>> otherwise let's use Bill's notation.)
>>>
>>> I don't have time right at this moment to answer the questions at the
>>> bottom
>>> of your message, but I'll try to get back to it later today, as I think
>>> we've got the foundation now to complete a semantically deeper and
>>> stronger
>>> definition of $has and $has$a, and with my improved understanding of the
>>> notation, we might even be able to make good progress on that in email.
>>>
>>> Thanks,
>>>
>>> =Drummond
>>>
>>> On Fri, Jan 15, 2010 at 5:51 AM, Barnhill, William [USA] <
>>> barnhill_william@bah.com> wrote:
>>>
>>>   Hi Drummond,
>>>>
>>>> Your email makes sense but I have a few things I'd like to ehar what you
>>>> think about:
>>>>
>>>> The terms I was using 'logical equivalence', 'implies/implication', '!=
>>>> for
>>>> not equal' are not terms I created. They're terms in wide adoption within
>>>> semantic community and in the case of !=, the development community.
>>>> Typically what you're calling inference is isually termed implication. I
>>>> haven't encountered the terms you used before, so what do you think about
>>>> using the following terms and notation instead:
>>>>
>>>>
>>>>            Implication                                   => (there's no
>>>> need for <= as A <= B is stated as B => A)
>>>>
>>>>            Logical equivalence                      <=>
>>>>
>>>>            Alias (ie resolution equivalence)     ==
>>>>
>>>>            Non-equivalence                           !=
>>>>
>>>>            And                                            ^
>>>>
>>>>            Or                                              ||
>>>>
>>>>            Not And (NAND)                          |
>>>>
>>>>            Exclusive Or (XOR)                      ~|
>>>>
>>>>            Nor (Not Or)                                !|
>>>>
>>>>
>>>>
>>>>           The notation for the boolean operations NAND, XOR, and NOR
>>>> above
>>>> are my own. The std notation is
>>>>
>>>>           not easily entered into plain text except as an image.
>>>>
>>>>
>>>>
>>>> You asked about the following
>>>>
>>>> :+a/$has/+b  =  (+a/+b)  =  +a+b
>>>>
>>>> Yes, I think so. Though this presents a hefty practicality concern that
>>>> =Bill+email now resolves to the graph of statements about the statement
>>>> (=Bill/+email), rather than a set of statements describing Bill's email.
>>>>
>>>>
>>>>
>>>> :+a/$has/+b  /=  +a/+b
>>>>
>>>> Yes the right one contains statements about the value of the left one,
>>>> the
>>>> left statements about the value of the +b property of +a.
>>>>
>>>>
>>>>
>>>> :+a/+b  /=  +a+b
>>>>
>>>> Yes the left one contains statements about the value of the +b property
>>>> of
>>>> +a, the right statements about the left one.
>>>>
>>>>
>>>>
>>>> :+a/$has$a/+b  =  +a/+b
>>>>
>>>> Don't know. I never got the feeling we nailed down $has$a enough for me
>>>> to
>>>> make this determination. I still do not like $a as an inverse operator,
>>>> would prefer defining inverse $ words for the key concepts like $has,
>>>> maybe
>>>> $hadby. I think we have to nail $has down real well before we can define
>>>> the
>>>> inverse of it.
>>>>
>>>>
>>>>
>>>> :+a/$has$a/+b  /=  (+a/+b)
>>>>
>>>> See above.
>>>>
>>>>
>>>>  Kind regards,
>>>>
>>>> Bill Barnhill
>>>> Booz Allen Hamilton - Rome, NY
>>>> 315-330-7386  | william.barnhill.ctr@rl.af.mil |
>>>> barnhill_william@bah.com
>>>>
>>>>  ------------------------------
>>>> *From:* drummond.reed@gmail.com [drummond.reed@gmail.com] On Behalf Of
>>>> Drummond Reed [drummond.reed@xdi.org]
>>>> *Sent:* Friday, January 15, 2010 5:12 AM
>>>> *To:* OASIS - XDI TC
>>>> *Subject:* [xdi] Inference and equivalence defintion/notation (and $has)
>>>>
>>>>  Bill,
>>>>
>>>>
>>>>
>>>> In thinking about your definitions of "logical equivalence" and
>>>> "resolution
>>>> equivalence" today, I'm wondering if we couldn't just use the terms
>>>> "inference" and "equivalence". In other words, it is easier for me to
>>>> understand the differences between the sentence "XDI statement A infers
>>>> XDI
>>>> statement B" and the sentence "XDI statement A is equivalent to XDI
>>>> statement B".
>>>>
>>>>
>>>>
>>>> My understanding of inference is that it can be unidirectional (A infers
>>>> B
>>>> but B does not infer A) or bi-directional (A infers B and B infers A).
>>>>
>>>>
>>>>
>>>> My understanding of equivalence is that it must be bidirectional, i.e.,
>>>> if
>>>> A is equivalent to B then B MUST be equivalent to A. This is the purpose
>>>> of
>>>> XDI $is statements – they express equivalence between an XDI subject and
>>>> and
>>>> XDI object.
>>>>
>>>>
>>>>
>>>> If you agree these terms will work, let's use your proposed notation:
>>>>
>>>>
>>>>
>>>>            Unidirectional inference               <=        or         =>
>>>>
>>>>            Bidirectional inference                <=>
>>>>
>>>>            Equivalence                               =
>>>>
>>>>            Non-equivalence             /=
>>>>
>>>>
>>>>
>>>> With that said I can restate the first part of the final conclusion of
>>>> today's telecon (see
>>>> http://lists.oasis-open.org/archives/xdi/201001/msg00046.html for the
>>>> chat
>>>> transcript) as saying that:
>>>>
>>>>
>>>>
>>>>            +a/$has/+b  <=>  (+a/+b)  <=>  +a/+b  <=>  +a+b
>>>>
>>>>
>>>>
>>>> This is true because each infers the other, and all these inferences are
>>>> bidirectional (I think).
>>>>
>>>>
>>>>
>>>> This would also imply that:
>>>>
>>>>
>>>>
>>>>            +a  <=>  (+a)
>>>>
>>>>
>>>>
>>>> Again, this is because each implies the other (I have one small
>>>> reservation
>>>> about this but I'll save that for another time).
>>>>
>>>>
>>>>
>>>> And part two of our final conclusion of the telecon was that:
>>>>
>>>>
>>>>
>>>>            +a  /=  (+a)
>>>>
>>>>
>>>>
>>>> This is true because as you described resolving (doing a $get) on +a and
>>>> (+a) respectively would not return the same graph.
>>>>
>>>>
>>>>
>>>> Lastly, we can finally get to the definition of $has and $has$a
>>>> statements.
>>>> What I have been saying (but not expressing with the proper notation) is:
>>>>
>>>>
>>>>
>>>>            +a/$has/+b  =  (+a/+b)  =  +a+b
>>>>
>>>>
>>>>
>>>>            +a/$has/+b  /=  +a/+b
>>>>
>>>>
>>>>
>>>>            +a/+b  /=  +a+b
>>>>
>>>>
>>>>
>>>>            +a/$has$a/+b  =  +a/+b
>>>>
>>>>
>>>>
>>>>            +a/$has$a/+b  /=  (+a/+b)
>>>>
>>>>
>>>>
>>>> Do you agree?
>>>>
>>>>
>>>>
>>>> =Drummond
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>
>>
>>
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