Re: Hypothetical possibilities --- Complete Proof

Op 02.aug.2024 om 13:24 schreef olcott:

On 8/2/2024 3:13 AM, Fred. Zwarts wrote:

Op 01.aug.2024 om 23:03 schreef olcott:

On 8/1/2024 2:30 PM, Fred. Zwarts wrote:

Op 01.aug.2024 om 18:32 schreef olcott:

On 8/1/2024 11:11 AM, joes wrote:

Am Thu, 01 Aug 2024 09:30:00 -0500 schrieb olcott:

On 8/1/2024 9:23 AM, Fred. Zwarts wrote:

Op 01.aug.2024 om 15:29 schreef olcott:

On 8/1/2024 8:12 AM, Fred. Zwarts wrote:

Op 01.aug.2024 om 14:20 schreef olcott:

On 8/1/2024 3:10 AM, Fred. Zwarts wrote:

Op 31.jul.2024 om 23:23 schreef olcott:

On 7/31/2024 3:01 PM, Fred. Zwarts wrote:

Op 31.jul.2024 om 17:14 schreef olcott:

On 7/31/2024 3:44 AM, Fred. Zwarts wrote:

Op 31.jul.2024 om 06:09 schreef olcott:

>

The trace stops and hides what happens when 000015d2 is called.
Olcott is hiding the conditional branch instructions in the
recursion.

These next lines conclusively prove that DDD is being correctly
emulated by HHH after DDD calls HHH(DDD).

It also shows that HHH when simulating itself, does not reach the end
of its own simulation.

If you weren't a clueless wonder you would understand that DDD correctly
emulated by HHH including HHH emulating itself emulated DDD has no end
of correct emulation.

>

It does if the simulated HHH aborts, but its simulating copy preempts
that. Indeed, it has no choice, but if it didn't abort, the simulation
wouldn't abort either. Therefore it can't simulate itself.
>

>
<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>

>
Sipser agreed only to a correct simulation.

>
of N steps.

>
Without skipping M steps of a halting program.
>

 THAT IS WRONG. IT IS MAKING SURE TO SKIP ALL THE STEPS AFTER

Are you shouting because you see my claim is true?

 H correctly determines that its simulated D would never
stop running unless aborted

It skips the last M steps of a halting program. The programmer was very confused when he thought that two recursions implied an infinite number of recursions. He did not notice that one cycle later the halting program would halt.
The 'never stop running unless aborted' is irrelevant, because the simulated program does abort and halt.
You are dreaming of a HHH that does not abort. Dreams are no substitute for fact, nor for logic.

 

>

>
I spent two years carefully composing the above before I even
asked professor Sipser to review it.
>
DDD is correctly emulated by HHH until HHH sees the same
never ending pattern that anyone else can see.

>
The never ending pattern is there only in your dreams. The HHH that halts after two cycles has a halting pattern.

>
In order for DDD correctly emulated by HHH to halt
DDD correctly emulated  must reach its emulated "ret"
instruction. This <is> impossible.

>
Indeed! HHH cannot possibly simulate itself correctly.

 You are a damned liar about how correct emulation is defined.
 

The fact that you need such language reveals how uncertain you feel about your claims.