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computers / comp.ai.philosophy / Re: What if a cat barks? [ sound deduction proves that I am correct ]

SubjectAuthor
* What if a cat barks?olcott
+* Re: What if a cat barks?Chris M. Thomasson
|+* Re: What if a cat barks?olcott
||+- Re: What if a cat barks?olcott
||`* Re: What if a cat barks?olcott
|| `* Re: What if a cat barks?olcott
||  `* Re: What if a cat barks?olcott
||   `* Re: What if a cat barks?olcott
||    `* Re: What if a cat barks? [ sound deduction is a proof ]olcott
||     `* Re: What if a cat barks? [ sound deduction is a proof ]olcott
||      +* Re: What if a cat barks? [ sound deduction is a proof ]olcott
||      |`- Re: What if a cat barks? [ sound deduction is a proof ]olcott
||      `* Re: What if a cat barks? [ sound deduction is a proof ](axiom)olcott
||       `* Re: What if a cat barks? [ sound deduction is a proof ](axiom)olcott
||        `* Re: What if a cat barks? [ sound deduction is a proof ](axiom)olcott
||         `* Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||          `* Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||           `* Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            +* Re: What if a cat barks? [ sound deduction is a proof ](infiniteChris M. Thomasson
||            |`* Re: What if a cat barks? [ sound deduction is a proof ](infiniteJeff Barnett
||            | +* Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            | |`* Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            | | +* Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            | | |+- Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            | | |`- Re: What if a cat barks? [ sound deduction is a proof ](infiniteChris M. Thomasson
||            | | +* Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||            | | |`* Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            | | | `* Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||            | | |  `* Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            | | |   `* Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            | | |    `* Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||            | | |     `- Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||            | | `- Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            | +- Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            | +* Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||            | |`* Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||            | | +- Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chaiolcott
||            | | +- Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||            | | `- Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||            | `- Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||            `* Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
||             `- Re: What if a cat barks? [ sound deduction is a proof ](infiniteolcott
|`* Re: What if a cat barks? [ sound deduction proves that I am correct ]olcott
| `* Re: What if a cat barks? [ sound deduction proves that I am correct ]olcott
|  +- Re: What if a cat barks? [ sound deduction proves that I am correct ]olcott
|  +* Re: What if a cat barks? [ sound deduction proves that I am correct ]olcott
|  |`- Re: What if a cat barks? [ sound deduction proves that I am correct ]olcott
|  `- Re: What if a cat barks? [ sound deduction proves that I am correct ]olcott
+- Re: What if a cat barks? [ How can a cat bark? ]olcott
+- Re: What if a cat barks?olcott
`* Re: What if a cat barks?esa 4me
 `* Re: What if a cat barks?esa 4me
  `* Re: What if a cat barks?Don Stockbauer
   `* Re: What if a cat barks?esa 4me
    `* Re: What if a cat barks?esa 4me
     `* Re: What if a cat barks?Don Stockbauer
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        `* Re: What if a cat barks?Don Stockbauer
         `* Re: What if a cat barks?assumed. identiy.3396
          `* Re: What if a cat barks?Don Stockbauer
           `* Re: What if a cat barks?assumed. identiy.3396
            `* Re: What if a cat barks?Don Stockbauer
             `* Re: What if a cat barks?assumed. identiy.3396
              `* Re: What if a cat barks?Don Stockbauer
               `* Re: What if a cat barks?Don Stockbauer
                `* Re: What if a cat barks?assumed. identiy.3396
                 +- Re: What if a cat barks?assumed. identiy.3396
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?assumed. identiy.3396
                 +- Re: What if a cat barks?assumed. identiy.3396
                 +- Re: What if a cat barks?assumed. identiy.3396
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?assumed. identiy.3396
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?assumed. identiy.3396
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?assumed. identiy.3396
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?assumed. identiy.3396
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?esa 4me
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 +- Re: What if a cat barks?Don Stockbauer
                 `- Re: What if a cat barks?Don Stockbauer

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What if a cat barks?

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From: NoO...@NoWhere.com (olcott)
Subject: What if a cat barks?
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 by: olcott - Mon, 21 Jun 2021 04:15 UTC

If you see an animal and test its DNA and confirm that it is definitely
a cat, what happens when the cat barks?

When we examine the behavior of the Peter Linz Ĥ applied to its own
Turing machine description: ⟨Ĥ⟩ and simply assume that the embedded halt
decider at its internal state of Ĥ.qx is a UTM then we find that this
machine has infinitely nested simulation.

SELF-EVIDENT-TRUTH
Every computation that never halts unless its simulation is aborted is a
computation that never halts.

SELF-EVIDENT-TRUTH
The <Ĥ> <Ĥ> input to the embedded halt decider at Ĥ.qx is a computation
that never halts unless its simulation is aborted.

∴ IMPOSSIBLY FALSE CONCLUSION
The embedded simulating halt decider at Ĥ.qx correctly decides its
input: <Ĥ> <Ĥ> is a computation that never halts.

The above three elements essentially provide the DNA of the cat.

Halting problem undecidability and infinitely nested simulation

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks?

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 by: Chris M. Thomasson - Mon, 21 Jun 2021 05:04 UTC

On 6/20/2021 9:15 PM, olcott wrote:
> If you see an animal and test its DNA and confirm that it is definitely
> a cat, what happens when the cat barks?
[...]

Have you been hearing cats bark lately? Wow.

Re: What if a cat barks? [ How can a cat bark? ]

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 by: olcott - Mon, 21 Jun 2021 15:07 UTC

On 6/20/2021 11:15 PM, olcott wrote:
> If you see an animal and test its DNA and confirm that it is definitely
> a cat, what happens when the cat barks?
>
> When we examine the behavior of the Peter Linz Ĥ applied to its own
> Turing machine description: ⟨Ĥ⟩ and simply assume that the embedded halt
> decider at its internal state of Ĥ.qx is a UTM then we find that this
> machine has infinitely nested simulation.
>
> SELF-EVIDENT-TRUTH
> Every computation that never halts unless its simulation is aborted is a
> computation that never halts.
>
> SELF-EVIDENT-TRUTH
> The <Ĥ> <Ĥ> input to the embedded halt decider at Ĥ.qx is a computation
> that never halts unless its simulation is aborted.
>
> ∴ IMPOSSIBLY FALSE CONCLUSION
> The embedded simulating halt decider at Ĥ.qx correctly decides its
> input: <Ĥ> <Ĥ> is a computation that never halts.
>
> The above three elements essentially provide the DNA of the cat.
>
>
> Halting problem undecidability and infinitely nested simulation
>
> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

So when we check the DNA of the cat [ review the sound deductive
inference chain of the embedded halt decider's decision to abort the
simulation of Ĥ(⟨Ĥ⟩) ] We find that even when the cat barks [ Ĥ(⟨Ĥ⟩)
halts ] that this does not contradict that the cat is indeed a cat [
that Ĥ(⟨Ĥ⟩) really does specify a computation that never halts ].

Here is how the cat barks [ non-halting computation Ĥ(⟨Ĥ⟩) halts ]:
Ĥ(⟨Ĥ⟩) specifies an infinite chain of invocation that is terminated
at its third invocation. The first call from Ĥ(⟨Ĥ⟩) to H(⟨Ĥ⟩, ⟨Ĥ⟩) is
the first element of the infinite chain of invocations.

Everyone knows that when any invocation of an infinite chain of
invocations is terminated that the whole chain terminates. That the
first element of this infinite chain terminates after its third element
has been terminated does not entail that this first element is an actual
terminating computation.

For the first element to be an actual terminating computation it must
terminate without any of the elements of the infinite chain of
invocations being terminated.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks?

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Mon, 21 Jun 2021 21:11 UTC

On 6/21/2021 1:24 PM, André G. Isaak wrote:
> On 2021-06-20 23:04, Chris M. Thomasson wrote:
>> On 6/20/2021 9:15 PM, olcott wrote:
>>> If you see an animal and test its DNA and confirm that it is
>>> definitely a cat, what happens when the cat barks?
>> [...]
>>
>> Have you been hearing cats bark lately? Wow.
>
> As long as he's just hearing them bark, we're probably fine.
>
> It's when the barking cats start telling him to do things (like kill
> neighbours, steal catnip, or conquer Liechtenstein) that we really need
> to worry.
>
> André
>

The point is that the mere intuition about the halting behavior of Ĥ
applied to ⟨Ĥ⟩ is superseded by meticulous sound deductive inference.

SELF-EVIDENT-TRUTH
Every computation that never halts unless its simulation is aborted is a
computation that never halts.

SELF-EVIDENT-TRUTH
The ⟨Ĥ⟩ ⟨Ĥ⟩ input to the embedded halt decider at Ĥ.qx is a computation
that never halts unless its simulation is aborted.

∴ IMPOSSIBLY FALSE CONCLUSION
The embedded simulating halt decider at Ĥ.qx correctly decides its
input: ⟨Ĥ⟩ ⟨Ĥ⟩ is a computation that never halts.

Ĥ.q0 ⟨Ĥ⟩ specifies an infinite chain of invocations that is terminated
at its third invocation. The first invocation of Ĥ.qx ⟨Ĥ⟩, ⟨Ĥ⟩ is the
first element of an infinite chain of invocations.

It is common knowledge that when any invocation of an infinite chain of
invocations is terminated that the whole chain terminates. That the
first element of this infinite chain terminates after its third element
has been terminated does not entail that this first element is an actual
terminating computation.

For the first element to be an actual terminating computation it must
terminate without any of the elements of the infinite chain of
invocations being terminated.

Halting problem undecidability and infinitely nested simulation
https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks?

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<sap6l7$130c$5@gioia.aioe.org> <saqlhk$860$1@dont-email.me>
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Mon, 21 Jun 2021 21:26 UTC

On 6/21/2021 4:11 PM, olcott wrote:
> On 6/21/2021 1:24 PM, André G. Isaak wrote:
>> On 2021-06-20 23:04, Chris M. Thomasson wrote:
>>> On 6/20/2021 9:15 PM, olcott wrote:
>>>> If you see an animal and test its DNA and confirm that it is
>>>> definitely a cat, what happens when the cat barks?
>>> [...]
>>>
>>> Have you been hearing cats bark lately? Wow.
>>
>> As long as he's just hearing them bark, we're probably fine.
>>
>> It's when the barking cats start telling him to do things (like kill
>> neighbours, steal catnip, or conquer Liechtenstein) that we really
>> need to worry.
>>
>> André
>>
>
> The point is that the mere intuition about the halting behavior of Ĥ
> applied to ⟨Ĥ⟩ is superseded by meticulous sound deductive inference.
>

Self-Evident-Truth (premise[1])
Every computation that never halts unless its simulation is aborted is a
computation that never halts.

Self-Evident-Truth (premise[2])
The ⟨Ĥ⟩ ⟨Ĥ⟩ input to the embedded halt decider at Ĥ.qx is a computation
that never halts unless its simulation is aborted.

∴ Sound Deductive Conclusion
The embedded simulating halt decider at Ĥ.qx correctly decides its
input: ⟨Ĥ⟩ ⟨Ĥ⟩ is a computation that never halts.

> Ĥ.q0 ⟨Ĥ⟩ specifies an infinite chain of invocations that is terminated
> at its third invocation. The first invocation of Ĥ.qx ⟨Ĥ⟩, ⟨Ĥ⟩ is the
> first element of an infinite chain of invocations.
>
> It is common knowledge that when any invocation of an infinite chain of
> invocations is terminated that the whole chain terminates. That the
> first element of this infinite chain terminates after its third element
> has been terminated does not entail that this first element is an actual
> terminating computation.
>
> For the first element to be an actual terminating computation it must
> terminate without any of the elements of the infinite chain of
> invocations being terminated.
>
>
>
>
>
> Halting problem undecidability and infinitely nested simulation
> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>
>
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks?

