Re: HHH maps its input to the behavior specified by it --- never reaches its halt state ---

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Sujet : Re: HHH maps its input to the behavior specified by it --- never reaches its halt state ---
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theory
Date : 10. Aug 2024, 14:55:22
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v97rg9$l4f4$3@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
User-Agent : Mozilla Thunderbird
Op 10.aug.2024 om 15:25 schreef olcott:
On 8/10/2024 8:08 AM, Fred. Zwarts wrote:
Op 10.aug.2024 om 13:34 schreef olcott:
On 8/10/2024 3:27 AM, Fred. Zwarts wrote:
Op 09.aug.2024 om 22:53 schreef olcott:
On 8/9/2024 2:35 PM, Fred. Zwarts wrote:
Op 09.aug.2024 om 18:19 schreef olcott:
>
void DDD()
{
   HHH(DDD);
   return;
}
>
Each HHH of every HHH that can possibly exist definitely
emulates zero to infinity instructions of DDD correctly.
Every expert in the C language sees that this emulated DDD
cannot possibly reaches its own "return" instruction halt state.
And you don't need to be an expert to see that this proves that all these simulations are incorrect.
>
In other words you are trying to get away with the lie that
Richard has been persistently pushing:
>
When N > 0 instructions of DDD are correctly emulated by HHH
then no instructions of DDD have been correctly emulated.
>
>
>
In other words, you believe that only the numbers zero and infinite exist.
>
 *The set of HHH x86 emulators are defined such that*
 Each element of this set corresponds to one element of
the set of positive integers indicating the number of
x86 instructions of DDD that it correctly emulates.
 When we can see that in none of these cases that the
correctly emulated DDD ever reaches its "return" instruction
halt state.
Proving that none of these simulations is correct. Because each of them has a reachable return (as proven by the direct execution and by another simulator, like HHH1), but the simulator failed to reach it.

 This entails that each HHH can take a wild guess that
its input does not reach this halt state and HHH would
necessarily be correct.
If each of these simulations is incorrect, they are all incorrect, so the wild guess that none of them reach the halt state is correct, proving that all of them failed to simulate correctly up to the halt state. So, each of them did an incomplete and therefore incorrect simulation.

 When all X has property Y then each X is necessarily
correct to state that is has property Y.
 From this it follows that:
When all HHH have the property that it fails to simulate itself correctly up to the halt state, then it is necessarily correct to state that each HHH fails to simulate itself correctly up to the halt state.
Everybody denying such simple facts should stop the discussion immediately.

Date Sujet#  Auteur
6 Jul 25 o 

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