- From: Drew McDermott <drew.mcdermott@yale.edu>
- Date: Mon, 18 Aug 2003 12:10:23 -0400 (EDT)
- To: www-ws@w3.org
Background: The DAML-S group has been having discussions on
streamlining its ontology by making processes (abstract) instances,
i.e., objects, rather than classes. The latter seemed at first like
the obvious way to go, because semantically a process is a collection
of all its possible execution traces. However, the resulting
formalism has come out to be cumbersome and bug-prone. If processes
are objects, then the relationship a process and its execution traces
still exists, but can't be identified with the OWL/DAML instanceOf
relation.
Monika Solanki said:
>> I am a little confused over the discussions we had today regarding
>> representation of process description in the process model and also a
>> representation of the execution trace of the process. As I understand
>> (Pls correct me if I am wrong). that Massimo wants to have a separate
>> ontology for expressing the trace, in terms of occurence of events,
>> data flow etc and instantiate this ontology for "a" process( for e.g
>> ExpressCongoBuy) and then provide a reference to this instantiation
>> in the core process model for that process. This means that we will
>> be having four Upper level ontology instantiations for a service:
>> Profile, Process, Grounding and Trace ?.
No, if I understand the phrase "ontology instantiations" correctly.
We don't need to say anything about traces in the DAML-S ontology if
we don't want to. A theory of traces would have axioms such as
"A trace of (if q a b) by agent $A$ starting in situation $s$ consists
of a trace of $a$ starting in $s$ if $A$ knows in $s$ that $q$ is
true; it consists of a trace of $b$ if $A$ knows in $s$ that $q$ is
false; otherwise there is no trace (it fails)." [I just made this up
so I don't endorse it particularly.]
Note that (1) a given process has many traces, in fact, infinitely
many; (2) in a trace of a process, some of its subprocesses need have
no traces (e.g., in a trace of (if q a b), one of $a$ or $b$ will have
no trace).
--
-- Drew McDermott
Received on Monday, 18 August 2003 12:10:25 UTC