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The present Newsletter issue contains a contribution by Jixin Ma to the discussion about Ontologies for time.

Debates

Jixin Ma:

Reply to Sergio Brandano (ENRAC 8.5.1998)

The only one "minor adjustment" I made consists in the first four lines of my contribution to ENRAC 3.5.1998, where no inequality appears at all. Concerning the hypothesis, I remind you what I wrote in ENRAC 24.4.1998:

When you presented the (classical) axiom of completeness (ENRAC 23.4.1998), you used   <   in the hypothesis (You are now still using it, see below). But for the case where elements in domain S are just intervals, you used   <   instead (otherwise, Pat's example is valid, see below).

The axiom of completeness imposes   <  , so no "alternation" is needed at all. The reason why I wrote   <   instead of   <   is simply due to my need to stress the example, since the case  <s1,t1> = <s2,t2>  is trivial. If you like to check, the reference is ENRAC 24.4.1998.

Again, if you re-claim that the axiom of completeness imposes   <  , then Pat's example is valid (see below).

...when you construct intervals out of points. In the case where intervals are taken as primitive, the need of such alternative is indeed more conceptually necessary). However, your adjustment is not enough, or you haven't reached the proper form for general treatments. In fact, you need address the issue regarding different cases. To see this, you may just consider the difference between the case where at least one of  <s1,t1>  and  <s2,t2>  is "closed" at  t1  (  = s2 ), and the case where both  <s1,t1>  and  <s2,t2>  are "open" at  t1  (  = s2 ). In the former case, you need use   <   in the hypothesis; otherwise, Pat's example will be a valid counterexample. In the latter case, you need

... at least one is closed. So we have, since  s2 = t1 :   [s1, t1]  <  [t1, t2]     [s1, t1]  <  (t1, t2]     [s1, t1)  <  [t1, t2]   where  xi =  [t1, t1] in S  in all cases. Note I used   <  , as required by the axiom of completeness. If I use $lt$, as you recommend, then all cases trivially fail. Pat's example: trivially fails when using   <  , trivially succeeds ( xi =  [q, q]  ) when using   <  .

This is exactly what I wanted to show and have shown a few times now. That is, in the case there your domain S contains intervals, to fulfill the axiom of completeness, S has to contains singletons (single points) as well, not as you specially claimed that S contains points or (exclusive-or) intervals. My observation that Pat's example would be valid is under your assumption that the domain S refuses to take both intervals and singletons (points). I think when Pat gave the example, he also followed this assumption of yours. (Actually, Pat did specially claim that "unless you allow intervals consisting of a single point" when he gave the example in ENRAC 24.4.1998).

use   <   in the hypothesis; otherwise, your axiom cannot not prevent a "gap" between   <s1, t1)   and   (s2, t2>  , that is, there is no guarantee that the singleton   [t1, t1]   is contained in  S  (Do you think this is consistent with the "classical" concept of contiunity?).

... the latter case. So we have, since  s2 = t1 : 4.   [s1, t1)  <  (t1, t2]   where  xi =  [t1, t1] in S . I used   <   here too.

This is exactly what I have suggested in my message to you (see above).

The problem about intervals is whether one needs to introduce them into the temporal domain, and the few argument-examples I encountered are far from being convincing. Furthermore, in this debate, you and Hayes proposed the DIP, and I refuted it.

The DIP was proposed much earlier in the literature. What you have done, is just a re-Writing of a model where the DIP arises. Could you please check carefully what exactly is the problem and if it can be solved by your formulation?

Jixin