Re: How to avoid that collections "break" relationships

On Mar 29, 2014, at 8:10 PM, Peter F. Patel-Schneider <pfpschneider@gmail.com> wrote:

> 
> On 03/29/2014 03:30 PM, Markus Lanthaler wrote:
>> On Wednesday, March 26, 2014 5:26 AM, Pat Hayes wrote:
>>> Hmm. I would be inclined to violate IRI opacity at this point and have
>>> a convention that says that any schema.org property schema:ppp can have
>>> a sister property called schema:pppList, for any character string ppp.
>>> So you ought to check schema:knowsList when you are asked to look for
>>> schema:knows. Then although there isn't a link in the conventional
>>> sense, there is a computable route from schema:knows to
>>> schema:knowsList, which as far as I am concerned amounts to a link.
>> Schema.org doesn't suffer from this issue as much as other vocabularies do
>> as it isn't defined with RDFS but uses its own, looser description
>> mechanisms such as schema:domainIncludes and schema:rangeIncludes. So what
>> I'm really looking for is a solution that would work in general, not just
>> for some vocabularies.
>> [...]
>> 
>> 
>> --
>> Markus Lanthaler
>> @markuslanthaler
>> 
> I would  like to see some firm definition of just how these looser description mechanisms actually work.

Yes, I agree. Let me put the question rather more sharply. What follows from knowing that 

ppp schema:domainIncludes ccc . ?

Suppose you know this and you also know that 

x ppp y .

Can you infer x rdf:type ccc? I presume not, since the domain might include other stuff outside ccc. So, what *can* be inferred about the relationship between x and ccc ? As far as I can see, nothing can be inferred. If I am wrong, please enlighten me. But if I am right, what possible utility is there in even making a schema:domainIncludes assertion? 

If "inference" is too strong, let me weaken my question: what possible utility **in any way whatsoever** is provided by knowing that schema:domainIncludes holds between ppp and ccc? What software can do what with this, that it could not do as well without this? 

Having a piece of formalism which claims to be a 'weak' assertion becomes simply ludicrous when it is so weak that it carries no content at all. This bears the same relation to axiom writing that miming does to wrestling. 

Pat

> 
> peter
> 
> 

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Received on Sunday, 30 March 2014 07:14:00 UTC