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From: Richard Heathfield <rjh@cpax.org.uk>
Newsgroups: comp.theory
Subject: =?UTF-8?Q?Re=3A_Analysis_of_Flibble=E2=80=99s_Latest=3A_Detecting_v?=
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Date: Wed, 21 May 2025 21:13:12 +0100
Organization: Fix this later
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On 21/05/2025 20:28, olcott wrote:
> On 5/21/2025 2:06 PM, Richard Heathfield wrote:
>> On 21/05/2025 19:48, olcott wrote:
>>
>> <snip>
>>
>>> Show how to define a D that actually does the opposite
>>> of what its termination analyzer reports.
>>
>> The whole point of the proof is that no algorithm can define a 
>> universal halt decider.
>>
> 
> I have NEVER been talking about that.

Who cares what *you're* talking about?

Um... let's see. You? Probably.

> I have only been talking about the ACTUAL
> conventional proof of the halting problem.

The ACTUAL conventional proof of the Halting Problem goes 
something like this:

1) assume that it is possible to devise an algorithm that can 
determine in finitely many steps ascertain whether an arbitrary 
program applied to arbitrary data does or does not stop.

2) given such an algorithm, imagine incorporating it into a 
program that ascertains whether a supplied program with supplied 
data halts, loops if it does, and halts if it doesn't.

3) imagine feeding the program to itself, and we arrive at the 
contradiction that the program would halt if it didn't but not if 
it did.

4) Our reasoning has led us to a contradiction, so we deduce that 
the only assumption we made, in 1) above, is false. QED.

That, highly paraphrased, is the ACTUAL conventional proof of the 
halting problem.

Note: no simulation required. There's nothing to simulate; it's a 
thought experiment, not something you actually do.

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within