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computers / comp.ai.philosophy / Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]

SubjectAuthor
* Re: Black box halt decider is NOT a partial decider [ H(P,P)==0 isolcott
`* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
 +* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
 |`* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
 | +- Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
 | `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
 |  `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
 |   +- Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
 |   `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
 |    +- Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
 |    `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
 |     `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
 |      `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
 |       `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
 |        `- Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
 `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
  `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
   `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
    `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
     +- Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
     +* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
     |`* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
     | `- Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
     +- Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn [ succinolcott
     +- Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
     `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
      `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
       +* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
       |`* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
       | `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
       |  `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
       |   `- Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
       `* Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
        +- Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.olcott
        +- Re: _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
        `* Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn isolcott
         `* Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn isolcott
          +* Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn isolcott
          |`- Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn isolcott
          +- Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn isolcott
          `- Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn isolcott

Pages:12
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you game_?_)
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Tue, 3 Aug 2021 21:50 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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NNTP-Posting-Date: Tue, 03 Aug 2021 16:50:59 -0500
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you g
ame_?_)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <87tukc12yh.fsf@bsb.me.uk>
<e6OdnW_rdvusk5n8nZ2dnUU7-dPNnZ2d@giganews.com> <875ywrmwsr.fsf@bsb.me.uk>
<GtmdnfBrysPrAZn8nZ2dnUU7-L_NnZ2d@giganews.com> <874kcakv3d.fsf@bsb.me.uk>
<-M-dnfsXl4WvWpj8nZ2dnUU7-IGdnZ2d@giganews.com> <87im0pa1pp.fsf@bsb.me.uk>
<6eOdnaoMUdZN_pr8nZ2dnUU7-cPNnZ2d@giganews.com> <87y29j6c5e.fsf@bsb.me.uk>
<p8SdnV8dJu3wq5X8nZ2dnUU7-a3NnZ2d@giganews.com> <875ywn5qfr.fsf@bsb.me.uk>
<hJmdnSWv8-zPCJX8nZ2dnUU7-T_NnZ2d@giganews.com>
<3df616af-1e70-413f-8534-fb4ca6204c28n@googlegroups.com>
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<f5026a04-cbe0-4809-82d9-27ecaf1fd1afn@googlegroups.com>
<3IadnZ2CTqZnw5T8nZ2dnUU7-VudnZ2d@giganews.com> <sebtb9$plb$1@dont-email.me>
<dPGdnfUV6vjsBZT8nZ2dnUU7-V3NnZ2d@giganews.com> <sec7nt$333$1@dont-email.me>
<Ea2dnYUpZJLzNpT8nZ2dnUU78LXNnZ2d@giganews.com> <secahd$l3q$1@dont-email.me>
<zqKdnXqSUdOSMpT8nZ2dnUU7-f_NnZ2d@giganews.com> <seccad$o4$1@dont-email.me>
From: NoO...@NoWhere.com (olcott)
Date: Tue, 3 Aug 2021 16:50:52 -0500
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On 8/3/2021 4:26 PM, André G. Isaak wrote:
On 2021-08-03 15:03, olcott wrote:
On 8/3/2021 3:56 PM, André G. Isaak wrote:
On 2021-08-03 14:47, olcott wrote:
On 8/3/2021 3:08 PM, André G. Isaak wrote:
On 2021-08-03 13:26, olcott wrote:
On 8/3/2021 12:11 PM, André G. Isaak wrote:
On 2021-08-03 09:21, olcott wrote:
On 8/3/2021 9:33 AM, Malcolm McLean wrote:
On Tuesday, 3 August 2021 at 14:00:28 UTC+1, olcott wrote:
On 8/3/2021 12:29 AM, Malcolm McLean wrote:

You're trying to claim that H_Hat<H_Hat> doens't really halt because
the recursion of H instances is terminated by H.
No I have never been saying that. I am claiming that the input to H(P,P)
never halts whether or not H terminates its simulation of this input.

We agree that P(P) halts.
So now you're drawing a distinction between P(P) and "the input to H(P,P)".
ThIs is nonsense..

Try and find any error in (a)(b)(c)(d) on page 6

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









The error, which has been pointed out repeatedly, is in your (b).

You claim "there are no control flow instructions in the execution trace that would escape the infinite recursion", but there *are* flow control instructions. But your trace is incomplete and skips over the call to B82 where these flow control instructions reside.


Whether H aborts its simulation of P or not P never reaches its final state.

But H's simulation of P is *not* a computation.


Sure it is.
The pure simulation of a computation on its input is computationally equivalent to the direct execution of this same computation on its input.

No. It isn't. A computation is a free-standing, self-contained piece of code. When you run H(P, P) the computation P(P) isn't being computed. It's being evaluated. H(P, P) is the only computation being run in this case.


So you don't understand how UTMs work.

I'm well aware of how UTMs work. I am also aware that an x86 simulator is not the same thing as a UTM.


It is equivalent. In any case my exactly same reasoning directly applies to the Linz proof.

