Re: Mike Terry Proves --- How the requirements that Professor Sipser agreed to are exactly met

On 5/22/25 6:42 PM, olcott wrote:

On 5/22/2025 3:10 AM, Mikko wrote:

On 2025-05-21 15:41:45 +0000, olcott said:
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On 5/21/2025 3:05 AM, Mikko wrote:

On 2025-05-20 18:31:01 +0000, Mr Flibble said:
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On Tue, 20 May 2025 19:51:59 +0200, Fred. Zwarts wrote:
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Op 20.mei.2025 om 16:22 schreef olcott:

On 5/20/2025 2:00 AM, Mikko wrote:

On 2025-05-20 04:10:54 +0000, olcott said:
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On 5/19/2025 5:12 AM, Mikko wrote:

On 2025-05-18 19:18:21 +0000, olcott said:
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On 5/18/2025 2:08 PM, joes wrote:

Am Sun, 18 May 2025 12:28:05 -0500 schrieb olcott:

On 5/18/2025 10:21 AM, Mike Terry wrote:

On 18/05/2025 10:09, Mikko wrote:

On 2025-05-17 17:15:14 +0000, olcott said:

>

HHH(DDD) does not base its decision on the actual behavior of
DDD after it has aborted its simulation of DDD, instead it
bases its decision on a different HHH/DDD pair that never
aborts.

>
This is why HHH does not satisfy "H correctly determines that
its simulated D would never stop running unless aborted". If
HHH bases its decision on anything else than what its actual
input actually specifies it does not decide correctly.
>

Right.  It seems to be a recent innovation in PO's wording that
he has started using the phrase "..bases its decision on a
different *HHH/DDD pair* ..".
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Thus SHD must report on a different SHD/Infinite_Loop pair where
this hypothetical instance of itself never aborts.

This, the simulator. The input still calls the same real aborting
HHH.
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If H always reports on the behavior of its simulated input after
it aborts then every input including infinite_loop would be
determined to be halting.

Yes, that is why H is wrong.
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Instead H must report on the hypothetical H/D input pair where
the very same H has been made to not abort its input.

Just no.
>

*H correctly determines that its simulated D*
*would never stop running unless aborted*
by a hypothetical instance of itself that never aborts.

H does stop running when simulated without aborting, because it
aborts.
>
>

H is required to report on the behavior of D in the case where a
hypothetical instance of itself never aborts its simulated D.
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When the hypothetical H never aborts its simulated D then:
(a) Simulated D  NEVER HALTS (b) Executed D() NEVER HALTS (c)
Executed H() NEVER HALTS (d) Everything that H calls NEVER HALTS

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You forgot (e) H does not report

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HHH is required to report, that is why it must always report on the
behavior of the hypothetical H/D pair and not the actual behavior of
the actual H/D pair for every non-terminating input.

>
Every decider is required to report. But your (c) above prevents the
hypothetical H from reporting. Therefore the hypothetical H is not a
decider.
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>

I wish that people would pay attention.
People only glance at a couple of words that I say then artificially
contrive a fake rebuttal.
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*We are ONLY measuring HHH/DDD against this criteria*
<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
>
>

We use the same criteria. We see that there is no correct simulation and
that H does not correctly determine that its simulated D would never
stop running. In fact the input specified to H contains code to abort,
so a simulation of this input without abort would lead to a natural
halt.
>
So, because the criteria are not met, we see that Sipser agreed to a
vacuous statement.
>
But you do not pay attention to what is said, because you stay in
rebuttal mode and, after seeing just a few words, keep repeating
statements that are proven to be irrelevant, without even touching the
fact that you are proven to be irrelevant.

>
The halting problem is defined in terms of UTMs with infinite tape so

>
It usually isn't. There are many variants of the problem but if you
have an oracle for one of the you can solve them all. Usually an UTM
is not mentioned in the problem statement. The tape is potentially
infinite but one execution of a decider never uses more than a finite
segment of the tape.
>

>
I think that I am the original inventor of
the notion of simulating halt decider as it
pertains to the halting problem proofs.
>
Computer Science professor Eric Hehner PhD
was the first one that noticed this:
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*Problems with the Halting Problem*
 From a programmer's point of view, if we apply
an interpreter to a program text that includes
a call to that same interpreter with that same
text as argument, then we have an infinite loop.
https://www.cs.toronto.edu/~hehner/PHP.pdf
>
Professor Hehner did not notice that the above
could be used to create a simulating halt decider
that rejects the conventional HP counter-example
input as non-halting.

>
It can't. A decider, unlike a simulator, cannot be infinitely looping.
>

 void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}
 _Infinite_Loop()
[0000212e] 55             push ebp
[0000212f] 8bec           mov ebp,esp
[00002131] ebfe           jmp 00002131
[00002133] 5d             pop ebp
[00002134] c3             ret
Size in bytes:(0007) [00002134]
 In other words (even though Mike proved otherwise)
No one and nothing can possibly know that an infinite
loop will never terminate until they wait until
the end of time and it didn't stop running yet.
 Why the Hell are you trying to get away with something
so moronically stupid?
 

No one is saying that HHH can't detect that this input is non-halting.
The problem is that when the input contains a call to HHH, but HHH isn't included as part of the input, then the input can't be correctly simulated.
And when you include the code of THAT HHH (the one that aborts and returns 0) you can't then put at that same location the HHH that correctly simuates the input to try to LIE that DDD is non-halting.
THe correct simulation of the DDD that calls the HHH that aborts and returns 0, shows it is HALTING.
Only by making a LIE, that the HHH that aborts is the same code as the HHH that doesn't, which is like saying that 1 is equal to 2, can you even attempt your argument.
Sorry, you are just proving that you logic is based on allowing lying.