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computers / comp.ai.philosophy / Re:_My_honest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_last_step_of_my_proof_]

SubjectAuthor
* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ keyolcott
+* Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
|`* Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
| +* Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
| |`- Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
| +- Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
| +* Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
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| |`* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
| | `* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
| |  +* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
| |  |`- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
| |  +- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
| |  `* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
| |   +* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
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| `- Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
+* Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
|`- Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
|`- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
|`- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
|+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
|`* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
| +* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
| |`* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
| | `* Re: _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
| |  +* Re: _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
| |  |+- Re: _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
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| |  ||`* Re: _My_honest_reviewers:_André,_Ben,_Mike,olcott
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| |  ||  `- Re: _My_honest_reviewers:_André,_Ben,_Mike,olcott
| |  |`- Re: _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
| |  `- Re: _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
| `- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
`- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott

Pages:123
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ self-evident truth ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Mon, 11 Apr 2022 02:53 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
self-evident truth ]
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 4/10/2022 9:47 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 10:44:25 PM UTC-4, olcott wrote:
On 4/10/2022 9:41 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 10:38:33 PM UTC-4, olcott wrote:
On 4/10/2022 9:32 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 10:27:20 PM UTC-4, olcott wrote:
On 4/10/2022 9:17 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 10:09:41 PM UTC-4, olcott wrote:
On 4/10/2022 9:06 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 10:02:56 PM UTC-4, olcott wrote:
On 4/10/2022 8:59 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 9:55:22 PM UTC-4, olcott wrote:
On 4/10/2022 8:53 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 9:43:47 PM UTC-4, olcott wrote:
On 4/10/2022 8:38 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 9:36:11 PM UTC-4, olcott wrote:
On 4/10/2022 8:30 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 9:26:57 PM UTC-4, olcott wrote:
On 4/10/2022 8:17 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 9:13:23 PM UTC-4, olcott wrote:
On 4/10/2022 8:10 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 9:07:18 PM UTC-4, olcott wrote:
On 4/10/2022 8:03 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 9:00:25 PM UTC-4, olcott wrote:
On 4/10/2022 7:44 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 8:40:34 PM UTC-4, olcott wrote:
On 4/10/2022 7:28 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 7:58:55 PM UTC-4, olcott wrote:
On 4/10/2022 6:26 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 7:20:44 PM UTC-4, olcott wrote:
On 4/10/2022 6:14 PM, André G. Isaak wrote:
On 2022-04-10 17:08, olcott wrote:
On 4/10/2022 5:59 PM, André G. Isaak wrote:
On 2022-04-10 16:40, olcott wrote:
On 4/10/2022 5:35 PM, André G. Isaak wrote:
On 2022-04-10 15:56, olcott wrote:
On 4/10/2022 4:49 PM, André G. Isaak wrote:
On 2022-04-10 15:00, olcott wrote:
On 4/10/2022 3:15 PM, olcott wrote:
On 4/10/2022 3:07 PM, André G. Isaak wrote:

I'm trying to get you to write using correct and coherent
notation. That's one of the things you'll need to be able to
do if you ever hope to publish. That involves remembering to
always include conditions and using the same terms in your
'equations' as in your text.

Not sure how that makes me a 'deceitful bastard'.

André


THAT you pretended to not know what I mean by embedded_H so
that you could artificially contrive a fake basis for rebuttal
when no actual basis for rebuttal exists makes you a deceitful
bastard.

IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or
⟨Ĥ0.qn⟩ under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is
proved to specify a non-halting sequence of configurations.

Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn

This is now the third reply you've made to the same post.

That post didn't make any arguments whatsoever about your claims.
It simply pointed out that you are misusing your notation and
urged you to correct it.


THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS
INCORRECT.

If the notation is junk, then the definition is also junk.

That's like "stipulating" that

+×yz÷² = ±z+³

It's meaningless because the notation is meaningless, much like
your notation above.

This is meaningless:

Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy // what's the condition?
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn // what's the condition?

With no conditions specified, the above is just nonsense.

André


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

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


This is still nonsense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its
own final state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
reach its own final state.

And again you're still being inconsistent. You can either use H or use
embedded_H, but you can't mix the two.

Sure I can. I just did.
This means that H pretends that it is only a UTM to see what its
simulated input would do in this case. If it would never reach its own
final state then H correctly rejects this input.

A Turing Machine cannot "pretend" to be some different Turing Machine.
It can perform a pure simulation of its input until this simulated input
matches a repeating behavior pattern that proves this input never
reaches its own final state.

If that's the case, why does an actual UTM applied to the *same* input halt?

Hint: Because the result of an actual UTM applied to the input defines the correct answer, so H answers wrong.
Intuitively that would seem to be true, this intuition is incorrect.

The ultimate definition of correct is the computation of the mapping of
the inputs to an accept or reject state on the basis of the behavior
that these inputs specify.

That simulated inputs to embedded_H would never reach their own final
state under any condition what-so-ever
IS THE ULTIMATE MEASURE OF THEIR HALTING BEHAVIOR
and conclusively proves they specify a non-halting sequence of
configurations.

Since embedded_H is the same as H (as you have yet to provide any evidence to the contrary), then the above can be applied to any simulating halt decider.

The topic is the single point that the simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
embedded_h cannot possibly reach its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩.

And that same logic gives us this:

That simulated inputs <N><5> to Ha3 would never reach their own final
state under any condition what-so-ever
IS THE ULTIMATE MEASURE OF THEIR HALTING BEHAVIOR
and conclusively proves they specify a non-halting sequence of
configurations.

THE SIMULATED INPUT CANNOT POSSIBLY REACH ITS OWN FINAL STATE
THIS SINGLE FACT BY ITSELF PROVES THAT THE INPUT IS CORRECTLY REJECTED

Try to find an error in the above.

That simulated inputs <N><5> to Ha3
are pure nonsense gibberish.

So you agree that Ha3 is correct to reject <N><5>?
It has no associated meaning

Sure it does. Ha3 is a simulating halt decider whose halt status criteria is to abort after 3 steps, and N takes <n> as input and runs for exactly n steps before halting in state N.qy. And Ha3 rejects <N><5>.

That simulated inputs <N><5> to Ha3 would never reach their own final
You are starting with a broken halt decider, my rules only apply to
correct halt deciders.

It is like you are saying that no one can possibly drive their car and
to prove this you drive your car into a tree.

So what criteria would you apply to Ha3 to determine that it's broken?
I would simply say that ridiculously stupid ideas such as Ha3 should
never be presented by someone with your top 0.04% reputation they
denigrate you. You must know how stupid this idea was before you first
mentioned it.

If it's so obvious to you that Ha3 is broken, then you should be able to explain exactly how to determine that.
Obviously a very lame attempt at a head game by a guy with your reputation.

Explain how you would determine Ha3 is broken. If you don't, I'll be forced to conclude that you agree that it's correct.
I will quit talking to you until you stop the head games.

That you have difficultly following my logic (actually, your own logic applied to a different halt decider) doesn't mean I'm playing head games.

Explain exactly how you would determine that Ha3 is broken, or admit that it is not.
You know that the question is mere denigration:
https://en.wikipedia.org/wiki/UFTP

This is how you know you're stuck, when you go completely off topic.

Explain exactly how you would determine that Ha3 is broken, or admit that it is not.
I will ask you a similar question in the same vein:

Exactly how do you know that driving a car into a tree is not a very
good way to see how a car drives, please provide every single detail.

Dishonest dodge.

If Ha3 is broken, explain exactly how you would determine that. Tell us how you would determine that the result that a simulating halt decider gives is correct.
I thought of a good way to answer this.
It does not meet the specs of a simulating halt decider.


A simulating halt decider must continue to simulate its input until it
has proof that this simulation would never end.

Since a given halt decider X has a fixed algorithm, that would mean the input would need to be passed to a different halt decider Y. So if Y showed that the input halts, would you agree that it shows that X was incorrect to report non-halting?
The given halt decider has a fixed algorithm that can be applied to any
input on its tape. The whole idea of Y seems pretty crazy.

Multiple candidate halt deciders H can exist, with each one potentially getting cases right that another might get wrong. Ha3 is one of these. The multiple halt deciders you refer to as "H" (and which Ha3 is actually a part of) are others. So if one simulating halt decider X reports a given input as non-halting, and if another simulating halt decider Y is given the same input and reports halting, would you agree that Y shows that X was incorrect because it didn't simulate for long enough?

Click here to read the complete article
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ correct halt deciding criteria ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Mon, 11 Apr 2022 04:15 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
correct halt deciding criteria ]
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<maednRn7IZ4_Cc7_nZ2dnUU7_8zNnZ2d@giganews.com> <t305f9$odp$1@dont-email.me>
<KcqdnXkO6ddoBM7_nZ2dnUU7_8xh4p2d@giganews.com> <t309pq$enf$1@dont-email.me>
Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 4/10/2022 11:11 PM, André G. Isaak wrote:
On 2022-04-10 21:01, olcott wrote:
On 4/10/2022 9:57 PM, André G. Isaak wrote:
On 2022-04-10 20:38, olcott wrote:

That is too far off topic. I have been talking circles with Ben for 17 years. We now must talk in hierarchies, cyclic paths are trimmed off of the decision tree.

Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn

And we're back to the meaningless notation again.

You are truly incapable of learning anything.

André


You can't remember the details between posts? Everyone else can.

I can remember. But that doesn't change the fact that the notation you write above is meaningless without a condition specified.

You claim you want to know how to present your ideas so you will be taken seriously. I'm trying to help with that. When someone points out an error in your notation, why insist on continuing to use the broken notation?

André

I have a single primary goal that supersedes and overrides all other goals, get mutual agreement on my current stage of progress.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its own final state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never reach its own final state.

embedded_H correctly rejects its input.

--
Copyright 2022 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 --- Version(11) [ correct halt deciding criteria ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Mon, 11 Apr 2022 20:04 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
correct halt deciding criteria ]
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On 4/10/2022 11:26 PM, André G. Isaak wrote:
On 2022-04-10 22:15, olcott wrote:
On 4/10/2022 11:11 PM, André G. Isaak wrote:
On 2022-04-10 21:01, olcott wrote:
On 4/10/2022 9:57 PM, André G. Isaak wrote:
On 2022-04-10 20:38, olcott wrote:

That is too far off topic. I have been talking circles with Ben for 17 years. We now must talk in hierarchies, cyclic paths are trimmed off of the decision tree.

Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn

And we're back to the meaningless notation again.

You are truly incapable of learning anything.