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Tue, 22 Jun 2021 01:39 UTC

On 6/21/2021 7:57 PM, André G. Isaak wrote:
> On 2021-06-21 15:11, olcott wrote:
>> On 6/21/2021 1:24 PM, André G. Isaak wrote:
>>> On 2021-06-20 23:04, Chris M. Thomasson wrote:
>>>> On 6/20/2021 9:15 PM, olcott wrote:
>>>>> If you see an animal and test its DNA and confirm that it is
>>>>> definitely a cat, what happens when the cat barks?
>>>> [...]
>>>>
>>>> Have you been hearing cats bark lately? Wow.
>>>
>>> As long as he's just hearing them bark, we're probably fine.
>>>
>>> It's when the barking cats start telling him to do things (like kill
>>> neighbours, steal catnip, or conquer Liechtenstein) that we really
>>> need to worry.
>>>
>>> André
>>>
>>
>> The point is that the mere intuition about the halting behavior of Ĥ
>> applied to ⟨Ĥ⟩ is superseded by meticulous sound deductive inference.
>>
>> SELF-EVIDENT-TRUTH
>> Every computation that never halts unless its simulation is aborted is
>> a computation that never halts.
>
> How can that possibly be "self-evident" when it doesn't even explain
> what "its simulation" means. Its simulation of/by what?
>

We could simplify this and say that any computation that never halts
unless this computation is aborted is a computation that never halts.

>> SELF-EVIDENT-TRUTH
>> The ⟨Ĥ⟩ ⟨Ĥ⟩ input to the embedded halt decider at Ĥ.qx is a
>> computation that never halts unless its simulation is aborted.
>
> If that were self-evident, you wouldn't have so many people pointing out
> to you that it is simply wrong. Things can't be both wrong and
> self-evident.
>

That people simply don't want to bother to pay enough attention to see
that I am right is not actually any rebuttal at all.

>> ∴ IMPOSSIBLY FALSE CONCLUSION
>
> "Impossibly false" is a meaningless expression. No journal is going to
> take you seriously if this phrase appears anywhere in your work.
>

I improved this. I now call it the conclusion of sound deduction, which
means exactly the same thing as impossibly false.

>> The embedded simulating halt decider at Ĥ.qx correctly decides its
>> input: ⟨Ĥ⟩ ⟨Ĥ⟩ is a computation that never halts.
>>
>> Ĥ.q0 ⟨Ĥ⟩ specifies an infinite chain of invocations that is terminated
>> at its third invocation. The first invocation of Ĥ.qx ⟨Ĥ⟩, ⟨Ĥ⟩ is the
>> first element of an infinite chain of invocations.
>>
>> It is common knowledge that when any invocation of an infinite chain
>> of invocations is terminated that the whole chain terminates. That the
>> first element of this infinite chain terminates after its third
>> element has been terminated does not entail that this first element is
>> an actual terminating computation.
>
> No. It isn't "common knowledge". It is simply false.
>

I will make a common knowledge concrete example:
Infinite recursion is an infinite sequence of invocations right?

If any element of the infinite sequence of invocations of infinite
recursion is aborted then the whole sequence stops right?

If the third invocation of the infinite sequence of invocations of
infinite recursion is aborted then the whole sequence stops right?

Was it really that hard to see the above three steps on the basis of my
claim of common knowledge?

> And if the simulation is terminated after the third call to Ĥ, the you
> don't have an infinite chain of calls. You have a chain of three calls.
>

When the halt decider is analyzing the behavior stipulated by a finite
string it is incorrect for the halt decider to conflate its own behavior
in this analysis.

The question that the halt decider is answering is whether or not a pure
simulation/execution of the input must be aborted to prevent the
infinite execution of this input.

>> For the first element to be an actual terminating computation it must
>> terminate without any of the elements of the infinite chain of
>> invocations being terminated.
>
> This is just plain silly.

See my infinite recursion example above.

> If some program H simulates another program Y
> along with an input string but has the ability to terminate a
> simulation, then there are three possibilities:
>
> A) The simulation is allowed to continue until Y reaches one of its
> final states. In such a case we can say that Y halts. Since Y halts, H
> can also halt.
>
> B) The simulation is allowed to continue forever, but it never reaches a
> final state. The simulation continues forever. In this case, Y doesn't
> halt. H therefore also doesn't halt. Of course, this option would be
> difficult to empirically verify since we can't actually observe
> something running for an infinite amount of time.
>
> C) H decides to discontinue the simulation. In this case the simulation
> neither halts nor runs forever. It may be that that Y is non-halting, or
> it may be that H simply discontinued the simulation prematurely. But in
> either of these two cases, H can halt.
>
> André
>

As I have said so very many hundreds of times now that you really should
not have made such a terrible mistake with part C

When H must terminate the simulation of its input to prevent the
infinite execution of P then H does necessarily infallibly correctly
decide that its input never halts.

Unlike Turing machines where we must simply imagine crucially important
details and have no way to infallibly ascertain that our imagination
accurately represent the truth we can examine the x86 execution trace of
P and know with 100% perfectly correct certainty that P would never ever
halt unless H aborts P.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks?

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 by: olcott - Tue, 22 Jun 2021 03:09 UTC

On 6/21/2021 9:29 PM, André G. Isaak wrote:
> On 2021-06-21 19:39, olcott wrote:
>> On 6/21/2021 7:57 PM, André G. Isaak wrote:
>>> On 2021-06-21 15:11, olcott wrote:
>>>> On 6/21/2021 1:24 PM, André G. Isaak wrote:
>>>>> On 2021-06-20 23:04, Chris M. Thomasson wrote:
>>>>>> On 6/20/2021 9:15 PM, olcott wrote:
>>>>>>> If you see an animal and test its DNA and confirm that it is
>>>>>>> definitely a cat, what happens when the cat barks?
>>>>>> [...]
>>>>>>
>>>>>> Have you been hearing cats bark lately? Wow.
>>>>>
>>>>> As long as he's just hearing them bark, we're probably fine.
>>>>>
>>>>> It's when the barking cats start telling him to do things (like
>>>>> kill neighbours, steal catnip, or conquer Liechtenstein) that we
>>>>> really need to worry.
>>>>>
>>>>> André
>>>>>
>>>>
>>>> The point is that the mere intuition about the halting behavior of Ĥ
>>>> applied to ⟨Ĥ⟩ is superseded by meticulous sound deductive inference.
>>>>
>>>> SELF-EVIDENT-TRUTH
>>>> Every computation that never halts unless its simulation is aborted
>>>> is a computation that never halts.
>>>
>>> How can that possibly be "self-evident" when it doesn't even explain
>>> what "its simulation" means. Its simulation of/by what?
>>>
>>
>> We could simplify this and say that any computation that never halts
>> unless this computation is aborted is a computation that never halts.
>
> That's no better. A computation which is aborted doesn't halt. 'Halt'
> means to reach one of the final states. If you abort something it
> doesn't reach a final state. But the simulator itself can.
>

When we are trying to determine whether or not an infinite loop is an
infinite loop we can debug step through this code and see that it
endlessly repeats and there is no escape from this endless repetition in
this code. It is not really that hard.

>>>> SELF-EVIDENT-TRUTH
>>>> The ⟨Ĥ⟩ ⟨Ĥ⟩ input to the embedded halt decider at Ĥ.qx is a
>>>> computation that never halts unless its simulation is aborted.
>>>
>>> If that were self-evident, you wouldn't have so many people pointing
>>> out to you that it is simply wrong. Things can't be both wrong and
>>> self-evident.
>>>
>>
>> That people simply don't want to bother to pay enough attention to see
>> that I am right is not actually any rebuttal at all.
>>
>>>> ∴ IMPOSSIBLY FALSE CONCLUSION
>>>
>>> "Impossibly false" is a meaningless expression. No journal is going
>>> to take you seriously if this phrase appears anywhere in your work.
>>>
>>
>> I improved this. I now call it the conclusion of sound deduction,
>> which means exactly the same thing as impossibly false.
>
> So where is the sound deduction from which you reach the above
> conclusion? You start with two premises, one of which is too vague to be
> interpreted and the other of which is simply false. That's not how
> deductively sound arguments work.

We really have to look at this in terms of H and P because there is no
other possible way to make sure that we examine all the details when we
try to imagine what a Turing machine might do.

>>>> The embedded simulating halt decider at Ĥ.qx correctly decides its
>>>> input: ⟨Ĥ⟩ ⟨Ĥ⟩ is a computation that never halts.
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ specifies an infinite chain of invocations that is
>>>> terminated at its third invocation. The first invocation of Ĥ.qx
>>>> ⟨Ĥ⟩, ⟨Ĥ⟩ is the first element of an infinite chain of invocations.
>>>>
>>>> It is common knowledge that when any invocation of an infinite chain
>>>> of invocations is terminated that the whole chain terminates. That
>>>> the first element of this infinite chain terminates after its third
>>>> element has been terminated does not entail that this first element
>>>> is an actual terminating computation.
>>>
>>> No. It isn't "common knowledge". It is simply false.
>>>
>>
>> I will make a common knowledge concrete example:
>> Infinite recursion is an infinite sequence of invocations right?
>
> Sure, but there is no infinite recursion (or recursion period) in the
> Linz example.
>
>> If any element of the infinite sequence of invocations of infinite
>> recursion is aborted then the whole sequence stops right?
>
> Not necessarily. That depends entirely on what is meant by 'abort'.
>
>> If the third invocation of the infinite sequence of invocations of
>> infinite recursion is aborted then the whole sequence stops right?
>
> No. The third invocation is aborted. The simulator itself continues to
> run and is able to halt.
>
>> Was it really that hard to see the above three steps on the basis of
>> my claim of common knowledge?
>
> Things that you believe and 'common knowledge' are not the same thing.
>
>>> And if the simulation is terminated after the third call to Ĥ, the
>>> you don't have an infinite chain of calls. You have a chain of three
>>> calls.
>>>
>>
>> When the halt decider is analyzing the behavior stipulated by a finite
>> string it is incorrect for the halt decider to conflate its own
>> behavior in this analysis.
>
> And I am not conflating them. When you abort the simulation that doesn't
> entail aborting the simulator. You conflate them by concluding that the
> topmost program doesn't halt based on what happens to the program it is
> simulating.
>
>> The question that the halt decider is answering is whether or not a
>> pure simulation/execution of the input must be aborted to prevent the
>> infinite execution of this input.
>>
>>>> For the first element to be an actual terminating computation it
>>>> must terminate without any of the elements of the infinite chain of
>>>> invocations being terminated.
>>>
>>> This is just plain silly.
>>
>> See my infinite recursion example above.
>
> As I said, the above is just plain silly.
>
>>> If some program H simulates another program Y along with an input
>>> string but has the ability to terminate a simulation, then there are
>>> three possibilities:
>>>
>>> A) The simulation is allowed to continue until Y reaches one of its
>>> final states. In such a case we can say that Y halts. Since Y halts,
>>> H can also halt.
>>>
>>> B) The simulation is allowed to continue forever, but it never
>>> reaches a final state. The simulation continues forever. In this
>>> case, Y doesn't halt. H therefore also doesn't halt. Of course, this
>>> option would be difficult to empirically verify since we can't
>>> actually observe something running for an infinite amount of time.
>>>
>>> C) H decides to discontinue the simulation. In this case the
>>> simulation neither halts nor runs forever. It may be that that Y is
>>> non-halting, or it may be that H simply discontinued the simulation
>>> prematurely. But in either of these two cases, H can halt.
>>>
>>> André
>>>
>>
>> As I have said so very many hundreds of times now that you really
>> should not have made such a terrible mistake with part C
>
> There is no mistake in C.

C) H decides to discontinue the simulation.
This is very terribly incorrect

H MUST stop its simulation of P or P never halts
H MUST stop its simulation of P or P never halts
H MUST stop its simulation of P or P never halts
H MUST stop its simulation of P or P never halts
H MUST stop its simulation of P or P never halts

This is not at all the same thing as H arbitrarily stops simulating P
for possibly no good reason at all.

This is not at all the same thing as H arbitrarily stops simulating P
for possibly no good reason at all.

This is not at all the same thing as H arbitrarily stops simulating P
for possibly no good reason at all.

This is not at all the same thing as H arbitrarily stops simulating P
for possibly no good reason at all.


