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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 16:26:50 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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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:

>>> When this is construed as non-halting criteria then simulating
>>> termination analyzer H0 is correct to reject these inputs as
>>> non-halting by returning 0 to its caller.
>> To the caller DDD, which then returns to its own caller H0, which
>> returns „halting” to main… hold on.
Where do you disagree?

>>> Simulating termination analyzers must report on the behavior that
>>> their finite string input specifies thus H0 must report that DDD
>>> correctly emulated by H0 remains stuck in recursive simulation.
>> 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))}.

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

> Deciders compute the mapping from
> their actual finite string input to an output by a sequence of finite
> string transformations.
And should get the right answer.

> 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 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.