André


You can't remember the details between posts? Everyone else can.

I can remember. But that doesn't change the fact that the notation you write above is meaningless without a condition specified.

You claim you want to know how to present your ideas so you will be taken seriously. I'm trying to help with that. When someone points out an error in your notation, why insist on continuing to use the broken notation?

André

I have a single primary goal that supersedes and overrides all other goals, get mutual agreement on my current stage of progress.

I thought your goal was to eventually publish,

My goal is to be understood to be essentially correct every other goal has 100,000-fold less priority.

It is self-evident that the actual behavior of the actual simulated
input is the ULTIMATE MEASURE of the correctness of any halt decider.

By this measure embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct
and H(P,P) correctly returns false.



--
Copyright 2022 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 --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Tue, 12 Apr 2022 00:07 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ][ back door ]
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On 4/11/2022 6:55 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/10/2022 7:05 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/10/2022 4:18 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/10/2022 10:52 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/9/2022 5:54 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/9/2022 7:20 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/8/2022 4:08 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/7/2022 8:14 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/7/2022 6:37 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/7/2022 10:51 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

THIS PROVES THAT I AM CORRECT
It is the case that the correctly simulated input to embedded_H can
never possibly reach its own final state under any condition at all.
Therefore embedded_H is necessarily correct to reject its input.

Yet you won't answer two simple questions!  Why?

Because I absolutely positively will not tolerate divergence from
validating my 17 years worth of work.

But you have no choice but to tolerate it.  If someone wants to talk
about why you are wrong, they will do so.

You are wrong (for the C version of H) because H(P,P) == false but P(P)
halts.  You are wrong about your TM H because H <Ĥ> <Ĥ> transitions to
qn, but Ĥ applied to <Ĥ> is a halting computation. (Feel free to deny
any of these facts if the mood takes you.)

If you believe (against the verified facts) that the simulated ⟨Ĥ0⟩
reaches its final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩...

I believe what you've told me: that you claim that H(P,P)==false is
correct despite the fact that P(P) halts.  That's wrong.

If the input to H(P,P) cannot possibly reach its final state then this
input is correctly rejected and nothing in the universe can possibly
contradict this.

Agreed facts: (1) H(P,P) == false, (2) P(P) halts.  You don't dispute
either (indeed they come from you).
At least you don't contend these facts.

Your new line in waffle is just an attempt to distract attention from a
very simple claim: that the wrong answer is the right one.

Even Linz got this wrong because it is counter-intuitive.

A halt decider must compute the mapping from its inputs (not any damn
thing else in the universe) to its own final state on the basis of the
behavior specified by these inputs

That's not counter intuitive, it's basic.  Everyone knows this, though
it took you a while to get round to it.  A halt decider accepts or
rejects a string based on the behaviour of the computation specified by
that string.  Of course, you never got as far in my exercises as
specifying any TM that decides something on the basis of behaviour, so
you really don't know how it's actually done.  That was, I thought, the
whole point of the exercises -- to see how TMs are specified to decide
properties of computations.

You have to actually pay attention to this,

Flip, flop!  Back to being wrong about TMs rather than being wrong about
your old C junk.  These uncontested facts: (1) H(P,P) == false, (2) P(P)
halts are why your H and P are wrong.

If you are able to break the problem down to it micro component parts
and carefully analyze each of these separately instead of simply
slipping down the slide of intuition then you can see that I am
correct.

If it is true that the correct simulation input to H(P,P) cannot
possibly reach its own final state then

The input to H(P,P) is non-halting then
There is no "input to H(P,P)".

The correct simulation of the input to H
Better.  I still would not call it "input" (since these are C functions)
but you've got the hang of what am saying.  Well done.

cannot possibly ever reach it final state thus is a non-halting
sequence of configurations even if everyone and everything in the
universe disagrees.

The truth is not determined by who does or does not agree with
something.  But to find the truth of the matter you must first stop
talking literal nonsense.  The arguments to H (what you call the
"input") are two pointers.  What does simulating two pointers mean?
What you mean, I hope, is simulating calling the first pointer with the
second as it's argument.  That simulation, according to you, will halt
(or "reach it's final state" in your flamboyant, sciencey, language).
It will halt because the direct call P(P) halts.  Everything here halts
(according to you).  That's why H is wrong.

You simply are ignoring the actual execution trace that conclusively
proves that the simulated input to H cannot possibly reach its final
own state.
The traces that matter are the one of P(P) halting (you made the mistake
of posting it once), and the one of H(P,P) return false (you posted that
as well).  You a free to retract any of these at any time, but until you
do, your H is wrong by your own supplied traces.


It is never the case that the simulated input to H(P,P) ever reaches
its own final state.

Waffle.  HP(P) halts so (P,P) == false is wrong.  You can retract these
facts (since they come from you in the first place).  Until then, you've
told us that your H is wrong.


It is the case that the simulated input never reaches its [00000970] machine address, no waffle there merely an easily verified fact.

_P()
[00000956](01)  55              push ebp
[00000957](02)  8bec            mov ebp,esp
[00000959](03)  8b4508          mov eax,[ebp+08]
[0000095c](01)  50              push eax
[0000095d](03)  8b4d08          mov ecx,[ebp+08]
[00000960](01)  51              push ecx
[00000961](05)  e8c0feffff      call 00000826 // H(P,P)

// The above (as simulated input) keeps repeating until aborted.

[00000966](03)  83c408          add esp,+08
[00000969](02)  85c0            test eax,eax
[0000096b](02)  7402            jz 0000096f
[0000096d](02)  ebfe            jmp 0000096d
[0000096f](01)  5d              pop ebp
[00000970](01)  c3              ret
Size in bytes:(0027) [00000970]

It is self-evident that the actual behavior of the actual simulated
input is the ULTIMATE MEASURE of the correctness of any halt decider.


--
Copyright 2022 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 --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Tue, 12 Apr 2022 01:17 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ][ back door ]
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On 4/11/2022 8:02 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/11/2022 6:55 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/10/2022 7:05 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/10/2022 4:18 PM, Ben wrote:

The truth is not determined by who does or does not agree with
something.  But to find the truth of the matter you must first stop
talking literal nonsense.  The arguments to H (what you call the
"input") are two pointers.  What does simulating two pointers mean?
What you mean, I hope, is simulating calling the first pointer with the
second as it's argument.  That simulation, according to you, will halt
(or "reach it's final state" in your flamboyant, sciencey, language).
It will halt because the direct call P(P) halts.  Everything here halts
(according to you).  That's why H is wrong.

You simply are ignoring the actual execution trace that conclusively
proves that the simulated input to H cannot possibly reach its final
own state.
The traces that matter are the one of P(P) halting (you made the mistake
of posting it once), and the one of H(P,P) return false (you posted that
as well).  You a free to retract any of these at any time, but until you
do, your H is wrong by your own supplied traces.

It is never the case that the simulated input to H(P,P) ever reaches
its own final state.

Waffle.  HP(P) halts so (P,P) == false is wrong.  You can retract
typo: "so H(P,P) == false is wrong"
these facts (since they come from you in the first place).  Until
then, you've told us that your H is wrong.

It is the case that the simulated input never reaches its [00000970]
machine address, no waffle there merely an easily verified fact.

You can verify a thousand more irrelevant facts.  The facts that matter
are already known: that P(P) halts and that H(P,P) == false.  Are you
presenting any verified facts that corrects this mistake?  If so, just
say and I'll stop quoting it.


The sequence of configurations specified by P(P) intuitively seems like it must be identical to the correct simulation of the input to H(P,P).
It turns out that intuition is incorrect.

It is the case that the input to H(P,P) never halts and it is the case the P(P) halts because these are different computations they need not have the same behavior.

When you actually examine the actual behavior of the correctly simulated input to H(P,P) then we can see that it is an easily verified fact that this input would never halt: AKA reach machine address [00000970].

_P()
[00000956](01)  55              push ebp
[00000957](02)  8bec            mov ebp,esp
[00000959](03)  8b4508          mov eax,[ebp+08]
[0000095c](01)  50              push eax       // push P
[0000095d](03)  8b4d08          mov ecx,[ebp+08]
[00000960](01)  51              push ecx       // push P
[00000961](05)  e8c0feffff      call 00000826  // call H(P,P)

The above keeps repeating until aborted

[00000966](03)  83c408          add esp,+08
[00000969](02)  85c0            test eax,eax
[0000096b](02)  7402            jz 0000096f
[0000096d](02)  ebfe            jmp 0000096d
[0000096f](01)  5d              pop ebp
[00000970](01)  c3              ret    // final state.
Size in bytes:(0027) [00000970]




--
Copyright 2022 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 --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Tue, 12 Apr 2022 02:03 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 24
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ][ back door ]
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From: NoO...@NoWhere.com (olcott)
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On 4/11/2022 8:36 PM, Dennis Bush wrote:
On Monday, April 11, 2022 at 9:21:59 PM UTC-4, olcott wrote:
On 4/11/2022 8:17 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/11/2022 6:52 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/10/2022 7:00 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/10/2022 4:41 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

The above means this:
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
That's funny! You really have no idea what this notation means, do you?

embedded_H is a simulating halt decider that has a full UTM embedded
within it. As soon as it sees that the pure UTM simulation of its
input would never reach the final state of this input it aborts this
simulation and rejects this non-halting input.

So you had no business writing those two junk lines, did you? Or do you
really think that they are in some way compatible with that last
paragraph? Probably neither. I really think you see it much like
poetry. Meanings are supposed to be intuited from unusual, often
metaphorical, juxtapositions of symbols.

Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
Still junk.


Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its
own final state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
reach its own final state.
Still junk. Mixing embedded_H and H. Also H.qy is a final state so
this is not the hat construction form Linz. Also uses triger word
"would". What matters is what is the case, not what would be the case.
But, much like a poem, I can a feeling for what you might mean -- it's
the same old reject is correct because of what would happen if H (and
it's embedded copy) where not the TMs that actually are.
To see why you are clearly wrong, just say what state H ⟨Ĥ⟩ ⟨Ĥ⟩
transitions to, and what string must be passed to H for H to tell us
whether Ĥ applied to ⟨Ĥ⟩ halts or not.

The fact that I do not express myself perfectly does not freaking mean
that the key essence of all my ideas is not exactly correct.

Absolutely right. The reason the key essence of all my ideas is known
to be wrong is because you have, occasionally, been clear.
I addressed this in my other reply to you.

I am only talking about H(P,P) now because if someone imagines that it
does differently that it does an actual execution trace proves that they
are incorrect as a matter of objective fact with zero room for debate.