Click here to read the complete article
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 by: olcott - Tue, 22 Jun 2021 17:05 UTC

On 6/22/2021 11:47 AM, André G. Isaak wrote:
> On 2021-06-21 21:09, olcott wrote:
>> On 6/21/2021 9:29 PM, André G. Isaak wrote:
>>> On 2021-06-21 19:39, olcott wrote:
>>>> On 6/21/2021 7:57 PM, André G. Isaak wrote:
>>>>> On 2021-06-21 15:11, olcott wrote:
>>>>>> On 6/21/2021 1:24 PM, André G. Isaak wrote:
>>>>>>> On 2021-06-20 23:04, Chris M. Thomasson wrote:
>>>>>>>> On 6/20/2021 9:15 PM, olcott wrote:
>>>>>>>>> If you see an animal and test its DNA and confirm that it is
>>>>>>>>> definitely a cat, what happens when the cat barks?
>>>>>>>> [...]
>>>>>>>>
>>>>>>>> Have you been hearing cats bark lately? Wow.
>>>>>>>
>>>>>>> As long as he's just hearing them bark, we're probably fine.
>>>>>>>
>>>>>>> It's when the barking cats start telling him to do things (like
>>>>>>> kill neighbours, steal catnip, or conquer Liechtenstein) that we
>>>>>>> really need to worry.
>>>>>>>
>>>>>>> André
>>>>>>>
>>>>>>
>>>>>> The point is that the mere intuition about the halting behavior of
>>>>>> Ĥ applied to ⟨Ĥ⟩ is superseded by meticulous sound deductive
>>>>>> inference.
>>>>>>
>>>>>> SELF-EVIDENT-TRUTH
>>>>>> Every computation that never halts unless its simulation is
>>>>>> aborted is a computation that never halts.
>>>>>
>>>>> How can that possibly be "self-evident" when it doesn't even
>>>>> explain what "its simulation" means. Its simulation of/by what?
>>>>>
>>>>
>>>> We could simplify this and say that any computation that never halts
>>>> unless this computation is aborted is a computation that never halts.
>>>
>>> That's no better. A computation which is aborted doesn't halt. 'Halt'
>>> means to reach one of the final states. If you abort something it
>>> doesn't reach a final state. But the simulator itself can.
>>>
>>
>> When we are trying to determine whether or not an infinite loop is an
>> infinite loop we can debug step through this code and see that it
>> endlessly repeats and there is no escape from this endless repetition
>> in this code. It is not really that hard.
>
> Which has nothing to do with what I wrote.
>

It explains the details of how a computation that must be aborted by the
halt decider to prevent the infinite execution of this computation <is>
a computation that never halts.

>>>>>> SELF-EVIDENT-TRUTH
>>>>>> The ⟨Ĥ⟩ ⟨Ĥ⟩ input to the embedded halt decider at Ĥ.qx is a
>>>>>> computation that never halts unless its simulation is aborted.
>>>>>
>>>>> If that were self-evident, you wouldn't have so many people
>>>>> pointing out to you that it is simply wrong. Things can't be both
>>>>> wrong and self-evident.
>>>>>
>>>>
>>>> That people simply don't want to bother to pay enough attention to
>>>> see that I am right is not actually any rebuttal at all.
>>>>
>>>>>> ∴ IMPOSSIBLY FALSE CONCLUSION
>>>>>
>>>>> "Impossibly false" is a meaningless expression. No journal is going
>>>>> to take you seriously if this phrase appears anywhere in your work.
>>>>>
>>>>
>>>> I improved this. I now call it the conclusion of sound deduction,
>>>> which means exactly the same thing as impossibly false.
>>>
>>> So where is the sound deduction from which you reach the above
>>> conclusion? You start with two premises, one of which is too vague to
>>> be interpreted and the other of which is simply false. That's not how
>>> deductively sound arguments work.
>>
>> We really have to look at this in terms of H and P because there is no
>> other possible way to make sure that we examine all the details when
>> we try to imagine what a Turing machine might do.
>
> Which doesn't answer my question. Where is your 'sound deductive argument'?
>

(a) Every computation P that never halts unless the halt decider H
aborts this computation is a computation that never halts.

(b) X is a computation that never halts unless it is aborted by its halt
decider.

∴ (c) X is a computation that is correctly decided to be a computation
that never halts.

>>>>>> The embedded simulating halt decider at Ĥ.qx correctly decides its
>>>>>> input: ⟨Ĥ⟩ ⟨Ĥ⟩ is a computation that never halts.
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ specifies an infinite chain of invocations that is
>>>>>> terminated at its third invocation. The first invocation of Ĥ.qx
>>>>>> ⟨Ĥ⟩, ⟨Ĥ⟩ is the first element of an infinite chain of invocations.
>>>>>>
>>>>>> It is common knowledge that when any invocation of an infinite
>>>>>> chain of invocations is terminated that the whole chain
>>>>>> terminates. That the first element of this infinite chain
>>>>>> terminates after its third element has been terminated does not
>>>>>> entail that this first element is an actual terminating computation.
>>>>>
>>>>> No. It isn't "common knowledge". It is simply false.
>>>>>
>>>>
>>>> I will make a common knowledge concrete example:
>>>> Infinite recursion is an infinite sequence of invocations right?
>>>
>>> Sure, but there is no infinite recursion (or recursion period) in the
>>> Linz example.
>>>
>>>> If any element of the infinite sequence of invocations of infinite
>>>> recursion is aborted then the whole sequence stops right?
>>>
>>> Not necessarily. That depends entirely on what is meant by 'abort'.
>>>
>>>> If the third invocation of the infinite sequence of invocations of
>>>> infinite recursion is aborted then the whole sequence stops right?
>>>
>>> No. The third invocation is aborted. The simulator itself continues
>>> to run and is able to halt.
>>>
>>>> Was it really that hard to see the above three steps on the basis of
>>>> my claim of common knowledge?
>>>
>>> Things that you believe and 'common knowledge' are not the same thing.
>>>
>>>>> And if the simulation is terminated after the third call to Ĥ, the
>>>>> you don't have an infinite chain of calls. You have a chain of
>>>>> three calls.
>>>>>
>>>>
>>>> When the halt decider is analyzing the behavior stipulated by a
>>>> finite string it is incorrect for the halt decider to conflate its
>>>> own behavior in this analysis.
>>>
>>> And I am not conflating them. When you abort the simulation that
>>> doesn't entail aborting the simulator. You conflate them by
>>> concluding that the topmost program doesn't halt based on what
>>> happens to the program it is simulating.
>>>
>>>> The question that the halt decider is answering is whether or not a
>>>> pure simulation/execution of the input must be aborted to prevent
>>>> the infinite execution of this input.
>>>>
>>>>>> For the first element to be an actual terminating computation it
>>>>>> must terminate without any of the elements of the infinite chain
>>>>>> of invocations being terminated.
>>>>>
>>>>> This is just plain silly.
>>>>
>>>> See my infinite recursion example above.
>>>
>>> As I said, the above is just plain silly.
>>>
>>>>> If some program H simulates another program Y along with an input
>>>>> string but has the ability to terminate a simulation, then there
>>>>> are three possibilities:
>>>>>
>>>>> A) The simulation is allowed to continue until Y reaches one of its
>>>>> final states. In such a case we can say that Y halts. Since Y
>>>>> halts, H can also halt.
>>>>>
>>>>> B) The simulation is allowed to continue forever, but it never
>>>>> reaches a final state. The simulation continues forever. In this
>>>>> case, Y doesn't halt. H therefore also doesn't halt. Of course,
>>>>> this option would be difficult to empirically verify since we can't
>>>>> actually observe something running for an infinite amount of time.
>>>>>
>>>>> C) H decides to discontinue the simulation. In this case the
>>>>> simulation neither halts nor runs forever. It may be that that Y is
>>>>> non-halting, or it may be that H simply discontinued the simulation
>>>>> prematurely. But in either of these two cases, H can halt.
>>>>>
>>>>> André
>>>>>
>>>>
>>>> As I have said so very many hundreds of times now that you really
>>>> should not have made such a terrible mistake with part C
>>>
>>> There is no mistake in C.
>>
>>
>> C) H decides to discontinue the simulation.
>> This is very terribly incorrect
>>
>> H MUST stop its simulation of P or P never halts
>
> Which is subsumed under option C. The above enumerates all logical
> possibilities.
>


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Re: What if a cat barks?

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From: NoO...@NoWhere.com (olcott)
Date: Tue, 22 Jun 2021 20:12:37 -0500
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 by: olcott - Wed, 23 Jun 2021 01:12 UTC

On 6/22/2021 7:31 PM, wij wrote:
> On Wednesday, 23 June 2021 at 01:22:47 UTC+8, olcott wrote:
>> On 6/22/2021 12:16 PM, wij wrote:
>>> On Wednesday, 23 June 2021 at 01:14:01 UTC+8, wij wrote:
>>>> On Wednesday, 23 June 2021 at 01:08:26 UTC+8, olcott wrote:
>>>>> On 6/22/2021 12:02 PM, wij wrote:
>>>>>> On Tuesday, 22 June 2021 at 22:06:42 UTC+8, olcott wrote:
>>>>>>> On 6/22/2021 6:52 AM, wij wrote:
>>>>>>>> On Monday, 21 June 2021 at 23:37:49 UTC+8, olcott wrote:
>>>>>>>>> On 6/21/2021 10:33 AM, wij wrote:
>>>>>>>>>> On Monday, 21 June 2021 at 21:47:51 UTC+8, olcott wrote:
>>>>>>>>>>> On 6/21/2021 2:46 AM, wij wrote:
>>>>>>>>>>>> On Monday, 21 June 2021 at 12:15:27 UTC+8, olcott wrote:
>>>>>>>>>>>>> If you see an animal and test its DNA and confirm that it is definitely
>>>>>>>>>>>>> a cat, what happens when the cat barks?
>>>>>>>>>>>>>
>>>>>>>>>>>>> When we examine the behavior of the Peter Linz Ĥ applied to its own
>>>>>>>>>>>>> Turing machine description: ⟨Ĥ⟩ and simply assume that the embedded halt
>>>>>>>>>>>>> decider at its internal state of Ĥ.qx is a UTM then we find that this
>>>>>>>>>>>>> machine has infinitely nested simulation.
>>>>>>>>>>>>>
>>>>>>>>>>>>> SELF-EVIDENT-TRUTH
>>>>>>>>>>>>> Every computation that never halts unless its simulation is aborted is a
>>>>>>>>>>>>> computation that never halts.
>>>>>>>>>>>>>
>>>>>>>>>>>>> SELF-EVIDENT-TRUTH
>>>>>>>>>>>>> The <Ĥ> <Ĥ> input to the embedded halt decider at Ĥ.qx is a computation
>>>>>>>>>>>>> that never halts unless its simulation is aborted.
>>>>>>>>>>>>>
>>>>>>>>>>>>> ∴ IMPOSSIBLY FALSE CONCLUSION
>>>>>>>>>>>>> The embedded simulating halt decider at Ĥ.qx correctly decides its
>>>>>>>>>>>>> input: <Ĥ> <Ĥ> is a computation that never halts.
>>>>>>>>>>>>>
>>>>>>>>>>>>> The above three elements essentially provide the DNA of the cat.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Halting problem undecidability and infinitely nested simulation
>>>>>>>>>>>>>
>>>>>>>>>>>>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> --
>>>>>>>>>>>>> Copyright 2021 Pete Olcott
>>>>>>>>>>>>>
>>>>>>>>>>>>> "Great spirits have always encountered violent opposition from mediocre
>>>>>>>>>>>>> minds." Einstein
>>>>>>>>>>>>
>>>>>>>>>>>> As I said the question is very simple:
>>>>>>>>>>>> You have to show a correct implement (pseudo-code is OK) of the function
>>>>>>>>>>>> "bool HaltDecider(Func f, Arg a)". This is a MUST.
>>>>>>>>>>>> Other things (paper/talk) are auxiliary.
>>>>>>>>>>> I have done that six months ago using different naming conventions.
>>>>>>>>>>
>>>>>>>>>> This is a very great achievement, deserves 3 Nobel Prizes.
>>>>>>>>>>
>>>>>>>>>>> Halting problem undecidability and infinitely nested simulation
>>>>>>>>>>>
>>>>>>>>>>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Quoting the paper makes me baffled completely. It to me just is like searching for a set of
>>>>>>>>>> codes using 'simulator', not a good strategy while static code analyzer is sufficient.
>>>>>>>>> This is my paper that I wrote that has the code that you asked for.
>>>>>>>>>
>>>>>>>>> // Simplified Linz Ĥ (Linz:1990:319)
>>>>>>>>> void P(u32 x)
>>>>>>>>> {
>>>>>>>>> u32 Input_Halts = H(x, x);
>>>>>>>>> if (Input_Halts)
>>>>>>>>> HERE: goto HERE;
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> int main()
>>>>>>>>> {
>>>>>>>>> u32 Input_Halts = H((u32)P, (u32)P);
>>>>>>>>> Output("Input_Halts = ", Input_Halts);
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> H is a simulating halt decider based on an x86 emulator. I spent nearly
>>>>>>>>> two years creating the x86utm operating system so that I could implement H.
>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Evading this 'simple' question is taken as "No, my proof can't stand such a test".
>>>>>>>>>>>> Therefore... everything you have said is.... you imagine it.
>>>>>>>>>>>>
>>>>>>>>>>> --
>>>>>>>>>>> Copyright 2021 Pete Olcott
>>>>>>>>>>>
>>>>>>>>>>> "Great spirits have always encountered violent opposition from mediocre
>>>>>>>>>>> minds." Einstein
>>>>>>>>> --
>>>>>>>>> Copyright 2021 Pete Olcott
>>>>>>>>>
>>>>>>>>> "Great spirits have always encountered violent opposition from mediocre
>>>>>>>>> minds." Einstein
>>>>>>>>
>>>>>>>> Your proof may be 100% correct. But it only valid for your instance P.
>>>>>>>> I think you mis-interpreted the conventional HP proof.
>>>>>>>>
>>>>>>> When we compare the conventional pseudo-code to my C code that statement
>>>>>>> seem ridiculously stupid.
>>>>>>>
>>>>>>> procedure compute_g(i):
>>>>>>> if f(i, i) == 0 then
>>>>>>> return 0
>>>>>>> else
>>>>>>> loop forever // (Wikipedia:Halting Problem)
>>>>>>> // Simplified Linz Ĥ (Linz:1990:319)
>>>>>>> void P(u32 x)
>>>>>>> {
>>>>>>> u32 Input_Halts = H(x, x);
>>>>>>> if (Input_Halts)
>>>>>>> HERE: goto HERE;
>>>>>>> }
>>>>>>>
>>>>>>> int main()
>>>>>>> {
>>>>>>> u32 Input_Halts = H((u32)P, (u32)P);
>>>>>>> Output("Input_Halts = ", Input_Halts);
>>>>>>> }
>>>>>>>> I have shown an instance P that simulates H in different way(H2) will make H
>>>>>>>> behave incorrectly. The conventional HP proof can be demonstrated in C-like
>>>>>>> If it is not a pure simulation then it is wrong and all pure simulations
>>>>>>> must be identical.
>>>>>>
>>>>>> H2 is designed to simulate H in different way.
>>>>>> Why anyone's simulation of H2 is not a pure simulation while your H is?
>>>>>>
>>>>> Every simulation that is not a pure simulation is a wrong simulation.
>>>>> If your simulation is not a pure simulation then it is wrong.
>>>>>
>>>>> If your simulation is a pure simulation then it cannot possibly differ
>>>>> from any other pure simulation. That you claim that it is different
>>>>> proves that it is wrong.
>>>> Your H does not do what P exactly does. That you claim that it 'simulate'
>>>>> proves that it is wrong.
>>>>
>>>>>>>> pseudo-code which is more useful, applicable, most people can comprehend
>>>>>>>> immediately. A refutation should be capable of being demonstrated in the same way.
>>>>>>>>
>>>>>>>> From software engineering point of view, your proof is 'optimized' too soon
>>>>>>>> to the lowest level (assembly, TM). Creating a x86utm operating system makes
>>>>>>>> no sense to refute HP. Beside, to refute, the 'x86utm operating system' (all) has to
>>>>>>>> be present in the paper for peer to reproduce the result.
>>>>>>>>
>>>>>>> It is enormously easier to analyze the ready made directed graphs of
>>>>>>> control flow that assembly language provides rather than have to build
>>>>>>> these directed graphs from scratch manually. Any unbroken cycle in a
>>>>>>> directed graph is infinite execution that must be aborted.
>>>>>>> --
>>>>>>> Copyright 2021 Pete Olcott
>>>>>>>
>>>>>>> "Great spirits have always encountered violent opposition from mediocre
>>>>>>> minds." Einstein
>>>>>>
>>>>>> You fabricated a halt-decider which only works in your head.
>>>>>>
>>>>> --
>>>>> Copyright 2021 Pete Olcott
>>>>>
>>>>> "Great spirits have always encountered violent opposition from mediocre
>>>>> minds." Einstein
>>>
>>> Your H does not do what P exactly does. That you claim that it 'simulate'
>>> proves that it is wrong.
>>>
>> H is a simulator and P is not a simulator therefore if H did exactly
>> what P does H would be wrong. H does show exactly what P does.
>> --
>> Copyright 2021 Pete Olcott
>>
>> "Great spirits have always encountered violent opposition from mediocre
>> minds." Einstein
>
> H2 would do functionally exactly the same H does (H2 can show exactly what H
> does), whatever you call H is (pure?). Manipulating descriptive words is not a
> good sign you honestly want to understand the issues of your proof.
>
> Since you made your refutation a real program (this is very good), but it
> can't stand real tests in my estimate.
> In any cases, reviewer need to duplicate your running program to reproduce
> your result and claim. Everything else is just talk.
>


