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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05
 --partial agreement--
Date: Sat, 9 Mar 2024 09:50:15 -0800
Organization: i2pn2 (i2pn.org)
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On 3/9/24 8:15 AM, olcott wrote:
> On 3/9/2024 9:33 AM, immibis wrote:
>> On 9/03/24 16:04, olcott wrote:
>>> On 3/9/2024 7:11 AM, immibis wrote:
>>>> On 8/03/24 20:08, olcott wrote:
>>>>> On 3/8/2024 12:58 PM, immibis wrote:
>>>>>> On 8/03/24 19:12, olcott wrote:
>>>>>>>
>>>>>>> Unless it is an extra parameter it has no basis for doing this.
>>>>>>> If it is an extra parameter then it is no longer the Linz proof.
>>>>>>>
>>>>>>
>>>>>> So it's impossible to make a Turing machine that writes 12345 onto 
>>>>>> its tape unless 12345 is a parameter?
>>>>>
>>>>> That is a valid point.
>>>>> It is impossible to get me to talk about that until
>>>>> we first have full closure that the Linz H correctly
>>>>> determines the halt status of the Linz Ĥ ⟨Ĥ⟩ when run
>>>>> in the Olcott master UTM.
>>>>>
>>>>
>>>> When run in the Olcott master UTM as Olcott machines, the Linz proof 
>>>> does not work because it is designed for Turing machines. A 
>>>> different proof works.
>>>
>>> Because Olcott machines are more powerful than Turing Machines they 
>>> can correctly determine the halt status of this input: H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>.
>>>
>>
>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> always calculates the same answer that H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> 
>> calculates.
>>
> 
> No it does not. Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ has some preceding steps that copy its input
> and H ⟨Ĥ⟩ ⟨Ĥ⟩ does not have these steps. Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see all of these
> steps.

HOW?

The only "memory" in the machine is the tape, and that doesn't have any 
of that information on it.

> 
> It turns out that this single difference is why Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort 
> its simulation and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort its simulation.

I.E, if you lie about the system, you can claim anything you want.

You just proved you are totally ignorant about how Turing Machines and 
computations work.

This makes you the pathetic ignorant pathoogical lying idiot that you are.

> 
> I only created Olcott machines so that it is easy to see how Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
> would detect that it must abort its simulation and transition to Ĥ.Hqn.

Except you don't explain HOW it does that,

Since the Olcott H^.H is pased the context <H> instead of <H^> it has no 
way to see what you are claiming.

> 
> Turing machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ might still be able to do this thus
> nullifying your counter-example.
> 
> Your reviews are often quite good. Some of Richard's reviews
> were very good and some of them are stuck in circular reasoning
> anchored in very persistently false assumptions.
> 

You are just proving you are too stupid to understand what yoare talking 
about.