Translation:

"People are finding so many holes in my logic on Turing machines

There is an inherent hole the the logic specified by Linz
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn

The Linz text basically says "magic happens here" ⊢*
at the second wild card state transition shown above:
https://www.liarparadox.org/Linz_Proof.pdf

With H(P,P) we see exactly what the x86 emulator sees and both what H should do and why it does it.

Pages 4-5
https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation


that I don't know how to respond them all without admitting I'm wrong, so I'm going to change the way I talk about the problem in hopes I can hide my errors better."


--
Copyright 2022 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 --- Version(11) [ key missing piece in dialogue ] H(P,P)
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 13 Apr 2022 02:49 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ] H(P,P)
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From: NoO...@NoWhere.com (olcott)
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On 4/12/2022 9:29 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/12/2022 9:02 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/12/2022 7:49 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/12/2022 8:06 AM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/11/2022 8:02 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/11/2022 6:55 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/10/2022 7:05 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/10/2022 4:18 PM, Ben wrote:

The truth is not determined by who does or does not agree with
something.  But to find the truth of the matter you must first stop
talking literal nonsense.  The arguments to H (what you call the
"input") are two pointers.  What does simulating two pointers mean?
What you mean, I hope, is simulating calling the first pointer with the
second as it's argument.  That simulation, according to you, will halt
(or "reach it's final state" in your flamboyant, sciencey, language).
It will halt because the direct call P(P) halts.  Everything here halts
(according to you).  That's why H is wrong.

You simply are ignoring the actual execution trace that conclusively
proves that the simulated input to H cannot possibly reach its final
own state.
The traces that matter are the one of P(P) halting (you made the mistake
of posting it once), and the one of H(P,P) return false (you posted that
as well).  You a free to retract any of these at any time, but until you
do, your H is wrong by your own supplied traces.

It is never the case that the simulated input to H(P,P) ever reaches
its own final state.

Waffle.  HP(P) halts so (P,P) == false is wrong.  You can retract
typo: "so H(P,P) == false is wrong"
these facts (since they come from you in the first place).  Until
then, you've told us that your H is wrong.

It is the case that the simulated input never reaches its [00000970]
machine address, no waffle there merely an easily verified fact.
You can verify a thousand more irrelevant facts.  The facts that matter
are already known: that P(P) halts and that H(P,P) == false.  Are you
presenting any verified facts that corrects this mistake?  If so, just
say and I'll stop quoting it.

The sequence of configurations specified by P(P) intuitively seems
like it must be identical to the correct simulation of the input to
H(P,P).  It turns out that intuition is incorrect.

So which fact are you retracting?  That P(P) halts or that H(P,P) ==
false?

As long as the correctly simulated input to H(P,P) cannot possibly
reach the final state of this input then we know that it never halts
even if everyone in the universe disagrees.

So you plan to keep posting the same sentence in an attempt to take
the focus off the fact that H is obviously wrong?

Then you must mean
WHOOP!  WHOOP!  WHOOP!  Danger Will Robinson.
You should avoid trying to paraphrase other people.  Your replies
suggest you don't often understand the various points being put to you,
so when you try to re-word them the results are usually bogus.

that a correctly simulated input that would never reaches its own
final state is still a computation that halts.

I meant what I said.  If you are not sure that I meant, asking
well-chosen questions about it is the way to go.

We know [by definition] that a correctly simulated input that would
never reach its own final state is not a halting computation.

Deception alert: "would" rather than "does".

A simulating halt decider cannot wait for an otherwise infinite sequence of configurations to end, so "would" is correct.

 Also, delete "correctly".

Without the word "correctly" People simply stick to the view that it must be incorrect. I would use a different language for academia.

You only need an adverb in the exceptional case: "incorrectly
simulated", "partially simulated".  Also the "input" to H are it's
parameters.  The parameters are both C pointers.  What does it mean to
simulate two pointers? 

We can't actually pass a finite string to a "C" function so we have to use pointers to strings.

And "final state" is a term from Turing machines
but H and P are C functions.


I mean in in a perfectly analogous way.

I'd re-word it in a sane way, but you'd reject my re-wording (I know
because I did in a few years ago).  This is all old hat.

Are you saying that the definition of halting is incorrect?

No.  You don't know what it is, but it's not incorrect.  You could try
to say what it is, but I doubt you can without the long list of category
errors above.


I have already quoted this very many dozens of times:

Are you saying that the correctly simulated input to H(P,P) does reach
its own final state?

Simulating P(P) and calling P(P) have the same halting status.  They
both halt or neither halts.


I never bring this up because it always causes a short-circuit in your brain: The execution of P(P) is not computationally equivalent to the correct simulation of the input to H(P,P).

That is why I focus on the X proves Y

(X) We can verify that the simulated input to H(P,P) fails to meet the Linz definition of

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

(Y) When H(P,P) returns false it is correct.

I CAN'T POSSIBLE BE WRONG UNLESS THERE IS AN ESSENTIAL MISTAKE IN X OR Y.

--
Copyright 2022 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 --- Version(11) [ key missing piece in dialogue ] H(P,P)
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 13 Apr 2022 20:15 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ] H(P,P)
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From: NoO...@NoWhere.com (olcott)
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On 4/13/2022 2:38 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/13/2022 10:44 AM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/12/2022 9:37 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/12/2022 8:55 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/12/2022 7:47 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/12/2022 8:09 AM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/11/2022 8:17 PM, Ben wrote:
I am only talking about H(P,P) now because if someone imagines that it
does differently that it does an actual execution trace proves that
they are incorrect as a matter of objective fact with zero room for
debate.

That's good because you have been 100% clear about H.  It's wrong
because P(P) halts (according to you) and H(P,P) == false (according to
you).

Why do you insist that a halt decider must compute the mapping from
non-inputs: P(P) when you know that it only computes the mapping from
inputs H(P,P) ?

I don't.  H takes arguments (what you insist on calling inputs) and maps
them to a result.  The correct result is defined by people who know what
the halting problem is -- you don't get to decide.  You may never
understand the specification, but you only need to know one fact about
it: mapping the "inputs" P and P to false is wrong.

So in other words you are saying that mere opinions carry more weight
than the following verified fact:

The facts that (a) H(P,P) == false and (b) P(P) halts are not in dispute
(so far as I know).  That H mapping P and P to false is wrong is a
matter of definition.  There are no opinions being expressed here at
all.

So you agree that an input that never halts is a halting computation.
Don't be silly.  As the culmination of 17 years of work, this looks
pathetic.  Either retract one of your previously asserted facts (that
P(P) halts or that H(P,P) == false) or say that your H is not deciding
halting, but something else.


That is why I focus on the X proves Y
You don't know what a proof is, remember?  You still think that if
{A,B,C} ⊦ X then {A,B,C,~X} ⊬ X.

(X) We can verify that the simulated input to H(P,P) fails to meet the
Linz definition of computation that halts … the Turing machine will
halt whenever it enters a final state. (Linz:1990:234)
No.

(Y) When H(P,P) returns false it is correct.
No.

I CAN'T POSSIBLY BE WRONG UNLESS THERE IS AN ESSENTIAL MISTAKE IN X OR
Y.
Both.  I give more detail (if you are interested) in another reply.


Since what I said is true by logical necessity correct rebuttals are
logically impossible.

Why are you posting then?  Time to publish your halt decider H that has
H(P,P) == false and P(P) halting.  Try the JACM.  Make sure you put
those undisputed facts at the top of the abstract.

The simulated input to H(P,P) is non halting.

Then you are either (a) doing it wrong, or (b) wrong to have said that
P(P) halts.  Oh, there is a third (c) you are using poetic license, and
simulating the input means something silly.  It's literal nonsense to
there's a lot of scope for you make up some silly meaning.


When mere rhetoric goes against easily verified facts rhetoric loses:
(perhaps you don't have the slightest clue how the x86 language works)

The simulated input to H(P,P) cannot possibly reach its own final state it keeps repeating [00000956] to [00000961] until aborted.

_P()
[00000956](01)  55              push ebp
[00000957](02)  8bec            mov ebp,esp
[00000959](03)  8b4508          mov eax,[ebp+08]
[0000095c](01)  50              push eax       // push P
[0000095d](03)  8b4d08          mov ecx,[ebp+08]
[00000960](01)  51              push ecx       // push P
[00000961](05)  e8c0feffff      call 00000826  // call H(P,P)
The above keeps repeating until aborted


[00000966](03)  83c408          add esp,+08
[00000969](02)  85c0            test eax,eax
[0000096b](02)  7402            jz 0000096f
[0000096d](02)  ebfe            jmp 0000096d
[0000096f](01)  5d              pop ebp
[00000970](01)  c3              ret    // final state.
Size in bytes:(0027) [00000970]



--
Copyright 2022 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 --- Version(11) [ key missing piece in dialogue ] H(P,P)
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 14 Apr 2022 14:36 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ] H(P,P)
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<b866e22d-423b-4c52-88fb-60b924930a1an@googlegroups.com>
From: NoO...@NoWhere.com (olcott)
Followup-To: comp.theory
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On 4/14/2022 3:56 AM, Malcolm McLean wrote:
On Thursday, 14 April 2022 at 01:37:11 UTC+1, olcott wrote:
On 4/13/2022 6:02 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/13/2022 2:38 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

The simulated input to H(P,P) is non halting.

Then you are either (a) doing it wrong, or (b) wrong to have said that
P(P) halts. Oh, there is a third (c) you are using poetic license, and
simulating the input means something silly. It's literal nonsense to
there's a lot of scope for you make up some silly meaning.

When mere rhetoric goes against easily verified facts rhetoric loses:

Your own claim: H(P,P) == false is "correct" even though P(P) halts.
That's not rhetoric. You've been too clear about this attempt. You
need to try a new ruse.

Because the input to H(P,P) is non-halting then nothing in the universe
can possibly contradict the fact that it is non-halting.
The simulated input to H(P,P) cannot possibly reach its own final state
it keeps repeating [00000956] to [00000961] until aborted.

