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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