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From: immibis <news@immibis.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: D correctly simulated by H cannot possibly halt --- templates and
 infinite sets
Date: Thu, 30 May 2024 12:11:24 +0200
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On 30/05/24 05:48, olcott wrote:
> On 5/29/2024 9:55 PM, Richard Damon wrote:
>> On 5/29/24 10:36 PM, olcott wrote:
>>> On 5/29/2024 9:25 PM, Richard Damon wrote:
>>>> On 5/29/24 9:55 PM, olcott wrote:
>>>>> When the category is examined all at once then there is no need
>>>>> to look at each individual element.
>>>>
>>>> So, which one or ones gave the correct answer for their input?
>>>>
>>>
>>> *Formalizing the Linz Proof structure*
>>> ∃H  ∈ Turing_Machines
>>> ∀x  ∈ *Turing_Machines_Descriptions*
>>> ∀y  ∈ Finite_Strings
>>> such that H(x,y) = Halts(x,y)
>>>
>>> When we formalize it that way then some simulating halt deciders
>>> get the correct answer.
>>>
>>> *Everyone else implicitly assumes this incorrect formalization*
>>> ∃H  ∈ Turing_Machines
>>> ∀x  ∈ *Turing_Machines*
>>> ∀y  ∈ Finite_Strings
>>> such that H(x,y) = Halts(x,y)
>>>
>>>
>>
>> Nope.
>>
>> You just don't understand the meaning of a "Description" in the problem.
>>
> 
> A deciders compute the mapping FROM ITS INPUTS
> to it own accept or reject state
> Deciders cannot take ACTUAL TURING MACHINES AS INPUTS
> Deciders can only take FINITE STRINGS AS INPUTS
> 

If you want to be pedantic, you made the mistake.
It's actually H(DescriptionOf(x),y) = Halts(x,y)
DescriptionOf is an injective function that converts Turing machines 
into finite strings.