_P()
[00000956](01) 55 push ebp
[00000957](02) 8bec mov ebp,esp
[00000959](03) 8b4508 mov eax,[ebp+08]
[0000095c](01) 50 push eax // push P
[0000095d](03) 8b4d08 mov ecx,[ebp+08]
[00000960](01) 51 push ecx // push P
[00000961](05) e8c0feffff call 00000826 // call H(P,P)
The above keeps repeating until aborted


[00000966](03) 83c408 add esp,+08
[00000969](02) 85c0 test eax,eax
[0000096b](02) 7402 jz 0000096f
[0000096d](02) ebfe jmp 0000096d
[0000096f](01) 5d pop ebp
[00000970](01) c3 ret // final state.
Size in bytes:(0027) [00000970]

A simulator is a machine code program. When you examine its "execution
trace" it looks nothing like the program it is simulating. You wouldn't
know that it was a simulator or which program it was simulating, except by
a most exhaustive analysis.

So it's not clear what you have done. However, in posts many months ago, you
mentioned removing the simulator code itself from the execution traces.

Since if you use a simulating halt decider and apply the H_Hat, H_Hat
construction to it, you get a nested series of simulations of simulations, which
are eventually halted by the simulator, removing the simulator code itself from
the execution traces makes it looks as though the code is caught in an infinite
loop, when in fact it isn't.

The simulated input does not need to be caught in an infinite loop, as long as it would never reach is own final state in an unlimited number of steps of simulation it fails to meet the Linz definition:

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

thus making it a non-halting sequence of configurations.

--
Copyright 2022 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 --- Version(11) [ key missing piece in dialogue ] H(P,P)
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 14 Apr 2022 14:54 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ] H(P,P)
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From: NoO...@NoWhere.com (olcott)
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On 4/14/2022 9:30 AM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/13/2022 6:02 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/13/2022 2:38 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

The simulated input to H(P,P) is non halting.

Then you are either (a) doing it wrong, or (b) wrong to have said that
P(P) halts.  Oh, there is a third (c) you are using poetic license, and
simulating the input means something silly.  It's literal nonsense to
there's a lot of scope for you make up some silly meaning.

When mere rhetoric goes against easily verified facts rhetoric loses:

Your own claim: H(P,P) == false is "correct" even though P(P) halts.
That's not rhetoric.  You've been too clear about this attempt.  You
need to try a new ruse.

Because the input to H(P,P) is non-halting then nothing in the
universe can possibly contradict the fact that it is non-halting.

Being generous, the "input" to H in the call H(P,P) is just two
pointers.  They are neither halting nor non-halting -- they are just
pointers.

Up until, now I was prepared to take your words metaphorically, but
since you duck the key question of what "the input to H(P,P) is
non-halting" means,

Sure when I make to to explain ever details many hundreds of times damned liars will say that I never mentioned any of this.

The input to H is the only way that finite strings can be passed to a "C" function and points to the finite string of the machine code of P.

The simulating halt decider H uses an x86 emulator to simulate its input (P,P) and finds that it would never reach its own final state in an unlimited number of simulated steps.

This conclusively proves that this simulated input fails to match the Linz definition: computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)

 I will have start to take you are your word, though
I have still guessed what you mean by "input to H(P,P)".  I may have to
start replying that "H does no I/O" if you keep using the silly term
input.

What we do know, for sure, is that H(P,P) == false even though P(P)
halts.  And that's wrong.  And I can keep saying that even if you never
explain your poetic use of language.



As long as the correctly simulated input to H(P,P) would never halt then we know it is non-halting.

Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.

--
Copyright 2022 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 --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ][ back door ]
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On 4/14/2022 3:28 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:
As long as the correctly simulated input to H(P,P) would never halt
then we know it is non-halting.

Readers beware!  Ask yourselves why PO does not say the simpler "never
halts", but instead says "/would/ never halt".  Why the implied
subjunctive mood? 

A simulating halt decider must correctly predict the future behavior of of its simulated input. Many of my readers get confused and believe that an aborted simulation means that the simulated input has halted.

To an academic audience of computer scientists I would simply say that the behavior of the simulated input to H(P,P) conclusively proves that it is non-halting.

They would recognize the functional notation of H(P,P) indicating its input as is standard functional notation for all functions.

_P()
[00000956](01)  55              push ebp
[00000957](02)  8bec            mov ebp,esp
[00000959](03)  8b4508          mov eax,[ebp+08]
[0000095c](01)  50              push eax       // push P
[0000095d](03)  8b4d08          mov ecx,[ebp+08]
[00000960](01)  51              push ecx       // push P
[00000961](05)  e8c0feffff      call 00000826  // call H(P,P)
The above keeps repeating until aborted

[00000966](03)  83c408          add esp,+08
[00000969](02)  85c0            test eax,eax
[0000096b](02)  7402            jz 0000096f
[0000096d](02)  ebfe            jmp 0000096d
[0000096f](01)  5d              pop ebp
[00000970](01)  c3              ret            // final state.
Size in bytes:(0027) [00000970]


--
Copyright 2022 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 --- Version(11) [ liar by definition ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 15 Apr 2022 00:14 UTC
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liar by definition ]
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On 4/14/2022 7:02 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/14/2022 3:28 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:
As long as the correctly simulated input to H(P,P) would never halt
then we know it is non-halting.

Readers beware!  Ask yourselves why PO does not say the simpler "never
halts", but instead says "/would/ never halt".  Why the implied
subjunctive mood?

A simulating halt decider must correctly predict the future behavior
of of its simulated input.

But that's not what the implied subjunctive is about.  We know what it's
about because you've been clear in the past.  You need to justify the
wrong answer (false) for a halting computation.

As long as the correctly simulated input to H(P,P) would never halt then we know it is non-halting.

Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.


--
Copyright 2022 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 --- Version(11) [ liar by definition ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 15 Apr 2022 00:17 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
liar by definition ]
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On 4/14/2022 7:12 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/14/2022 3:54 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/14/2022 11:40 AM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/14/2022 9:30 AM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/13/2022 6:02 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/13/2022 2:38 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

The simulated input to H(P,P) is non halting.

Then you are either (a) doing it wrong, or (b) wrong to have said that
P(P) halts.  Oh, there is a third (c) you are using poetic license, and
simulating the input means something silly.  It's literal nonsense to
there's a lot of scope for you make up some silly meaning.

When mere rhetoric goes against easily verified facts rhetoric loses:

Your own claim: H(P,P) == false is "correct" even though P(P) halts.
That's not rhetoric.  You've been too clear about this attempt.  You
need to try a new ruse.

Because the input to H(P,P) is non-halting then nothing in the
universe can possibly contradict the fact that it is non-halting.

Being generous, the "input" to H in the call H(P,P) is just two
pointers.  They are neither halting nor non-halting -- they are just
pointers.

Up until, now I was prepared to take your words metaphorically, but
since you duck the key question of what "the input to H(P,P) is
non-halting" means,

Sure when I make to to explain ever details many hundreds of times
damned liars will say that I never mentioned any of this.
Just use the right terms.  H(P,P) has not input.  The call has
arguments.  They are just pointers.  Pointers are not halting nor are
they non halting.  Given that this mantra is the core of what you are
now claiming, I would have thought you would want to avoid it being
patentent nonsense.

The input to H is the only way that finite strings can be passed to a
"C" function and points to the finite string of the machine code of P.
H has no input.  Do you mean the two pointer arguments?

The simulating halt decider H uses an x86 emulator to simulate its
input (P,P) and finds that it would never reach its own final state in
an unlimited number of simulated steps.

(P,P) is too vague.  What needs to be simulated is the first pointer
being called as a function with the second as it's argument.  I.e. the
call P(P) is what should be simulated.

That the correctly simulated input to H(P,P) cannot possibly reach its
own final state proves that this input is not-halting.

Adding all of the tedious details that you suggest does not change
this fact.

If you add all the corrections, sorry, "tedious details", it contradicts
what you've said in the past.  With the errors left in place, the naive
reader won't knowing exactly what's being said -- and I think that's
deliberate.

For example, why talk about simulation at all since simulations of
computations halt or don't halt if and only if the computations do
themselves?  Well, it adds yet another puff of smoke to the mirrors
you've got in there already like what that "its" refers to (since
nothing here has a final state), or what non-halting pointers are.
"The input to H(P,P)" should mean the two pointers, P and P.  Simulating
them should mean simulating the call P(P) and the simulation "not
reaching its own final state" should mean that the simulation of P(P)
does not halt.  And that happens if, and only if, the call P(P) itself
does not halt.

I honestly have no idea if that is what you mean, but if it is, it's
wrong because P(P) halts.  You are probably just trying to cover that
up.

That a non input halts, converts the world to Fascism or opens a very
popular brothel is totally unrelated to the easily confirmed fact that
the input to H(P,P)* does not halt.

The correct value of H(P,P) is determined by the halting status of what
you call a non input:

LIAR LIAR PANTS ON FIRE

As long as the correctly simulated input to H(P,P) would never halt then we know it is non-halting.

Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.
Anyone that disagrees with this is a liar by definition.

--
Copyright 2022 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 --- Version(11) [ liar by definition ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 15 Apr 2022 00:21 UTC
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liar by definition ]
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On 4/14/2022 6:55 AM, Dennis Bush wrote:
On Wednesday, April 13, 2022 at 11:40:55 PM UTC-4, olcott wrote:
On 4/13/2022 10:33 PM, Dennis Bush wrote:
On Wednesday, April 13, 2022 at 11:27:49 PM UTC-4, olcott wrote:
On 4/13/2022 10:16 PM, Dennis Bush wrote:
On Wednesday, April 13, 2022 at 11:14:05 PM UTC-4, olcott wrote:
On 4/13/2022 10:09 PM, Dennis Bush wrote:
On Wednesday, April 13, 2022 at 11:08:02 PM UTC-4, olcott wrote:
On 4/13/2022 10:03 PM, Dennis Bush wrote:
On Wednesday, April 13, 2022 at 10:56:03 PM UTC-4, olcott wrote:
On 4/13/2022 9:50 PM, Dennis Bush wrote:
On Wednesday, April 13, 2022 at 10:48:25 PM UTC-4, olcott wrote:
On 4/13/2022 9:36 PM, Dennis Bush wrote:
On Wednesday, April 13, 2022 at 10:34:37 PM UTC-4, olcott wrote:
On 4/13/2022 9:32 PM, Dennis Bush wrote:
On Wednesday, April 13, 2022 at 10:28:54 PM UTC-4, olcott wrote:
On 4/13/2022 9:14 PM, Dennis Bush wrote:
On Wednesday, April 13, 2022 at 10:06:03 PM UTC-4, olcott wrote:
On 4/13/2022 6:45 AM, Dennis Bush wrote:
On Tuesday, April 12, 2022 at 11:35:34 PM UTC-4, olcott wrote:
On 4/12/2022 10:25 PM, Dennis Bush wrote:
On Tuesday, April 12, 2022 at 11:23:46 PM UTC-4, olcott wrote:
On 4/12/2022 10:20 PM, Dennis Bush wrote:
On Tuesday, April 12, 2022 at 11:17:52 PM UTC-4, olcott wrote:
On 4/12/2022 10:15 PM, Dennis Bush wrote:
On Tuesday, April 12, 2022 at 11:09:11 PM UTC-4, olcott wrote:
On 4/12/2022 10:05 PM, Dennis Bush wrote:
On Tuesday, April 12, 2022 at 10:59:47 PM UTC-4, olcott wrote:
On 4/12/2022 9:56 PM, Dennis Bush wrote:
On Tuesday, April 12, 2022 at 10:53:05 PM UTC-4, olcott wrote:
On 4/12/2022 9:13 PM, Dennis Bush wrote:
On Tuesday, April 12, 2022 at 10:06:02 PM UTC-4, olcott wrote:
On 4/12/2022 9:02 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/12/2022 7:49 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/12/2022 8:06 AM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/11/2022 8:02 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/11/2022 6:55 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/10/2022 7:05 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/10/2022 4:18 PM, Ben wrote:

