Issue rdfs-transitive-subSubProperty and AP 2001-08-02#18

Is a subproperty of rdfs:subPropertyOf necessarily transitive? No.

The action point asks something slightly different, viz:
  Propose an explanation of why a subproperty of a transitive property
  need not be transititive.

I don't have much of an explanation, except to say: transitivity and
subpropertyness aren't particularly closely related.

In particular, if A is a subproperty of B, then just because B is
transitive doesn't mean that A has to be (and vice versa). I'll give
examples below, at which point I suggest this issue be properly closed.
Oh, and rdfs:subpropertyOf is no exception; while it is transitive itself,
it has nontransitive subproperties.


1. A subproperty of a transitive property is not necessarily transitive.

	ancestorOf	is a transitive relationship
	parentOf	is not; it is, however, a subPropertyOf ancestorOf

There are lots of examples like these. Conceptually, many of the obvious
ones arise from having the subproperty be a "single step" (eg, one place
up a family tree) and the superproperty be something akin to the Kleene
closure (ie, zero or more) of that step.


2. A nontransitive property can have transitive subproperties.

The example here is: "is less than" is a transitive subproperty of
"is not equal to".

In fact, _every_ nontransitive property has at least one (trivial)
transitive subproperty.


3. Subproperties of rdfs:subPropertyOf need not be transitive.

Below, P, Q and R stand for properties. a, b, c etc. are things related
by those properties. I use a handwaving ntriples-alike that includes
those variables*

We note that
	P rdfs:subPropertyOf Q .

means that

	forall x, y
		x P y .
		->
		x Q y .

(that is the definition of what it means for one property to be a
subproperty of another).

Now we define a new property, SP/2, as follows:

  forall x, y
    x SP/2 y .
    iff
    x, y are properties (relationships) and
      if <(a_1, b_1), (a_2, b_2), ...> is the sequence of pairs of
        a, b such that a y b . arranged alphabetically
      then a x b. iff (a,b) occurs in the sequence at an
	odd-numbered position.

informally, a x b . holds true in half the cases that a y b . holds
true.

We're now done; it's pretty straightforward to show that
1. SP/2 rdfs:subPropertyOf rdfs:subPropertyOf .
2. SP/2 is not transitive.


jan

* which I propose we leave out of the MT for the moment :-)

-- 
jan grant, ILRT, University of Bristol. http://www.ilrt.bris.ac.uk/
Tel +44(0)117 9287163 Fax +44 (0)117 9287112 RFC822 jan.grant@bris.ac.uk
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Received on Saturday, 11 August 2001 11:59:05 UTC