Re: Decision from the Semantics TF: liberal baseline

> Am 08.01.2025 um 18:14 schrieb Franconi Enrico <franconi@inf.unibz.it>:
> 
> With this in mind:
> 
> if the triple structure appears in S then S RDF(S) entails
> reif0 sss aaa ooo sss rdf:type rdfs:Resource . 
> ooo rdf:type rdfs:Resource . 
> aaa rdf:type rdf:Property .
> reif1 sss aaa <<(xxx yyy zzz)>> <<(xxx yyy zzz)>> rdf:type rdfs:Proposition .
> reif2 <<(xxx yyy zzz)>> aaa ooo <<(xxx yyy zzz)>> rdf:type rdfs:Proposition .
> reif3 sss rdf:reifies ooo ooo rdf:type rdfs:Proposition .
> 
> On 8 Jan 2025, at 18:07, Doerthe Arndt <doerthe.arndt@tu-dresden.de> wrote:
> 
>> General:
>> - Niklas made an interesting point: If you  derive from 
>>> sss aaa <<(xxx yyy zzz)>>
>> 
>> that
>> xxx a rdfs:Resource. and zzz a rdfs:Resource. 
>> then we do not need the  „if the triple structure appears in S“ for rdf:Resource and can stick to „if S contains“.
> 
> What if this triple structure is deeply embedded?


I propose an alternative (based on Niklas idea):


> if  S contains then S RDF(S) entails
> reif0 sss aaa ooo sss rdf:type rdfs:Resource . 
> ooo rdf:type rdfs:Resource . 
> aaa rdf:type rdf:Property .
> reif1 sss aaa <<(xxx yyy zzz)>> <<(xxx yyy zzz)>> rdf:type rdfs:Proposition .
> xxx rdf:type rdfs:Resource.
> zzz rdf:type rdfs:Resource.
> yyy rdf:type rdf:Property.
> reif2 <<(xxx yyy zzz)>> aaa ooo <<(xxx yyy zzz)>> rdf:type rdfs:Proposition .
> xxx rdf:type  rdfs:Resource.
> zzz rdf:type  rdfs:Resource.
> yyy rdf:type rdf:Property.
> reif3 sss rdf:reifies ooo ooo rdf:type rdfs:Proposition .
> 


If I now have

s p <<(a b <<(x y z)>>)>>.

I get with reif1:

<<(a b <<(x y z)>>)>> rdf:type rdfs:Resource.

But from that, I get with reif2:

<<(x y z)>>  rdf:type rdfs:Resource.

I can again apply reif2 and get: 

x a rdfs:Resource. 
z a rdfs:Resource. 
y a rdf:Property. 

