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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: H(D,D) cannot even be asked about the behavior of D(D) V3
 ---IGNORING ALL OTHER REPLIES
Date: Sat, 15 Jun 2024 20:13:27 -0400
Organization: i2pn2 (i2pn.org)
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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.

If we simulate THAT input for one more step, it will see the FIRST H 
decide to abort and return.

If you don't bind the first H, yo can't do the first step.

> 
> _D()
> [00000cfc](01) 55          push ebp
> [00000cfd](02) 8bec        mov ebp,esp
> [00000cff](03) 8b4508      mov eax,[ebp+08]
> [00000d02](01) 50          push eax       ; push D
> [00000d03](03) 8b4d08      mov ecx,[ebp+08]
> [00000d06](01) 51          push ecx       ; push D
> [00000d07](05) e800feffff  call 00000b0c  ; call H
> [00000d0c](03) 83c408      add esp,+08
> [00000d0f](02) 85c0        test eax,eax
> [00000d11](02) 7404        jz 00000d17
> [00000d13](02) 33c0        xor eax,eax
> [00000d15](02) eb05        jmp 00000d1c
> [00000d17](05) b801000000  mov eax,00000001
> [00000d1c](01) 5d          pop ebp
> [00000d1d](01) c3          ret
> Size in bytes:(0034) [00000d1d]
> 
> 
>> You haven't shown the required pieces for an inductive proof.
>>
>> I doubt you even know what you need to do, let alone be able to do it.
> 
>