# Re: Apparent inconsistency regarding equality in RIF-PRD

From: Christian De Sainte Marie <csma@fr.ibm.com>
Date: Fri, 6 Jul 2012 17:11:13 +0200
To: Jesse Weaver <weavej3@rpi.edu>

Message-ID: <OF66EC62A1.DCD0DC1E-ONC1257A33.0050F59A-C1257A33.0053700D@fr.ibm.com>
Hi Jesse,

When I wrote that I would examine your comments, I meant that I would
check wrt the spec, which I did not read since a couple months, and think
a little bit before responding; but I expect to have a full answer today,
or Monday at the latest...

And, anyway, I committed to produce a draft of the revised edition by
Tuesday, so, in the worst case, you will not have to wait for a very long
time :-)

As far as I can see from where I stand now, you are right re the
membership assertion (although your example rule is not well-formed :-)
and re the case of the unbindable action variable; and you are wrong re
equality (although I have to check whether and how the value of a constant
is defined, since matching an equality is defined with respect to it in
PRD, see definition of matching substitution, in 2.2.1).

Re membership, I have no doubt that it is pure oversight and that the
intent was that a class membership assertion would assert all the class
membership according to the class hierarchy. I just have to find out how
that was supposed to work, and repair the text accordingly.

Re the unbindable action variable, I have to check the discussions that we

I hope this helps. More later...

Cheers,

Christian

IBM
9 rue de Verdun
94253 - Gentilly cedex - FRANCE
Tel./Fax: +33 1 49 08 29 81

From:   Jesse Weaver <weavej3@rpi.edu>
To:     Christian De Sainte Marie/France/IBM@IBMFR
cawelty@gmail.com, public-rif-wg@w3.org, sandro@w3.org,
team-rif-chairs@w3.org
Date:   06/07/2012 16:40
Subject:        Re: Apparent inconsistency regarding equality in RIF-PRD

Hi Christian.

Thanks for responding.  I am working on my dissertation in which I use the
operational semantics of RIF-PRD as the underpinnings for some
formalization.  Are there any brief answers that can be given at this time
for the three problems?  Otherwise, I will have to resolve it in my own
way, but I would prefer to align with whatever the revised standard will
specify.

Jesse Weaver
Ph.D. Student, Patroon Fellow
Tetherless World Constellation
Rensselaer Polytechnic Institute
http://www.cs.rpi.edu/~weavej3/index.xhtml

On Jul 6, 2012, at 5:14 AM, Christian De Sainte Marie wrote:

Hi Jesse,

Thanx for the feedback (which could not be more appropriately timed:
indeed, we are preparing a revised edition to correct the errors found in
the first edition :-)

I will examine the three problems that you have found re equality,
subclasses and action variables and let you know.

Cheers,

Christian

IBM
9 rue de Verdun
94253 - Gentilly cedex - FRANCE
Tel./Fax: +33 1 49 08 29 81

From:        Jesse Weaver <weavej3@rpi.edu>
To:        team-rif-chairs@w3.org
Cc:        cawelty@gmail.com, Christian De Sainte Marie/France/IBM@IBMFR,
sandro@w3.org, Ankesh <ankesh@cs.rpi.edu>, public-rif-wg@w3.org,
Date:        05/07/2012 21:22
Subject:        Apparent inconsistency regarding equality in RIF-PRD

RIF Chairs,

In short, if a=b and b=c, then intuitively, it should hold that a=c, but
the operational semantics do not seem to ensure such.

By definition of atomic formula (section 2.1.2) and informal definition of
fact (section 2.2.2), a set of facts may include ground, equality, atomic
formulas.  Consider then the following set of facts where t_1, t_2, and
t_3 are different ground terms: \Phi = {t_1=t_2, t_2=t_3}.  By
(operational) definition of state of the fact base (section 2.2.2), \Phi
represents a state of the fact base.  According to the interpretation of
condition formulas (section 12.2), I_{truth}(I(x=y)) iff I(x)=I(y).
Intuitively, then, it should hold that since t_1=t_2 \in \Phi implies
I(t_1)=I(t_2), and since t_2=t_3 \in \Phi implies I(t_2)=I(t_3), by
transitivity of mathematical equality, I(t_1)=I(t_3).  However, there
exists no ground substitution that matches t_1=t_3 to \Phi, and since
ground, equality, atomic formulas cannot be inferred by rules (they are
not syntactically allowed as the target of an assert action; see
definition of atomic action, section 3.1.1), then there is also no
subsequent state in which there exists a ground substitution that matches
t_1=t_3 to the state's associated set of facts.  This seems like an
inconsistency.

A similar argument can be made for the simpler (but perhaps less
interesting and less obvious) case of when \Phi = {t_1=t_2}, where t_1 and
t_2 are different ground terms.  Then no ground substitution exists that
matches t_2=t_1 to \Phi or to any subsequently inferred sets of facts.

I can think of a number of ways to resolve this apparent inconsistency,
but I wish to know: have I correctly determined an inconsistency, and if
so, what were the intended (consistent) semantics (or definitions)?

Jesse Weaver
Ph.D. Student, Patroon Fellow
Tetherless World Constellation
Rensselaer Polytechnic Institute
http://www.cs.rpi.edu/~weavej3/index.xhtml

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Compagnie IBM France
Siège Social : 17 avenue de l'Europe, 92275 Bois-Colombes Cedex
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Sauf indication contraire ci-dessus:/ Unless stated otherwise above:
Compagnie IBM France
Siège Social : 17 avenue de l'Europe, 92275 Bois-Colombes Cedex
RCS Nanterre 552 118 465
Forme Sociale : S.A.S.
Capital Social : 645.605.931,30 ?
SIREN/SIRET : 552 118 465 03644 - Code NAF 6202A

Received on Friday, 6 July 2012 15:12:05 GMT

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