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computers / comp.ai.philosophy / Re: Concise refutation of halting problem proofs V5 [Linz version] (Flibble)

SubjectAuthor
* Concise refutation of halting problem proofs V5olcott
+- Re: Concise refutation of halting problem proofs V5 [Linz version]olcott
+* Re: Concise refutation of halting problem proofs V5 [Linz version]olcott
|`- Re: Concise refutation of halting problem proofs V5 [Linz version]olcott
+- Re: Concise refutation of halting problem proofs V5 [ logicalolcott
`- Re: Concise refutation of halting problem proofs V5 [Linz version]olcott

1
Subject: Concise refutation of halting problem proofs V5
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Tue, 9 Nov 2021 14:52 UTC
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Subject: Concise refutation of halting problem proofs V5
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// Simplified Linz Ĥ (Linz:1990:319)
// Strachey(1965) CPL translated to C
void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
}

_P()
[00000c36](01)  55          push ebp
[00000c37](02)  8bec        mov ebp,esp
[00000c39](03)  8b4508      mov eax,[ebp+08] // 2nd Param
[00000c3c](01)  50          push eax
[00000c3d](03)  8b4d08      mov ecx,[ebp+08] // 1st Param
[00000c40](01)  51          push ecx
[00000c41](05)  e820fdffff  call 00000966    // call H
[00000c46](03)  83c408      add esp,+08
[00000c49](02)  85c0        test eax,eax
[00000c4b](02)  7402        jz 00000c4f
[00000c4d](02)  ebfe        jmp 00000c4d
[00000c4f](01)  5d          pop ebp
[00000c50](01)  c3          ret
Size in bytes:(0027) [00000c50]

Begin Local Halt Decider Simulation at Machine Address:c36

  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[00000c36][002117ca][002117ce] 55          push ebp
[00000c37][002117ca][002117ce] 8bec        mov ebp,esp
[00000c39][002117ca][002117ce] 8b4508      mov eax,[ebp+08]
[00000c3c][002117c6][00000c36] 50          push eax       // push P
[00000c3d][002117c6][00000c36] 8b4d08      mov ecx,[ebp+08]
[00000c40][002117c2][00000c36] 51          push ecx       // push P
[00000c41][002117be][00000c46] e820fdffff  call 00000966  // call H(P,P)

Because P calls H(P,P) P specifies infinitely nested simulation to every simulating halt decider H.

This has been verified with the execution trace of a correct pure simulation of P.

Whether or not H aborts its simulation P never reaches its final state therefore the simulated P never halts.

If the simulated P never halts then H(P,P)==0 is always correct for every simulating halt decider H.

To refute the claim that the direct execution of P(P) halts thus its simulation is incorrect H(P,P) directly executes its input instead of simulating it. The result is the infinitely nested simulation becomes infinite recursion.

In no case does the following directly executed P ever reach its final state of c50.

// call void P(I);
int H(u32 P, u32 I)
{
   if (!Halts(P,I)
     return 0;
   ((void(*)(int))P)(I);
   return 1;
}

int main()
{
   H((u32)P, (u32)P);
}


--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V5 [Linz version]
From: olcott
Newsgroups: comp.theory, sci.logic, sci.math, comp.ai.philosophy
Followup: comp.theory
Date: Wed, 10 Nov 2021 15:09 UTC
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Subject: Re: Concise refutation of halting problem proofs V5 [Linz version]
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On 11/10/2021 8:36 AM, Malcolm McLean wrote:
On Wednesday, 10 November 2021 at 13:37:01 UTC, Ben Bacarisse wrote:

I, for one, am fed up with trying to explain ever more silly
consequences of the initial assumption about H. The proof is simple and
the contradiction entailed is obvious.

There seems to be an oscillating between a pure simulator and a halt
decider. Then recently there's been a distinction between a decider which
simulates its input and one which executes its input directly.

But the core problem remains that the halt decider returns "false" for a
machine / program which halts. I haven't see any real advance on that, and
as you say, it become repetitive pointing it out.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

If the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts then the transition to ⊢* Ĥ.qn is necessarily correct no matter what Ĥ ⟨Ĥ⟩ does. A halt decider is only accountable for correctly deciding the halt status of its actual input.


Indeed. So tell PO that there is no string <H^>. If there were, there
wold be a TM H^ halts if it does not and does not halt if it does.

There might be a assumption that H is a correct halt decider built into
PO's argument somewhere, but I haven't actually found it. However I
haven't been paying such close attention.



