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computers / comp.ai.philosophy / Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual agreement ]

SubjectAuthor
* Refuting the Peter Linz Halting Problem Proof V5 [ without anolcott
+- Re: Refuting the Peter Linz Halting Problem Proof V5 [ without anolcott
+* Re: Refuting the Peter Linz Halting Problem Proof V5 [ without anolcott
|+* Re: Refuting the Peter Linz Halting Problem Proof V5 [ without anolcott
||+* Re: Refuting the Peter Linz Halting Problem Proof V5 [ without anolcott
|||`- Re: Refuting the Peter Linz Halting Problem Proof V5 [ without anolcott
||`* Re: Refuting the Peter Linz Halting Problem Proof V5 [ without anolcott
|| +* Re: Refuting the Peter Linz Halting Problem Proof V5 [ without anolcott
|| |`* Re: Refuting the Peter Linz Halting Problem Proof V5 [ bald facedolcott
|| | `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ correctolcott
|| |  `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ correctolcott
|| |   `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ correctolcott
|| |    `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ correctolcott
|| |     `- Re: Refuting the Peter Linz Halting Problem Proof V5 [ correctolcott
|| `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||  `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||   `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||    `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||     `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||      +* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||      |+- Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||      |`* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||      | `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||      |  `- Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||      `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||       `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||        `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutualolcott
||         `- Re: Refuting the Peter Linz Halting Problem Proof V8olcott
|`- Re: Refuting the Peter Linz Halting Problem Proof V5 [ without anolcott
`* Re: Refuting the Peter Linz Halting Problem Proof V5 [ disingenuous ]olcott
 `* Re: Refuting the Peter Linz Halting Problem Proof V5 [ disingenuous ]olcott
  `- Re: Refuting the Peter Linz Halting Problem Proof V5 [ accurateolcott

Pages:12
Subject: Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual agreement ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 24 Mar 2022 16:09 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual
agreement ]
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 3/24/2022 10:58 AM, wij wrote:
On Thursday, 24 March 2022 at 23:35:56 UTC+8, richar...@gmail.com wrote:
...[cut]
I would challenge you to try to design a Turing Machine (not just an
'equivalent, but an ACTUAL Turing Machine) that even trys to detect that
its input is a representation of itself. (note, *A*, not *the* as
machines have MANY (infinite) representations of themselves).

That's the point. Not only PO don't have a complete H, he don't have a 'correct'
P either. At most we can see his 'invalid' C implement:
1. H does not exist. P can't exist.
2. The H(P,P) in P should be a 'description' to be a formal proof.

IMO, what PO need is basic logic, otherwise all is meaningless to him.

When you simply assume that H does not exist then it logically follows that P does not exist. When you look at the actual x86 execution trace (on pages 4-5) you see that H does correctly determine that its input never halts and does correctly report this.

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

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual agreement ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 24 Mar 2022 17:00 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
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Subject: Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual
agreement ]
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From: NoO...@NoWhere.com (olcott)
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On 3/24/2022 11:53 AM, wij wrote:
On Friday, 25 March 2022 at 00:09:19 UTC+8, olcott wrote:
On 3/24/2022 10:58 AM, wij wrote:
On Thursday, 24 March 2022 at 23:35:56 UTC+8, richar...@gmail.com wrote:
...[cut]
I would challenge you to try to design a Turing Machine (not just an
'equivalent, but an ACTUAL Turing Machine) that even trys to detect that
its input is a representation of itself. (note, *A*, not *the* as
machines have MANY (infinite) representations of themselves).

That's the point. Not only PO don't have a complete H, he don't have a 'correct'
P either. At most we can see his 'invalid' C implement:
1. H does not exist. P can't exist.
2. The H(P,P) in P should be a 'description' to be a formal proof.

IMO, what PO need is basic logic, otherwise all is meaningless to him.
When you simply assume that H does not exist then it logically follows
that P does not exist. When you look at the actual x86 execution trace
(on pages 4-5) you see that H does correctly determine that its input
never halts and does correctly report this.

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

Try replace the call 'H(x,x)' (or TM's codes) in your P with 'description', this
is required by a formal proof.

void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
}

The x86 machine code of P is a 100% complete machine description of x.
It is self-evident that the x86 emulation of the input to H cannot possibly ever reach its final state of [00000c50].

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual agreement ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 24 Mar 2022 17:30 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
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Subject: Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual
agreement ]
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 3/24/2022 12:21 PM, wij wrote:
On Friday, 25 March 2022 at 01:00:56 UTC+8, olcott wrote:
On 3/24/2022 11:53 AM, wij wrote:
On Friday, 25 March 2022 at 00:09:19 UTC+8, olcott wrote:
On 3/24/2022 10:58 AM, wij wrote:
On Thursday, 24 March 2022 at 23:35:56 UTC+8, richar...@gmail.com wrote:
...[cut]
I would challenge you to try to design a Turing Machine (not just an
'equivalent, but an ACTUAL Turing Machine) that even trys to detect that
its input is a representation of itself. (note, *A*, not *the* as
machines have MANY (infinite) representations of themselves).

