# Re: Existential Quantification [Re: New RDF model theory]

From: pat hayes <phayes@ai.uwf.edu>
Date: Mon, 27 Aug 2001 12:12:44 -0700
Message-Id: <v04210106b7b0464ab8ec@[130.107.66.237]>
To: Jeremy Carroll <jjc@hplb.hpl.hp.com>

```>In danger of making another mistake ....
>
>I didn't quite see how the interpolation lemma worked when the LHS has
>anonymous nodes.
>
>How does it get the following entailment
>
>
>_:x <b> <c> .
>
>
>entails
>
>_:y <b> <c> .

Ive been assuming that these actually describe the *same* graph, ie
the one with a single anonymous node in the subject position. (Those
'bound variables' _:x and _:y aren't actually there in the graphs,
right? ) The labels of anonymous nodes play no role in the model
theory. All that matters about anonymous nodes is that they are in a
particular graph and they are different from one another.

Thanks for noticing this. This all needs to be written out more
carefully to emphasise what is going on. Its rather hard to know
quite what needs explaining. Things that seem obvious to someone with
a logical background tend to seem opaque to those without, and vice
versa.

Actually this raises a very interesting point about graph syntax. The
graphs contain only anonymous nodes; but (by definition) an anonnode
in one graph isn't the same node as an exactly similar anonnode in
another graph. What counts as being 'another' graph, however? Eg what
if we infer a graph from itself? Do the anonnodes in the inferred
graph retain their identity *as nodes* from the graph considered as
an antecedent to the 'same' graph considered as a consequent? (Surely
yes.)  What if we copy the graph, creating a new (isomorphic but
distinct) graph; do the anonnodes in this new graph retain their
identity? (Probably not.)  Now what if we take a graph with an
anonnode in it, and add something to it, creating a 'new' graph? Do
the anonnodes in this 'new' graph retain their identity from the
'old' graph? I really am not sure what the proper answer should be.
Anyone used to thinking about data structures would say yes, but that
thinks of graphs as having a state, and that's not the standard
mathematical way of thinking.

>Sorry if I'm being stupid.

You aren't. I was stupid to write 'Lemma: ..' before actually writing
the proof out in detail.

Pat

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```
Received on Monday, 27 August 2001 15:11:42 UTC

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