--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V5 [Linz version]
From: olcott
Newsgroups: comp.theory, sci.logic, sci.math, comp.ai.philosophy
Date: Wed, 10 Nov 2021 16:09 UTC
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Subject: Re: Concise refutation of halting problem proofs V5 [Linz version]
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On 11/10/2021 9:44 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

In this Linz machine:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Remember to add that this must be the case if, and only if, Ĥ applied to
⟨Ĥ⟩ does not halt.  Then it all becomes clear to the average reader.


If the actual input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ would never halt then the transition to ⊢* Ĥ.qn is necessarily correct no matter what Ĥ ⟨Ĥ⟩ does. A halt decider is only accountable for correctly deciding the halt status of its actual input.


--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V5 [ logical necessity ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 10 Nov 2021 20:36 UTC
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Subject: Re: Concise refutation of halting problem proofs V5 [ logical
necessity ]
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On 11/10/2021 2:18 PM, Andy Walker wrote:
On 10/11/2021 13:36, Ben Bacarisse wrote:
His removal of the key condition on H^ is crucial to his attempt to keep
the discussion going.

     Disagree.  What is crucial to the attempt to keep the
discussion going is the fact that every man and his dog [other
sexes and animals are available] feels the need to reply.


It can be objectively verified that the correct pure simulation
of the input to H(P,P) never halts. (pages 3-4 of the new paper).

It is known on the basis of logical necessity that when-so-ever
the correctly simulated input to a halt decider never halts
that this halt decider would always be correct when it reports
that its input never halts.

All of the current rebuttals are entirely based on denying this
logical necessity.

Halting problem undecidability and infinitely nested simulation (V2)
https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2 --
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V5 [Linz version]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
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Subject: Re: Concise refutation of halting problem proofs V5 [Linz version]
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On 11/10/2021 2:49 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 11/10/2021 1:20 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 11/10/2021 11:02 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 11/10/2021 9:44 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

In this Linz machine:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Remember to add that this must be the case if, and only if, Ĥ applied to
⟨Ĥ⟩ does not halt.  Then it all becomes clear to the average reader.

If the actual input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ would never halt
Inputs don't halt or not halt.  You've been told this many times.  You
also know how to say what you are trying to say correctly, but I think
you want to avoid being clear.
Anyway, just make sure you keep the correct condition: if, and only if,
Ĥ applied to ⟨Ĥ⟩ does not halt.  Any other "facts" care to add are
immaterial since the correct condition shows that there are not such
TMs.

then the transition to ⊢* Ĥ.qn is necessarily correct no matter what Ĥ
⟨Ĥ⟩ does.
What Ĥ ⟨Ĥ⟩ does is what makes the line above apply or not -- it's there
in the part you deliberately keep omitting.  Ignoring (and not stating)
what Ĥ ⟨Ĥ⟩ does is central to why you are wrong.

A halt
decider is only accountable for correctly deciding the halt status of
its actual input.

There is no halt decider present in the line you keep misquoting.  There
is a "half-decider" at Ĥ.qx and we know what it does.  You just keep
omitting the key statements so that you can add some waffle instead.

If the pure simulation of the actual input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ would never
reach a final state of Ĥ then the transition to ⊢* Ĥ.qn is necessarily
correct no matter what Ĥ ⟨Ĥ⟩ does. A halt decider is only accountable
for correctly deciding the halt status of its actual input.

As I say, you can add any waffle you like provided you keep the correct
condition in place.  You can prove that for every TM H that behaves as
Linz specifies, the string ⟨Ĥ⟩ is even.  And you can prove that it's odd
as well.  You can prove anything from a contradiction.

If it is necessarily true that input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts when
it is correctly simulated then it is necessarily correct for Ĥ.qx to
transition to Ĥ.qn on this input.

You don't seem to be listening.  You certainly don't have anything
pertinent to say.


If an X is a Y then Z is always correct when it reports that an X is a Y.

If the correctly simulated input to any halt decider never halts then it is always correct for this halt decider to report that this input never halts.


You don't seem to be able to comprehend the concept of logical
necessity or that disagreeing with logical necessity is woefully
foolish.

I think it's rather foolish of you to think I might care about your
opinion of me.  In case you have forgotten, I don't.  I happy to stand
by my posts and let others decide who is being logical.



--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V5 [Linz version] (Flibble)
From: olcott
Newsgroups: comp.theory, sci.logic, sci.math, comp.ai.philosophy
Followup: comp.theory
Date: Thu, 11 Nov 2021 02:33 UTC
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Subject: Re: Concise refutation of halting problem proofs V5 [Linz version]
(Flibble)
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On 11/10/2021 8:08 PM, Ben Bacarisse wrote:
Richard Damon <Richard@Damon-Family.org> writes:

On 11/10/21 8:36 AM, Ben Bacarisse wrote:
Richard Damon <Richard@Damon-Family.org> writes:

On 11/10/21 6:35 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

Here is the same thing using Peter Linz notation:
Oh dear, back to talking about Turing machines...

