From: Seth Russell <seth@robustai.net>

Date: Tue, 1 May 2001 12:51:34 -0700

Message-ID: <004801c0d278$25fcb200$b17ba8c0@c1457248a.sttls1.wa.home.com>

To: "pat hayes" <phayes@ai.uwf.edu>

Cc: <www-rdf-interest@w3.org>

Date: Tue, 1 May 2001 12:51:34 -0700

Message-ID: <004801c0d278$25fcb200$b17ba8c0@c1457248a.sttls1.wa.home.com>

To: "pat hayes" <phayes@ai.uwf.edu>

Cc: <www-rdf-interest@w3.org>

Well, sorry, I'm still not getting this. Could I impose on you again to refer to a new mentograph: [1] http://robustai.net/mentography/TransitiveProperties.gif See you say ... "What you are talking about is meta-language statements (statements about other statements), not higher-order statements." Yet I have been able to transform the two examples of "higher-order statements" that you gave below into RDF statements merely about other statements - which I take to mean that statements can be objects of other statements. So I am at a loss to make the distinction you require. A couple of notes on my diagram. * I use a short-hand notation for RDF reification - explained at [2] http://robustai.net/mentography/reification.gif * I had to change your example slightly away from unary relations so as to correspond with the RDF way of doing things. But I was able to duplicate the problem to which you referred and then resolve it with a arc labeled "not" between the variable class and the designated class. So, what am I missing ?? Thanks for your patience with this troublesome student ... Seth .... in response to your examples below ... > > >.. i have these concepts all smushed together in > >my mind ... and am playing catch up with my education. But I still don't > >get what makes logic higher order. I have tried to depict my understanding > >of your description in the graph at > >http://robustai.net/mentography/higherOrder.gif which I have also put on > >the Public CMap server under the SemanticWeb Project. If you do find the > >time to answer me, maybe you could show where I have gone wrong by mutating > >my graph. > > Hmm, not sure I follow the graph , I'm afraid. Sorry, I tend to work > with words better. > The higher-order/firstorder distinction is rather a subtle one to get > exactly right. Let me sketch it first and then correct the sketch > later, OK? > > Sketch > First-order logic asserts relations between things, so you can say things like > (IsBiggerThan Bill Fred) > ie relation IsBiggerThan holds between things Bill and Fred, and it > quantifies over the things, so you can say > (forall (?x) (exists (?y)(IsBiggerThan ?y ?x))) > ie for any thing x there is something y which is bigger than it, ie > everything has something bigger than it. (I didnt say it was true, > only that you can say it.) > > OK. In second-order logic, you can also quantify over the > (first-order) relations and have (second-order) relations on > relations, so for example you could say that IsBiggerThan is > transitive: > (Transitive IsBiggerThan) > and define Transitive: > (forall (?R) > (iff (Transitive ?R) > (forall (?x ?y ?z)(implies (and (?R ?x ?y)(?R ?y ?z)) > (?r ?x ?z))) > )) > Notice that the ?R ranges over (first-order) relations, not just things. > In third-order, you can have relations on second-order relations, and so on... > Higher-order means you can go as high up the ladder of relations of > relations of... as you want. In practice nobody much wants to go > beyond second-order, most of the time, but you never know. > > Real Story > > What *really* makes a logic higher-order is that when you quantify > over 'all relations', that really does mean ALL relations, not just > the ones you happen to mention. There are a hell of a lot of > relations; more than you probably ever want to think about. For > example, consider the property (unary relation, ie relation with one > argument) of being further north than the oldest plumber born in > Philadelphia. Hey, its a perfectly good property; but when you said > (forall (?p)...) did you really have that in mind as a possibility? > Answer: if you are a mathematical logician, yes, you did. The moral > of which is that real higher-order logic is probably more use to > mathematicians than anyone else. For another example, suppose you > wanted to say that two people had something in common, and thought of > using a second-order sentence like > (exists (?P) (and (?P Bill)(?P Joe))) > to say it (ie there is some property true of Bill and of Joe), and > you were thinking of ?P's like 'eye-color' or 'watches baseball'. It > wouldnt do the job for you, since in real higher-order logic, this is > trivially true of any two things, since the property of 'being either > Bill or Joe' satisfies it. Written using lambda this would be > (lambda (?x) (or (= ?x Bill)(= ?x Joe))). Obviously this is true of > Bill (who is equal to Bill) and also of Joe, so it works for ?P. No > good saying "that's not a real property": in real second-order logic > it is, tough luck. > The connection with lambda-calculus is that any lambda-expression > with a sentence body defines a relation. ANY lambda-expression. So > higher-order logic has an inference rule (called 'comprehension', > sometimes its phrased as an axiom) which allows you to make any > sentence into a lambda-expression. If you can say it, its can be used > to define a relation, is the idea. > > ------ > > As you can see, none of this has got anything to do with sentences > about sentences: its all to do with sentences about relations. > > Hope this helps. > > Pat Hayes > > --------------------------------------------------------------------- > IHMC (850)434 8903 home > 40 South Alcaniz St. (850)202 4416 office > Pensacola, FL 32501 (850)202 4440 fax > phayes@ai.uwf.edu > http://www.coginst.uwf.edu/~phayes > >Received on Tuesday, 1 May 2001 15:56:51 UTC

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