# Re: test case A revisited

From: pat hayes <phayes@ai.uwf.edu>
Date: Mon, 8 Jul 2002 16:48:50 -0700
Message-Id: <p05111b24b94fd3f893ed@[192.168.0.146]>
To: Brian McBride <bwm@hplb.hpl.hp.com>

```
>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. Put another way: if all
literals are tidy, then all literals in containers must be tidy.
Well, right. And if all literals are not tidy, the ones in container
need not be either. So... what has been demonstrated, exactly?

>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.

I fail to see why.

Pat

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```
Received on Monday, 8 July 2002 19:48:39 UTC

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