Re: test case A revisited

At 16:48 08/07/2002 -0700, pat hayes wrote:

>>Friday's telecon reminded me that I had left test case A in for a 
>>reason.  There was more I had to say about it, and writing that message 
>>the following occurred to me.
>>
>>Test case A says:
>>
>>   <s1> <p> "lit" .
>>   <s2> <p> "lit" .
>>
>>can we conclude that value of both properties are the same.
>>
>>Consider
>>
>>   _:b1 rdf:type rdf:Seq .
>>   _:b1 rdf:_1   "10" .
>>   _:b2 rdf:type rdf:Seq .
>>   _:b2 rdf:_1   "10" .
>>
>>This would require that the first member of each sequence is the same.
>>
>>Given that we also have a common superproperty of the rdf:_xxx 
>>properties, this essentially means that all literals which are members of 
>>any container must all have the same dataype, i.e. all literals in 
>>containers must be tidy.
>>
>>I suggest this is incompatible with the untidy literals and a yes to test 
>>case A above.
>
>?? I fail to follow your reasoning here. It seems circular.
>
>There are two cases to consider, right? We can have (semantically) tidy 
>literals, where each literal node labelled with the same literal denotes 
>the same thing; or we can not. Call these the ST and NST cases. Test A is 
>'yes' for ST, 'no' for NST. Now consider your container example. In an ST 
>reading, b1 and b2 have the same first element; in an NST reading, they 
>need not.

True, but in the f2f proposal it was suggested that they would because the 
denotation is a function of the literal and the property, i.e.

>>   <s1> <p> "lit" .
>>   <s2> <p> "lit" .

|=
   <s1> <p> _:l .
   <s2> <p> _:l .



>  Put another way: if all literals are tidy, then all literals in 
> containers must be tidy.

Sure.

>  Well, right. And if all literals are not tidy, the ones in container 
> need not be either. So... what has been demonstrated, exactly?

Ok, given:

   <a> rdf:_1 "10" .

what does the "10" denote?  What is in the container?

Brian

Received on Tuesday, 9 July 2002 11:04:00 UTC