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## Re: [ontolog-forum] HOL decidability [Was: using SKOS forcontrolled valu

 To: "'[ontolog-forum] '" "Rich Cooper" Thu, 14 Oct 2010 12:06:53 -0700 <20101014190656.20185138CFD@xxxxxxxxxxxxxxxxx>
 Thanks Michael,   Very interesting page!  But it seems to contradict your statement where it says:   It is known no non-constant polynomial function P(n) with integer coefficients exists that evaluates to a prime number for all integers n. The proof is: Suppose such a polynomial existed. Then P(1) would evaluate to a prime p, so $P(1) \equiv 0 \pmod p$. But for any k, $P(1+kp) \equiv 0 \pmod p$also, so P(1 + kp) = p (as it is prime and divisible by p), but the only way P(1 + kp) = P(1) for all k is if the polynomial function is constant.   If I had been a mathematician instead of an engineer, I might have known that for presenting earlier.  Your suggested set of functions are no “non-constant polynomials” if they iterate the primes.  Whatever that “non-constant polynomials” means!  Again, the precision of my original statement could have been greatly improved, and thanks for your addition to the precise constraints on the which no functions exist might be useful for other mathemticians to read, if interested.     End of soap box.   Sincerely, Rich Cooper EnglishLogicKernel.com Rich AT EnglishLogicKernel DOT com 9 4 9 \ 5 2 5 - 5 7 1 2 From: ontolog-forum-bounces@xxxxxxxxxxxxxxxx [mailto:ontolog-forum-bounces@xxxxxxxxxxxxxxxx] On Behalf Of Michael Gruninger Sent: Thursday, October 14, 2010 11:36 AM To: ontolog-forum@xxxxxxxxxxxxxxxx Subject: Re: [ontolog-forum] HOL decidability [Was: using SKOS forcontrolled values for controlledvocabulary]   On 10-10-14 1:15 PM, Rich Cooper wrote: Hi Doug,   Thanks for you post, you seem to be honestly trying to understand what I meant by the statement "there is no function that can iterate the primes", and perhaps I should have originally said "directly, without iterating other types", which seems to have set off this mess.  But I expected Menzel to make an honest answer instead of an ad hominem attack.     The point is that there is no function which takes primes as its domain and ranges over the primes.  Actually, there is indeed such a function. Matiyasevich's theorem says: Every recursively enumerable set is Diophantine, that is, every recursively enumerable set is equivalent to the integer solutions of some polynomial. Since the set of prime numbers is recursively enumerable, there exists a polynomial function whose integer solutions are exactly the prime numbers. You might even check out the page http://en.wikipedia.org/wiki/Prime_number_formulas for examples of such functions. As a wise person said earlier in this thread "we've made progress if you get off the soap box. " - michael

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