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On 6/7/24 11:56 AM, olcott wrote:I provide conclusive proof otherwise and your "rebuttal" isOn 6/7/2024 10:46 AM, Richard Damon wrote:And for this, "Correct Simulation" means a simulation that accurated reflects that actual behavior of the dirrectly executed machine,On 6/7/24 11:29 AM, olcott wrote:>On 6/7/2024 10:14 AM, Richard Damon wrote:>On 6/7/24 9:09 AM, olcott wrote:I conclusively prove my point and you finally admit that your wholeOn 6/6/2024 10:29 PM, Richard Damon wrote:>>>
If the essence of your life's work is that you came up with a way to not-prove the thing you were trying to prove
No you are just a Liar
Then try to show it.
>
CHANGE-THE-SUBJECT strawman deception fake rebuttal has always simply
ignored the proof that I am correct shown below:
>
Try to show how this DD correctly simulated by any HH ever
stops running without having its simulation aborted by HH.
>
_DD()
[00001e12] 55 push ebp
[00001e13] 8bec mov ebp,esp
[00001e15] 51 push ecx
[00001e16] 8b4508 mov eax,[ebp+08]
[00001e19] 50 push eax ; push DD
[00001e1a] 8b4d08 mov ecx,[ebp+08]
[00001e1d] 51 push ecx ; push DD
[00001e1e] e85ff5ffff call 00001382 ; call HH
>
A {correct simulation} means that each instruction of the
above x86 machine language of DD is correctly simulated
by HH and simulated in the correct order.
>
Anyone claiming that HH should report on the behavior
of the directly executed DD(DD) is requiring a violation
of the above definition of correct simulation.
>
And your last statement proves why you have the problem.
>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D
until H correctly determines that its simulated D would never
stop running unless aborted then
>
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
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