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Re: What if a cat barks? [ sound deduction proves that I am correct ]

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Subject: Re: What if a cat barks? [ sound deduction proves that I am correct ]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 23 Jun 2021 02:15 UTC

On 6/22/2021 8:47 PM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>
>> On 6/21/2021 4:16 AM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> On 6/20/2021 9:15 PM, olcott wrote:
>>>>> If you see an animal and test its DNA and confirm that it is
>>>>> definitely a cat, what happens when the cat barks?
>>>> [...]
>>>>
>>>> Have you been hearing cats bark lately? Wow.
>>> He's making a bad analogy. The correct analogy is that /assuming/ that
>>> a cat barks leads to a contradiction so we must reject the assumption.
>>> He can see the contradiction that assuming a halt decider leads to (at
>>> least he claimed to be able to see it) so what to do? He has to state
>>> that he has a barking cat -- a halt decider that works at least for the
>>> confounding case. Of course he doesn't, but he has to find some way to
>>> keep the discussion going (he only cares about keeping people talking).
>>
>> Ahhh. Good. Okay, well I am still wondering why, when I tell him to
>> run his "halt decider" against an unknown, black box program created
>> by somebody else... Well, he seems to get pissed off. Afaict, his
>> decider only works on programs that he already knows are,
>> decided. Cheating 101? Or what? ;^o
>
> Well he flip-flops on this. He keeps saying that he has a halt decider
> (sometimes without realising that he's said it) but when this is pointed
> out he claims he only cares about the one case -- the "hat" construction
> given in so many proofs.
>
> That, alone, would be noteworthy because it's impossible. Way back in
> Dec 2018, he was waffling about a decider for one case (which is
> trivial), when someone (I think it was Mike Terry) spotted that he was
> actually claiming to have a TM H that gave the correct result for the
> computation <[H^], [H^]> (my notation -- I'll explain it if you really
> care about the details). Once it was pointed out that everyone knows
> this is impossible PO was delighted and wrote things like:
>
> "Everyone has claimed that H on input pair (Ĥ, Ĥ) meeting the Linz
> specs does not exist. I now have a fully encoded pair of Turing
> Machines H / Ĥ proving them wrong."
>
> "I now have an actual H that decides actual halting for an actual (Ĥ,
> Ĥ) input pair. I have to write the UTM to execute this code, that
> should not take very long. The key thing is the H and Ĥ are 100%
> fully encoded as actual Turing machines."
>
> "I am waiting to encode the UTM in C++ so that I can actually execute
> H on the input pair: (Ĥ, Ĥ). This should take a week or two. It is not
> a universal halt decider, yet it is exactly and precisely the Peter
> Linz H and Ĥ, with H actually deciding input pair: (Ĥ, Ĥ) on the basis
> of Ĥ actually deciding itself."
>
> I refer to this as the Big Lie because, of course, he never had any such
> pair of TMs, but as you can see, it was getting one (impossible) case
> right that was the jumping-off point for the last two an half years of
> nonsense.
>

Yes and of course when I make a C program that is computationally
equivalent to the standard relation between the HP halt decider and its
impossible input and show all of the steps of how this halt decider
correctly decides this "known" to be impossible input that does not
count at all because it is not 100% perfectly an actual TM.

The C program is implemented in the x86utm operating system using an
actual x86 emulator. This work began in earnest at the end of 2019 after
the huge false start of trying to implement an interpreter for a tiny
subset of C from scratch.

Halting problem undecidability and infinitely nested simulation

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

Ben: You (and others) have affirmed premise (1).

You might be able to see that the conclusion logically follows from
premise (1) and premise (2).

There was one very highly qualified person that has affirmed that
premise (2) is a verified fact.

Premise(1) Every computation that never halts unless its simulation is
aborted is a computation that never halts. This verified as true on the
basis of the meaning of its words.

Premise(2) The simulation of the input to H(P,P) never halts without
being aborted is a verified fact on the basis of its x86 execution
trace. (shown below).

From the above true premises it necessarily follows that simulating
halt decider H correctly reports that its input: (P,P) never halts.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction proves that I am correct ]

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Subject: Re: What if a cat barks? [ sound deduction proves that I am correct ]
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From: NoO...@NoWhere.com (olcott)
Date: Tue, 22 Jun 2021 22:21:43 -0500
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 by: olcott - Wed, 23 Jun 2021 03:21 UTC

On 6/22/2021 10:00 PM, Richard Damon wrote:
> On 6/22/21 10:15 PM, olcott wrote:
>> On 6/22/2021 8:47 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> On 6/21/2021 4:16 AM, Ben Bacarisse wrote:
>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>
>>>>>> On 6/20/2021 9:15 PM, olcott wrote:
>>>>>>> If you see an animal and test its DNA and confirm that it is
>>>>>>> definitely a cat, what happens when the cat barks?
>>>>>> [...]
>>>>>>
>>>>>> Have you been hearing cats bark lately? Wow.
>>>>> He's making a bad analogy.  The correct analogy is that /assuming/ that
>>>>> a cat barks leads to a contradiction so we must reject the assumption.
>>>>> He can see the contradiction that assuming a halt decider leads to (at
>>>>> least he claimed to be able to see it) so what to do?  He has to state
>>>>> that he has a barking cat -- a halt decider that works at least for the
>>>>> confounding case.  Of course he doesn't, but he has to find some way to
>>>>> keep the discussion going (he only cares about keeping people talking).
>>>>
>>>> Ahhh. Good. Okay, well I am still wondering why, when I tell him to
>>>> run his "halt decider" against an unknown, black box program created
>>>> by somebody else... Well, he seems to get pissed off. Afaict, his
>>>> decider only works on programs that he already knows are,
>>>> decided. Cheating 101?  Or what? ;^o
>>>
>>> Well he flip-flops on this.  He keeps saying that he has a halt decider
>>> (sometimes without realising that he's said it) but when this is pointed
>>> out he claims he only cares about the one case -- the "hat" construction
>>> given in so many proofs.
>>>
>>> That, alone, would be noteworthy because it's impossible.  Way back in
>>> Dec 2018, he was waffling about a decider for one case (which is
>>> trivial), when someone (I think it was Mike Terry) spotted that he was
>>> actually claiming to have a TM H that gave the correct result for the
>>> computation <[H^], [H^]> (my notation -- I'll explain it if you really
>>> care about the details).  Once it was pointed out that everyone knows
>>> this is impossible PO was delighted and wrote things like:
>>>
>>>    "Everyone has claimed that H on input pair (Ĥ, Ĥ) meeting the Linz
>>>    specs does not exist. I now have a fully encoded pair of Turing
>>>    Machines H / Ĥ proving them wrong."
>>>
>>>    "I now have an actual H that decides actual halting for an actual (Ĥ,
>>>    Ĥ) input pair.  I have to write the UTM to execute this code, that
>>>    should not take very long.  The key thing is the H and Ĥ are 100%
>>>    fully encoded as actual Turing machines."
>>>
>>>    "I am waiting to encode the UTM in C++ so that I can actually execute
>>>    H on the input pair: (Ĥ, Ĥ). This should take a week or two. It is not
>>>    a universal halt decider, yet it is exactly and precisely the Peter
>>>    Linz H and Ĥ, with H actually deciding input pair: (Ĥ, Ĥ) on the basis
>>>    of Ĥ actually deciding itself."
>>>
>>> I refer to this as the Big Lie because, of course, he never had any such
>>> pair of TMs, but as you can see, it was getting one (impossible) case
>>> right that was the jumping-off point for the last two an half years of
>>> nonsense.
>>>
>>
>> Yes and of course when I make a C program that is computationally
>> equivalent to the standard relation between the HP halt decider and its
>> impossible input and show all of the steps of how this halt decider
>> correctly decides this "known" to be impossible input that does not
>> count at all because it is not 100% perfectly an actual TM.
>
> Except that it isn't because H^ doesn't make copies as it is supposed to
> so you don't end up with a true equivalent.