The truth is not determined by who does or does not agree with
something. But to find the truth of the matter you must first stop
talking literal nonsense. The arguments to H (what you call the
"input") are two pointers. What does simulating two pointers mean?
What you mean, I hope, is simulating calling the first pointer with the
second as it's argument. That simulation, according to you, will halt
(or "reach it's final state" in your flamboyant, sciencey, language).
It will halt because the direct call P(P) halts. Everything here halts
(according to you). That's why H is wrong.

You simply are ignoring the actual execution trace that conclusively
proves that the simulated input to H cannot possibly reach its final
own state.
The traces that matter are the one of P(P) halting (you made the mistake
of posting it once), and the one of H(P,P) return false (you posted that
as well). You a free to retract any of these at any time, but until you
do, your H is wrong by your own supplied traces.

It is never the case that the simulated input to H(P,P) ever reaches
its own final state.

Waffle. HP(P) halts so (P,P) == false is wrong. You can retract
typo: "so H(P,P) == false is wrong"
these facts (since they come from you in the first place). Until
then, you've told us that your H is wrong.

It is the case that the simulated input never reaches its [00000970]
machine address, no waffle there merely an easily verified fact.
You can verify a thousand more irrelevant facts. The facts that matter
are already known: that P(P) halts and that H(P,P) == false. Are you
presenting any verified facts that corrects this mistake? If so, just
say and I'll stop quoting it.

The sequence of configurations specified by P(P) intuitively seems
like it must be identical to the correct simulation of the input to
H(P,P). It turns out that intuition is incorrect.

So which fact are you retracting? That P(P) halts or that H(P,P) ==
false?

As long as the correctly simulated input to H(P,P) cannot possibly
reach the final state of this input then we know that it never halts
even if everyone in the universe disagrees.

So you plan to keep posting the same sentence in an attempt to take
the focus off the fact that H is obviously wrong?

Then you must mean

WHOOP! WHOOP! WHOOP! Danger Will Robinson.

You should avoid trying to paraphrase other people. Your replies
suggest you don't often understand the various points being put to you,
so when you try to re-word them the results are usually bogus.

that a correctly simulated input that would never reaches its own
final state is still a computation that halts.

I meant what I said. If you are not sure that I meant, asking
well-chosen questions about it is the way to go.

We know [by definition] that a correctly simulated input that would
never reach its own final state is not a halting computation.

Are you saying that the definition of halting is incorrect?

Are you saying that the correctly simulated input to H(P,P) does reach
its own final state?

As a matter of fact it does.

The simulated input to H(P,P) cannot possibly reach its own final state
it keeps repeating [00000956] to [00000961] until aborted.

_P()
[00000956](01) 55 push ebp
[00000957](02) 8bec mov ebp,esp
[00000959](03) 8b4508 mov eax,[ebp+08]
[0000095c](01) 50 push eax // push P
[0000095d](03) 8b4d08 mov ecx,[ebp+08]
[00000960](01) 51 push ecx // push P
[00000961](05) e8c0feffff call 00000826 // call H(P,P)
The above keeps repeating until aborted


[00000966](03) 83c408 add esp,+08
[00000969](02) 85c0 test eax,eax
[0000096b](02) 7402 jz 0000096f
[0000096d](02) ebfe jmp 0000096d
[0000096f](01) 5d pop ebp
[00000970](01) c3 ret // final state.
Size in bytes:(0027) [00000970]

And when Hb simulates this input, it reaches a final state. Therefore H is wrong to report non-halting.
A dishonest attempt at the strawman fallacy.

Bill up the street did not rob the liquor store I know this because the
other guy that I know named Bill was watching TV at the time of the
robbery.

So in other words you have no rebuttal because you know that Hb(P,P) == true is correct and proves that H(P,P)==false is incorrect.
Hb(P,P) is off topic because it proves nothing and you know that it
proves nothing and is just a disgusting attempt at a head game.

It is not off topic because it directly contradicts your desired result. That you haven't explained why it's wrong is a confirmation of this.

That is why I focus on the X proves Y

(X) We can verify that the simulated input to H(P,P) fails to meet the
Linz definition of

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

FALSE, because an incorrect simulation is not the same as a turing machine.

(Y) When H(P,P) returns false it is correct.

FALSE, as proved by Hb(P,P) returning true for the same input.


I CAN'T POSSIBLY BE WRONG UNLESS THERE IS AN ESSENTIAL MISTAKE IN X OR Y.

If the simulated input to H(P,P) fails to meet the definition of halting, then Ha3(N,5) also fails to meet that same definition of halting. So that along shows that your criteria is bogus.
I can see that you don't want this dialogue to continue.
I will not tolerate head games.

Translation:

"I will not tolerate arguments that conclusively prove me wrong because I can't bear the though of having wasted the last 17 years".
You are trying to prove that Bill Smith is not guilty because Bill Jones
didn't do it.

Another bad analogy. Further proof that you have no rebuttal.
That H(P,P)==false is correct is true by logical necessity.

No, that Hb(P,P)==true is correct by logical necessity.
I will dumb it down for you: (Do not not believe in tautologies ?)
Because the input to H(P,P) is non-halting

Which is isn't as Hb demonstrates
Liar

I see you didn't bother to explain why my explanation is wrong.
If an X <is a> Y and Dennis disagrees then Dennis is a liar.

Still no explanation.

It is a verified fact that the input to H(P,P) is non-halting and Dennis
denies this therefore Dennis is a liar. Is it fun being a liar?

It is a verified fact that the input to H(P,P) is *halting* as follows:
Liar !

This just shows everyone reading not only that you're wrong, but that you know it.
If an X is a Y and you deny it then you are a liar.
The simulated input to H(P,P) is non-halting and you know it.
Is it fun being a liar?

You're really struggling with this aren't you?

I am not struggling with the fact that you are having fun being a liar.

Unlike everyone else I have a direct measure of your competence, so you
can't fool me about what you don't understand.

Then you should have no problem explaining EXACTLY why this is wrong:


The simulated input does reach a final state when simulated by Hb.

_P()
[00000956](01) 55 push ebp
[00000957](02) 8bec mov ebp,esp
[00000959](03) 8b4508 mov eax,[ebp+08]
[0000095c](01) 50 push eax // push P
[0000095d](03) 8b4d08 mov ecx,[ebp+08]
[00000960](01) 51 push ecx // push P
[00000961](05) e8c0feffff call 00000826 // call H(P,P)
//The above returns false
[00000966](03) 83c408 add esp,+08
[00000969](02) 85c0 test eax,eax

Click here to read the complete article
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ Why lie ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Sun, 17 Apr 2022 02:17 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
Why lie ? ]
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<3776f609-1a71-4286-89fa-067190c12134n@googlegroups.com>
Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 4/16/2022 8:57 PM, Dennis Bush wrote:
On Saturday, April 16, 2022 at 9:33:22 PM UTC-4, olcott wrote:
On 4/16/2022 8:24 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/15/2022 5:10 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/15/2022 6:39 AM, Ben wrote:

Anyway, you don't get to say what the correct answer is.

Computer science says that...
WHOOP! WHOOP! WHOOP! **wonky paraphrase alert**

On 4/15/2022 10:02 AM, olcott wrote:
Computer science says that any input to a halt decider that
would never reach its own final state is non-halting.

The above is the rest of the quote that you despicably removed.
You are such a complete Jackass.

I'll remember to keep the wonky paraphrase intact in future. I was not
accusing you of lying about what "computer science" says, just of not
understanding what it says. This is true, I'm sure, of all your "so
what you are saying is..." catastrophes. Hanlon's Razor and all that.

The key point is that because the input to H(P,P) is non-halting
H(P,P)==false is necessary correct.
If I was wrong then the correct simulation of the 27 bytes of machine
code at machine address [000009d6] by H would show some correct
execution trace from machine address [000009d6] ending at machine
address [000009f0].

_P()
[000009d6](01) 55 push ebp
[000009d7](02) 8bec mov ebp,esp
[000009d9](03) 8b4508 mov eax,[ebp+08]
[000009dc](01) 50 push eax // push P
[000009dd](03) 8b4d08 mov ecx,[ebp+08]
[000009e0](01) 51 push ecx // push P
[000009e1](05) e840feffff call 00000826 // call H
[000009e6](03) 83c408 add esp,+08
[000009e9](02) 85c0 test eax,eax
[000009eb](02) 7402 jz 000009ef
[000009ed](02) ebfe jmp 000009ed
[000009ef](01) 5d pop ebp
[000009f0](01) c3 ret // Final state
Size in bytes:(0027) [000009f0]

That everyone refuses this challenge proves that it is beyond their
technical capacity or that they are liars.
Halting problem undecidability and infinitely nested simulation (V5)

https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5

You claim that because Ha(Pa,Pa) cannot simulate its input to a final state then it is correct to return false.

By the same logic, because Ha3(N,5) cannot simulate its input to a final state then it is correct to return false.

Agreed?

I agreed that you dodged my challenge.
I will stop responding until you meet my challenge.

--
Copyright 2022 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 --- Version(11) [ liar by definition ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Mon, 18 Apr 2022 00:00 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
liar by definition ]
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Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math
References: <v6idnaCJifSVTtT_nZ2dnUU7_8zNnZ2d@giganews.com>
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On 4/17/2022 6:43 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/17/2022 10:20 AM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/16/2022 8:08 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/15/2022 5:03 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/15/2022 6:48 AM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/14/2022 7:12 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

That a non input halts, converts the world to Fascism or opens a very
popular brothel is totally unrelated to the easily confirmed fact that
the input to H(P,P)* does not halt.