In this Linz machine:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
it is true that the correct pure simulation of the
input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
There is no string ⟨Ĥ⟩, so there is nothing that can be simulated.  You
keep removing the clause that defines Ĥ's behaviour:
     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
With that clause put back, you (well, other people at least) can see
that no TM can behave like this.

But, if you step back a bit and ask about the H^ based on a machine H
that CLAIMS to meet the requirements, then you can have a H^ and a
<H^>.
I disagree (though it's largely philosophical at this point).  From the
claim that H meets Linz's spec we can deduce all sort of things.  One of
which is that there is no string <H^>.  But can also deduce any other
contradiction you like.
And that's the trouble.  Anything PO says about such an H^ is validly
entailed by the assumption.  He can even claim that 1 == 0 follows from
"Linz's specs", and he'd be right.  At some stage you have to point out
the contradictions and force PO to abandon the initial assumption.

I, for one, am fed up with trying to explain ever more silly
consequences of the initial assumption about H.  The proof is simple and
the contradiction entailed is obvious.

The difference is that PO HAS an H, so H does exist, and we can thus
build a H^ from. (This assumes we can convert is 'C Program' into some
sort of Turing Machine).

Not in this sub-thread.  He starts: "In this Linz machine:" and then
gives the line he say not yet understood.  He calls everything H, but
when it's Linz's we must assume what Linz assumes.

When H refers to his code, it's wrong for the reasons you and André and
I have been saying.  If it's Linz H, it does not exist (or the class of
TMs referred to by H is empty).

Anyway, I'm butting out.  If I've written out the proof using TMs more
than once, and the world has told him that his C code is wrong by
definition, but there is no way he will ever see any of this.  It's not
that he's stupid, it's just that he's invested a significant proportion
of his life, and most of his self-esteem, in this ill thought out idea.
The mind will play extraordinary tricks in that sort of situation.


Within the definition of the x86 language H(P,P)==0 is a necessary consequence for every simulating halt decider H.

// Simplified Linz Ĥ (Linz:1990:319)
// Strachey(1965) CPL translated to C
void P(u32 x)
{
   if (H(x, x))
     HERE: goto HERE;
}

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

_P()
[00000c36](01)  55          push ebp
[00000c37](02)  8bec        mov ebp,esp
[00000c39](03)  8b4508      mov eax,[ebp+08] // 2nd Param
[00000c3c](01)  50          push eax
[00000c3d](03)  8b4d08      mov ecx,[ebp+08] // 1st Param
[00000c40](01)  51          push ecx
[00000c41](05)  e820fdffff  call 00000966    // call H
[00000c46](03)  83c408      add esp,+08
[00000c49](02)  85c0        test eax,eax
[00000c4b](02)  7402        jz 00000c4f
[00000c4d](02)  ebfe        jmp 00000c4d
[00000c4f](01)  5d          pop ebp
[00000c50](01)  c3          ret
Size in bytes:(0027) [00000c50]

Within the definition of the x86 language H(P,P)==0 is a necessary consequence for every simulating halt decider H.


Indeed.  So tell PO that there is no string <H^>.  If there were, there
wold be a TM H^ halts if it does not and does not halt if it does.

Except that you can create an machine H that you can (incorrectly)
claim to meet the requirements, and thus a machine H^ can be created
and thus the string <H^> exists.

If he starts "my H" then yes, but he started "this Linz machine".  Of
course calling everything H is just another silly thing he does, but
there is "Linz's H" which does not exist, and "his H" which does not
meet Linz's specification.


Everyone simply leaps to the conclusion that I must be wrong yet can't point to a single error in the actual inference steps that prove my point. Perhaps this simply means that everyone here is mostly clueless about the x86 language.

Flibble seems to understand me quite well. He even understood the last required step to convert my H into a pure function so well that he presented it before I did.

[Olcott wrong about infinite recursion] comp.theory
On 11/10/2021 3:53 PM, Mr Flibble wrote:
 > Olcott is barking up the wrong tree re infinite recursion: there
 > is only a need to detect a single recursive call into the halt decider
 > and signal an exception.  Why?  Because infinite recursion is valid
 > program behavior.
 >
 > /Flibble




Halting problem undecidability and infinitely nested simulation (V2)

https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2 --
Copyright 2021 Pete Olcott

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


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