Others know that it is sufficiently equivalent.

procedure compute_g(i):
if f(i, i) == 0 then
return 0
else
loop forever // (Wikipedia:Halting Problem)

Its pretty obvious that the C code and the pseudo-code are saying
exactly the same thing.

// Simplified Linz Ĥ (Linz:1990:319)
void P(u32 x)
{ u32 Input_Halts = H(x, x);
if (Input_Halts)
HERE: goto HERE;
}

> This lets you 'cheat' to
> detect the recursion that doesn't really exist in the original problem.

Because others know that it is sufficiently equivalent they can also see
the infinite recursion in the Linz proof as soon as they sufficiently
understand my x86/C proof.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction proves that I am correct ]

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Subject: Re: What if a cat barks? [ sound deduction proves that I am correct ]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 23 Jun 2021 16:12 UTC

On 6/23/2021 11:00 AM, Ben Bacarisse wrote:
> Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
>
>> On Wednesday, 23 June 2021 at 11:57:06 UTC+1, Richard Damon wrote:
>>> On 6/22/21 11:21 PM, olcott wrote:
>>>> On 6/22/2021 10:00 PM, Richard Damon wrote:
>>>>>
>>>>>> Yes and of course when I make a C program that is computationally
>>>>>> equivalent to the standard relation between the HP halt decider and its
>>>>>> impossible input and show all of the steps of how this halt decider
>>>>>> correctly decides this "known" to be impossible input that does not
>>>>>> count at all because it is not 100% perfectly an actual TM.
>>>>>
>>>>> Except that it isn't because H^ doesn't make copies as it is supposed to
>>>>> so you don't end up with a true equivalent.
>>>>
>>>> Others know that it is sufficiently equivalent.
>>> There is no 'sufficiently' it only is or it isn't.
>>>
>>> Is 2.0001 sufficiently equivalent to 2? Its pretty close, but is it
>>> equivalent.
>>>
>> PO's system uses x86 code instead of Turing machines. That enables him
>> to detect when the halt decider is being called on a program including itself.
>> That's not the root of the error (the halting problem also applies to x86
>> programs) but it is the root of the confusion.
>> PO obviously thinks that he has found a counter example to Linz's
>> impossible H.
>
> I find this hard to believe but I accept that there are minds out there
> that I have little understanding of. Even so, how can anyone think they
> have a decider M that is correct about <[M^],[M^]> if
>
> a. M rejects <[M^],[M^]>, but
> b. M^([M^]) halts?
>

// Simplified Linz Ĥ (Linz:1990:319)
void P(u32 x)
{ u32 Input_Halts = H(x, x);
if (Input_Halts)
HERE: goto HERE;
}

int main()
{ u32 Input_Halts = H((u32)P, (u32)P);
Output("Input_Halts = ", Input_Halts);
}

Premise(1) Every computation that never halts unless its simulation is
aborted is a computation that never halts. This verified as true on the
basis of the meaning of its words.

Premise(2) The simulation of the input to H(P,P) never halts without
being aborted is a verified fact on the basis of its x86 execution
trace. (shown below).

Conclusion (3) From the above true premises it necessarily follows that
simulating halt decider H correctly reports that its input: (P,P) never
halts.

When (1) is true and (2) is true and (1) & (2) logically entails that
(3) is true, nothing can possibly contradict that (3) is true.

Make sure you carefully ignore all those words because that is the only
way that any fake ruse of a rebuttal can continue.

Halting problem undecidability and infinitely nested simulation

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

> It seems impossible to me that anyone can accept these facts and claim
> to have "a counter example" to the proof's hat construction. Surely
> even PO can see that such a TM is not correct? It think he know.
> That's why he was, briefly, honest about redefining what "halting"
> means. It was only after he realised that everyone would ignore him if
> he was talking about some other notion of "halting" that he stopped
> being clear about this redefinition.
>
> Further, PO recently posted this:
>
> | Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
> | if M applied to wM halts, and
> |
> | Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
> | if M applied to wM does not halt
>
> which is (roughly) Linz's notation for what H^, built from a correct H,
> should do. I showed the last tiny step which is to set wM = <Ĥ> (so M
> is Ĥ) to get
>
> Ĥ.q0 <Ĥ> ⊢* Ĥ.qx <Ĥ> <Ĥ> ⊢* Ĥ.qy ⊢* ∞
> if Ĥ applied to <Ĥ> halts, and
>
> Ĥ.q0 <Ĥ> ⊢* Ĥ.qx <Ĥ> <Ĥ> ⊢* Ĥ.qn
> if Ĥ applied to <Ĥ> does not halt
>
> which, in words is, "Ĥ run (from state q0) on input <Ĥ> does not halt if
> Ĥ applied to <Ĥ> halts", and "Ĥ run (from state q0) on input <Ĥ> halts
> (in state qn) if Ĥ applied to <Ĥ> does not halt".
>
> In reply, PO said he understood that, but then went on to say more
> things as if anything, true or false, can alter the fact that the
> assumption about H leads to a contradiction. Do you think he genuinely
> believes that saying more things, even true things, can "undo" a
> contradiction? That is a strange mind indeed.
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks?

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 23 Jun 2021 20:07 UTC

On 6/23/2021 12:29 PM, André G. Isaak wrote:
> On 2021-06-22 13:17, olcott wrote:
>> On 6/22/2021 1:38 PM, André G. Isaak wrote:
>>> On 2021-06-22 11:05, olcott wrote:
>>>> On 6/22/2021 11:47 AM, André G. Isaak wrote:
>
> <snip>
>
>> In the case of the simulation of P it is verified on the basis of the
>> x86 execution trace of B, thus an established verified fact.
>
> Except that an execution trace doesn't 'verify' anything. No one doubts
> that your program decides that P(P) == 0. But unless the logic
> underlying your program is actually correct, this result tells us
> nothing. And you haven't provided any sort of proof that the logic
> underlying your program is actually correct.
>

Halting problem undecidability and infinitely nested simulation

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

Sure I have. When we look as page 3 and 4 of the above paper it is very
easy to see that H does faithfully simulate the execution of P and that
P has no escape from infinite execution in its own code.

Premise(1) Every computation that never halts unless its simulation is
aborted is a computation that never halts. This verified as true on the
basis of the meaning of its words.

Premise(2) The simulation of the input to H(P,P) never halts without
being aborted is a verified fact on the basis of its x86 execution
trace. (shown below).

Conclusion(3) From the above true premises it necessarily follows that
simulating halt decider H correctly reports that its input: (P,P) never
halts.

Only by very diligently making sure to not pay attention is it possible
to maintain the ruse of a fake rebuttal.

> And it isn't correct. Most significantly, your trace skips over certain
> portions of the P being simulated which completely invalidates its results.
>

The question is:

Does the simulation of P have to be aborted to prevent its infinite
execution?

As long as the answer to this question is [YES] then Premise(2) is
perfectly satisfied and your objection is utterly pointless.

As long as the answer to this question is [YES] then Premise(2) is
perfectly satisfied and your objection is utterly pointless.

As long as the answer to this question is [YES] then Premise(2) is
perfectly satisfied and your objection is utterly pointless.

As long as the answer to this question is [YES] then Premise(2) is
perfectly satisfied and your objection is utterly pointless.

> <snip>
>
>>>>> Option C never said anything about why it aborts the simulation.
>>>>
>>>> That is its horribly terrible error. To simply ignore the key most
>>>> important halt deciding criteria is a terribly awful mistake.
>>>
>>> Clearly you have reading comprehension problems. The halting status
>>> of the *simulator* has nothing to do with the halting status of its
>>> input computation if the simulator is capable of aborting the
>>> simulation of that input. That applies equally to your (C) and your
>>> (D). In either case, the simulator can halt.
>>>
>>
>> An intentional fallacy of equivocation error that very persistently
>> tries to hide the correct halt deciding criteria.
>>
>> Key to refuting all the conventional HP undecidability proofs
>>
>> -- H MUST stop its simulation of P or P never halts
>
> I'm not hiding anything. Whether the halt deciding criterion is correct
> or not has no bearing on my claim, which is that once a simulation is
> aborted the simulator can still halt. That's true of a simulator using a
> 'correct' criterion just as much as its true of one which simply aborts
> its simulation at random.
>
> André
>
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction is a proof ]

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Thu, 24 Jun 2021 00:01 UTC

On 6/23/2021 6:13 PM, André G. Isaak wrote:
> On 2021-06-23 14:07, olcott wrote:
>> On 6/23/2021 12:29 PM, André G. Isaak wrote:
>>> On 2021-06-22 13:17, olcott wrote:
>>>> On 6/22/2021 1:38 PM, André G. Isaak wrote:
>>>>> On 2021-06-22 11:05, olcott wrote:
>>>>>> On 6/22/2021 11:47 AM, André G. Isaak wrote:
>>>
>>> <snip>
>>>
>>>> In the case of the simulation of P it is verified on the basis of
>>>> the x86 execution trace of B, thus an established verified fact.
>>>
>>> Except that an execution trace doesn't 'verify' anything. No one
>>> doubts that your program decides that P(P) == 0. But unless the logic
>>> underlying your program is actually correct, this result tells us
>>> nothing. And you haven't provided any sort of proof that the logic
>>> underlying your program is actually correct.
>>>
>>
>> Halting problem undecidability and infinitely nested simulation
>>
>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>>
>>
>> Sure I have. When we look as page 3 and 4 of the above paper it is
>> very easy to see that H does faithfully simulate the execution of P
>> and that P has no escape from infinite execution in its own code.
>
> No. First, it *doesn't* faithfully simulate the code. If H(P, P) is
> simulating P(P), then P(P) can't call the main H -- it can only invoke a
> copy of H *within the simulation*.
>

It is actually and quite factually not a copy.
I have had to say this hundreds of times now.

It is the same freaking machine language at the
same freaking machine language address.

> Second, even if we ignore the above, the decision criterion you use is
> flawed because it completely ignores all conditional branches contained
> within P's copy of H, and it's these conditional branches which prevent
> P(P) from being an 'infinitely recursive' computation.
>

Unless we keep the independent variable separate from the dependent
variable our analysis is incorrect.

The independent variable is the cause. Its value is independent of other
variables in your study.

The dependent variable is the effect. Its value depends on changes in
the independent variable.

https://www.scribbr.com/methodology/independent-and-dependent-variables/

When we are asking whether or not H must abort its simulation of P we
examine the simulation of P. We must not examine the behavior of H.

> Once again, traces do not constitute proofs. If you want to be taken
> seriously, you need to provide some actual *proof* that your algorithm
> works.

That my proof does not conform to academic conventions does not show
that it is not a proof. As long as I start with a set of premises that
can be verified as true and my conclusion logically follows from these
true premises then I definitely have a proof even if everyone in the
universe disagrees.

>> Premise(1) Every computation that never halts unless its simulation is
>> aborted is a computation that never halts. This verified as true on
>> the basis of the meaning of its words.
>>
>> Premise(2) The simulation of the input to H(P,P) never halts without
>> being aborted is a verified fact on the basis of its x86 execution
>> trace. (shown below).
>>
>> Conclusion(3) From the above true premises it necessarily follows that
>> simulating halt decider H correctly reports that its input: (P,P)
>> never halts.
>>
>> Only by very diligently making sure to not pay attention is it
>> possible to maintain the ruse of a fake rebuttal.
>>
>>> And it isn't correct. Most significantly, your trace skips over
>>> certain portions of the P being simulated which completely
>>> invalidates its results.
>>>
>>
>> The question is:
>>
>> Does the simulation of P have to be aborted to prevent its infinite
>> execution?
>>
>> As long as the answer to this question is [YES] then Premise(2) is
>> perfectly satisfied and your objection is utterly pointless.
>
> You should cut it out with the asinine repetition.It just makes you look
> like a three-year old.
>

If I don't get ridiculous with the repetition in a single reply then I
need twenty replies to make the same point because this point is
perpetually ignored. I usually don't repeat the same sentence five times
until after a key point has been ignored in several replies.

> And no, the objection is *not* pointless. You can't claim that there is
> "no escape from infinite execution" when you ignore the conditional
> statements which would allow P(P) to abort the simulation it is
> conducting. Those are the conditional branches which provide that escape.
>
> André
>

The question is whether or not H must abort its simulation of P.
The behavior of H is totally irrelevant to this question.

