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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH cannot possibly halt
Date: Thu, 11 Jul 2024 23:18:08 -0400
Organization: i2pn2 (i2pn.org)
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On 7/11/24 10:28 PM, olcott wrote:
> On 7/11/2024 9:08 PM, Richard Damon wrote:
>> On 7/11/24 10:51 AM, olcott wrote:
>>> On 7/10/2024 8:50 PM, Richard Damon wrote:
>>>> On 7/10/24 9:21 PM, olcott wrote:
>>>>> On 7/10/2024 8:11 PM, Richard Damon wrote:
>>>>>> On 7/10/24 9:01 PM, olcott wrote:
>>>>>>> On 7/10/2024 7:37 PM, Richard Damon wrote:
>>>>>>>> On 7/10/24 8:24 PM, olcott wrote:
>>>>>>>>> On 7/10/2024 7:01 PM, Richard Damon wrote:
>>>>>>>>>> On 7/10/24 9:41 AM, olcott wrote:
>>>>>>>>>>> On 7/10/2024 8:27 AM, joes wrote:
>>>>>>>>>>>> Am Tue, 09 Jul 2024 23:19:25 -0500 schrieb olcott:
>>>>>>>>>>>>> On 7/9/2024 11:01 PM, joes wrote:
>>>>>>>>>>>>>   > That means that HHH doesn't return, in particular that 
>>>>>>>>>>>>> it doesn't
>>>>>>>>>>>>>   > abort.
>>>>>>>>>>>>> DDD correctly emulated by any pure function HHH that 
>>>>>>>>>>>>> correctly emulates
>>>>>>>>>>>>> 1 to ∞ steps of DDD can't make it past the above line of 
>>>>>>>>>>>>> code no matter
>>>>>>>>>>>>> what.
>>>>>>>>>>>> That line being the call to itself -> it can't simulate itself.
>>>>>>>>>>>>
>>>>>>>>>>>>> *DDD NEVER HALTS*
>>>>>>>>>>>> DDD ONLY calls HHH...
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> void DDD()
>>>>>>>>>>> {
>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>    return;
>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>>> DDD correctly emulated by any pure function HHH that
>>>>>>>>>>> correctly emulates 1 to ∞ lines of DDD can't make it
>>>>>>>>>>> to the second line of DDD no matter what.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Nope, DDD does if HHH(DDD) returns.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> You have a dead cat in your driveway does not mean that
>>>>>>>>> you have a peanut butter sandwich on your front porch.
>>>>>>>>> It has taken you at least 1000 messages to see that.
>>>>>>>>>
>>>>>>>>> DDD correctly emulated by any pure function HHH that
>>>>>>>>> correctly emulates 1 to ∞ lines of DDD can't make it
>>>>>>>>> to the second line of DDD no matter what.
>>>>>>>>
>>>>>>>> WRONG, you don't seem to understand the difference between DDD 
>>>>>>>> and HHH's emualtion of it.
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>> Would you bet your immortal soul that DDD simulated
>>>>>>> by HHH (as provided above) would terminate normally?
>>>>>>>
>>>>>>
>>>>>> That is a ambiguous statement, showing your attempt at deciet.
>>>>>>
>>>>>
>>>>> We stipulate that the only measure of a correct emulation
>>>>> is the semantics of the x86 programming language. By this
>>>>> measure when 1 to ∞ steps of DDD are correctly emulated
>>>>> by each pure function x86 emulator HHH (of the infinite
>>>>> set of every HHH that can possibly exist) then DDD cannot
>>>>> possibly reach its own machine address of 00002174 and halt.
>>>>
>>>> And thus you stipulate that you are a LIAR.
>>>>
>>>> By the semantic of the x86 programming language, the only correct 
>>>> simulation is a FULL simulation
>>> In other words you are trying to get away with the lie that
>>> when 1 step of DDD is correctly emulated that 0 steps of DDD
>>> are correctly emulated.
>>>
>>> Repent of this lie or risk damnation.
>>>
>>
>> WHAT LIE?
>>
> 
> When 1,2,3... ∞ steps of DDD are correctly emulated by
> HHH it is a lie to say that this many instructions were
> not correctly emulated and you know it.
> 

But only N instructions "correctly emulated" is NOT a CORRECT emulaition 
of the instructions of DDD/HHH

WHERE in the Intel x86 documenation does it say that the processor might 
just stop after one of the normal instructions.

They ALL have the definition that after they execute, then next one to 
follow will also be executed.

You logic says that you can call a road as having infinite length by 
riving on it one mile and then gett ing off it, since hta tshowed that 
you didn't drive it to the end.

Remember, by the time that HHH can start to emulate the input, teh full 
program has been determined by pairing it with that HHH, so if you talk 
about "changing" HHH to something different, you can't change the copy 
that DDD is calling, or you change the input and thus invalidate you claims.

You are just painting yourself into the corner with your lies.

This HHH, that aborts its simulation did NOT do a "Correct Emulation" 
per the definition of the x86 processor, but only a PARTIAL emulation, 
that BY DEFINITION doesn't define the behavior after the emulation stopped.

And the actual CORRECT emulation of that exact input shows that it will 
emulate that code of HHH that aborts it emulation of it copy of DDD (in 
a different process space) and returning to the DDD in its process space 
and that DDD returning, thus showing that this HHH didn't actually need 
to abort its emulation, as the HHH in the other process space will and 
that makes the input halt.

You are just stuck with bad definitions that just show you to be an 
igmorant liar.