Re: Liar detector: Peter Olcott

On 7/7/24 9:52 AM, olcott wrote:

On 7/7/2024 2:29 AM, Mikko wrote:

On 2024-07-06 12:54:59 +0000, olcott said:
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On 7/6/2024 1:33 AM, Mikko wrote:

On 2024-07-05 12:43:45 +0000, olcott said:
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On 7/5/2024 7:26 AM, joes wrote:

Am Fri, 05 Jul 2024 07:20:04 -0500 schrieb olcott:

On 7/5/2024 4:49 AM, joes wrote:

Am Wed, 03 Jul 2024 13:57:40 -0500 schrieb olcott:

On 7/3/2024 1:40 PM, Fred. Zwarts wrote:

Op 03.jul.2024 om 20:20 schreef olcott:

>

DDD correctly emulated by any element of the infinite set of every
pure function HHH cannot possibly reach its own ret instruction and
halt. That HHH aborts its emulation at some point or never aborts
its emulation cannot possibly change this.
>

Ad hominem attacks always try to hide a lack of argumentation.
It has been proved that HHH cannot possibly correctly simulate
itself.

>
That is false and you know it. That might not be a flat out lie as it
is an sloppy use of language.
>
HHH does correctly simulate itself simulating DDD one time, then it
stops correctly simulating itself because this criteria is met:
HHH correctly simulates its input DDD until HHH correctly
determines that its simulated DDD would never stop running
unless aborted

But it would stop running.

Not if not aborted.

But it is aborted!
>

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*It is not aborted when HHH makes its decision to abort*

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If it will be aborted in future it will not run forever.
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>
<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>
>
HHH(DDD) does meet this criteria. Professor Sipser is not wrong.

>
You have not proven that it does meet both criteria.
>

 Knowledge of arithmetic proves that 2 + 3 = 5. If you lack
this knowledge that does not mean that 2 + 3 = 5 is not proven.
 _DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
 Sufficient knowledge of the x86 language conclusively proves
that the call from DDD correctly simulated by HHH to HHH(DDD)
cannot possibly return for any pure function HHH.
 That you continue to disagree only having your own ignorance
as a fake basis is ridiculously stupid.
 

Nope, knowledge of the x86 language tells us that a given set of instructions, when executed with the same data, will always behave the same, and thus if HHH(DDD) is known to return 0 to main, it will also return to DDD and thus a CORRECT SIMULATION of DDD, will show that.
If HHH's simulation doesn't it just wasn't a correct simulation, perhaps incorrect by being just partial, thus your premise of DDD correctly simulated gy HHH and HHH(DDD) correctly returning 0 and answer are just inconsistant.
yes, your HHH will not reach the return instruction of HHH in its simulation of DDD, but that doesn't mean that a correct simulation will not.
Only if HHH never aborts, (and that mean NEVER) will DDD be non-halting.
But, since that isn't the HHH you talk about, we can state that a correct simulation of DDD will return, and that you HHH(DDD) just fails to meet either its requirement to correctly simulate the input or to correct decide on the bahavior of its input.