That's the point. Not only PO don't have a complete H, he don't have a 'correct'
P either. At most we can see his 'invalid' C implement:
1. H does not exist. P can't exist.
2. The H(P,P) in P should be a 'description' to be a formal proof.

IMO, what PO need is basic logic, otherwise all is meaningless to him.
When you simply assume that H does not exist then it logically follows
that P does not exist. When you look at the actual x86 execution trace
(on pages 4-5) you see that H does correctly determine that its input
never halts and does correctly report this.

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

Try replace the call 'H(x,x)' (or TM's codes) in your P with 'description', this
is required by a formal proof.

void P(u32 x)
{
if (H(x, x))
HERE: goto HERE;
}
The x86 machine code of P is a 100% complete machine description of x.

I mean replace the call 'H(x,x)' with the description (x86 machine code) of H.

That is many hundreds of pages, all that we need to know is that it implements an x86 emulator that emulates one x86 instruction at a time and as soon as it detects an in infinite pattern it aborts and returns 0.

It is self-evident on pages 4-5 that H does do this correctly.


It is self-evident that the x86 emulation of the input to H cannot
possibly ever reach its final state of [00000c50].

Yes, I believe a faithful implement cannot reach [00000c50].


That proves that H(P,P) == 0 is correct.

-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer


--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual agreement ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 24 Mar 2022 18:00 UTC
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 3/24/2022 12:51 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 3/23/2022 11:21 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 3/23/2022 9:31 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 3/23/2022 8:47 PM, Ben Bacarisse wrote:

As far as I can tell, when you write H and Ĥ we must assume that these
refer to some kind of as yet unspecified machine that has "machine
addresses" (so definitely not what Linz is taking about).  And that
these H and Ĥ have the property that H applied to some string can
transition to a different state to an exact copy of H if it's embedded
in some other machine.  I suppose they don't actually have to be
literally magic to do this, but they are certainly not Turing machines.
Do you have anything to say about Linz's proof?

All halt deciders must map their input to an accept or reject state
based on the actual behavior actually specified by this input as
measured by a finite number of N steps of the correct UTM simulation
of this input.

No.  You'll find what a halt decider must do in any good text book.
It appears that after 17 years of work in this you are:

(a) trying to redefine what that halting problem is;

Someone that understands these things deeper than learned by rote
would agree with me.

No one agrees with the above.  It's not the halting problem.  What it
is, ironically, is an admission that halting is undecidable!  If you
really thought it weren't, there would be no need to keep writing the
condition for accepting an input incorrectly.

Certainly you can state this dogmatically, yet what you cannot do is
anchor this rebuttal in any sort of correct reasoning.

You are impervious to reasoning so I resorted to quoting you.  You
explicitly accept reject as the correct "mapping" for a string
representing a halting computation.  I'm not sure if there is a
reasoning that could help you here.

In computability theory, the halting problem is the problem of
determining, from a description of an arbitrary computer program and
an input, whether the program will finish running, or continue to run
forever. https://en.wikipedia.org/wiki/Halting_problem

A halt decider must map its finite string input pair to an accept or
reject state on the basis of the actual halting / non-halting behavior
specified by this finite string pair.

Exactly.  Yet you pretend you can't see that declaring reject to be the
correct mapping for a string representing a halting computation is
wrong.  Do you really think there is anything I could say that would
help you understand?

It is the case that all deciders compute the mapping from their input
finite strings to their own accept or reject state.

Silly wording, but it's not wrong.

It is the case that all deciders compute the mapping from their input
finite strings to their own accept or reject state on the basis of the
behavior specified by these finite strings.

No, but only because you missed out a work.  (A decider that accepts
those string the encode prime numbers does not care about the
"behaviour" specified by those strings.)

It is the case that the behavior specified by these finite input
strings is correctly measured by the correct UTM simulation of these
strings by the simulating halt decider.

Indeed.  A pointless thing to do but you are obsessed with simulation so
you prefer to write it that way.

It is the case that the correctly simulated input: ⟨Ĥ⟩ ⟨Ĥ⟩ to
embedded_H would never reach its final state.

That's a trick question. 

Not at all. If embedded_H would simulate its input ⟨Ĥ⟩ ⟨Ĥ⟩ as if embedded_H was a UTM then this simulated input would never reach its final state.


--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual agreement ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 25 Mar 2022 01:00 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual
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On 3/24/2022 7:24 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 3/24/2022 12:51 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

It is the case that the correctly simulated input: ⟨Ĥ⟩ ⟨Ĥ⟩ to
embedded_H would never reach its final state.

That's a trick question.

