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Re: [ontolog-forum] Foundation ontology, CYC, and Mapping

To: "'[ontolog-forum] '" <ontolog-forum@xxxxxxxxxxxxxxxx>
From: "Matthew West" <dr.matthew.west@xxxxxxxxx>
Date: Thu, 4 Feb 2010 18:07:14 -0000
Message-id: <4b6b0cd4.0a04d00a.7896.ffff843f@xxxxxxxxxxxxx>

Dear Duane,

 

That looks fine. There will be an elegant way to say that in Common Logic (KIF is dead long live CL). I will defer to Chris M on what that is.

 

Regards

 

Matthew West                           

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From: ontolog-forum-bounces@xxxxxxxxxxxxxxxx [mailto:ontolog-forum-bounces@xxxxxxxxxxxxxxxx] On Behalf Of Duane Nickull
Sent: 04 February 2010 17:32
To: [ontolog-forum]
Subject: Re: [ontolog-forum] Foundation ontology, CYC, and Mapping

 

Matthew/John

Thank you.  This speaks to me as a developer.

I remember writing equations using the membership operator (the one that looks like a rounded “E”).  Is this symbol representative of the same type of membership?  For example, please see the attached diagram to tell me if it makes sense.  I tried to express that “x” is a member of either “P” or “S” but cannot be a member of both.  Is there a better way to write this in KIF?

Duane


On 2/4/10 5:17 AM, "John F. Sowa" <sowa@xxxxxxxxxxx> wrote:

Dear Matthew and Duane,

MW> It is the most basic thing about sets that they are defined
 > by their membership, which does not change.

Yes.  If you write something like the following,

    S1 = S union {x}

mathematicians don't say that S has changed, but that S1 is
a different set.

In programming languages, it is common to write

    x = x + 1;

but when that operation is carefully analyzed and described,
it's described as "the value stored at the location designated
by x has been replaced or updated with a new value."

There are also some programming languages, such as ML or Haskell,
called *functional* or *single-assignment* languages in which
no variable can be reassigned a different value -- i.e., no
statement of the form x=x+1 is permitted.

MW> So if you have something that has members, but the membership
 > can change, then what you know for certain is that it is not a set.
 > Some people use the word type for such things. A type will have,
 > at a point in time, a set which is its membership.

The usual distinction:

   The identity conditions for a set are *extensional*:  if S1 and S2
   have the same members, they are identical; otherwise they are two
   different sets.

   The identity conditions for a type are *intensional*; two types
   T1 and T2 are identical if their definitions are equivalent;
   their set of instances in a particular world w is called their
   *denotation* in w.  Those worlds may be, for example, the physical
   world or a particular computer system at different times.

John


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