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[protege-owl] SWRL rule query

Joseph Thomas-Kerr joetk at elec.uow.edu.au
Wed Jan 2 17:41:26 PST 2008


But I don't want a global maximum, I want a maximum for the given set  
selected by the rest of the rule.

As it stands it looks like I can't use some property of the resultant set  
(ie a maximum value) to further constrain that set. Is that correct?

Regards,Joe.

On Thu, 03 Jan 2008 06:47:14 +1100, Martin O'Connor  
<martin.oconnor at stanford.edu> wrote:

> Query and inference are deliberately separated in SQWRL. OWL's open
> world assumption will prevent you from inferring the maximum of your
> sequence numbers because there may another number out there that is  
> greater.
>
> Martin
>
> Joseph Thomas-Kerr wrote:
>
>> Hi,
>>
>> I want to be able to express the following rule:
>>
>> avc:pps(?pps) ^ avc:id(?pps,?spsID)
>> ^ avc:sps(?sps) ^ avc:id(?sps,?spsID)
>> ^ avc:seqNo(?pps,?ppsSeqNo) ^ avc:seqNo(?sps,?spsSeqNo)
>> ^ swrlb:lessThan(?spsSeqNo,?ppsSeqNo)
>> ^ (?spsSeqNo = MAXIMUM(?spsSeqNo))
>> 	-> rdo:dependsOn(?pps,?sps)}
>>
>> That is, I want to infer a relationship between a avc:pps and the  
>> avc:sps
>> with the matching spsID that MOST RECENTLY precedes it. An example might
>> help. Given the following individuals
>>
>> sps(SeqNo=0,id=0)
>> sps(SeqNo=1,id=1)
>> sps(SeqNo=2,id=2)
>> sps(SeqNo=3,id=0)
>> PPS(SeqNo=4,id=0)
>> sps(SeqNo=5,id=3)
>> sps(SeqNo=6,id=2)
>> sps(SeqNo=7,id=0)
>> sps(SeqNo=8,id=2)
>> PPS(SeqNo=9,id=2)
>>
>> I want to infer that
>>
>> rdo:dependsOn(pps(SeqNo=4,id=0),sps(SeqNo=3,id=0))
>> rdo:dependsOn(pps(SeqNo=9,id=2),sps(SeqNo=8,id=2))
>>
>> and nothing else. sqwrl has a max function, but I don't think that this
>> will get me what I am after. I tried using this function in the  
>> consequent
>> along with my inferred property, but it doesn't seem to work; query and
>> inference seem to have been deliberately separated.
>>
>> I am not a logic expert; I suspect that this inference is nonmonotonic,
>> but I'm not certain.
>>
>> can anyone provide some insight?
>>
>> Regards,
>> Joe.
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