The correct value of H(P,P) is determined by the halting status of what
you call a non input:

LIAR LIAR PANTS ON FIRE

No.  The correct value of H(P,P) is not for you to choose.

I have proven that the correctly simulated input to H(P,P) never halts
therefore H would be necessarily correct to report this?

H(M,I) should be false if and only if M(I) does not halt.

The fact that the input H(P,P) is non-halting completely refutes
anything and everyone in the universe that says otherwise.

It's a mantra now.

If I was wrong then the correct simulation of the 27 bytes of machine
code...
If you were right you'd publish the code.

This is the only relevant code to the question does the correctly
simulated input to H(P,P) halt?

Clearly not.  The code for P is not in doubt.


So then you must agree that when H correctly simulates the input to H(P,P) that it would never reach its own final state.

We can determine the correct simulation of this input entirely on the basis of the following machine code and the definition of the x86 language.

_P()
[000009d6](01) 55         push ebp
[000009d7](02) 8bec       mov ebp,esp
[000009d9](03) 8b4508     mov eax,[ebp+08]
[000009dc](01) 50         push eax         // push P
[000009dd](03) 8b4d08     mov ecx,[ebp+08]
[000009e0](01) 51         push ecx         // push P
[000009e1](05) e840feffff call 00000826    // call H
[000009e6](03) 83c408     add esp,+08
[000009e9](02) 85c0       test eax,eax
[000009eb](02) 7402       jz 000009ef
[000009ed](02) ebfe       jmp 000009ed
[000009ef](01) 5d         pop ebp
[000009f0](01) c3         ret              // Final state
Size in bytes:(0027) [000009f0]

Begin Local Halt Decider Simulation
....[000009d6][00211368][0021136c] 55         push ebp         // enter P
....[000009d7][00211368][0021136c] 8bec       mov ebp,esp
....[000009d9][00211368][0021136c] 8b4508     mov eax,[ebp+08]
....[000009dc][00211364][000009d6] 50         push eax         // Push P
....[000009dd][00211364][000009d6] 8b4d08     mov ecx,[ebp+08]
....[000009e0][00211360][000009d6] 51         push ecx         // Push P
....[000009e1][0021135c][000009e6] e840feffff call 00000826    // Call H
....[000009d6][0025bd90][0025bd94] 55         push ebp         // enter P
....[000009d7][0025bd90][0025bd94] 8bec       mov ebp,esp
....[000009d9][0025bd90][0025bd94] 8b4508     mov eax,[ebp+08]
....[000009dc][0025bd8c][000009d6] 50         push eax         // Push P
....[000009dd][0025bd8c][000009d6] 8b4d08     mov ecx,[ebp+08]
....[000009e0][0025bd88][000009d6] 51         push ecx         // Push P
....[000009e1][0025bd84][000009e6] e840feffff call 00000826    // Call H


--
Copyright 2022 Pete Olcott

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


Subject: Re:_My_Dishonest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_continue_to_lie_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 20 Apr 2022 17:35 UTC
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Subject: Re:_My_Dishonest_reviewers:_André,_Ben,_Mike
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From: NoO...@NoWhere.com (olcott)
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On 4/19/2022 9:49 AM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/18/2022 4:53 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/17/2022 6:43 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/17/2022 10:20 AM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

If I was wrong then the correct simulation of the 27 bytes of machine
code...

If you were right you'd publish the code.

This is the only relevant code to the question does the correctly
simulated input to H(P,P) halt?

Clearly not.  The code for P is not in doubt.

So then you must agree that when H correctly simulates the input to
H(P,P) that it would never reach its own final state.
Deflection.  As I said, if you were right you'd publish the code.  You
pointlessly showed P which is not in doubt.  I can only assume you know
that if you publish H the game will be well and truly up.  You need to
trot out one of your excuses for keeping the flawed code secret.

I am going to continue to present this same point until everyone
reading this forum realizes that the only reason that you dodge it is
because you already know that it is correct.

Everyone here has seen me address it many times and has seen you ignore
what I've said about it many times.  Do you really think anyone will
change their option of me just because you keep typing the same vague
sentence?  And, more to the point, why do you care what people think of
me?

What you need is someone you trust to tell you to do something useful
with your time, not some way to get "everyone reading this forum" to
form your preferred opinion of me.  I am not that important.

When the input to H(P,P) is non halting then it is necessarily correct
for H to report that the input to H(P,P) is non halting.

On more time: H(P,P) has no input.

According to functional notation the inputs to a function are its parameters.


Function notation is a way to write functions that is easy to read and understand. Functions have dependent and independent variables, and when we use function notation the independent variable is commonly x, and the dependent variable is F(x).
https://www.brightstorm.com/math/algebra/graphs-and-functions/function-notation/#:~:text=Function%20notation%20is%20a%20way,variable%20is%20F(x

When we construe H(P,P) as a computable function then H computes the mapping from its inputs/parameters to its own accept of reject state, thus you "rebuttal" is merely double-talk misdirection.

 If you mean the two pointer
parameters, say so.  That "input" -- those two pointers -- are neither
halting nor non halting.  Maybe you mean calling the first with the
second as its only argument is non halting?

The finite string of machine code pointed to by P.

 If you do, you are
(according to other posts of yours) wrong.  Say what you mean, or accept
that honest commentators will have to explain your words back to you.


I expressed it accurately in the functional notation which specifies that the parameters are the inputs to a function and the way that finite strings are passed to any computable function must be something like the address of some memory location. In the RASP, RAM and x86 models this is simply a memory address.

So although you incorrectly nitpick at my terminology you continue to dodge the key point.

A halt decider computable function computes the mapping from its finite string parameters to its own accept or reject state based on the actual behavior specified by these parameters.
Example: The correctly simulated input to H(P,P) by H.

You continue to insist that
A halt decider computable function computes the mapping from non finite string non-parameters to its own accept or reject state based on the actual behavior specified by these non finite string non-parameters.
Example: The behavior of P(P).

At least you've removed the "would" and you have stopped using the
unclear "it" and "its".



--
Copyright 2022 Pete Olcott

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


Subject: Re:_My_Dishonest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_continue_to_lie_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
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On 4/20/2022 3:24 PM, Ben wrote:
Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:

You say that P(P) halts, but the correctly simulated input to H(P,P)
does not halt. You've been pretty consistent on this point.  But I'm
still mystified. I've made a suggestion that might resolve the obvious
contradiction, but it hasn't been accepted.

PO has reached the stage where his only option is to be unclear.

THAT YOU SAY THIS IS UNCLEAR IS A FALSEHOOD IT HAS ZERO AMBIGUITY:
The technical computer science term "halt" means that a program will reach its last instruction technically called its final state. For P this would be its machine address [000009f0].

The function named H continues to simulate its input using an x86 emulator until this input either halts on its own or H detects that it would never halt. If its input halts H returns 1. If H detects that its input would never halt H returns 0.

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

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

_P()
[000009d6](01) 55         push ebp
[000009d7](02) 8bec       mov ebp,esp
[000009d9](03) 8b4508     mov eax,[ebp+08]
[000009dc](01) 50         push eax         // push P
[000009dd](03) 8b4d08     mov ecx,[ebp+08]
[000009e0](01) 51         push ecx         // push P
[000009e1](05) e840feffff call 00000826    // call H
[000009e6](03) 83c408     add esp,+08
[000009e9](02) 85c0       test eax,eax
[000009eb](02) 7402       jz 000009ef
[000009ed](02) ebfe       jmp 000009ed
[000009ef](01) 5d         pop ebp
[000009f0](01) c3         ret              // Final state
Size in bytes:(0027) [000009f0]

The simulated input to H(P,P) cannot possibly reach its own final state of [000009f0] it keeps repeating [000009d6] to [000009e1] until aborted.

Begin Local Halt Decider Simulation
....[000009d6][00211368][0021136c] 55         push ebp         // enter P
....[000009d7][00211368][0021136c] 8bec       mov ebp,esp
....[000009d9][00211368][0021136c] 8b4508     mov eax,[ebp+08]
....[000009dc][00211364][000009d6] 50         push eax         // Push P
....[000009dd][00211364][000009d6] 8b4d08     mov ecx,[ebp+08]
....[000009e0][00211360][000009d6] 51         push ecx         // Push P
....[000009e1][0021135c][000009e6] e840feffff call 00000826    // Call H
....[000009d6][0025bd90][0025bd94] 55         push ebp         // enter P
....[000009d7][0025bd90][0025bd94] 8bec       mov ebp,esp
....[000009d9][0025bd90][0025bd94] 8b4508     mov eax,[ebp+08]
....[000009dc][0025bd8c][000009d6] 50         push eax         // Push P
....[000009dd][0025bd8c][000009d6] 8b4d08     mov ecx,[ebp+08]
....[000009e0][0025bd88][000009d6] 51         push ecx         // Push P
....[000009e1][0025bd84][000009e6] e840feffff call 00000826    // Call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

Because the correctly simulated input to H(P,P) cannot possibly reach its own final state at [000009f0] it is necessarily correct for H to reject this input as non-halting.

The above verifies the fact that the input to H(P,P) is non-halting.
Anyone insisting on disagreeing with verified facts that they know are verified facts is a liar by definition.


--
Copyright 2022 Pete Olcott

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


Subject: Re:_My_Dishonest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_continue_to_lie_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 20 Apr 2022 20:43 UTC
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Subject: Re:_My_Dishonest_reviewers:_André,_Ben,_Mike
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From: NoO...@NoWhere.com (olcott)
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On 4/20/2022 3:33 PM, Malcolm McLean wrote:
On Wednesday, 20 April 2022 at 20:49:35 UTC+1, olcott wrote:
On 4/20/2022 2:44 PM, olcott wrote:
On 4/20/2022 2:30 PM, Malcolm McLean wrote:
On Wednesday, 20 April 2022 at 19:10:37 UTC+1, olcott wrote:
On 4/20/2022 12:51 PM, Malcolm McLean wrote:
On Wednesday, 20 April 2022 at 18:35:43 UTC+1, olcott wrote:
On 4/19/2022 9:49 AM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/18/2022 4:53 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/17/2022 6:43 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/17/2022 10:20 AM, Ben wrote:
olcott <No...@NoWhere.com> writes:

If I was wrong then the correct simulation of the 27 bytes
of machine
code...