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

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

When Ĥ is applied to its own Turing Machine description ⟨Ĥ⟩ the embedded halt decider at Ĥ.qx correctly decides that its input ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.

And your H *doesn't* perform a pure simulation of its input, so we can't even talk about the simulation as being 'computationally equivalent' to some computation.


The fact that the first seven instructions of P keep repeating while H is in pure simulation mode proves that H is doing pure simulation mode correctly.

The problem is that your 'pure simulation mode' is a cheat.


That the execution trace of the simulation of P precisely corresponds to its source-code proves beyond all possible doubt that the pure simulation of P by H is correct. There is no way to wiggle out of this.

Your H and P *need* to be independent, self-contained programs. They need to both be compiled and linked separately resulting in two different executables. This means that the copy of H contained in P will be *distinct* from program H. It will be at an entirely different address and any code inside H which tells it to ignore its 'own code' will not result in it ignoring the H contained in P.

That also means that any change in the "mode" in which the main H runs *cannot* have any effect on what the H contained in P does.

Your 'pure simulation mode' is only valid as a test if it applies *only* to the behaviour of H, and not to the behaviour of the H contained inside P.

If you change what the H inside P does, you are then talking about an *entirely different* computation which has no bearing on the halting behaviour of P.

You can't evaluate whether P halts by considering what some modified version of P does, which is in effect what you are doing.

André


That the execution trace of the simulation of P precisely corresponds to its source-code proves beyond all possible doubt that the pure simulation of P by H is correct. There is no way to wiggle out of this.

--
Copyright 2021 Pete Olcott

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


Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you game_?_)
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 4 Aug 2021 00:42 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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NNTP-Posting-Date: Tue, 03 Aug 2021 19:42:48 -0500
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] ( Are you g
ame_?_)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <875ywrmwsr.fsf@bsb.me.uk>
<GtmdnfBrysPrAZn8nZ2dnUU7-L_NnZ2d@giganews.com> <874kcakv3d.fsf@bsb.me.uk>
<-M-dnfsXl4WvWpj8nZ2dnUU7-IGdnZ2d@giganews.com> <87im0pa1pp.fsf@bsb.me.uk>
<6eOdnaoMUdZN_pr8nZ2dnUU7-cPNnZ2d@giganews.com> <87y29j6c5e.fsf@bsb.me.uk>
<p8SdnV8dJu3wq5X8nZ2dnUU7-a3NnZ2d@giganews.com> <875ywn5qfr.fsf@bsb.me.uk>
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<3df616af-1e70-413f-8534-fb4ca6204c28n@googlegroups.com>
<_ZudnRCgbJV5oJT8nZ2dnUU7-VXNnZ2d@giganews.com>
<f5026a04-cbe0-4809-82d9-27ecaf1fd1afn@googlegroups.com>
<3IadnZ2CTqZnw5T8nZ2dnUU7-VudnZ2d@giganews.com> <sebtb9$plb$1@dont-email.me>
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<Ea2dnYUpZJLzNpT8nZ2dnUU78LXNnZ2d@giganews.com> <secahd$l3q$1@dont-email.me>
<zqKdnXqSUdOSMpT8nZ2dnUU7-f_NnZ2d@giganews.com> <seccad$o4$1@dont-email.me>
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From: NoO...@NoWhere.com (olcott)
Date: Tue, 3 Aug 2021 19:42:48 -0500
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On 8/3/2021 5:35 PM, André G. Isaak wrote:
On 2021-08-03 15:50, olcott wrote:
On 8/3/2021 4:26 PM, André G. Isaak wrote:
On 2021-08-03 15:03, olcott wrote:
On 8/3/2021 3:56 PM, André G. Isaak wrote:
On 2021-08-03 14:47, olcott wrote:
On 8/3/2021 3:08 PM, André G. Isaak wrote:
On 2021-08-03 13:26, olcott wrote:
On 8/3/2021 12:11 PM, André G. Isaak wrote:
On 2021-08-03 09:21, olcott wrote:
On 8/3/2021 9:33 AM, Malcolm McLean wrote:
On Tuesday, 3 August 2021 at 14:00:28 UTC+1, olcott wrote:
On 8/3/2021 12:29 AM, Malcolm McLean wrote:

You're trying to claim that H_Hat<H_Hat> doens't really halt because
the recursion of H instances is terminated by H.
No I have never been saying that. I am claiming that the input to H(P,P)
never halts whether or not H terminates its simulation of this input.

We agree that P(P) halts.
So now you're drawing a distinction between P(P) and "the input to H(P,P)".
ThIs is nonsense..

Try and find any error in (a)(b)(c)(d) on page 6

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











The error, which has been pointed out repeatedly, is in your (b).

You claim "there are no control flow instructions in the execution trace that would escape the infinite recursion", but there *are* flow control instructions. But your trace is incomplete and skips over the call to B82 where these flow control instructions reside.


Whether H aborts its simulation of P or not P never reaches its final state.

