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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: Bad faith and dishonesty
Date: Fri, 30 May 2025 15:01:11 -0400
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On 5/30/2025 3:00 PM, olcott wrote:
> On 5/30/2025 1:48 PM, dbush wrote:
>> On 5/30/2025 2:40 PM, olcott wrote:
>>> On 5/30/2025 1:20 PM, Richard Heathfield wrote:
>>>> On 30/05/2025 18:45, dbush wrote:
>>>>> On 5/30/2025 1:27 PM, olcott wrote:
>>>>>> On 5/30/2025 12:06 PM, Richard Heathfield wrote:
>>>>>>> There aren't many ways to invalidate a proof. Demonstrating that 
>>>>>>> the conclusion is false is insufficient (because you now have two 
>>>>>>> proofs, each of which claims that 'I'm right so you're wrong'); 
>>>>>>> one must attack the reasoning or the assumptions (or both) and 
>>>>>>> show how a flawed step or a flawed assumption invalidates the 
>>>>>>> method (and perhaps the conclusion).
>>>>>>>
>>>>>>> As it happens, Olcott accepts anyway that Turing's conclusion is 
>>>>>>> correct, so his only beef can be with an assumption or a step.
>>>>>>>
>>>>>>
>>>>>> Turing's conclusion *is correct within a false assumption*
>>>>>
>>>>> Specifically, the assumption that the following requirements can be 
>>>>> met:
>>>>>
>>>>>
>>>>> Given any algorithm (i.e. a fixed immutable sequence of 
>>>>> instructions) X described as <X> with input Y:
>>>>>
>>>>> A solution to the halting problem is an algorithm H that computes 
>>>>> the following mapping:
>>>>>
>>>>> (<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed 
>>>>> directly
>>>>
>>>> Yes, that is precisely the assumption Turing makes, and he makes it 
>>>> explicitly, and he makes it with the express intent of showing that 
>>>> it cannot be true.
>>>>
>>>>>> YOU MUST PAY ATTENTION TO ALL THE WORDS THAT I SAY.
>>>>
>>>> Noise on the line again, I see. I must call the broadband people.
>>>>
>>>>>>
>>>>>>> Turing's only assumption is overturned by reductio within the 
>>>>>>> proof itself, so that can't be it... which only leaves steps.
>>>>>>>
>>>>>>> As far as I can recall, Olcott's ramblings never go within 
>>>>>>> discus- throwing distance of a potentially erroneous step.
>>>>>>>
>>>>>>
>>>>>> There is no *INPUT* D to termination analyzer H
>>>>>> that can possibly do the opposite of whatever
>>>>>> value that H returns.
>>>>>
>>>>> False. "DDD" is a description/specification of algorithm DDD 
>>>>> consisting of the fixed code of the function DDD, the fixed code 
>>>>> function HHH, and the fixed code of everything that HHH calls down 
>>>>> to the OS level.
>>>>
>>>> HHH is not the computation Turing assumed could exist (for the sole 
>>>> purpose of showing that it could not). HHH is a hodgepodge of shit C 
>>>> and what looks like more line noise in assembly mnemonics. It is not 
>>>> a universal computation such as Turing envisaged:
>>>>
>>>> +++++
>>>> Let us suppose that there is such a process; that is to say, that we 
>>>> can invent a machine <D- which, when supplied with the S.D of any 
>>>> computing machine i l will test this S.D and if i l is circular will 
>>>> mark the S.D with the symbol "u" and if it is circle-free will mark 
>>>> it with " s ".
>>>> +++++
>>>>
>>>> By "the S.D. of any computing machine" he means the 'standard 
>>>> description' of >>>>any<<<< Turing machine.
>>>>
>>>> HHH is not that process, and thus HHH has no bearing whatsoever on 
>>>> the Turing proof.
>>>>
>>>
>>> It is a verified fact that the
>>> *input input input input input input*
>>> *input input input input input input*
>>> *input input input input input input*
>>> *input input input input input input*
>>>
>>> to HHH(DDD)
>>
>>
>> i.e. a description of algorithm DDD consisting of the fixed code of 
>> the function DDD, the fixed code of the function HHH, and the fixed 
>> code of everything that HHH calls down to the OS level.
>>
> 
> Never stops running unless HHH aborts its emulation

In other words, if you change the input so that HHH doesn't abort.

Changing the input is not allowed.