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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Computable Functions --- finite string transformation rules
Date: Fri, 25 Apr 2025 21:50:19 -0400
Organization: i2pn2 (i2pn.org)
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On 4/25/25 5:52 PM, olcott wrote:
> On 4/25/2025 4:42 PM, joes wrote:
>> Am Fri, 25 Apr 2025 16:21:30 -0500 schrieb olcott:
>>> On 4/25/2025 8:56 AM, joes wrote:
>>>> Am Thu, 24 Apr 2025 19:03:34 -0500 schrieb olcott:
>>
>>>>> Mathematical induction proves that DD emulated by HHH cannot possibly
>>>>> reach its own final state in an infinite number of steps and it does
>>>>> this with one recursive emulation.
>>>>> There is a repeating pattern that every C programmer can see.
>>>
>>>> Like Fred wrote months ago, that has nothing to do with the
>>>> contradictory part of DD,
>>> Sure it does. The contradictory part of DD has always been unreachable
>>> thus only a ruse.
> 
>> Then it can't be due to that. It even works with HHH(HHH)!
>>
> 
> When the finite string transformation rules of any
> concrete (thus fully specified) computer language
> are applied to the input to any conventional Halting
> Problem proof the contradictory part is always unreachable.

No, because the input defined in that Halting Problem proof is always 
built on the one specific decider that it is designed to prove wrong.

The UTM machine is not one of those machines, so there is nothing that 
makes that machine unable to simulate the input.

> 
> This works for C, for x86, and for Turing Machines.
> 

Right, and they ALL show that if the Halt Decider that the Input was 
built on returns a value, (and if it doesn't, it isn't a decideer), then 
the UTM or other complete simulator for the input will get to that final 
decision and see the proof program get its answer from its copy of that 
decider, and then act the opposite.

Your problem is your computer model is just incorrect, and what you call 
the input, isn't actually valid, but then you fix that problem by your 
implied inclusion of Halt7.c which then defines the one decider that the 
input will be built on and makes wrong.

You are just to ignorant and stupid to understand this issue, The input 
doesn't (and can't) call whatever decider it trying to decide it, it 
MUST call the one specific decider it was built on, and thus it HAS a 
actual correct simulation, and if the decider returns 0 for that input, 
that behavior WILL halt.