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On 7/13/2024 3:48 AM, joes wrote:But every DDD based on an HHH that stops its emulation and returns, will itself return when fully correctly emulated or run. It is only HHH INCORRECT because it is only PARTIAL emulation of that DDD that doesn't make it to the end.Am Fri, 12 Jul 2024 22:00:08 -0500 schrieb olcott:*This proves that every rebuttal is wrong somewhere*On 7/12/2024 6:41 PM, Richard Damon wrote:>On 7/12/24 7:19 PM, olcott wrote:On 7/12/2024 5:56 PM, Richard Damon wrote:On 7/12/24 10:56 AM, olcott wrote:Where does it disagree?!You try to cut my airtight proof up in little pieces and fail. EveryWhich is just your double-talk to try to redefine what halting means.>Thus each HHH element of the above infinite set of HHH/DDD pairs isNope.
necessarily correct to reject its DDD as non-halting.
>
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are
shown to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because
THEIR DDD will halt. and the HHH that doen't abort never get around
to rejecting its DDD as non-halting.
When no DDD of every HHH/DDD that can possibly exist halts then each
HHH that rejects its DDD as non-halting is necessarily correct.
*No double-talk and weasel words can overcome that*
>
>
rebuttal that you make has disagreeing with the semantics of the x86
language as its basis.
>
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
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