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From: joes <noreply@example.com>
Newsgroups: comp.theory
Subject: Re: 197 page execution trace of DDD correctly simulated by HHH
Date: Fri, 28 Jun 2024 17:41:17 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Thanks for leaving the unanswered questions in place, though I’d rather
have you answer them.

Am Fri, 28 Jun 2024 12:05:18 -0500 schrieb olcott:
> On 6/28/2024 11:26 AM, joes wrote:
>> Am Fri, 28 Jun 2024 10:25:36 -0500 schrieb olcott:
>>> On 6/28/2024 8:14 AM, joes wrote:
>>>> Am Thu, 27 Jun 2024 12:30:38 -0500 schrieb olcott:

>>>> To the caller DDD, which then returns to its own caller H0, which
>>>> returns „halting” to main… hold on.
>> Where do you disagree?

>>>> H0 must not report on itself, only on DDD. Which you’ve proven halts.
>>>> We don’t care how H0 deviates (i.e. is incorrect) in its simulation.
>>>> That would be main {H0(H0(DDD))}.
Do you see what I mean?

>>> The behavior of the directly executed DDD() is irrelevant because that
>>> is not the behavior of the input.
>> What is the difference here?
Isn’t the input DDD?

>>> In this case the sequence is the line-by-line execution trace of the
>>> behavior of DDD correctly emulated by HHH.
>> No, the sequence is the behaviour of DDD, period.
The input is not HHH(DDD). See above.

>>> The behavior of this input must include and cannot ignore the
>>> recursive emulation specified by the fact that DDD is calling its own
>>> emulator.
>> Yes, and the behaviour of H0 is that it produces the exact same
>> behaviour as DDD.
Because it is a simulator.

> The call from DDD to HHH(DDD) when N steps of DDD are correctly emulated
> by any pure function x86 emulator HHH cannot possibly return.
> That you assume that it does against the facts is ridiculous.
I don’t. A simulator doesn’t even need to return. That’s not in question.
A decider however must.