But H's simulation of P is *not* a computation.


Sure it is.
The pure simulation of a computation on its input is computationally equivalent to the direct execution of this same computation on its input.

No. It isn't. A computation is a free-standing, self-contained piece of code. When you run H(P, P) the computation P(P) isn't being computed. It's being evaluated. H(P, P) is the only computation being run in this case.


So you don't understand how UTMs work.

I'm well aware of how UTMs work. I am also aware that an x86 simulator is not the same thing as a UTM.


It is equivalent. In any case my exactly same reasoning directly applies to the Linz proof.

No, it isn't equivalent. A UTM takes a description of a *Turing Machine* as its input. A description of an x86 program is not a Turing Machine. How can two computations which take entirely distinct inputs be equivalent?


Apparently you don't understand computational equivalence either.

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

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

When Ĥ is applied to its own Turing Machine description ⟨Ĥ⟩ the embedded halt decider at Ĥ.qx correctly decides that its input ⟨Ĥ⟩ ⟨Ĥ⟩

Specifies a computation that halts.

never halts.

Inputs neither halt nor don't halt. Only computations halt. >
And Ĥ.qx isn't an 'embedded halt decider'. It is a state.


Linz stipulates that it <is> Turing machine H.

And your H *doesn't* perform a pure simulation of its input, so we can't even talk about the simulation as being 'computationally equivalent' to some computation.


The fact that the first seven instructions of P keep repeating while H is in pure simulation mode proves that H is doing pure simulation mode correctly.

The problem is that your 'pure simulation mode' is a cheat.


That the execution trace of the simulation of P precisely corresponds to its source-code proves beyond all possible doubt that the pure simulation of P by H is correct. There is no way to wiggle out of this.

No. It doesn't 'precisely correspond' to its source code because your trace *omits* a big chunk of code.


*What details of this paragraph do you feel are incorrect (and why)*

Because H only acts as a pure simulator of its input until after its halt status decision has been made it has no behavior that can possibly effect the behavior of its input. Because of this H screens out its own address range in every execution trace that it examines. This is why we never see any instructions of H in any execution trace after an input calls H.

Your H and P *need* to be independent, self-contained programs. They need to both be compiled and linked separately resulting in two different executables. This means that the copy of H contained in P will be *distinct* from program H. It will be at an entirely different address and any code inside H which tells it to ignore its 'own code' will not result in it ignoring the H contained in P.

That also means that any change in the "mode" in which the main H runs *cannot* have any effect on what the H contained in P does.

Your 'pure simulation mode' is only valid as a test if it applies *only* to the behaviour of H, and not to the behaviour of the H contained inside P.

If you change what the H inside P does, you are then talking about an *entirely different* computation which has no bearing on the halting behaviour of P.

You can't evaluate whether P halts by considering what some modified version of P does, which is in effect what you are doing.

André


That the execution trace of the simulation of P precisely corresponds to its source-code proves beyond all possible doubt that the pure simulation of P by H is correct. There is no way to wiggle out of this.

No. It doesn't 'precisely correspond' to its source code because your trace *omits* a big chunk of code.


*What details of this paragraph do you feel are incorrect (and why)*

Because H only acts as a pure simulator of its input until after its halt status decision has been made it has no behavior that can possibly effect the behavior of its input. Because of this H screens out its own address range in every execution trace that it examines. This is why we never see any instructions of H in any execution trace after an input calls H.

Come back and post a complete trace once you've actually set up your H and P *properly*. i.e. as two *separate* programs which have been compiled and linked two form two separate, *complete* executables where P does not call any code contained in H. If there is code in common, it must be *duplicated* so that each executable contains its own copy of this code.

There is no point discussing your implementation where your H and P are contained in a single executable. That's not how 'TM equivalents' work, nor is it how the Linz Proof constructs H_Hat, and no conclusions which you draw from such a mangled implementation can possibly shed any light on the halting problem.


So when your reasoning fails to show that my H is incorrect you change the subject. It is true that the input to H(P,P) never halts therefore H(P,P)==0 is correct.

If you don't grasp why what you are doing is illegitimate, then I suggest you go and actually learn what the term 'computation' means.

The standard formal definition of computation, repeated in all the major textbooks, derives from these early ideas. Computation is defined as the execution sequences of halting Turing machines (or their equivalents). An execution sequence is the sequence of total configurations of the machine, including states of memory and control unit.

https://dl.acm.org/doi/pdf/10.1145/1880066.1880067

That would involve actually reading Linz and/or Sipser *from the beginning*. Your notion than you can somehow overcome a well-established proof without even grasping the fundamentals of the subject matter is mystifying to say the least.

