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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is correctly rejected as
 non-halting V2
Date: Tue, 16 Jul 2024 20:43:10 -0000 (UTC)
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Am Tue, 16 Jul 2024 13:18:07 -0500 schrieb olcott:
> 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.
>>>>>>> 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.
>>>>>>> ...
>>>>>>> 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.
You have been saying for a while that HHH returns what it would report
on its input DDD. Glad to see you come around.

> Only because DDD calls HHH(DDD) in recursive emulation it is impossible
> for DDD correctly emulated by HHH to reach past its own machine address
> of 0000216b.
It is not impossible. The simulated HHH aborts just the same and returns
to DDD.

>> But we can see more, in particuar that DDD() halts if HHH(DDD) does.
> It is still a fact that HHH(DDD) was required to abort its emulation.
Just no.

>> Anyway, if the function DDD calls is not a part of the input then the
>> question whether DDD halts is not well-posed and can only be ansered
>> with a conditional.
> We are analyzing whether or not DDD halts.
> We are NOT analyzing whether or not HHH halts.
Whether DDD halts depends entirely on HHH, because DDD does nothing
else but call it.

-- 
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.