Not at all. If embedded_H would simulate its input ⟨Ĥ⟩ ⟨Ĥ⟩ as if
embedded_H was a UTM then this simulated input would never reach its
final state.

The fact that you keep using ever more vague and hand waving ways to
avoid stating the correct condition should alert everyone to what you
are doing.  The correct condition is not about what "would happen if"
things were not as they actually are.

You've been trying this on for months: your Halts function was correct
to reject a halting computation because of what "would happen if" line
15 were commented out (line 15 was the abort that stopped halts from
being a UTM).


I think that the problem is that you really want to disagree at all costs and that you understood that I am correct many months ago.
No recent rebuttals by anyone have been coherent.

Because the only fake rebuttal that you have is pointing out that I made a mistake (many months ago) on something not related to my current proof I take this as evidence that you can't find anything wrong with my current proof.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

It is the case that the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to simulating halt decider embedded_H cannot reach its final state. This conclusively proves that this simulated input fails to meet the Linz definition of halting.

computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)

This conclusively proves the if embedded_H rejects its input it is correct.

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof V5 [ mutual agreement ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 25 Mar 2022 03:23 UTC
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On 3/24/2022 9:19 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 3/24/2022 7:24 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 3/24/2022 12:51 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

It is the case that the correctly simulated input: ⟨Ĥ⟩ ⟨Ĥ⟩ to
embedded_H would never reach its final state.

That's a trick question.

Not at all. If embedded_H would simulate its input ⟨Ĥ⟩ ⟨Ĥ⟩ as if
embedded_H was a UTM then this simulated input would never reach its
final state.

The fact that you keep using ever more vague and hand waving ways to
avoid stating the correct condition should alert everyone to what you
are doing.  The correct condition is not about what "would happen if"
things were not as they actually are.
You've been trying this on for months: your Halts function was correct
to reject a halting computation because of what "would happen if" line
15 were commented out (line 15 was the abort that stopped halts from
being a UTM).

I think that the problem is that you really want to disagree at all
costs and that you understood that I am correct many months ago.  No
recent rebuttals by anyone have been coherent.
  I'll leave that for others to decide, but /you/ are not competent to
assess what I've said.  Every time you try to paraphrase my words the
result has been junk (though that might have been deliberate).

This is what I am talking about now.
I will not respond to anything else.

This is what I am talking about now.
I will not respond to anything else

This is what I am talking about now.
I will not respond to anything else

This is what I am talking about now.
I will not respond to anything else

This is what I am talking about now.
I will not respond to anything else

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

It is the case that the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to simulating halt decider embedded_H cannot reach its final state. This conclusively proves that this simulated input fails to meet the Linz definition of halting.

computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)

This conclusively proves the if embedded_H rejects its input it is correct.

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof V8
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
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On 3/25/2022 3:13 PM, Ben Bacarisse wrote:
Which Ĥ is this? 
A Simulating Halt Decider (SHD) computes the mapping from its input to its own accept or reject state based on whether or not the pure simulation of its input could reach the final state of this input in a finite number of simulated steps.

The following simplifies the syntax for the definition of the Linz Turing machine Ĥ, it is now a single machine with a single start state. A copy of Linz H is embedded at Ĥ.qx

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its final state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach its final state.

When Ĥ is applied to ⟨Ĥ⟩
   Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then embedded_H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩

Then these steps would keep repeating:
   Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
   Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
   Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H2 simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩...

The above shows that the simulated input to embedded_H never reaches its own final state whether or not its simulation is aborted.
(a) If the simulation is not aborted the above sequence never ends.
(b) If the simulation is aborted the entire chain of recursive simulations immediately stops.

In no case does the simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ ever reach its final state ⟨Ĥ⟩.qn thus never meets the Linz definition of halting:

computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234) Thus if embedded_H rejects its input it is necessarily correct.

Because all halt deciders are deciders they compute the mapping from their input finite strings inputs to their own accept or reject state. Halt deciders (because they are deciders) do not compute any mappings from non-finite string non-inputs.

No halt decider ever determines the halt status of the computation that contains its actual self thus embedded_H does not compute the mapping from Ĥ ⟨Ĥ⟩ because it is neither an input nor a finite string.

Even Linz was confused by this. embedded_H is not supposed to report on itself or the computation that it is contained within.

As long as it is verified that the simulated input to embedded_H cannot reach its final state then we know that this simulated input cannot meet the Linz definition of halting.

As long as we know that this simulated input cannot meet the Linz definition of halting we know that this input specifies a non-halting sequence of configurations.

As long as we know that this input  specifies a non-halting sequence of configurations then we know that embedded_H would be correct to reject this input.


Halting problem undecidability and infinitely nested simulation (V4)

https://www.researchgate.net/publication/359349179_Halting_problem_undecidability_and_infinitely_nested_simulation_V4




--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


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