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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Proof that DDD specifies non-halting behavior --- point by point
Date: Wed, 14 Aug 2024 13:25:48 -0000 (UTC)
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Am Wed, 14 Aug 2024 08:14:42 -0500 schrieb olcott:
> On 8/14/2024 2:43 AM, joes wrote:
>> Am Tue, 13 Aug 2024 21:38:07 -0500 schrieb olcott:
>>> On 8/13/2024 9:29 PM, Richard Damon wrote:
>>>> On 8/13/24 8:52 PM, olcott wrote:
>> 
>>>>> A simulation of N instructions of DDD by HHH according to the
>>>>> semantics of the x86 language is necessarily correct.
>>>> Nope, it is just the correct PARTIAL emulation of the first N
>>>> instructions of DDD, and not of all of DDD,
>>> That is what I said dufuss.
>> You were trying to label an incomplete/partial/aborted simulation as
>> correct.
> When one instruction of DDD is correctly emulated then one instruction
> was correctly emulated.
Only one, so not all.

>>>>> A correct simulation of N instructions of DDD by HHH is sufficient
>>>>> to correctly predict the behavior of an unlimited simulation.
>>>> Nope, if a HHH returns to its caller,
>>> *Try to show exactly how DDD emulated by HHH returns to its caller*
>> how *HHH* returns
> Changing the question is the strawman error or reasoning.
Richard was talking about HHH returning.

>>> (the first one doesn't even have a caller)
>>> Use the above machine language instructions to show this.
>> HHH simulates DDD	        enter the matrix
>>    DDD calls HHH(DDD)	Fred: could be eliminated
>>    HHH simulates DDD         second level
>>       DDD calls HHH(DDD)	recursion detected
>>    HHH aborts, returns	outside interference
>>    DDD halts		    	voila
>> HHH halts
>> 
> That is the strawman error of reasoning.
This is not a misrepresentation of your position, this is mine.

> DDD correctly emulated by HHH never reaches its own "return"
> instruction. Show how it does or admit that I am correct.
See above. Show the error.

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.