André



--
Copyright 2021 Pete Olcott

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


Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs_Count_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 4 Aug 2021 03:23 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
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NNTP-Posting-Date: Tue, 03 Aug 2021 22:23:54 -0500
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inpu
ts_Count_]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <87pmv4ab6r.fsf@bsb.me.uk>
<JNadnQD-Ofr-SJz8nZ2dnUU7-XHNnZ2d@giganews.com> <871r7i6n2u.fsf@bsb.me.uk>
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<1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk>
<7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk>
<j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk>
<Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk>
<qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
Date: Tue, 3 Aug 2021 22:23:54 -0500
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On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/2/2021 2:10 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/1/2021 5:54 AM, Ben Bacarisse wrote:

I am happy you have a notation you like.  Are you prepared to address
that fact that your H^ is not "as in Linz"?

My Ĥ is exactly the Linz Ĥ with the additional elaboration that the
second wildcard state transition ⊢* is defined to be a simulating halt
decider.
No.  I've explained many times now why your Ĥ is not at all "the Linz
Ĥ".  Do you see any point in my doing so again?

I suspect not.  You certainly have not asked a single question that
could help you to understand why your Ĥ is irrelevant.

We all know you are declaring that to be correct.  Here's why your Ĥ is
not "as in Linz".  Linz requires that

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

Any Ĥ that eventually transitions to Ĥ.qn on input wM must do so
if, and only if, the encoded M applied to wM does not halt.  But you've
given us a case where your Ĥ is not like this:

     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

And when we eliminate the fallacy of equivocation error we have

Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
The only thing I think you are equivocating on is pretending that Ĥ[0]
is not Ĥ, and it doesn't look as if you've removed that error.  I think
you /should/ remove it so that you can be saying something of value
about Ĥ.

It is clear from the text that Linz does specify at least three
different instances of Ĥ, The TM the TMD input ⟨Ĥ⟩ and a copy of this
TMD input.

Still equivocating.  Either Ĥ[0] = Ĥ or you are wasting everyone's time.

(This is mathematical equality.  Ĥ is a tuple of sets.  If Ĥ[0] is not
exactly identical in every way to Ĥ then I don't care about it.  Note
that I'm not disputing your right, for ease of explanation, to give
identical things more than one name.  But the same permission allows me
to use any of the names because they name the same thing.)

You are wrong because

   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn

where Linz requires that

  Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
  if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.


You have the above incorrectly in these two
M refers to the Turing Machine described by wM.

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

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

This is *NOT* the Ĥ that is executed.
This is only the Turing Machine description ⟨Ĥ⟩ input.

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

The key point is that only the input to the halt decider is within the scope of the halting problem.

As long as the halt decider correctly decides its input what these same programs do what they are not inputs to the halt decider make no difference to the actual halting problem.


--
Copyright 2021 Pete Olcott

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


Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs_Count_](_Honest_Dialogue_)
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 4 Aug 2021 15:11 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 25
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NNTP-Posting-Date: Wed, 04 Aug 2021 10:11:33 -0500
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs_Count_](_Honest_Dialogue_)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <87sfzw3ao1.fsf@bsb.me.uk> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk> <w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk> <1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <sed35j$od8$1@dont-email.me> <KcOdnSwOTcFujZf8nZ2dnUU7-W_NnZ2d@giganews.com> <sed4b4$tqd$1@dont-email.me> <PKGdnRIK9KVMi5f8nZ2dnUU78V3NnZ2d@giganews.com> <sed5fo$31j$1@dont-email.me> <INCdnRriUbgihJf8nZ2dnUU7-YmdnZ2d@giganews.com> <Ysedne93LONKgJf8nZ2dnUU7-XvNnZ2d@giganews.com> <594cc8cb-cfac-41d4-b77b-b33278758c4an@googlegroups.com>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 10:11:33 -0500
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On 8/4/2021 9:13 AM, Malcolm McLean wrote:
On Wednesday, 4 August 2021 at 05:55:59 UTC+1, olcott wrote:

I am not willing to talk about anything besides page 6 until we have
mutual agreement on page 6.

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

We agree that P(P) halts and that H(P, P) reports false (non-halting).

I've suggested that maybe you are trying to argue that P(P)'s halting
doesn't count, because it is generated by the copy of H inside P.

You've rejected that idea, and insist that H(P,P) has correctly determined
  that P(P) is non-halting.

I don't really think there's anything useful left to say.


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

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation Because page 6 does prove that H(P,P) does correctly decide that its input never halts and the halting problem only requires that H decide its input correctly the fact that H(P,P)==0 is correct does directly refute the halting problem counter-example conclusion.

That people are unwilling to critique page 6 only shows that they are unwilling to have an actual honest dialogue.