If you were right you'd publish the code.

This is the only relevant code to the question does the
correctly
simulated input to H(P,P) halt?

Clearly not. The code for P is not in doubt.

So then you must agree that when H correctly simulates the
input to
H(P,P) that it would never reach its own final state.
Deflection. As I said, if you were right you'd publish the code.
You
pointlessly showed P which is not in doubt. I can only assume
you know
that if you publish H the game will be well and truly up. You
need to
trot out one of your excuses for keeping the flawed code secret.

I am going to continue to present this same point until everyone
reading this forum realizes that the only reason that you dodge
it is
because you already know that it is correct.

Everyone here has seen me address it many times and has seen you
ignore
what I've said about it many times. Do you really think anyone will
change their option of me just because you keep typing the same vague
sentence? And, more to the point, why do you care what people
think of
me?

What you need is someone you trust to tell you to do something useful
with your time, not some way to get "everyone reading this forum" to
form your preferred opinion of me. I am not that important.

When the input to H(P,P) is non halting then it is necessarily
correct
for H to report that the input to H(P,P) is non halting.

On more time: H(P,P) has no input.
According to functional notation the inputs to a function are its
parameters.


Function notation is a way to write functions that is easy to read and
understand. Functions have dependent and independent variables, and
when
we use function notation the independent variable is commonly x,
and the
dependent variable is F(x).
https://www.brightstorm.com/math/algebra/graphs-and-functions/function-notation/#:~:text=Function%20notation%20is%20a%20way,variable%20is%20F(x


When we construe H(P,P) as a computable function then H computes the
mapping from its inputs/parameters to its own accept of reject state,
thus you "rebuttal" is merely double-talk misdirection.
If you mean the two pointer
parameters, say so. That "input" -- those two pointers -- are neither
halting nor non halting. Maybe you mean calling the first with the
second as its only argument is non halting?
The finite string of machine code pointed to by P.

Machine code is tree-like in structure. I particular, in your
system, the "string of
machine code pointed to by P" contains a call to H.
Now are you excluding that code in H from "the input to H(P,P)?".
The only relevant point is that the correctly simulated P simulated by H
cannot possibly reach its own final state under any condition
what-so-ever thus conclusively fails to meet the Linz criteria of a
halting computation thus is definitively determined to be a non-halting
sequence of configurations.

If {an X is a Y} "the input to H(P,P) is non-halting"

then it is necessarily correct for Z to report that {an X is a Y} "H
reports that its input is non-halting"

You say that P(P) halts, but the correctly simulated input to H(P,P) does
not halt. You've been pretty consistent on this point.


But I'm still mystified. I've made a suggestion that might resolve the
obvious
contradiction, but it hasn't been accepted.

It is an empirically proved verified fact that the input to H(P,P) does
not halt so anyone and anything disagreeing is necessarily incorrect.

If I smash a Boston cream pie in someones face and they have a
fundamental religious belief that there is no such thing as pies the pie
dripping from their face conclusively proves that they are incorrect.

The verified fact that the input to H(P,P) is non-halting is the pie
dripping from the face.

You have never published H, so we can't verify anything.

This right here provides complete proof that the input to H(P,P) is non-halting. If you are clueless about the x86 language that does not mean that it is not complete proof.

The technical computer science term "halt" means that a program will reach its last instruction technically called its final state. For P this would be its machine address [000009f0].

The function named H continues to simulate its input using an x86 emulator until this input either halts on its own or H detects that it would never halt. If its input halts H returns 1. If H detects that its input would never halt H returns 0.

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

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

_P()
[000009d6](01) 55         push ebp
[000009d7](02) 8bec       mov ebp,esp
[000009d9](03) 8b4508     mov eax,[ebp+08]
[000009dc](01) 50         push eax         // push P
[000009dd](03) 8b4d08     mov ecx,[ebp+08]
[000009e0](01) 51         push ecx         // push P
[000009e1](05) e840feffff call 00000826    // call H
[000009e6](03) 83c408     add esp,+08
[000009e9](02) 85c0       test eax,eax
[000009eb](02) 7402       jz 000009ef
[000009ed](02) ebfe       jmp 000009ed
[000009ef](01) 5d         pop ebp
[000009f0](01) c3         ret              // Final state
Size in bytes:(0027) [000009f0]

The simulated input to H(P,P) cannot possibly reach its own final state of [000009f0] it keeps repeating [000009d6] to [000009e1] until aborted.

Begin Local Halt Decider Simulation
....[000009d6][00211368][0021136c] 55         push ebp         // enter P
....[000009d7][00211368][0021136c] 8bec       mov ebp,esp
....[000009d9][00211368][0021136c] 8b4508     mov eax,[ebp+08]
....[000009dc][00211364][000009d6] 50         push eax         // Push P
....[000009dd][00211364][000009d6] 8b4d08     mov ecx,[ebp+08]
....[000009e0][00211360][000009d6] 51         push ecx         // Push P
....[000009e1][0021135c][000009e6] e840feffff call 00000826    // Call H
....[000009d6][0025bd90][0025bd94] 55         push ebp         // enter P
....[000009d7][0025bd90][0025bd94] 8bec       mov ebp,esp
....[000009d9][0025bd90][0025bd94] 8b4508     mov eax,[ebp+08]
....[000009dc][0025bd8c][000009d6] 50         push eax         // Push P
....[000009dd][0025bd8c][000009d6] 8b4d08     mov ecx,[ebp+08]
....[000009e0][0025bd88][000009d6] 51         push ecx         // Push P
....[000009e1][0025bd84][000009e6] e840feffff call 00000826    // Call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

Because the correctly simulated input to H(P,P) cannot possibly reach its own final state at [000009f0] it is necessarily correct for H to reject this input as non-halting.

I do believe you
when you say that P(P) halts and H(P,P) reports "non-halting". The question
which everyone is asking is how you get from that fact to the claim that
"The input to H(P,P) is non-halting".

As I said, I suspect that the reason is confusion, and the root of the confusion
is that you are using an x86 emulator to make statements about Turing machines.
Using an x86 emulator instead a Turing machine isn't of itself fatal to your
project, but it opens the way for all sorts of errors that using Turing machines
themselves would avoid. So really I urge you to learn more about Turing
machines.

The exact same reasoning applies to
embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would reach its own final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

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

Click here to read the complete article
Subject: Re:_My_Dishonest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_continue_to_lie_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 21 Apr 2022 19:45 UTC
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From: NoO...@NoWhere.com (olcott)
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On 4/21/2022 2:22 PM, olcott wrote:
On 4/21/2022 2:10 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/21/2022 1:08 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/20/2022 7:17 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/20/2022 4:46 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/20/2022 3:24 PM, Ben wrote:
Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:

You say that P(P) halts, but the correctly simulated input to H(P,P)
does not halt. You've been pretty consistent on this point.  But I'm
still mystified. I've made a suggestion that might resolve the obvious
contradiction, but it hasn't been accepted.
PO has reached the stage where his only option is to be unclear.

THAT YOU SAY THIS IS UNCLEAR IS A FALSEHOOD IT HAS ZERO AMBIGUITY:
Of course it's ambiguous.  Some people have determined (by guessing)
that you mean that P(P) is non halting.  But you reject that and anyway
you tell us that P(P) halts so that can't be what you mean.  I prefer
not to guess because I think you are deliberately not saying what you
mean by simulating two pointers.  Go on, explain it -- what does it mean
to simulate the two pointers P and P?

The simulated input to H(P,P) never reaches its last instruction at
machine address [000009f0].
So you won't say?  That figures.  Not saying what you mean is all you
have left...

I have already fully proved my point to everyone that is very familiar
with the interface between C and x86 which is apparently no one here
besides me.

So what now?  Publish?  Fame?  Fortune?  Or keep posting here to chat
with people you think are liars?  Let me guess...

That everyone here claims that I am wrong knowing full well that they
have no idea what I am saying because they don't know the first thing
about the x86 language is still a lie.

It seems you won't address even a simple question put to you like "what
now?".  I know you think everyone here is ignorant and deliberately
lying, but since that is your view I wonder what your plan is.


Continue to refine my words until one or more qualified reviewers accept that I am correct.

HERE IS THE INTRO:
To fully understand this paper a software engineer must be an expert in the C programming language, the x86 programming language, exactly how C translates into x86 and what an x86 process emulator is. No knowledge of the halting problem is required.

Unbelievably it appears to be to continue to chat away with this bunch
of ignorant liars.  How do you think that will help other than satisfy
your apparent desire which is to be chat about your "discoveries"?


I was aiming for the point where my words become so clear that even my lying reviewers won't lie because it will be far too obvious that they are lying making their lie unbearably denigrating to themselves.

You pretend to want to publish and be famous, but even you should be
able to see that talking to a bunch of ignorant lairs isn't going to
help you.  In fact, it makes your supposed intent seem less that honest.


The fact that it has helped tremendously proves that it was always a good idea.


If we merely eliminate all of the reviews of my work on comp.theory my current paper or anything remotely as good would have been infeasible.

Most of the reviewers did initially have a much better understanding of the halting problem than I had. In the last month or two places have switched.

A halt decider computes the mapping from its input finite strings to its own accept or reject state on the basis of the actual behavior of its correctly simulated input.

--
Copyright 2022 Pete Olcott

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


Subject: Re:_My_Dishonest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_continue_to_lie_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 21 Apr 2022 23:32 UTC
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Subject: Re:_My_Dishonest_reviewers:_André,_Ben,_Mike
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 4/21/2022 6:10 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/21/2022 4:33 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

Read my addendum.
You haven't cited anyone without their permission have you?

THIS IS THE ADDENDUM

On 4/21/2022 2:45 PM, olcott wrote:
If we merely eliminate all of the reviews of my work on comp.theory my
current paper or anything remotely as good would have been infeasible.

Most of the reviewers did initially have a much better understanding of
the halting problem than I had. In the last month or two places have
switched.

A halt decider computes the mapping from its input finite strings to its
own accept or reject state on the basis of the actual behavior of its
correctly simulated input.

Unfortunate that you boast like that, but it's to be expected.  At least
you don't name anyone as having helped you.

Still, I expect there's a few years more chatting to be done before you
get to an editor's day.


People here have gotten the conventional halting problem dogma so ingrained in their psyche that they believe that even when it is proven to be a verified fact that the input to H(P,P) specifies a non-halting sequence of configurations it is somehow incorrect for H to report this.



