[FOM] Is current computability theory intuitionistic?
Craig Smorynski
smorynski at sbcglobal.net
Tue Jun 25 15:59:13 EDT 2013
Am I missing something? PA and HA have the same provably computable functions, so the two theories would have to have different strengths. Otherwise the answer is yes even for both theories classical, provided one has more provably computable functions, e.g. ZF vs. PA.
Or am I missing something?
On Jun 19, 2013, at 4:36 PM, WILLIAM TAIT wrote:
> Maybe an interesting question in this connection is whether there is an e such that
>
> \forall x \exists y T(e,x, y)
>
> is classically provable (by whatever means) but there is no e' such that
>
> \forall x \exists y T(e',x, y)
>
> is constructively provable and {e} = {e'}.
>
> Bill Tait
>
>
> On Jun 18, 2013, at 4:41 PM, Alasdair Urquhart <urquhart at cs.toronto.edu> wrote:
>
>> Current texts certainly do present computability in terms of classical
>> logic. I recall some example in Hartley Rogers along the lines
>> of a function defined by: f(n) = 1 if the Riemann hypothesis is true,
>> otherwise 0 (I don't have the book at hand right now). Rogers
>> says that f is a well defined function.
>>
>> Whether this matters or not, I am not sure. I think a lot of computability theory can be easily reworked in a constructive context, since the arguments are generally constructive. On the other hand,
>> the classical arithmetical hierarchy presupposes a classical
>> reading of the quantifiers.
>>
>> On Tue, 18 Jun 2013, Steve Stevenson wrote:
>>
>>> I didn't know to ask this question when I was learning and now I'm too
>>> old to read all the standard books. My recollection though is that
>>> computability texts use classical logic. Is that true? Does it matter?
>>>
>>> --
>>> D. E. (Steve) Stevenson, PhD, Emeritus Associate Professor, Clemson University
>>> "Those that know, do. Those that understand, teach," Aristotle.
>>
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Craig
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