Re: Semantic Predication: 5 - model theory

Dear Enrico,

Am 24.02.2023 um 16:19 schrieb Franconi Enrico <franconi@inf.unibz.it<mailto:franconi@inf.unibz.it>>:

I’m happy that somebody checks for potential errors in my formalisation!

I always like checking these things, so please keep posting you proposals :) And so far my assumption is that I simply do not fully understand, so maybe no errors at all.


Extend the mapping [I+A] as follows:
[I+A](<<<s,p,o>>>) = IT( ([I+A](s),[I+A](p),[I+A](o)) ).

What exactly is that b? Should it be a blank node or an IRI?

Yeah, sorry, I should have been more precise: b is indeed a bnode symbol.

I ask because I am bothered by the last line,

Extend the semantic conditions for an RDF-star graph as follows:
[I+A](b,unstar-sem:subject,s)=true   iff ∃p,o. [I+A](<<<s,p,o>>>)=[I+A](b)
[I+A](b,unstar-sem:predicate,p)=true iff ∃s,o. [I+A](<<<s,p,o>>>)=[I+A](b)
[I+A](b,unstar-sem:object,o)=true    iff ∃s,p. [I+A](<<<s,p,o>>>)=[I+A](b)
[I+A](s,p,o)=true                    iff ∃b.   [I+A](<<<s,p,o>>>)=[I+A](b)

This is always true since I know that [I+A](<<<s,p,o>>>)=[I+A](<<<s,p,o>>>), so there exists alway a b, namely <<<s,p,o>>>, or was that the idea? That the quoted triple is true if we have an interpretation for it, that is, if it is not malformed? Mmm, but what if

([I+A](s),[I+A](o)) not in IEXT([I+A](p))?

Then, the triple is true and false at the same time? That can’t be it.

So, could you please explain here? (And sorry for the „writing while-thinking-style“ you see some representation of my thinking process here ;) )

Yeah, sorry, I should have been more precise: IT is a partial function defined only for the embedded triples appearing in the graph.
And I could shorten the semantic condition, making it more precise, as follows:

∃s,p,o,b.
( [I+A](b,unstar-sem:subject,s)=true ⋀
  [I+A](b,unstar-sem:predicate,p)=true ⋀
  [I+A](b,unstar-sem:object,o)=true ⋀
  [I+A](s,p,o)=true )
iff [I+A](<<<s,p,o>>>)=[I+A](b)

Is this more convincing?

Not yet. So, let’s assume that [I+A](<<<s,p,o>>>)=[I+A](b) for some blank node b and some triple <<<s,p,o>>> both appearing in the graph. Then, I still cannot conclude that  [I+A](s,p,o)=true because there is the „normal“ interpretation of the triple (s,p,o) which says, that (s,p,o) is true iff ([I+A](s), [I+A](o)) \in IEXT([I+A](p)), so let’s assume, that IEXT([I+A](p))=\emptyset. Then the implication is not true. So, what is missing? Where is my misunderstanding?


Kind regards,
Dörthe

Thanks for double checking.
—e.

Add to the "5.3 Properties of simple entailment (Informative)" section:

The following holds: given the mapping LS from RDF-star graphs to RDF-1.1 graphs defined below, a RDF-star graph G entails a RDF-star graph E under the "sem" interpretation if and only if LS(G) entails LS(E) under RDF-1.1 simple interpretation.

LS(G) is defined as follows.
While G contains sem-quoted triples:
1) Pick an RDF-star sem-quoted triple (s, p, o) in the constituents of G such that neither s nor o is a sem-quoted triple.
2) Mint a fresh blank node b (i.e., such that b is not in the constituents of G).
3) Replace with b all occurrences of (s, p, o) in the subject or object position of an asserted or sem-quoted triple of G.
4) Add the following asserted triples to G:
   (b, unstar-sem:subject, s)
   (b, unstar-sem:predicate, p)
   (b, unstar-sem:object, o)
   (s, p, o)

This semantics has the following properties:

  *   Semantic embedded triples are fully transparent.
  *   Semantic embedded triples are always asserted.
  *   Semantic embedded triples entail their reification *and* vice-versa (logical equivalence).



  *   Monotonically extends RDF 1.1 and full backward compatible with RDF 1.1: if a RDF-star graph does not contain semantically quoted triples, then then the RDF-star "sem" semantics coincides with the RDF 1.1 semantics.
  *   The semantics is compositional.