--
Copyright 2021 Pete Olcott

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


Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inputs_Count_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 4 Aug 2021 15:14 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 25
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NNTP-Posting-Date: Wed, 04 Aug 2021 10:14:26 -0500
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ Only Inpu
ts_Count_]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <87sfzw3ao1.fsf@bsb.me.uk>
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<j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk>
<Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk>
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<Ysedne93LONKgJf8nZ2dnUU7-XvNnZ2d@giganews.com> <sed904$iae$1@dont-email.me>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 10:14:25 -0500
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On 8/4/2021 12:36 AM, André G. Isaak wrote:
On 2021-08-03 22:55, olcott wrote:
On 8/3/2021 11:38 PM, olcott wrote:
On 8/3/2021 11:36 PM, André G. Isaak wrote:
On 2021-08-03 22:25, olcott wrote:
On 8/3/2021 11:16 PM, André G. Isaak wrote:
On 2021-08-03 22:00, olcott wrote:
On 8/3/2021 10:56 PM, André G. Isaak wrote:
On 2021-08-03 21:23, olcott wrote:

The key point is that only the input to the halt decider is within the scope of the halting problem.

As long as the halt decider correctly decides its input what these same programs do what they are not inputs to the halt decider make no difference to the actual halting problem.

You have this completely back-assward.

How the TM described by the input to a halt decider behaves when they are run as *independent* computations (i.e. not as inputs to the halt decider) is the question a halt decider is supposed to answer by the very definition of a 'halt decider'. If the answer provided by the decider does not match the behavior of the *independent* computation, then that answer is wrong.

Talking about how the description behaves "as an input" isn't even coherent. Inputs don't behave like anything. They are just data.

André


    In computability theory, the halting problem is the
    problem of determining,

from a description of an arbitrary computer program and an input,
from a description of an arbitrary computer program and an input,
from a description of an arbitrary computer program and an input,
from a description of an arbitrary computer program and an input,

    whether the program will
    finish running, or continue to run forever.
    https://en.wikipedia.org/wiki/Halting_problem

I have proved that the input to H(P,P) never halts whether or not it is aborted by H, are you going to have an honest dialogue about the steps of the proof or not?

Yes, and when run H(P, P) the arbitrary program and input which it is being given described P(P).

It is supposed to answer how that *program* will behave. And you have acknowledged that that program *halts*. That behaviour is what *defines* the correct answer to the question 'does P(P) halt'. And your H gets this *wrong*.

The halting question is not concerned with how the input string behaves inside your partial simulator. It is *only* concerned with the actual, *independent* program P(P).


So in other words you insist on not going through my steps (a)(b)(c)(d) on page 6 and trying to find an error?


I already pointed out that your (b) has major errors in an earlier post which you disregarded. I see no reason to repeat that.


I am not willing to talk about anything else.

I am not willing to talk about anything besides page 6 until we have mutual agreement on page 6.

That's not going to happen since your argument on page 6 is fallacious.

Read my earlier message <sebtb9$plb$1@dont-email.me>

I see no reason to repeat it.


You simply ignored my correction of your error.
I responded to this message again with words that are more clear.

Simply ignoring this response is sufficient proof that you are unmotivated to have an actual honest dialogue.


--
Copyright 2021 Pete Olcott

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


Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ][ GIGO ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 4 Aug 2021 15:48 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
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NNTP-Posting-Date: Wed, 04 Aug 2021 10:48:23 -0500
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ][ GIGO ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <87pmv4ab6r.fsf@bsb.me.uk> <JNadnQD-Ofr-SJz8nZ2dnUU7-XHNnZ2d@giganews.com> <871r7i6n2u.fsf@bsb.me.uk> <OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk> <1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <53d47ab9-818c-4f40-8e72-bdb76fa416een@googlegroups.com> <87y29l8hhp.fsf@bsb.me.uk> <LZOdnR5aLooNKpv8nZ2dnUU7-SnNnZ2d@giganews.com> <87h7g988a6.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 10:48:23 -0500
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On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/1/2021 7:41 AM, Ben Bacarisse wrote:
Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:

On Sunday, 1 August 2021 at 11:54:57 UTC+1, Ben Bacarisse wrote:

Here we can see that Ĥ applied to ⟨Ĥ⟩ halts. You can call your Ĥ's
behaviour "correct". You can call it anything you like. But it's not
"as in Linz". It does not say anything about Linz's proof. It does not
do anything people would call impossible or even interesting.

It seems to be established that H(H_Hat, H_Hat) returns "non-halting"
whilst H_Hat(H_Hat) halts. So all is as Linz says it must be and no
theorems are refuted. Which you would expect. If results were consistent
it would have to be some cheap trick.
I case there is some confusion, I mean that PO's Ĥ is not an Ĥ as
specified in Linz.  Yes, everything is in accordance with the truth as
laid out in Linz and, indeed, in any textbook.
I point this out to PO because he brings it up.  He keeps posting the
specification of what an Ĥ, as Linz specifies it, would do:
    Ĥ.q0 wM ⊢* Ĥ.qx wM wM  ⊢* Ĥ.qn
    if (and only if) M applied to wM does not halt.
He claims (or used to claim) that his Ĥ meets this specification for at
least the one case where wM == ⟨Ĥ⟩:
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩  ⊢* Ĥ.qn
    if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
To remain relevant, he /must/ keep insisting that his Ĥ meets the
requirements laid out in Linz, if only for this one key input.


Ĥ[0].q0 is taken to mean Ĥ<sub>0</sub>.q0 which is the Turing machine.

