From Newsgroup: comp.ai.philosophy

On 7/29/2025 8:48 PM, Richard Damon wrote:

On 7/29/25 9:36 PM, olcott wrote:

On 7/29/2025 8:28 PM, Richard Damon wrote:

On 7/29/25 5:34 PM, olcott wrote:

On 7/29/2025 3:37 AM, Fred. Zwarts wrote:

Many times already olcott has claimed to have a proof.
Each time again it turned out that there was no proof.

Op 28.jul.2025 om 18:13 schreef olcott:

<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>

To make this easier to understand we replace line three
with the more conventional terminology this line:
"cannot possibly reach its own simulated final halt state then"

Which changes the meaning of the sentence, so you can no longer
claim that Sipser agreed to it.

Yes

In addition not-reaching the halt state can have more reasons than
a non-halting sequence of instructions. E.g.: a computer being
switched off, a simulation aborted.

This is you not paying close enough attention to the exact
words that I exactly said.

I did not say: DOES NOT REACH ITS FINAL STATE (might have been aborted) >>>> I said: CANNOT POSSIBLY REACH ITS FINAL STATE (aborted or not)

On 10/14/2022 7:44 PM, Ben Bacarisse wrote:

I don't think that is the shell game.  PO really /has/ an H (it's >>>>>>> trivial to do for this one case) that correctly determines that P(P) >>>>>>> *would* never stop running *unless* aborted.  He knows and
accepts that
P(P) actually does stop.  The wrong answer is justified by what >>>>>>> would
happen if H (and hence a different P) where not what they
actually are.

Saying that H is required report on the behavior of
machine M is a category error.

Turing machines cannot directly report on the behavior
of other Turing machines they can at best indirectly
report on the behavior of Turing machines through the
proxy of finite string machine descriptions such as ⟨M⟩.

Thus the behavior specified by the input finite string
overrules and supersedes the behavior of the direct
execution.

No, because there is no difference between the specification of the >>>>> input and the direct execution.

*I have proven this to be counter-factual*

HHH(DDD) must simulate itself simulating DDD because DDD calls HHH(DDD) >>>>
HHH1(DDD) must NOT simulate itself simulating DDD because DDD DOES
NOT CALL HHH1(DDD)   --- [same behavior as direct execution]

The problem is since you HHH *DOES* abort,

void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}

void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}

As it always does as soon as it correctly determines
that there exists no N steps of correct simulation
that can possibly reach the "return" instruction final
halt state.

But their is a N number of steps of the correct simulation of DDD that
cause it to halt,

Again you are a fucking liar. You know that halting is
reaching a final halt state and nothing else is halting.
When 0 to infinity steps of DDD are correctly simulated
by HHH no simulated DDD ever halts.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
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From Newsgroup: comp.ai.philosophy

On 7/29/25 11:02 PM, olcott wrote:

On 7/29/2025 8:48 PM, Richard Damon wrote:

On 7/29/25 9:36 PM, olcott wrote:

On 7/29/2025 8:28 PM, Richard Damon wrote:

On 7/29/25 5:34 PM, olcott wrote:

On 7/29/2025 3:37 AM, Fred. Zwarts wrote:

Many times already olcott has claimed to have a proof.
Each time again it turned out that there was no proof.

Op 28.jul.2025 om 18:13 schreef olcott:

<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>

To make this easier to understand we replace line three
with the more conventional terminology this line:
"cannot possibly reach its own simulated final halt state then"

Which changes the meaning of the sentence, so you can no longer
claim that Sipser agreed to it.

Yes

In addition not-reaching the halt state can have more reasons than >>>>>> a non-halting sequence of instructions. E.g.: a computer being
switched off, a simulation aborted.

This is you not paying close enough attention to the exact
words that I exactly said.

I did not say: DOES NOT REACH ITS FINAL STATE (might have been
aborted)
I said: CANNOT POSSIBLY REACH ITS FINAL STATE (aborted or not)

On 10/14/2022 7:44 PM, Ben Bacarisse wrote:

I don't think that is the shell game.  PO really /has/ an H (it's >>>>>>>> trivial to do for this one case) that correctly determines that >>>>>>>> P(P)
*would* never stop running *unless* aborted.  He knows and
accepts that
P(P) actually does stop.  The wrong answer is justified by what >>>>>>>> would
happen if H (and hence a different P) where not what they
actually are.

Saying that H is required report on the behavior of
machine M is a category error.

Turing machines cannot directly report on the behavior
of other Turing machines they can at best indirectly
report on the behavior of Turing machines through the
proxy of finite string machine descriptions such as ⟨M⟩.

Thus the behavior specified by the input finite string
overrules and supersedes the behavior of the direct
execution.

No, because there is no difference between the specification of
the input and the direct execution.

*I have proven this to be counter-factual*

HHH(DDD) must simulate itself simulating DDD because DDD calls
HHH(DDD)

HHH1(DDD) must NOT simulate itself simulating DDD because DDD DOES
NOT CALL HHH1(DDD)   --- [same behavior as direct execution]

The problem is since you HHH *DOES* abort,

void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}

void Infinite_Loop()
{
   HERE: goto HERE;
   return;
}

As it always does as soon as it correctly determines
that there exists no N steps of correct simulation
that can possibly reach the "return" instruction final
halt state.

But their is a N number of steps of the correct simulation of DDD that
cause it to halt,

Again you are a fucking liar. You know that halting is
reaching a final halt state and nothing else is halting.
When 0 to infinity steps of DDD are correctly simulated
by HHH no simulated DDD ever halts.

Right, it is *THE PROGRAM* reaching a final state. Dropping that
qualifier is just your standard method of operation.

And more imporantly, non-halting, is THE PROGRAM never reaching a final
state, even after running an unbounded number of steps.

A partial simulation not reaching the final state of the input is NOT non-halting.

And, different input are different and might behave differently,

and programs include ALL the code they use, so DDD includes the code of HHH,

so different HHHs are given different DDDs since they are each built on
the HHH that will try to simulate them.

Sorry, you are just proving you are just a liar who doesn't know what he
is talking about.
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