Of the two hypothetical possibilities: (independent variable)
(1) H aborts the simulation of P
(2) H never aborts the simulation of P

We ask what is the effect on the behavior of P? (dependent variable)

(1) P halts
(2) P never halts

The above answer applies to every possible encoding of any simulating
halt decoder H when applied to P or any computational equivalent to P
such as the Linz ⟨Ĥ⟩.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction is a proof ]

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Subject: Re: What if a cat barks? [ sound deduction is a proof ]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Thu, 24 Jun 2021 03:03 UTC

On 6/23/2021 7:49 PM, André G. Isaak wrote:
> On 2021-06-23 18:01, olcott wrote:
>> On 6/23/2021 6:13 PM, André G. Isaak wrote:
>>> On 2021-06-23 14:07, olcott wrote:
>>>> On 6/23/2021 12:29 PM, André G. Isaak wrote:
>>>>> On 2021-06-22 13:17, olcott wrote:
>>>>>> On 6/22/2021 1:38 PM, André G. Isaak wrote:
>>>>>>> On 2021-06-22 11:05, olcott wrote:
>>>>>>>> On 6/22/2021 11:47 AM, André G. Isaak wrote:
>>>>>
>>>>> <snip>
>>>>>
>>>>>> In the case of the simulation of P it is verified on the basis of
>>>>>> the x86 execution trace of B, thus an established verified fact.
>>>>>
>>>>> Except that an execution trace doesn't 'verify' anything. No one
>>>>> doubts that your program decides that P(P) == 0. But unless the
>>>>> logic underlying your program is actually correct, this result
>>>>> tells us nothing. And you haven't provided any sort of proof that
>>>>> the logic underlying your program is actually correct.
>>>>>
>>>>
>>>> Halting problem undecidability and infinitely nested simulation
>>>>
>>>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>>>>
>>>>
>>>> Sure I have. When we look as page 3 and 4 of the above paper it is
>>>> very easy to see that H does faithfully simulate the execution of P
>>>> and that P has no escape from infinite execution in its own code.
>>>
>>> No. First, it *doesn't* faithfully simulate the code. If H(P, P) is
>>> simulating P(P), then P(P) can't call the main H -- it can only
>>> invoke a copy of H *within the simulation*.
>>>
>>
>> It is actually and quite factually not a copy.
>> I have had to say this hundreds of times now.
>
> But if it isn't a copy, then your P isn't derived from your H in the
> manner prescribed by the Linz proof. If you want to claim to have
> provided a 'counterexample' to the 'standard proofs' then you need to
> actually construct your P in the same manner that H_Hat is constructed
> in those proofs.
>
>> It is the same freaking machine language at the
>> same freaking machine language address.
>
> If P is calling H directly, then in what sense is P being 'simulated'?
>
>>> Second, even if we ignore the above, the decision criterion you use
>>> is flawed because it completely ignores all conditional branches
>>> contained within P's copy of H, and it's these conditional branches
>>> which prevent P(P) from being an 'infinitely recursive' computation.
>>>
>>
>> Unless we keep the independent variable separate from the dependent
>> variable our analysis is incorrect.
>>
>> The independent variable is the cause. Its value is independent of
>> other variables in your study.
>>
>> The dependent variable is the effect. Its value depends on changes in
>> the independent variable.
>>
>> https://www.scribbr.com/methodology/independent-and-dependent-variables/
>>
>> When we are asking whether or not H must abort its simulation of P we
>> examine the simulation of P. We must not examine the behavior of H.
>
> This is simply wrong. You can ignore the code in the *decider*, but not
> code inside the computation which the decider is simulating. The call to
> H inside P is an integral part of the computation performed by P. You
> can't ignore this anymore than you can ignore any other instructions
> inside P.
>
> André
>

The independent variable is the cause. Its value is independent of other
variables in your study.

The dependent variable is the effect. Its value depends on changes in
the independent variable.

https://www.scribbr.com/methodology/independent-and-dependent-variables/

The question is whether or not H must abort its simulation of P.
The behavior of H is totally irrelevant to this question.

As I have proven below.
As I have proven below.
As I have proven below.
As I have proven below.

Of the two hypothetical possibilities: (independent variable)
(1) H aborts the simulation of P
(2) H never aborts the simulation of P

We ask what is the effect on the behavior of P? (dependent variable)

(1) P halts
(2) P never halts

It does not freaking matter at all what-ever-the-Hell H does except what
the behavior of P is under the above two hypothetical scenarios.

If the result of purely hypothetical H that never aborts its
simulation/execution of P would result in the infinite execution of P
then we know axiomatically that H decides that P never halts correctly.

Premise(1) Every computation that never halts unless its simulation is
aborted is a computation that never halts. This verified as true on the
basis of the meaning of its words.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction proves that I am correct ]

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Subject: Re: What if a cat barks? [ sound deduction proves that I am correct ]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Thu, 24 Jun 2021 15:25 UTC

On 6/24/2021 4:40 AM, Malcolm McLean wrote:
> On Wednesday, 23 June 2021 at 17:00:57 UTC+1, Ben Bacarisse wrote:
>>
>> In reply, PO said he understood that, but then went on to say more
>> things as if anything, true or false, can alter the fact that the
>> assumption about H leads to a contradiction. Do you think he genuinely
>> believes that saying more things, even true things, can "undo" a
>> contradiction? That is a strange mind indeed.
>>
> We've had quite a bit of flip-flopping. We had a period when it was
> acknowledged that Linz's proof was sound but he had a slightly
> different property.

// Simplified Linz Ĥ (Linz:1990:319)
void P(u32 x)
{ u32 Input_Halts = H(x, x);
if (Input_Halts)
HERE: goto HERE;
}

int main()
{ u32 Input_Halts = H((u32)P, (u32)P);
Output("Input_Halts = ", Input_Halts);
}

That H(P,P) cannot return a correct halt status to P in the above
computation is well proven.

That H(P,P) ever needs to return to P at all in the above code is the
key false assumption that unravels the whole proof.

> Defining a slightly different property to halting
> is not an inherently bad thing to do, it could be the starting point
> of some significant work. But that's very hard to do, and you and I
> pointed out that PO's different problem wasn't one of the "interesting"
> ones. You needed to solve the original halting problem first to decide
> it, for example.
>
> So there might have been a row back from that when he realised that
> the response would be "well that might be true but it doesn't get us
> anywhere".
>
> However I do think that he genuinely thinks that he's got something
> significant. I'm not a mind reader, but a troll would have got bored by
> now, whilst the posts go off at too many tangents to be a fraudster.
> When you dry run his simulating halt decider on itself, it is confusing.
> Take out the halt detector and it runs forever. Put the halt detector in,
> and it detects non-halting behaviour (according to PO, no real reason
> to disbelieve him). So surely it's got that right?

> But in fact simulate
> one more step, and the halt detection in the simulated machine will
> kick in and halt it.
>

I had the false assumption myself for a couple of days a year or two
ago. H can wait any fixed number of recursions before aborting its
input. If H simply waits for a later recursion to abort its input then
its input is never aborted because the code to abort this input does not
exist.

Halting problem undecidability and infinitely nested simulation
https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction proves that I am correct ]

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Subject: Re: What if a cat barks? [ sound deduction proves that I am correct ]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Thu, 24 Jun 2021 15:39 UTC

On 6/24/2021 5:49 AM, Ben Bacarisse wrote:
> Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:
>
>> On Wednesday, 23 June 2021 at 17:00:57 UTC+1, Ben Bacarisse wrote:
>>>
>>> In reply, PO said he understood that, but then went on to say more
>>> things as if anything, true or false, can alter the fact that the
>>> assumption about H leads to a contradiction. Do you think he genuinely
>>> believes that saying more things, even true things, can "undo" a
>>> contradiction? That is a strange mind indeed.
>>>
>> We've had quite a bit of flip-flopping. We had a period when it was
>> acknowledged that Linz's proof was sound but he had a slightly
>> different property. Defining a slightly different property to halting
>> is not an inherently bad thing to do, it could be the starting point
>> of some significant work. But that's very hard to do, and you and I
>> pointed out that PO's different problem wasn't one of the "interesting"
>> ones. You needed to solve the original halting problem first to decide
>> it, for example.
>>
>> So there might have been a row back from that when he realised that
>> the response would be "well that might be true but it doesn't get us
>> anywhere".
>>
>> However I do think that he genuinely thinks that he's got something
>> significant. I'm not a mind reader, but a troll would have got bored by
>> now, whilst the posts go off at too many tangents to be a fraudster.
>
> I don't think he's a troll, and I don't think he's a fraudster, but the
> changes in his claims suggest to me that he is aware of being wrong and
> having to row-back the wildest ones.

// Simplified Linz Ĥ (Linz:1990:319)
void P(u32 x)
{ u32 Input_Halts = H(x, x);
if (Input_Halts)
HERE: goto HERE;
}

int main()
{ u32 Input_Halts = H((u32)P, (u32)P);
Output("Input_Halts = ", Input_Halts);
}

That H(P,P) cannot return a correct halt status to P in the above
computation is well proven.

That H(P,P) ever needs to return to P at all in the above code is the
key false assumption that unravels the whole proof.

> He certainly knows, but has not
> admitted, that he never had "two actual Turing machines" "fully encoded"
> back in Dec 2018, waiting only for a TM simulator to be able to run them

Virtual machines that are "computationally equivalent" to a Turing
machines are sufficiently Turing machines.

> (he claimed that would take about a week). I don't think these
> deceptions are fraudulent. I think his self-image can't cope with the
> concept of being wrong.
>
>> When you dry run his simulating halt decider on itself, it is confusing.
>> Take out the halt detector and it runs forever. Put the halt detector in,
>> and it detects non-halting behaviour (according to PO, no real reason
>> to disbelieve him). So surely it's got that right?
>
> I don't know what you mean. How do you run anything of his? We've
> never even seen the key function.
>
>> But in fact simulate
>> one more step, and the halt detection in the simulated machine will
>> kick in and halt it.
>
> I don't see the value in discussing all these details, though others
> clearly do. He's stated in various round-about ways that his "decider"
> (the hidden code we will never see) is wrong as far as "conventional"
> halting is concerned. There is zero dispute on the facts that the code
> indicates false for a halting computation.
>

When a simulating halt decider decides that an actual infinite loop
would never halt and it stops this simulation of this infinite loop it
correctly reports that the input never halts, even though the input does
indeed halt.

The same reasoning applies to infinite recursion and infinitely nested
simulation.

You always dodge this challenge because you know that you will fail
You always dodge this challenge because you know that you will fail
You always dodge this challenge because you know that you will fail
You always dodge this challenge because you know that you will fail

Premise(1) Every computation that never halts unless its simulation is
aborted is a computation that never halts. This verified as true on the
basis of the meaning of its words.

Premise(2) The simulation of the input to H(P,P) never halts without
being aborted is a verified fact on the basis of its x86 execution
trace. (pages 3-4 of the paper).

Conclusion(3) From the above true premises it necessarily follows that
simulating halt decider H correctly reports that its input: (P,P) never
halts.

Halting problem undecidability and infinitely nested simulation

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction proves that I am correct ]

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Subject: Re: What if a cat barks? [ sound deduction proves that I am correct ]
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From: NoO...@NoWhere.com (olcott)
Date: Thu, 24 Jun 2021 11:03:33 -0500
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 by: olcott - Thu, 24 Jun 2021 16:03 UTC

On 6/24/2021 10:47 AM, Malcolm McLean wrote:
> On Thursday, 24 June 2021 at 16:25:00 UTC+1, olcott wrote:
>> On 6/24/2021 4:40 AM, Malcolm McLean wrote:
>>
>> That H(P,P) cannot return a correct halt status to P in the above
>> computation is well proven.
>>
>> That H(P,P) ever needs to return to P at all in the above code is the
>> key false assumption that unravels the whole proof.
>>
> H has to be a Turing machine.

In the pseudo-code proofs H is always pseudo-code and never a Turing
machine. It is not true that H must be a Turing machine as long as the
H/P relationship is sufficiently equivalent to the H/⟨Ĥ⟩ the C/x86
version applied.

>>
>>> But in fact simulate
>>> one more step, and the halt detection in the simulated machine will
>>> kick in and halt it.
>>>
>> I had the false assumption myself for a couple of days a year or two
>> ago. H can wait any fixed number of recursions before aborting its
>> input. If H simply waits for a later recursion to abort its input then
>> its input is never aborted because the code to abort this input does not
>> exist.
>> Halting problem undecidability and infinitely nested simulation
>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>>
> It's not really recursion. It's nested simulation, which is similar to recursion,
> but different from it in that control never passes to the child function.