Ĥ[1].q0 is taken to mean Ĥ<sub>1</sub>.q0 which is the Turing machine
description input to Ĥ[0].q0

Ĥ[2].q0 is taken to mean Ĥ<sub>2</sub>.q0 which is first copy of the
Turing machine description input to Ĥ[0].q0

Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn

Ĥ[0] is Ĥ so you are confirming, yet again, that

    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn

It is neither a contradiction nor a paradox because there are three
different instances of Ĥ.

I agree that this is neither a paradox nor a contradiction.  It's just a
fact derived form the logic of how your Ĥ is written (the majority of
which you are keeping hidden from us).

Because the only reason that the first instance halts is that Ĥ[0].qx
correctly determines that its input cannot possibly ever reach its
final state of Ĥ[1].qn or Ĥ[1].qy whether or not the simulating halt
decider aborts its simulation of this input, we know with 100%
perfectly justified logical certainty that the input to Ĥ[0].qx never
halts.

We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn.  This clearly
shows that Ĥ applied to ⟨Ĥ⟩ halts.  You can see the final state right

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

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

You are using the wrong Ĥ. Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of this ⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩. We can see that M never reaches its final state.

Because the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ does specify infinitely nested simulation we can know for sure that M never reaches its final state whether or not Ĥ.qx aborts its simulation of M.

there in the line you keep posting again and again.  As far as I know,
you have never disputed this fact.

As you say, this is neither a paradox nor a contradiction.  It just
shows that Ĥ does not behave as Linz says it should, in the one case you
have obsessed about for 17 years.  Having an H (and thus an Ĥ) that is
wrong (i.e. not "exactly and precisely as on Linz" as you once claimed)
is trivial.  It is not something that anyone (except Malcolm,
apparently) would care about.



--
Copyright 2021 Pete Olcott

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


Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 4 Aug 2021 16:15 UTC
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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <87k0la542c.fsf@bsb.me.uk> <1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk> <w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk> <xLednaPs_ZSXCZX8nZ2dnUU7-YnNnZ2d@giganews.com> <87o8af47y0.fsf@bsb.me.uk> <NsudnY99rthDOJX8nZ2dnUU7-c_NnZ2d@giganews.com> <87zgtx2wxn.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 11:15:57 -0500
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On 8/4/2021 7:53 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/2/2021 8:45 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/2/2021 7:11 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:


(I'm going to ignore errors that I've already pointed out.)

If you really do sincerely want an actual honest dialogue you would
carefully work through all the steps to confirm that the input ⟨Ĥ⟩ ⟨Ĥ⟩
to Ĥ.qx never halts.
If you want me to comment on that, write it without the errors.  It has
two, both of which I've commented on before.  If you are sincere, you
will want to write clearly and without errors.

Once we have mutual agreement that Ĥ.qx correctly decides that its
input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts and Ĥ does halt, then we have the basis to
go to the next step and resolve the actual paradox.
There is no paradox.  That Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn when Ĥ applied to ⟨Ĥ⟩
clearly halts is not paradoxical.  It's just not the sort of TM the
proof is talking about.  And how could it be?  The "Linz Ĥ" is as
illogical as a cat that is a dog.

As long as it is simply dismissed out-of-hand as a contradiction the
paradox remains unresolved.
There is no contradiction or paradox.  You Ĥ is just the wrong sort of
TM.  The proof you want to "refute" is talking about this sort of Ĥ:
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
    if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

Maybe saying it a couple more times will help.  After four times I can
tell you that it's still wrong.  Maybe about a dozen more?

Whether what happens after Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correct or not is determined
by Linz, not by you.  And you are clear that

   Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn.


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

As explained in complete detail below:
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

because M applied to wM does not halt
where M is Machine_of(⟨Ĥ⟩) (1st param) above
and wM is ⟨Ĥ⟩ the second param above.

Because wM is referring to ⟨Ĥ⟩ and M is referring to the underlying machine of ⟨Ĥ⟩ the last line above is translated to:
if Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ does not halt

We can know that Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ never reaches its final state whether or not the embedded halt decider at Ĥ.qx aborts it simulation of Machine_of(⟨Ĥ⟩), therefore we know that Machine_of(⟨Ĥ⟩) never halts. Therefore we know that M applied to wM does not halt.

Linz is equally clear that this is wrong.  You even helpfully keep
quoting Linz telling you it's wrong:

     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.

You can repeat, again and again, that "Ĥ.qx correctly decides"
something, but you've already told the world that the computation halts
when is should not.  Your Ĥ is not doing what it should to show a fault
in Linz's proof.  It's doing something else that is of no interest to
anyone.

(It's possible that just can't see why it is that those two lines from
Linz are telling you that your Ĥ does is wrong.  You've struggled with
what this notation really means, so it's possible.  But reasonable
student would ask if they could not see that Linz is telling you your Ĥ
is wrong.)

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

Really?  Well scan my tape and call me halting!

If you had had the two TMs you lied about having (sorry, used "poetic
license" to say you had) you could have run

   Ĥ.q0 ⟨Ĥ⟩

and

   H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩

and you would have seen, right away, that H(⟨Ĥ⟩ ⟨Ĥ⟩) is wrong about
Ĥ(⟨Ĥ⟩).  Maybe you did and that's why you've spent 30 months waking back
that lie^H^H^H poetic license.



--
Copyright 2021 Pete Olcott

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


Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ][ GIGO ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 4 Aug 2021 22:31 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
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Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ][ GIGO ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <871r7i6n2u.fsf@bsb.me.uk> <OqKdnROLKJ9CdJz8nZ2dnUU7-avNnZ2d@giganews.com> <87k0la542c.fsf@bsb.me.uk> <1NidnVPZ-NHDl5_8nZ2dnUU7-enNnZ2d@giganews.com> <87sfzw3ao1.fsf@bsb.me.uk> <7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <53d47ab9-818c-4f40-8e72-bdb76fa416een@googlegroups.com> <87y29l8hhp.fsf@bsb.me.uk> <LZOdnR5aLooNKpv8nZ2dnUU7-SnNnZ2d@giganews.com> <87h7g988a6.fsf@bsb.me.uk> <j8OdneamG91aK5f8nZ2dnUU7-fvNnZ2d@giganews.com> <87im0l2gc0.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 17:31:01 -0500
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On 8/4/2021 1:51 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/1/2021 11:00 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/1/2021 7:41 AM, Ben Bacarisse wrote:
Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:

On Sunday, 1 August 2021 at 11:54:57 UTC+1, Ben Bacarisse wrote:

Here we can see that Ĥ applied to ⟨Ĥ⟩ halts. You can call your Ĥ's
behaviour "correct". You can call it anything you like. But it's not
"as in Linz". It does not say anything about Linz's proof. It does not
do anything people would call impossible or even interesting.

It seems to be established that H(H_Hat, H_Hat) returns "non-halting"
whilst H_Hat(H_Hat) halts. So all is as Linz says it must be and no
theorems are refuted. Which you would expect. If results were consistent
it would have to be some cheap trick.
I case there is some confusion, I mean that PO's Ĥ is not an Ĥ as
specified in Linz.  Yes, everything is in accordance with the truth as
laid out in Linz and, indeed, in any textbook.
I point this out to PO because he brings it up.  He keeps posting the
specification of what an Ĥ, as Linz specifies it, would do:
     Ĥ.q0 wM ⊢* Ĥ.qx wM wM  ⊢* Ĥ.qn
     if (and only if) M applied to wM does not halt.
He claims (or used to claim) that his Ĥ meets this specification for at
least the one case where wM == ⟨Ĥ⟩:
     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩  ⊢* Ĥ.qn
     if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
To remain relevant, he /must/ keep insisting that his Ĥ meets the
requirements laid out in Linz, if only for this one key input.


Ĥ[0].q0 is taken to mean Ĥ<sub>0</sub>.q0 which is the Turing machine.

Ĥ[1].q0 is taken to mean Ĥ<sub>1</sub>.q0 which is the Turing machine
description input to Ĥ[0].q0

Ĥ[2].q0 is taken to mean Ĥ<sub>2</sub>.q0 which is first copy of the
Turing machine description input to Ĥ[0].q0

Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
Ĥ[0] is Ĥ so you are confirming, yet again, that
     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn

It is neither a contradiction nor a paradox because there are three
different instances of Ĥ.
I agree that this is neither a paradox nor a contradiction.  It's just a
fact derived form the logic of how your Ĥ is written (the majority of
which you are keeping hidden from us).

Because the only reason that the first instance halts is that Ĥ[0].qx
correctly determines that its input cannot possibly ever reach its
final state of Ĥ[1].qn or Ĥ[1].qy whether or not the simulating halt
decider aborts its simulation of this input, we know with 100%
perfectly justified logical certainty that the input to Ĥ[0].qx never
halts.
We know, since you keep telling us, that Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn.  This clearly
shows that Ĥ applied to ⟨Ĥ⟩ halts.  You can see the final state right

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

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

You are using the wrong Ĥ.

First of all, let's be 100% clear: I am talking about what /your/ Ĥ
does, based in the facts you have let slip about it.

Linz stipulates that wM is ⟨Ĥ⟩ and M is the underlying machine of this
⟨Ĥ⟩ therefore M applied to wM means ⟨Ĥ⟩ applied to ⟨Ĥ⟩.

No.  How many years have you been staring at this one page from Linz?
You still don't know what it says.  Do ask me questions, if you'd like
to know what the text you've been sure is wrong for 17 years really
says.


....Turing machine halting problem.
Simply stated, the problem is:
given the description of a Turing machine M
given the description of a Turing machine M
given the description of a Turing machine M
given the description of a Turing machine M
given the description of a Turing machine M

and an input w, does M, when started in the initial configuration q0w, perform a computation that eventually halts?

http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

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

When ⟨M⟩ = ⟨Ĥ⟩: I have proved that Ĥ ⟨M⟩ transitions to Ĥ.qn because M never reaches a final state.