Kind regards,
Dörthe

> 
> It seems I fixed your comment below with my latest proposal above.
> Comments welcome!
> —e.
> 
>> Problem keeps being rdf entailment and the property. 
>> 
>> RDF:
>> - I guess aaa in reif1 and reif2 should be yyy?
>> 
>> RDFS:
>> - we do not need  
>>> <<(xxx yyy zzz)>> rdf:type rdfs:Resource . 
>> in reif1 and reif 2 because we get that with the existing rules from RDFS.
>> We do need 
>> xxx a rdfs:Resource.
>> and
>> yyy a rdfs:Resource.  
>> instead.
>> 
>> 
>> Kind regards,
>> Dörthe
>> 
>> 
>>> 
>>> 
>>>> On 8 Jan 2025, at 17:35, Franconi Enrico <franconi@inf.unibz.it> wrote:
>>>> 
>>>> Option 1 (the current option) adds metamodelling inference only for asserted triples.:
>>>> Option 1 (shallow metamodelling)
>>>> 
>>>> ⏩ <[I+A](r), [I+A](rdf:Proposition)> ∈ IEXT([I+A](rdf:type))
>>>>           if r is a triple term and ∃ x,y . (<x,[I+A](r)> ∈ IEXT(y)) ⋁ (<[I+A](r),x> ∈ IEXT(y))
>>>>           or if ∃ x . <x,[I+A](r)> ∈ IEXT([I+A](rdf:reifies)) ⏪️
>>>> 
>>>> Note that this is just wrong since in this case we have 
>>>> [I+A](rdfs:Resource) ≠ IR
>>>> [I+A](rdfs:Property) ≠ IP
>>>> Option 2 (true metamodelling)
>>>> 
>>>> ⏩ <r, [I+A](rdf:Proposition)> ∈ IEXT([I+A](rdf:type))
>>>>           if r ∈ range(RE) or 
>>>>           if ∃ x,y . RE(x,[I+A](rdf:reifies),r)=y ⏪️
>>>> ⏩ <r, [I+A](rdfs:Resource)> ∈ IEXT([I+A](rdf:type))
>>>>           if r ∈ range(RE) or 
>>>>           if ∃ x,y,z . RE(x,z,r)=y or 
>>>>           if ∃ x,y,z . RE(r,z,x)=y ⏪️
>>>> ⏩ <r, [I+A](rdfs:Property)> ∈ IEXT([I+A](rdf:type))
>>>>           if ∃ x,y,z . RE(x,r,z)=y ⏪️
>>>> 
>>>> 
>>>> Option 2 adds new metamodelling conditions, which implies that
>>>> [I+A](rdfs:Resource) = IR
>>>> [I+A](rdfs:Property) = IP
>>>> as it should.
>>>> The entailment pattern for option 2 will have "if the triple structure appears in S”.
>>>> 
>>>> —e.
>>>> 
>>>>> On 8 Jan 2025, at 17:17, Doerthe Arndt <doerthe.arndt@tu-dresden.de> wrote:
>>>>> 
>>>>> Dear Niklas,
>>>>> 
>>>>>> 
>>>>>> I think that it should be derived. And I agree that the triple constituents are resources (due to transparency).
>>>>>> 
>>>>>> I believe the following rule does that (given the existing RDF 1.1 entailment):
>>>>>> 
>>>>>> If S contains:
>>>>>> 
>>>>>>     sss aaa <<(xxx yyy zzz)>> .
>>>>>> 
>>>>>> or S contains (in symmetric RDF):
>>>>>> 
>>>>>>     <<(xxx yyy zzz)>> aaa ooo .
>>>>>> 
>>>>>> then S RDF(1.2)-entails (in symmetric RDF):
>>>>>> 
>>>>>>     <<(xxx yyy zzz)>> rdf:type rdf:Proposition .
>>>>>>     <<(xxx yyy zzz)>> rdf:propositionSubject xxx .
>>>>>>     <<(xxx yyy zzz)>> rdf:propositionPredicate yyy .
>>>>>>     <<(xxx yyy zzz)>> rdf:propositionObject zzz .
>>>>>> 
>>>>>> Then define:
>>>>>> 
>>>>>>     rdf:propositionPredicate rdfs:range rdf:Property .
>>>>>> 
>>>>>> To make yyy a property. (Which I think makes sense, even though weird triple terms misusing e.g. classes as properties would have weird consequences.)
>>>>>> 
>>>>>> 
>>>>> 
>>>>> It is a little bit more complicated because of the nesting. We could have
>>>>> 
>>>>> :a :b <<( :s :p  <<( :x :y :z )>> )>>.
>>>>> 
>>>>> we would want to derive that
>>>>> 
>>>>> :y a rdf:Property.
>>>>> 
>>>>> But that could still be done with a detailed version of Enrico’s "triple structure appears in“ notation. We could still get your triples. 
>>>>> 
>>>>> Another problem I see with your approach here is that we depend on RDFS while the properties are already derived in RDF and I assume that we want to keep it that way.
>>>>> 
>>>>> Another question is whether or not we want the proposition subject, predicate and object, but they could serve the purpose.
>>>>> 
>>>>> Kind regards,
>>>>> Dörthe
>>>>> 
>>>>> 
>>>>>  
>>>>> 
>>>> 
>>> 
>> 
>> 
>