It is sufficiently computationally equivalent.

> Instead the child function is simulated, then simulated by a simulated function,
> then simulated by a simulated simulated function and so on, with control
> always remaining in the root function.
>
> If H doesn't abort, then this goes on forever. However if it does abort, it always
> aborts one step before the child would have performed the identical abort.

This is not true. If we simply wait for the child to do it then it never
happens.

> So you've kind of created your own paradox problem here. The abort / don't
> abort decision is always wrong.
>

Of the two hypothetical possibilities: (independent variable)
(1) H aborts the simulation of P
(2) H never aborts the simulation of P

We ask what is the effect on the behavior of P? (dependent variable)

(1) P halts
(2) P never halts

If the result of purely hypothetical H that never aborts its
simulation/execution of P would result in the infinite execution of P
then we know axiomatically that H decides that P never halts correctly.

This is the axiom:
Premise(1) Every computation that never halts unless its simulation is
aborted is a computation that never halts. This verified as true on the
basis of the meaning of its words.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction is a proof ]

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Thu, 24 Jun 2021 21:40 UTC

On 6/24/2021 4:13 PM, André G. Isaak wrote:
> On 2021-06-23 21:03, olcott wrote:
>> On 6/23/2021 7:49 PM, André G. Isaak wrote:
>>> On 2021-06-23 18:01, olcott wrote:
>>>> On 6/23/2021 6:13 PM, André G. Isaak wrote:
>>>>> On 2021-06-23 14:07, olcott wrote:
>>>>>> On 6/23/2021 12:29 PM, André G. Isaak wrote:
>>>>>>> On 2021-06-22 13:17, olcott wrote:
>>>>>>>> On 6/22/2021 1:38 PM, André G. Isaak wrote:
>>>>>>>>> On 2021-06-22 11:05, olcott wrote:
>>>>>>>>>> On 6/22/2021 11:47 AM, André G. Isaak wrote:
>>>>>>>
>>>>>>> <snip>
>>>>>>>
>>>>>>>> In the case of the simulation of P it is verified on the basis
>>>>>>>> of the x86 execution trace of B, thus an established verified fact.
>>>>>>>
>>>>>>> Except that an execution trace doesn't 'verify' anything. No one
>>>>>>> doubts that your program decides that P(P) == 0. But unless the
>>>>>>> logic underlying your program is actually correct, this result
>>>>>>> tells us nothing. And you haven't provided any sort of proof that
>>>>>>> the logic underlying your program is actually correct.
>>>>>>>
>>>>>>
>>>>>> Halting problem undecidability and infinitely nested simulation
>>>>>>
>>>>>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>>>>>>
>>>>>>
>>>>>> Sure I have. When we look as page 3 and 4 of the above paper it is
>>>>>> very easy to see that H does faithfully simulate the execution of
>>>>>> P and that P has no escape from infinite execution in its own code.
>>>>>
>>>>> No. First, it *doesn't* faithfully simulate the code. If H(P, P) is
>>>>> simulating P(P), then P(P) can't call the main H -- it can only
>>>>> invoke a copy of H *within the simulation*.
>>>>>
>>>>
>>>> It is actually and quite factually not a copy.
>>>> I have had to say this hundreds of times now.
>>>
>>> But if it isn't a copy, then your P isn't derived from your H in the
>>> manner prescribed by the Linz proof. If you want to claim to have
>>> provided a 'counterexample' to the 'standard proofs' then you need to
>>> actually construct your P in the same manner that H_Hat is
>>> constructed in those proofs.
>>>
>>>> It is the same freaking machine language at the
>>>> same freaking machine language address.
>>>
>>> If P is calling H directly, then in what sense is P being 'simulated'?
>>>
>>>>> Second, even if we ignore the above, the decision criterion you use
>>>>> is flawed because it completely ignores all conditional branches
>>>>> contained within P's copy of H, and it's these conditional branches
>>>>> which prevent P(P) from being an 'infinitely recursive' computation.
>>>>>
>>>>
>>>> Unless we keep the independent variable separate from the dependent
>>>> variable our analysis is incorrect.
>>>>
>>>> The independent variable is the cause. Its value is independent of
>>>> other variables in your study.
>>>>
>>>> The dependent variable is the effect. Its value depends on changes
>>>> in the independent variable.
>>>>
>>>> https://www.scribbr.com/methodology/independent-and-dependent-variables/
>>>>
>>>>
>>>> When we are asking whether or not H must abort its simulation of P
>>>> we examine the simulation of P. We must not examine the behavior of H.
>>>
>>> This is simply wrong. You can ignore the code in the *decider*, but
>>> not code inside the computation which the decider is simulating. The
>>> call to H inside P is an integral part of the computation performed
>>> by P. You can't ignore this anymore than you can ignore any other
>>> instructions inside P.
>>>
>>> André
>>>
>>
>> The independent variable is the cause. Its value is independent of
>> other variables in your study.
>>
>> The dependent variable is the effect. Its value depends on changes in
>> the independent variable.
>>
>> https://www.scribbr.com/methodology/independent-and-dependent-variables/
>
> Is there some reason you always insist on posting a webpage link to
> explain really basic concepts? In the unlikely event that someone here
> doesn't understand what this is they can always look it up themselves.
>
>>
>>
>> The question is whether or not H must abort its simulation of P.
>> The behavior of H is totally irrelevant to this question.
>>
>> As I have proven below.
>> As I have proven below.
>> As I have proven below.
>> As I have proven below.
>>
>> Of the two hypothetical possibilities: (independent variable)
>> (1) H aborts the simulation of P
>> (2) H never aborts the simulation of P
>
>> We ask what is the effect on the behavior of P? (dependent variable)
>>
>> (1) P halts
>
> You are abusing terms here. A Turing Machine halts only when it reaches
> one of its final states. If H aborts its simulation of P, then P does
> *not* halt (regardless of whether it is a halting or non-halting
> computation). Its simulation simply ends prematurely at which point H
> halts.
>
>> (2) P never halts
>>
>> It does not freaking matter at all what-ever-the-Hell H does except
>> what the behavior of P is under the above two hypothetical scenarios.
>
> P *is derived* from H, so the behaviour of P depends on the behaviour of
> H. Therefore it *does* matter what H does.
>
> Because of the above H and P are *interdependent* variables. If you want
> to treat H as an independent variable, then you must insure that P
> remains invariant, i.e. that the copy of H within P remains the same
> regardless of any changes you make to the outermost H.
>
> If the outermost H doesn't halt the simulation of P (i.e. if you disable
> the ability of the outermost H to abort simulations, and *only* the
> outermost H), then the copy of H inside P will abort the simulation of
> its input, at which point P *will* halt.
>

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
if M applied to wM halts, and

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt

The question can be rephrased this way in the hypothetical case where
the H/P is replaced with an x86 emulator and the hypothetical case where
the halt decider embedded at Ĥ.qx is replaced with a UTM

we can know that the original P(P) and Ĥ(⟨Ĥ⟩) are correctly decided as
computations that never halt by a simulating halt decider if the above
computations never halt because a simulating halt acts only as a x86
emulator / UTM until after its input demonstrates an infinitely
repeating behavior pattern.

That people have difficulty comprehending the necessary truth of this
does not count as any rebuttal what-so-ever.

>> If the result of purely hypothetical H that never aborts its
>> simulation/execution of P would result in the infinite execution of P
>> then we know axiomatically that H decides that P never halts correctly.
>>
>> Premise(1) Every computation that never halts unless its simulation is
>> aborted is a computation that never halts. This verified as true on
>> the basis of the meaning of its words.
>
> No, it isn't, because it is entirely unclear what 'its simulation'
> refers to. No claim with such sloppy wording can be 'verified as true on
> the basis of the meaning of its words.'
>
> And in actual proofs, we show things to be true based on the fact that
> they can be derived from axioms using accepted rules of inference, not
> 'on the basis of the meaning of [their] words'. What you're engaging in
> is simply word-games.
>
> André
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction is a proof ]

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Fri, 25 Jun 2021 13:53 UTC

On 6/24/2021 7:51 PM, André G. Isaak wrote:
> On 2021-06-24 15:40, olcott wrote:
>
> <snip>
>
>> The question can be rephrased this way in the hypothetical case where
>> the H/P is replaced with an x86 emulator and the hypothetical case
>> where the halt decider embedded at Ĥ.qx is replaced with a UTM
>
> If you want to test your claim that the halt decider *MUST* terminate
> its input, you change the behaviour of the halt decider so it can't
> terminate its input. (Or, better yet, you just run P(P) directly without
> making any changes at all -- your aversion to this option is truly
> mystifying).
>

It is silly that I have to repeat my words like this so that you pay
attention, you should just pay attention.

hypothetical case
hypothetical case
hypothetical case
hypothetical case

> What you *CAN'T* do is change the behaviour of the input to the decider
> which is what you are doing if you also change the H which is contained
> inside P. By doing this, you change P(P) into an entirely different
> computation.
>
> You claim that H is an independent variable and P is a dependent
> variable. If you change both P and H, then P is not an independent
> variable. P and H are interdependent variables, which means any test you
> perform is worthless.
>
>> we can know that the original P(P) and Ĥ(⟨Ĥ⟩) are correctly decided as
>> computations that never halt by a simulating halt decider if the above
>> computations never halt because a simulating halt acts only as a x86
>> emulator / UTM until after its input demonstrates an infinitely
>> repeating behavior pattern.
>>
>> That people have difficulty comprehending the necessary truth of this
>> does not count as any rebuttal what-so-ever.
>
> People comprehend things just fine. They comprehend that to test you
> claim we must change *only* the topmost H, not the copy of H embedded in P.
>
> André
>
>

In the computation P(P) if any invocation of the infinite chain of
invocations must be terminated to prevent the infinite execution of this
chain then we know that P(P) does specify an infinite chain of invocations.

I have finally gotten my words clear enough that any disagreement would
look quite foolish. On the above key point this took six months and
thousands of reviews.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction is a proof ](axiom)

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From: NoO...@NoWhere.com (olcott)
Date: Fri, 25 Jun 2021 11:33:53 -0500
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 by: olcott - Fri, 25 Jun 2021 16:33 UTC

On 6/24/2021 9:39 PM, Richard Damon wrote:
> On 6/24/21 11:42 AM, olcott wrote:
>> On 6/24/2021 6:29 AM, Richard Damon wrote:
>>> On 6/23/21 11:03 PM, olcott wrote:
>>>> If the result of purely hypothetical H that never aborts its
>>>> simulation/execution of P would result in the infinite execution of P
>>>> then we know axiomatically that H decides that P never halts correctly.
>>>
>>> Right, The P based on the never aborting H has infinite execution, but
>>> that isn't the P we are looking at,
>>
>> That hypothetical H/P proves beyond all possible doubt that the in the
>> actual H(P,P) the input to H never halts on the basis of this axiom:
>>
>> Premise(1) Every computation that never halts unless its simulation is
>> aborted is a computation that never halts. This verified as true on the
>> basis of the meaning of its words.
>>
>
> Except that that the two P's are different. You are saying that the
> Black cat is white because you are looking at the white dog instead.
>
> Remember, EVERY time you change the properties of H, you get a DIFFERENT P.

Not when the H and the P are both hypothetical.

Of the two hypothetical possibilities: (independent variable)
(1) H aborts the simulation of P
(2) H never aborts the simulation of P

We ask what is the effect on the behavior of P? (dependent variable)

(1) P halts
(2) P never halts

If the result of purely hypothetical H that never aborts its
simulation/execution of P would result in the infinite execution of P
then we know axiomatically that H decides that P never halts correctly.

