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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is correctly rejected as non-halting V2
Date: Thu, 1 Aug 2024 10:49:32 +0300
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On 2024-07-31 17:28:38 +0000, olcott said:

> On 7/31/2024 2:36 AM, Mikko wrote:
>> On 2024-07-16 18:18:07 +0000, olcott said:
>> 
>>> On 7/16/2024 2:57 AM, Mikko wrote:
>>>> On 2024-07-15 13:43:34 +0000, olcott said:
>>>> 
>>>>> On 7/15/2024 3:17 AM, Mikko wrote:
>>>>>> On 2024-07-14 14:50:47 +0000, olcott said:
>>>>>> 
>>>>>>> On 7/14/2024 5:09 AM, Mikko wrote:
>>>>>>>> On 2024-07-12 14:56:05 +0000, olcott said:
>>>>>>>> 
>>>>>>>>> We stipulate that the only measure of a correct emulation is the
>>>>>>>>> semantics of the x86 programming language.
>>>>>>>>> 
>>>>>>>>> _DDD()
>>>>>>>>> [00002163] 55         push ebp      ; housekeeping
>>>>>>>>> [00002164] 8bec       mov ebp,esp   ; housekeeping
>>>>>>>>> [00002166] 6863210000 push 00002163 ; push DDD
>>>>>>>>> [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
>>>>>>>>> [00002170] 83c404     add esp,+04
>>>>>>>>> [00002173] 5d         pop ebp
>>>>>>>>> [00002174] c3         ret
>>>>>>>>> Size in bytes:(0018) [00002174]
>>>>>>>>> 
>>>>>>>>> When N steps of DDD are emulated by HHH according to the
>>>>>>>>> semantics of the x86 language then N steps are emulated correctly.
>>>>>>>>> 
>>>>>>>>> When we examine the infinite set of every HHH/DDD pair such that:
>>>>>>>>> HHH₁ one step of DDD is correctly emulated by HHH.
>>>>>>>>> HHH₂ two steps of DDD are correctly emulated by HHH.
>>>>>>>>> HHH₃ three steps of DDD are correctly emulated by HHH.
>>>>>>>>> ...
>>>>>>>>> HHH∞ The emulation of DDD by HHH never stops running.
>>>>>>>>> 
>>>>>>>>> The above specifies the infinite set of every HHH/DDD pair
>>>>>>>>> where 1 to infinity steps of DDD are correctly emulated by HHH.
>>>>>>>> 
>>>>>>>> You should use the indices here, too, e.g., "where 1 to infinity steps of
>>>>>>>> DDD₁ are correctly emulated by HHH₃" or whatever you mean.
>>>>>>>> 
>>>>>>> 
>>>>>>> DDD is the exact same fixed constant finite string that
>>>>>>> always calls HHH at the same fixed constant machine
>>>>>>> address.
>>>>>> 
>>>>>> If the function called by DDD is not part of the input then the input does
>>>>>> not specify a behaviour and the question whether DDD halts is ill-posed.
>>>>>> 
>>>>> 
>>>>> We don't care about whether HHH halts. We know that
>>>>> HHH halts or fails to meet its design spec.
>>>>> 
>>>>> We are only seeing if DDD correctly emulated by HHH
>>>>> can can possibly reach its own final state.
>>>> 
>>>> HHH does not see even that. It only sees whther that it does not emulate
>>>> DDD to its final state.
>>> 
>>> No. HHH is not judging whether or not itself is a correct
>>> emulator. The semantics of the x86 instructions that emulates
>>> prove that its emulation is correct.
>> 
>> Semantics of x86 language alone doesn't prove anything. Only a detailed
>> comparison of the emulator code to the x86 semantics may prove that.
> 
> A proof is any sequence of steps such that the conclusion
> is a necessary consequence of its basis.

Only if every "step" is a sentence.

Your traces are not such sequences.

> Proving that DDD correctly emulated by HHH matches the
> infinite recursion behavior pattern.
> (a) The semantics of the x86 language.
> (b) the design of HHH provided below.
> (c) The definition of infinite recursion provided below.
> 
> *Infinite recursion behavior pattern*
> An emulated sequence of instructions that has no conditional
> branch instructions in this sequence is exactly repeated when
> it calls the same function with the same parameters again.
> As long as the called function can be determined to never
> return this proves infinite recursion.

The recursive simulation does not satisfy the "no conditional branch"
requirement so does not match the pattern.

-- 
Mikko