Re: N3 contexts vs RDF reification

Well, sorry,  I'm still not getting this.  Could I impose on you again to
refer to a new mentograph:


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

* 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 ...


.... 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
> >of your description in the graph at
> >  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
> >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
> (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
> 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
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Received on Tuesday, 1 May 2001 15:56:51 UTC