Re: Analysis of Flibble’s Latest: Detecting vs. Simulating Infinite Recursion ZFC

On 21/05/2025 18:47, olcott wrote:

On 5/21/2025 12:16 PM, Richard Heathfield wrote:

On 21/05/2025 17:51, olcott wrote:

On 5/21/2025 11:09 AM, Richard Heathfield wrote:

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That an algorithm for ascertaining whether an arbitrary program with arbitrary input halts cannot actually exist is precisely what the Halting Problem proves.

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If you think that it can exist then prove that it
exists by encoding it in C.

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It can't exist. The Halting Problem proves that it can't. I said that already.
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 *PAY ATTENTION*
I am saying that a key element of the halting problem
proof cannot exist, thus the proof itself cannot exist.

Yes, it can, and it does.
Watch.
Definition: a prime number is an integer >= 2 with no divisors >= 2 except itself.
Hypothesis: there is no largest prime number.
Proof:
Assume that a largest prime number Pn exists.
Itemise all the prime numbers from P1(=2) up to Pn :
P1 P2 P3 ... Pn
Insert x symbols all the way along.
P1 x P2 x P3 ... x Pn
Add 1.
The number thus calculated is not divisible by any prime in our list (there being a remainder of 1 in each case), so the number calculated is (a) prime, and (b) larger than Pn. Thus Pn is not the largest prime. This contradicts the assumption made at the beginning, which must therefore be false. Proof by contradiction.
The proof that no largest prime exists despite its assumption that such a prime /does/ exist - an assumption that turns out to be false.
*PAY ATTENTION*
I am saying that a key element of the Halting Problem proof is that it proves that a universal halting decider cannot exist.

Do you understand that a non-existent proof cannot
prove one damn thing?

The Halting Proof exists, whether you want it to or not.
I'm finding it hard to believe that you can really be this stupid. Perhaps you just get off yanking people's chains.
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
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