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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: H(D,D) cannot even be asked about the behavior of D(D) V3
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Date: Sat, 15 Jun 2024 19:48:24 -0500
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On 6/15/2024 7:13 PM, Richard Damon wrote:
> On 6/15/24 8:05 PM, olcott wrote:
>> On 6/15/2024 6:37 PM, Richard Damon wrote:
>>> On 6/15/24 7:30 PM, olcott wrote:
>>>> On 6/15/2024 6:01 PM, Richard Damon wrote:
>>>>> On 6/15/24 5:56 PM, olcott wrote:
>>>>>> On 6/15/2024 11:33 AM, Richard Damon wrote:
>>>>>>> On 6/15/24 12:22 PM, olcott wrote:
>>>>>>>> On 6/13/2024 8:24 PM, Richard Damon wrote:
>>>>>>>>  > On 6/13/24 11:32 AM, olcott wrote:
>>>>>>>>  >>
>>>>>>>>  >> It is contingent upon you to show the exact steps of how H 
>>>>>>>> computes
>>>>>>>>  >> the mapping from the x86 machine language finite string 
>>>>>>>> input to
>>>>>>>>  >> H(D,D) using the finite string transformation rules 
>>>>>>>> specified by
>>>>>>>>  >> the semantics of the x86 programming language that reaches the
>>>>>>>>  >> behavior of the directly executed D(D)
>>>>>>>>  >>
>>>>>>>>  >
>>>>>>>>  > Why? I don't claim it can.
>>>>>>>>
>>>>>>>> The first six steps of this mapping are when instructions
>>>>>>>> at the machine address range of [00000cfc] to [00000d06]
>>>>>>>> are simulated/executed.
>>>>>>>>
>>>>>>>> After that the behavior of D correctly simulated by H diverges
>>>>>>>> from the behavior of D(D) because the call to H(D,D) by D
>>>>>>>> correctly simulated by H cannot possibly return to D.
>>>>>>>
>>>>>>> Nope, the steps of D correctly simulated by H will EXACTLY match 
>>>>>>> the steps of D directly executed, until H just gives up and guesses.
>>>>>>>
>>>>>>
>>>>>> When we can see that D correctly simulated by H cannot possibly
>>>>>> reach its simulated final state at machine address [00000d1d]
>>>>>> after one recursive simulation and the same applies for 2,3,...N
>>>>>> recursive simulations then we can abort the simulated input and
>>>>>> correctly report that D correctly simulated by H DOES NOT HALT.
>>>>>
>>>>> Nope. Because an aborted simulation doesn't say anything about 
>>>>> Halting,
>>>>>
>>>>
>>>> It is the mathematical induction that says this.
>>>>
>>> WHAT "Mathematical Induction"?
>>>
>>
>> A proof by induction consists of two cases. The first, the base
>> case, proves the statement for n = 0 without assuming any knowledge
>> of other cases. The second case, the induction step, proves that
>> if the statement holds for any given case n = k then it must also
>> hold for the next case n = k + 1 These two steps establish that the
>> statement holds for every natural number n.
>> https://en.wikipedia.org/wiki/Mathematical_induction
> 
> Ok, so you can parrot to words.
> 
>>
>> It is true that after one recursive simulation of D correctly
>> simulated by H that D does not reach its simulated final state
>> at machine address [00000d1d].
> 
> Which means you consider that D has been bound to that first H, so you 
> have instruciton to simulate in the call H.
> 
>>
>> *We directly see this is true for every N thus no assumption needed*
>> It is true that after N recursive simulations of D correctly
>> simulated by H that D does not reach its simulated final state
>> at machine address [00000d1d].
> 
> Nope, because to do the first step, you had to bind the definition of 
> the first H to D, and thus can not change it.

So infinite sets are permanently beyond your grasp.
The above D simulated by any H has the same property
of never reaching its own simulated machine address
at [00000d1d].

What I mistook for dishonestly is simply a lack
of comprehension.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer