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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: H(D,D) cannot even be asked about the behavior of D(D) V3
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Date: Mon, 17 Jun 2024 07:57:00 -0500
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On 6/17/2024 2:19 AM, Mikko wrote:
> On 2024-06-16 12:48:56 +0000, olcott said:
> 
>> On 6/16/2024 3:00 AM, Mikko wrote:
>>> On 2024-06-16 01:42:29 +0000, olcott said:
>>>
>>>> On 6/15/2024 8:19 PM, Richard Damon wrote:
>>>>> On 6/15/24 8:48 PM, olcott wrote:
>>>>>> 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.
>>>>>>
>>>>>
>>>>>
>>>>> But it isn't an infinite set.
>>>>>
>>>>
>>>> Sure it is you are just clueless.
>>>> I mistook your ignorance for deception.
>>>>
>>>>> We don't ask an infinite set a question, or give a decider an 
>>>>> infinite set of inputs.
>>>>>
>>>>
>>>> Yes we do and this is simply over your head.
>>>>
>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>> The second ⊢* wildcard specifies this infinite set.
>>>
>>> As you should already know, ⊢* as used by Linz is not a wildcard.
>>> It is a repeated application of ⊢ without showing intermediate steps.
>>>
>>
>> It *is* a wild card such that the Linz template simultaneously
>> specifies an infinite set of machines.
> 
> No, it is not. In Linz' book an expression containing ⊢* (or just ⊢) does
> not specify anything. It merely expresses something about a computation.
> 

No you are wrong.

The Linz term “move” means a state transition and its corresponding
tape head action {move_left, move_right, read, write}.
⊢* indicates an arbitrary number of moves.



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