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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Wed, 2 Apr 2025 23:01:38 -0400
Organization: i2pn2 (i2pn.org)
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On 4/2/25 10:41 PM, olcott wrote:
> On 4/2/2025 6:18 AM, Richard Damon wrote:
>> On 4/1/25 10:40 PM, olcott wrote:
>>> On 4/1/2025 8:15 PM, Richard Damon wrote:
>>>> On 4/1/25 7:31 PM, olcott wrote:
>>>>> On 4/1/2025 1:25 AM, Mikko wrote:
>>>>>> On 2025-03-31 18:06:35 +0000, olcott said:
>>>>>>
>>>>>>> On 3/31/2025 3:47 AM, Mikko wrote:
>>>>>>>> On 2025-03-30 20:32:07 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 3/30/2025 1:52 PM, Richard Damon wrote:
>>>>>>>>>> On 3/30/25 2:27 PM, olcott wrote:
>>>>>>>>>>> On 3/30/2025 3:12 AM, joes wrote:
>>>>>>>>>>>> Am Sat, 29 Mar 2025 16:46:26 -0500 schrieb olcott:
>>>>>>>>>>>>> On 3/29/2025 3:14 PM, dbush wrote:
>>>>>>>>>>>>>> On 3/29/2025 4:01 PM, olcott wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>>>> We can know that when this adapted UTM simulates a finite 
>>>>>>>>>>>>>>> number of
>>>>>>>>>>>>>>> steps of its input that this finite number of steps were 
>>>>>>>>>>>>>>> simulated
>>>>>>>>>>>>>>> correctly.
>>>>>>>>>>>>>> And therefore does not do a correct UTM simulation that 
>>>>>>>>>>>>>> matches the
>>>>>>>>>>>>>> behavior of the direct execution as it is incomplete.
>>>>>>>>>>>>> It is dishonest to expect non-terminating inputs to complete.
>>>>>>>>>>>> A complete simulation of a nonterminating input doesn't halt.
>>>>>>>>>>>>
>>>>>>>>>>>>>>>> 2) changing the input is not allowed
>>>>>>>>>>>>>>> The input is unchanged. There never was any indication 
>>>>>>>>>>>>>>> that the input
>>>>>>>>>>>>>>> was in any way changed.
>>>>>>>>>>>>>> False, if the starting function calls UTM and UTM changes, 
>>>>>>>>>>>>>> you're
>>>>>>>>>>>>>> changing the input.
>>>>>>>>>>>>> When UTM1 is a UTM that has been adapted to only simulate a 
>>>>>>>>>>>>> finite
>>>>>>>>>>>>> number of steps
>>>>>>>>>>>> So not an UTM.
>>>>>>>>>>>>
>>>>>>>>>>>>> and input D calls UTM1 then the behavior of D simulated
>>>>>>>>>>>>> by UTM1 never reaches its final halt state.
>>>>>>>>>>>>> When D is simulated by ordinary UTM2 that D does not call 
>>>>>>>>>>>>> Then D reaches
>>>>>>>>>>>>> its final halt state.
>>>>>>>>>>>> Doesn't matter if it calls it, but if the UTM halts.
>>>>>>>>>>>>
>>>>>>>>>>>>>> Changing the input is not allowed.
>>>>>>>>>>>>> I never changed the input. D always calls UTM1.
>>>>>>>>>>>>> thus is the same input to UTM1 as it is to UTM2.
>>>>>>>>>>>> You changed UTM1, which is part of the input D.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> UTM1 simulates D that calls UTM1
>>>>>>>>>>> simulated D NEVER reaches final halt state
>>>>>>>>>>>
>>>>>>>>>>> UTM2 simulates D that calls UTM1
>>>>>>>>>>> simulated D ALWAYS reaches final halt state
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Only because UTM1 isn't actually a UTM, but a LIE since it 
>>>>>>>>>> only does a partial simulation, not a complete as REQURIED by 
>>>>>>>>>> the definition of a UTM.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> _DDD()
>>>>>>>>> [00002172] 55         push ebp      ; housekeeping
>>>>>>>>> [00002173] 8bec       mov  ebp,esp  ; housekeeping
>>>>>>>>> [00002175] 6872210000 push 00002172 ; push DDD
>>>>>>>>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>>>>>>>>> [0000217f] 83c404     add  esp,+04
>>>>>>>>> [00002182] 5d         pop  ebp
>>>>>>>>> [00002183] c3         ret
>>>>>>>>> Size in bytes:(0018) [00002183]
>>>>>>>>>
>>>>>>>>> DDD EMULATED BY HHH DOES SPECIFY THAT IT
>>>>>>>>> CANNOT POSSIBLY REACH ITS OWN FINAL HALT STATE.
>>>>>>>>
>>>>>>>> No, it does not. HHH misintepretes, contrary to the semantics of 
>>>>>>>> x86,
>>>>>>>> the specification to mean that.
>>>>>>>
>>>>>>> It is a truism that a correct x86 emulator
>>>>>>> would emulate itself emulating DDD whenever
>>>>>>> DDD calls this emulator with itself.
>>>>>>
>>>>>> Irrelevant. You didn't say anything about a correct emulator or 
>>>>>> emulation.
>>>>>>
>>>>>
>>>>> Sure all trolls would agree that when-so-ever a statement
>>>>> is made many dozens of time this proves that this statement
>>>>> was never said.
>>>>>
>>>>> When-so-ever a finite number of steps of x86 machine
>>>>> code have been emulated according to the semantics
>>>>> of the x86 language then these steps have been
>>>>> emulated correctly even if God himself disagrees.
>>>>>
>>>>
>>>> And the number of times you have INCORRECTLY claimed that you 
>>>> partial simulator did a correct emulation just shows how much of a 
>>>> liar you are.
>>>>
>>>
>>> Anyone that knows the C language knows that one recursive
>>> emulation proves that DDD emulated by HHH cannot possibly
>>> reach its final halt state in an infinite number of steps.
>>>
>>
>> No it doesn't, as if HHH actually emulated that path, it would see 
>> that the HHH that it emulated will eventually abort its emulation and 
>> return.
>>
> 
> This stumped me for abut four days several years ago.
> It turns out that unless the outer HHH aborts none
> of them ever abort. This proves that HHH is correct.
> 

But since the outer one DOES, as that is what the code does, all the 
inner ones do too.

All you are doing is admitting that your concept of programs is based on 
lies and deception and that a "given program" might not always be the 
same one.

Sorry, you are just proving that you ar ejust a pathologial liar that 
doesn't understand what he is talking about, and doesn't care.