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Re: [ontolog-forum] Inconsistent Theories

To: "[ontolog-forum] " <ontolog-forum@xxxxxxxxxxxxxxxx>
From: Christopher Menzel <cmenzel@xxxxxxxx>
Date: Tue, 9 Feb 2010 13:20:57 -0600
Message-id: <60FAB9AF-EE4C-44C6-A1B5-7238F0AB67C5@xxxxxxxx>
On Feb 9, 2010, at 11:37 AM, Rich Cooper wrote:
> ...
>>> On Feb 9, 2010, at 12:56 AM, Rich Cooper wrote:
>>>        Define( F(x), (x^3+27*x^2) )
>>  
>> I think you are referring to the LISP operator "defun", in which case you 
>want:
>>  
>>          (defun F (x)
>>             (* (+ (expt x 3) 27) (expt x 2))))
>  
> Yes, thanks for completing the lisp expr in proper format.  
>  
>> > Which states that the identify of the symbol F(x) is defined to be equal to
>> > the evaluable structure (x^3+27*x^2) which can be used to calculate the
>> > value of F(x) bound to any x.
>> >
>> > Those are two different views of '=' - by value and by defined structure
>> > (expression or object, perhaps). 
>  
>> No, they aren't.  Equality and defun are two entirely different things in 
>LISP. 
>  
> Isn't that what I just said above?      (01)

I wasn't clear.  My point was that defun isn't any sort of notion of equality.  
It is a procedural operation.  It *establishes* a certain class of equalities 
when it is called.  There simply *is* nothing like defun in ordinary 
first-order logic (though I supposed one could come up with something like it 
if one formalized first-order metatheory).    (02)

> Defun puts the structure on an association list with the decomposition above 
>as its 'value' propery.  That means, its value is defined by evaluating the 
>'value' property.  But there is still a significant difference between the 
>definition and the value of (F x).    (03)

Obviously, but my whole point is that NONE of this is apposite to the original 
discussion.  You have injected all sorts of issues from the semantics of 
programming languages into a simple discussion about the semantics of identity 
in first-order logic that was spawned by a related point about incompatible 
theories.  If you want to discuss the semantics of programming languages, fine 
and dandy, but start another thread or something.  You might as well have 
started injecting all sort of facts about quantum electrodynamics or 
agribusiness.  It's just not relevant. Start another thread.    (04)

> If you're going to appeal to LISP, you might be able to make your point more 
>profitably by appealing to its several related but distinct notions: -- "eq", 
>"eql", "equal", and "equalp".  But the general problem, again, is that you are 
>conflating the semantics of first-order logic with the semantics of 
>programming languages.
>  
> Its not Lisp per se which I am trying to use in explaining the property 
>concept, but the fact that any lisp atom or expr can have as many properties 
>as you want to associate with it.  The various versions of '=' in Lisp are 
>there for various programming reasons.  But the particular reason I am 
>bringing up in this discussion is not Lisp, itís the multiplicity of different 
>kinds of properties that an object in a database can exhibit.  Don't get hung 
>up on Lisp - I wanted to use something we all know to make the point about how 
>there are many properties, one of which is definition, another value, ... etc. 
> 
>  
> > The interpretation can be calculated to get the value, but the 
>interpretation is distinct from the value it gets at any given time.
>  
> You illustrate my point nicely.  LISP is a programming language; it is not a 
>first-order language. 
>  
> Actually, no.    (05)

Actually, yes.  You are confusing the fact that you can write S-expressions 
that look like sentences of a first-order language, and the fact that you can 
do some first-order theorem proving in LISP, with the idea that it's a 
first-order language or (as you say below) a "superset" of first-order logic.  
It isn't.  Really. Programming language models are vastly more complex and 
*also* far more constrained than models of first-order languages.  You can of 
course *transform* a first-order language into a programming language by giving 
it a programming language semantics -- that's basically what PROLOG does 
(though this claim needs to be qualified in numerous ways).  But, after doing 
so, it is no longer a first-order language.  In the context of this thread, a 
language is first-order in virtue of having a standard first-order model theory.    (06)

> Lisp has often been used to record FOL functionality.  Any programming 
>language can perform any FOL calculation short of infinite.    (07)

You are confusing first-order logic with automated theorem proving.    (08)

> For example, the conjunct:
>  
> (And  (Box 'A)
>       (Rectangle 'B)
>       (OnTopOf 'A 'B))
>  
> Is a perfectly feasible FOL situation description.    (09)

Vide the point above.  The fact that you can write S-expressions that look like 
FOL sentences does not mean LISP is a first-order language.     (010)

> And in fact, not only lisp, but any programming language incorporates full 
>FOL functionality (short of infinities).    (011)

I think you have something like expressiveness in mind but there is no 
straightforward comparison to be made between logical languages and programming 
languages in those terms.    (012)

> The semantics of programming languages is far more specialized and, 
>generally, far more complex than the semantics of first-order languages; 
>consider, e.g., the mathematics of domain theory that underlies the 
>denotational semantics of Scott and Strachey.  The semantics of LISP, in 
>particular, is a very rich and difficult subject that involves not only 
>general issues in the semantics of programming languages such as support for 
>recursion but also issues related to list processing and the semantics of 
>LISP's distinctive operators like "defun", "cons", "car", "cdr", "setq", etc.  
>These issues are light years removed from vanilla first-order model theory.
>  
> Wrong, as described above.  FOL is a SUBSET of modern programming languages 
>without infinity.     (013)

*sigh* This statement is confused along so many dimensions that one scarcely 
knows where to begin.  I don't plan to start.    (014)

-cm    (015)


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