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On 6/28/2024 12:41 PM, joes wrote:I rest my case.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:
Why doesn’t the first recursive H return?To the caller DDD, which then returns to its own caller H0, which
returns „halting” to main… hold on.
Can you see the difference?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))}.
What is the input?The behavior of the directly executed DDD() is irrelevant because
that is not the behavior of the input.
The input is not HHH(DDD). See above.In this case the sequence is the line-by-line execution trace of theNo, the sequence is the behaviour of DDD, period.
behavior of DDD correctly emulated by HHH.
The behavior of this input must include and cannot ignore theYes, and the behaviour of H0 is that it produces the exact same
recursive emulation specified by the fact that DDD is calling its
own emulator.
behaviour as DDD.
WTF? I just agreed with you.That you keep trying to ignore the fact that DDD calls HHH(DDD)The call from DDD to HHH(DDD) when N steps of DDD are correctlyI don’t. A simulator doesn’t even need to return. That’s not in
emulated by any pure function x86 emulator HHH cannot possibly return.
That you assume that it does against the facts is ridiculous.
question. A decider however must.
in recursive simulation is your huge mistake.
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