Axiom(1) Every computation that never halts unless its simulation is
aborted is a computation that never halts. This verified as true on the
basis of the meaning of its words.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction is a proof ](axiom)

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](axiom)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com> <sap6l7$130c$5@gioia.aioe.org> <saqlhk$860$1@dont-email.me> <tZOdneTmHpHNnUz9nZ2dnUU7-V3NnZ2d@giganews.com> <sarcip$jgd$1@dont-email.me> <zZOdnUG2JPuxokz9nZ2dnUU7-UPNnZ2d@giganews.com> <sarhu7$f83$1@dont-email.me> <E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com> <sat46g$afu$1@dont-email.me> <4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com> <satao3$gjk$1@dont-email.me> <YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me> <48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me> <6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com> <sb0kra$ert$1@dont-email.me> <kaudnZS1vPJiaE79nZ2dnUU7-QHNnZ2d@giganews.com> <C0_AI.793830$2A5.649020@fx45.iad> <udKdnabaTsZvOkn9nZ2dnUU7-dvNnZ2d@giganews.com> <OlbBI.605613$J_5.348305@fx46.iad> <fO6dnQEYd73PmEv9nZ2dnUU7-TmdnZ2d@giganews.com> <u9oBI.267517$lyv9.157656@fx35.iad>
From: NoO...@NoWhere.com (olcott)
Date: Fri, 25 Jun 2021 13:50:54 -0500
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 by: olcott - Fri, 25 Jun 2021 18:50 UTC

On 6/25/2021 12:14 PM, Richard Damon wrote:
> On 6/25/21 12:33 PM, olcott wrote:
>> On 6/24/2021 9:39 PM, Richard Damon wrote:
>>> On 6/24/21 11:42 AM, olcott wrote:
>>>> On 6/24/2021 6:29 AM, Richard Damon wrote:
>>>>> On 6/23/21 11:03 PM, olcott wrote:
>>>>>> If the result of purely hypothetical H that never aborts its
>>>>>> simulation/execution of P would result in the infinite execution of P
>>>>>> then we know axiomatically that H decides that P never halts
>>>>>> correctly.
>>>>>
>>>>> Right, The P based on the never aborting H has infinite execution, but
>>>>> that isn't the P we are looking at,
>>>>
>>>> That hypothetical H/P proves beyond all possible doubt that the in the
>>>> actual H(P,P) the input to H never halts on the basis of this axiom:
>>>>
>>>> Premise(1) Every computation that never halts unless its simulation is
>>>> aborted is a computation that never halts. This verified as true on the
>>>> basis of the meaning of its words.
>>>>
>>>
>>> Except that that the two P's are different. You are saying that the
>>> Black cat is white because you are looking at the white dog instead.
>>>
>>> Remember, EVERY time you change the properties of H, you get a
>>> DIFFERENT P.
>>
>> Not when the H and the P are both hypothetical.
>
> WRONG. P is DEFINED based on H. If you Hypothetically create a P that
> doesn't follow that form, then you are hypothetically creating nonsense
> and can't use it to for anything logical.
>

Of every possible encoding of simulating partial halt decider H that can
possibly exist H*, if H* never aborts the simulation of its input
results in the infinite execution of the invocation of of P(P) then a
simulating halt decider H that does abort its simulation of this input
does correctly decide that this input does specify the never halting
behavior of an infinite chain of invocations.

>>
>> Of the two hypothetical possibilities: (independent variable)
>> (1) H aborts the simulation of P
>> (2) H never aborts the simulation of P
>>
>> We ask what is the effect on the behavior of P? (dependent variable)
>>
>> (1) P halts
>> (2) P never halts
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction is a proof ](axiom)

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](axiom)
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From: NoO...@NoWhere.com (olcott)
Date: Fri, 25 Jun 2021 17:56:50 -0500
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 by: olcott - Fri, 25 Jun 2021 22:56 UTC

On 6/25/2021 4:59 PM, Richard Damon wrote:
> On 6/25/21 4:40 PM, olcott wrote:
>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>> On 6/25/21 2:50 PM, olcott wrote:
>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>
>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create a P that
>>>>> doesn't follow that form, then you are hypothetically creating nonsense
>>>>> and can't use it to for anything logical.
>>>>>
>>>>
>>>> Of every possible encoding of simulating partial halt decider H that can
>>>> possibly exist  H*, if H* never aborts the simulation of its input
>>>> results in the infinite execution of the invocation of of P(P) then a
>>>> simulating halt decider H that does abort its simulation of this input
>>>> does correctly decide that this input does specify the never halting
>>>> behavior of an infinite chain of invocations.
>>>
>>> Yes, if H* is an element of the set of non-aborting deciders (Hn), P
>>> will result in infinite recursion,
>>
>> Which logically entails beyond all possible doubt that the set of
>> encodings of simulating partial halt deciders H2* that do abort the
>> simulation of the (P,P) input would correctly report that this input
>> never halts.
>
> WHY?
>

Axiom(1) Every computation that never halts unless its simulation is
aborted is a computation that never halts. This verified as true on the
basis of the meaning of its words.

> In what universe does the properties of a Dog define the properties of a
> Cat.
>
> The logic shows that Ha can correctly decide that Pn(Pn) is non-Halting.
> Since your logic NEVER looked at Pa and determined anything about its
> behavior, it makes no determination about them. You are ignoring that P
> is a function of H, and thus if you change H then you have changed P so
> this whole argument isn't valid. You CAN'T decide the properties of one
> P based on another, which means you can't argue about one H based on
> another H.
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
invocation chain)
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References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Sat, 26 Jun 2021 01:01 UTC

On 6/25/2021 7:40 PM, Richard Damon wrote:
> On 6/25/21 6:56 PM, olcott wrote:
>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>> On 6/25/21 4:40 PM, olcott wrote:
>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>
>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create a P that
>>>>>>> doesn't follow that form, then you are hypothetically creating
>>>>>>> nonsense
>>>>>>> and can't use it to for anything logical.
>>>>>>>
>>>>>>
>>>>>> Of every possible encoding of simulating partial halt decider H
>>>>>> that can
>>>>>> possibly exist  H*, if H* never aborts the simulation of its input
>>>>>> results in the infinite execution of the invocation of of P(P) then a
>>>>>> simulating halt decider H that does abort its simulation of this input
>>>>>> does correctly decide that this input does specify the never halting
>>>>>> behavior of an infinite chain of invocations.
>>>>>
>>>>> Yes, if H* is an element of the set of non-aborting deciders (Hn), P
>>>>> will result in infinite recursion,
>>>>
>>>> Which logically entails beyond all possible doubt that the set of
>>>> encodings of simulating partial halt deciders H2* that do abort the
>>>> simulation of the (P,P) input would correctly report that this input
>>>> never halts.
>>>
>>> WHY?
>>>
>>
>> Axiom(1) Every computation that never halts unless its simulation is
>> aborted is a computation that never halts. This verified as true on the
>> basis of the meaning of its words.
>
> WRONG.
>
> Your test does not match the plain meaning of the words, as has been
> explained many times.
>

Those words may be over your head, yet several others understand that
they are necessarily correct.

> Note, it is easy to show that your interpretation is wrong since even
> you admit that Linz H^, now called P by you will come to its end and
> halt when given it own representation as its input, and thus is BY
> DEFINITION a Halting Computation, Thus the H deciding it didn't need to
> abort its execution to get the wrong answer of Non-Halting.
>

Because at least one invocation of the infinite invocation chain
specified by P(P) had to be terminated to prevent the infinite execution
of this infinite invocation chain it is confirmed beyond all possible
doubt that P(P) specifies an invocation chain.

That terminating a single invocation of this infinite invocation chain
terminates the whole chain is to be expected when any infinitely
recursive chain is broken.

To a guy that truly believes that functions called in infinite recursion
must return a value to their caller this all may be way over your head.

> You FAULTY language is thus shown to be incorrect and leads to
> inconsistent logic, as you whole argument has shown.
>
> I will again repeat, that statement does NOT need to be made an Axiom,
> it can actually be formally proven with the right interpretation of the
> words, you only need to try to make is an (incorrect) axiom because you
> can't show your false version to be true any other way.
>
> Face it, your logic generates so many contradictions that it gets hard
> to find anything that actually has a real factual basis.
>
>>
>>
>>> In what universe does the properties of a Dog define the properties of a
>>> Cat.
>>>
>>> The logic shows that Ha can correctly decide that Pn(Pn) is non-Halting.
>>> Since your logic NEVER looked at Pa and determined anything about its
>>> behavior, it makes no determination about them. You are ignoring that P
>>> is a function of H, and thus if you change H then you have changed P so
>>> this whole argument isn't valid. You CAN'T decide the properties of one
>>> P based on another, which means you can't argue about one H based on
>>> another H.
>>>
>>
>>
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
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From: NoO...@NoWhere.com (olcott)
Date: Fri, 25 Jun 2021 20:46:12 -0500
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 by: olcott - Sat, 26 Jun 2021 01:46 UTC

On 6/25/2021 8:37 PM, Richard Damon wrote:
> On 6/25/21 9:01 PM, olcott wrote:
>> On 6/25/2021 7:40 PM, Richard Damon wrote:
>>> On 6/25/21 6:56 PM, olcott wrote:
>>>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>>>> On 6/25/21 4:40 PM, olcott wrote:
>>>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>>>
>>>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create a P
>>>>>>>>> that
>>>>>>>>> doesn't follow that form, then you are hypothetically creating
>>>>>>>>> nonsense
>>>>>>>>> and can't use it to for anything logical.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Of every possible encoding of simulating partial halt decider H
>>>>>>>> that can
>>>>>>>> possibly exist  H*, if H* never aborts the simulation of its input
>>>>>>>> results in the infinite execution of the invocation of of P(P)
>>>>>>>> then a
>>>>>>>> simulating halt decider H that does abort its simulation of this
>>>>>>>> input
>>>>>>>> does correctly decide that this input does specify the never halting
>>>>>>>> behavior of an infinite chain of invocations.
>>>>>>>
>>>>>>> Yes, if H* is an element of the set of non-aborting deciders (Hn), P
>>>>>>> will result in infinite recursion,
>>>>>>
>>>>>> Which logically entails beyond all possible doubt that the set of
>>>>>> encodings of simulating partial halt deciders H2* that do abort the
>>>>>> simulation of the (P,P) input would correctly report that this input
>>>>>> never halts.
>>>>>
>>>>> WHY?
>>>>>
>>>>
>>>> Axiom(1) Every computation that never halts unless its simulation is
>>>> aborted is a computation that never halts. This verified as true on the
>>>> basis of the meaning of its words.
>>>
>>> WRONG.
>>>
>>> Your test does not match the plain meaning of the words, as has been
>>> explained many times.
>>>
>>
>> Those words may be over your head, yet several others understand that
>> they are necessarily correct.
>
> I have seen NO ONE agree to your interpretation of it. The plain meaning
> is that if it can be shown that if the given instance of the simulator
> simulating a given input doesn't stop its simulation that this
> simulation will run forevr, then the machine that is being simulated can
> be corrected decided as non-Halting.
>
> An more formal way to say that is if UTM(P,I) is non-halting then it is
> correct for H(P,I) to return the non-halting result.
>
> This actually follows since UTM(P,I) will be non-halting if and only if
> P(I) is non-halting by the definition of a UTM, so that statement is
> trivially proven.
>
> Your interpretation, where even if a copy of the algorithm of H is
> included in P and that included copy needs to abort the simulation of
> the copy of the machine that it was given, can be PROVEN wrong, as even
> you have shown that P(P) in this case does Halt, thus your claimed
> correct answer is wrong by the definition of the problem.
>
> Only if you define that your answer isn't actually supposed to be the
> answer to the halting problem can you justify your answer to be correct,
> but then you proof doesn't achieve the goal you claim.
>
>>
>>> Note, it is easy to show that your interpretation is wrong since even
>>> you admit that Linz H^, now called P by you will come to its end and
>>> halt when given it own representation as its input, and thus is BY
>>> DEFINITION a Halting Computation, Thus the H deciding it didn't need to
>>> abort its execution to get the wrong answer of Non-Halting.
>>>
>>
>> Because at least one invocation of the infinite invocation chain
>> specified by P(P) had to be terminated to prevent the infinite execution
>> of this infinite invocation chain it is confirmed beyond all possible
>> doubt that P(P) specifies an invocation chain.
>
> WRONG. Given that we have an H that can answer H(P,P) because it knows
> at least enough to terminate the pattern you describe, then when we run
> P(P) then because the H within it also knows to abort this sequence
> (since it is built on the same algorithm) this P is NOT part of an
> infinite chain of execution, and thus its H can return its (wrong)
> answer to it and that P can then Halt.

P(P) specifies in infinite invocation sequence that is terminated on its
third invocation of H(P,P).

P(P) specifies in infinite invocation sequence that is terminated on its
third invocation of H(P,P).

P(P) specifies in infinite invocation sequence that is terminated on its
third invocation of H(P,P).

P(P) specifies in infinite invocation sequence that is terminated on its
third invocation of H(P,P).

P(P) specifies in infinite invocation sequence that is terminated on its
third invocation of H(P,P).

Now I have told this this several hundred times.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

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