--
Copyright 2022 Pete Olcott

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


Subject: Re:_My_Dishonest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_last_step_of_my_proof_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 22 Apr 2022 00:53 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:_My_Dishonest_reviewers:_André,_Ben,_Mike
,_Dennis,_Richard_[_last_step_of_my_proof_]
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 4/21/2022 7:12 PM, Malcolm McLean wrote:
On Friday, 22 April 2022 at 00:19:50 UTC+1, olcott wrote:
On 4/21/2022 6:10 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/21/2022 4:33 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

Read my addendum.
You haven't cited anyone without their permission have you?

THIS IS THE ADDENDUM

On 4/21/2022 2:45 PM, olcott wrote:
If we merely eliminate all of the reviews of my work on comp.theory my
current paper or anything remotely as good would have been infeasible.

Most of the reviewers did initially have a much better understanding of
the halting problem than I had. In the last month or two places have
switched.

A halt decider computes the mapping from its input finite strings to its
own accept or reject state on the basis of the actual behavior of its
correctly simulated input.

Unfortunate that you boast like that, but it's to be expected. At least
you don't name anyone as having helped you.

Still, I expect there's a few years more chatting to be done before you
get to an editor's day.

I am writing it up so that software engineers can validate it without
any knowledge of the halting problem. I translate the key computer
science terminology into software engineering terms.
To fully understand this paper a software engineer must be an expert in
the C programming language, the x86 programming language, exactly how C
translates into x86 and what an x86 process emulator is. No knowledge of
the halting problem is required.
The computer science term “halting” means that a Turing Machine
terminated normally reaching its last instruction known as its “final
state”. This is the same idea as when a function returns to its caller
as opposed to and contrast with getting stuck in an infinite loop or
infinite recursion.

The halting problem applies to x86 programs as well as to Turing machines,
so it's not inherently an error to work with x86 code. However it can cause
confusion, and in fact it is causing confusion. For instance, if you make
H a simulating halt decider, there is no infinite recursion and no simple
infinite loop in Linz's H_Hat<H_Hat> system . What you get instead is nested
emulation, which may or may not be infinite. Since a Turing machine has no
subroutines, if you stick to Turing machines, you are not in danger of making
this mistake.


HERE IS THE ACTUAL PROBLEM:
People here have gotten the conventional halting problem dogma so ingrained in their psyche that they believe that even when it is proven to be a verified fact that the input to H(P,P) specifies a non-halting sequence of configurations it is somehow incorrect for H to report this.

AND

No one here knows the x86 language well enough to be able to verify the fact that the input to H(P,P) really does specify a non-halting sequence of configurations and in their ignorance they simply assume that I am wrong about this.


--
Copyright 2022 Pete Olcott

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


Subject: Re:_My_honest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_last_step_of_my_proof_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 22 Apr 2022 14:52 UTC
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Subject: Re:_My_honest_reviewers:_André,_Ben,_Mike,
_Dennis,_Richard_[_last_step_of_my_proof_]
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From: NoO...@NoWhere.com (olcott)
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On 4/22/2022 2:05 AM, Malcolm McLean wrote:
On Friday, 22 April 2022 at 02:33:17 UTC+1, olcott wrote:
On 4/21/2022 8:23 PM, Ben wrote:
olcott <No...@NoWhere.com> writes:

On 4/21/2022 6:10 PM, Ben wrote:

Still, I expect there's a few years more chatting to be done before you
get to an editor's day.

People here have gotten the conventional halting problem dogma so
ingrained in their psyche that they believe that even when it is
proven to be a verified fact that the input to H(P,P) specifies a
non-halting sequence of configurations it is somehow incorrect for H
to report this.

H(P,P) == false is wrong because P(P) halts. The problem of exhibiting
an H such that H(X,Y) == false if, and only if, X(Y) does not halt has
not gone away. It's called the halting problem. There are famous
theorems about it. At one time you were interested in addressing it.

If the correctly simulated input to H(P,P) specifies a non-halting
sequence of configurations on the basis of its actual behavior then this
actual behavior supersedes and overrules anything any everything else
that disagrees.

You keep maintaining the false assumption that the simulated input to
H(P,P) must be computationally equivalent to the direct execution of
P(P). It is easy to see that they specify a semantically different
sequence of configurations.

With H(P,P) H is called before P(P) is simulated.
With P(P) P is called before H(P,P) is invoked.

In x86 assembly, it is indeed not too difficult to arrange that P(P) has
one behaviour when called directly, and another when simulated.
However that doesn't seem to be your problem. If it was, then P(P)
and H(P,P) would be consistent, and it would look, superficially, as though
you had found a counter-example to Linz's proof. In reality, of course,
some global flag would be buried away in the x86 code, maybe very
cleverly disguised.

P(P) halts, H(P,P) reports "non-halting". So everyone says "what is there to
see, clearly H get P(P) wrong, which is exactly what Linz said must happen?".
The answer is the execution trace. When you look at that, it does indeed
look as though the program might go on forever, and H is right to describe it
as "non-halting".
However the execution trace misses the call to H.

It misses nothing, 264 pages of extraneous detail that says nothing more than the fact that H does emulate its input one instruction at a time with an x86 emulator until it recognizes the same infinitely repeating pattern that we can all see.

People here that know the x86 language (only me) can see that the execution trace that H provides of P is correct on the basis of the x86 source code for P.

Now plenty of people,
without ever seeing the source for H, have diagnosed what the problem
must be. Somewhere in the call to H, there is a test which breaks the
series of nested simulations.


Aborting an input does not help this now dead input to continue to execute until it reaches its own final state. Richard could never begin to understand this so I finally totally gave up on him.

The point is not whether or not the simulated P stops running the point is that it never halts even if it does stop running. It has to reach its own final state to halt. Infinite recursion prevents this.

P(P) specifies the first invocation of what would otherwise be infinite recursion and H(P,P) terminates this otherwise infinite recursion at its second call.

H(P,P) the infinite recursion is not hidden from H in this case so it terminates it at its first call.

They're almost certainly right.

If the actual behavior of the correctly simulated input to H(P,P) specifies a non-halting sequence of configurations

then this actual behavior supersedes and overrules anything and everything else that disagrees.

It does not matter at all that P(P) halts when we have proven that the input to H(P,P) specifies a non-halting sequence of configurations. The actual behavior of the actual input is the ultimate measure of its behavior.

It is necessary to examine H to be absolutely
sure. You say that's too much code to reasonably expect anyone to analyse,

(a) H continues to emulate its input with an x86 emulator until it sees the same infinitely repeating pattern that we can all see.

(b) We can verify this fact by simply looking at the x86 source code for P and the execution trace of P that H produces.

(c) 264 pages of additional execution trace would not help to see this, it would only provide enormous clutter that must be sifted through.

and you might well be right on that. So where you go is up to you. I urge to
to learn how to write Turing machines. If you are interested in Turing machines,
it's an essential skill.


--
Copyright 2022 Pete Olcott

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


Subject: Re:_My_honest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_last_step_of_my_proof_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Fri, 22 Apr 2022 19:57 UTC
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Subject: Re:_My_honest_reviewers:_André,_Ben,_Mike,
_Dennis,_Richard_[_last_step_of_my_proof_]
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From: NoO...@NoWhere.com (olcott)
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On 4/22/2022 1:52 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

This is clearer:
With H(P,P) H is invoked before P(P) is simulated.
With P(P) P is invoked before H(P,P) is invoked.

Because H and P are invoked in a different order this changes their
behavior.

Everyone has been quite sure that this is the trick you have been trying
to pull ever since you flatly denied it to Mike Terry more than three
years ago.  What's puzzling is why you don't use this trick to have H
return the correct answer!



It is a verified fact The actual behavior of the correctly simulated input to H(P,P) specifies a non-halting sequence of configurations.

Everyone simply assumes that I am wrong about this entirely on the basis of their woefully inadequate understanding of the x86 language.


Halting problem undecidability and infinitely nested simulation (V5)

This is an explanation of a key new insight into the halting problem provided in the language of software engineering. Technical computer science terms are explained using software engineering terms.

To fully understand this paper a software engineer must be an expert in the C programming language, the x86 programming language, exactly how C translates into x86 and what an x86 process emulator is. No knowledge of the halting problem is required.

https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5 --
Copyright 2022 Pete Olcott

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


Subject: Re:_My_honest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_last_step_of_my_proof_]_(_Mind_Reader_? )
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
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On 4/22/2022 4:42 PM, André G. Isaak wrote:
On 2022-04-22 14:58, olcott wrote:
On 4/22/2022 3:42 PM, André G. Isaak wrote:
On 2022-04-22 14:09, olcott wrote:
On 4/22/2022 2:37 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/21/2022 8:23 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/21/2022 6:10 PM, Ben wrote:

Still, I expect there's a few years more chatting to be done before you
get to an editor's day.

People here have gotten the conventional halting problem dogma so
ingrained in their psyche that they believe that even when it is
proven to be a verified fact that the input to H(P,P) specifies a
non-halting sequence of configurations it is somehow incorrect for H
to report this.

H(P,P) == false is wrong because P(P) halts.  The problem of exhibiting
an H such that H(X,Y) == false if, and only if, X(Y) does not halt has
not gone away.  It's called the halting problem.  There are famous
theorems about it.  At one time you were interested in addressing it.

If the correctly simulated input to H(P,P) specifies a non-halting
sequence of configurations on the basis of its actual behavior then
this actual behavior supersedes and overrules anything any everything
else that disagrees.

The halting problem is about H(P,P) being right about P(P), not about
anything else you might find to waffle about.

The halting problem *is not* about P(P) when P(P) is not computationally equivalent to the correct simulation of the input to H(P,P).

This is simply an ignorant statement. Rather, it was initially an ignorant statement but since there have been many attempts made to remedy your ignorance, it has since graduated to a willfully-ignorant-grasping-at-straws-statement.

A halt decider is a Turing Machine which computes the halting *function*.

The halting function is a mathematical function. it is not defined in terms of 'inputs' or 'simulators'. It is not defined in terms of halt deciders at all since a function is logically prior to any algorithm for computing that function.

The halting *function* is simply a mathematical mapping from computations to {yes, no} based on whether they halt.


No. A decider computes the mapping from (finite string) inputs to an accept or reject state.

I was defining the halting *function*. That is an entirely different animal from a halt *decider*. Do you still not grasp the distinction between a Turing Machine and the function which it computes?

So the halt decider computes the halting *function* on some other basis than what its input actually specifies (What is it a mind reader?)
You really don't think these things through do you?



--
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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