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Re: [ontolog-forum] type free logic and higher order quantification

To: "[ontolog-forum]" <ontolog-forum@xxxxxxxxxxxxxxxx>
From: Pat Hayes <phayes@xxxxxxx>
Date: Thu, 18 Aug 2011 19:53:15 -0500
Message-id: <34C4A73D-75B9-482D-8CE2-BFC0BF946216@xxxxxxx>

On Aug 18, 2011, at 5:52 PM, Christopher Menzel wrote:    (01)

> On Aug 18, 2011, at 5:15 PM, Pat Hayes wrote:
>> ...
>> Here is a test case to hone your intuitions. Consider these axioms:
>> 
>> (R a)
>> (Q b)
> 
> Pat, you obviously meant to write:
> 
> (R a)
> (R b)
> 
>> And ask yourself whether these entail the following:
>> 
>> (exists (p)(and (p a)(p b) )
>> 
>> i.e. that there is a property that applies both to a and to b. 
>> 
>> If you intuitively answer "no", then you are thinking first-order, and would 
>likely find CL congenial. If it seems obviously "yes", then you are thinking 
>in a genuinely higher-order way.
> 
> I agree that's a good place to start for testing your intuitions, but 
>answering "yes" here does not necessarily mean that you are thinking in a 
>genuinely higher-order way, i.e., in a way that commits you to full 
>second-order logic. For we can in fact add all of the comprehension axioms
> 
>  (exists (p) (forall (x1 ... xn) (iff (p x1 ... xn) φ)))
> 
> to the usual axioms for second-order quantification and the resulting logic 
>can still be given a Henkin-style general model theory that renders the logic 
>semantically first-order.  The comprehension axiom with φ = "(R x)" and some 
>simple logic give us the implication in your example.    (02)

Well, I would say that this is best stated by saying that my two axioms 
*together with the comprehension axioms* entail the conclusion. Of course, we 
can haver back and forth as to whether these comprehension axioms are part of a 
theory expressed in the logic or are part of the logic itself :-)    (03)

Pat    (04)

> 
> This, of course, only strengthens your initial point to Rick that "you need 
>to be achingly precise about what you count as 'higher order'".
> 
> -chris
> 
> 
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