test case A revisited

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.

The approach to untidiness suggested at the face to face was a 
compromise.  I think this example demonstrates that compromise does not 
work and we have to choose between tidiness or a stronger notion of untidyness.

Brian

Received on Monday, 1 July 2002 07:40:45 UTC