--
Copyright 2021 Pete Olcott

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


Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 4 Aug 2021 22:38 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
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NNTP-Posting-Date: Wed, 04 Aug 2021 17:38:26 -0500
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[
Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct
]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <87sfzw3ao1.fsf@bsb.me.uk>
<7oKdnTjx4IC20p78nZ2dnUU7-TvNnZ2d@giganews.com> <875yws36vt.fsf@bsb.me.uk>
<j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk>
<Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk>
<qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk>
<K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk>
<woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk>
<w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk>
<xLednaPs_ZSXCZX8nZ2dnUU7-YnNnZ2d@giganews.com> <87o8af47y0.fsf@bsb.me.uk>
<NsudnY99rthDOJX8nZ2dnUU7-c_NnZ2d@giganews.com> <87zgtx2wxn.fsf@bsb.me.uk>
<Rr6dnWKhH4ejIJf8nZ2dnUU7-Q3NnZ2d@giganews.com> <87czqt2ewt.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 4 Aug 2021 17:38:26 -0500
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On 8/4/2021 2:22 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/4/2021 7:53 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/2/2021 8:45 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

As long as it is simply dismissed out-of-hand as a contradiction the
paradox remains unresolved.

There is no contradiction or paradox.  You Ĥ is just the wrong sort of
TM.  The proof you want to "refute" is talking about this sort of Ĥ:

     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
     if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts
Ĥ.qx correctly decides that its input: ⟨Ĥ⟩ ⟨Ĥ⟩ never halts

Maybe saying it a couple more times will help.  After four times I can
tell you that it's still wrong.  Maybe about a dozen more?

Whether what happens after Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is correct or not is determined
by Linz, not by you.  And you are clear that
    Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn.

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

As explained in complete detail below:
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Yes, please don't tell me the final state yet again.  This is not been
in dispute for some time.

because M applied to wM does not halt
where M is Machine_of(⟨Ĥ⟩) (1st param) above
and wM is ⟨Ĥ⟩ the second param above.

Because wM is referring to ⟨Ĥ⟩ and M is referring to the underlying
machine of ⟨Ĥ⟩ the last line above is translated to: if
Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ does not halt

That's convoluted.  ⟨Ĥ⟩ is the encoding of Ĥ so to find out what Linz
expects from Ĥ applied to ⟨Ĥ⟩ we just substitute M = Ĥ and wM = ⟨Ĥ⟩ into
the above:

   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt.


It is not the first Ĥ that is being referred to it is *only* the machine represented by the input ⟨Ĥ⟩ that is being referred to. That machine never reaches its final state.

Ĥ.q0 ⟨Ĥ[1]⟩ ⊢* Ĥ.qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ.qn
if Ĥ[1] applied to ⟨Ĥ[2]⟩ does not halt.

That you persistently ignore this distinction really seems to be a diverge from an honest dialogue.

We can know that Machine_of(⟨Ĥ⟩) applied to ⟨Ĥ⟩ never reaches its

Machine_of(⟨Ĥ⟩) = Ĥ.

final state whether or not the embedded halt decider at Ĥ.qx aborts it

You keep telling me that Ĥ applied to ⟨Ĥ⟩ reaches a final state.  You
even tell me the final state.

simulation of Machine_of(⟨Ĥ⟩), therefore we know that Machine_of(⟨Ĥ⟩)
never halts. Therefore we know that M applied to wM does not halt.

You keep telling me that Ĥ applied to ⟨Ĥ⟩ reaches the final state Ĥ.qn.
Are you changing your mind after all this time?  No, you are searching
for some form of words that will make the wrong answer sound right.



--
Copyright 2021 Pete Olcott

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


Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Thu, 5 Aug 2021 12:07 UTC
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NNTP-Posting-Date: Thu, 05 Aug 2021 07:07:05 -0500
Subject: Re:_Black_box_halt_decider_is_NOT_a_partial_decider_[ Ĥ.qx(⟨Ĥ⟩,⟨Ĥ⟩) == Ĥ.qn ] [ succinct ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <875yws36vt.fsf@bsb.me.uk> <j66dnbdHrpV8_p78nZ2dnUU7-aXNnZ2d@giganews.com> <87im0s0ydp.fsf@bsb.me.uk> <Brqdnfehrf0Kj5n8nZ2dnUU7-X3NnZ2d@giganews.com> <87tukblgjy.fsf@bsb.me.uk> <qtGdnfuXs4nFOZn8nZ2dnUU7-cnNnZ2d@giganews.com> <871r7ekugt.fsf@bsb.me.uk> <K5-dndGZo_-VmJv8nZ2dnUU78QvNnZ2d@giganews.com> <87czqxa0zk.fsf@bsb.me.uk> <woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk> <w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk> <xLednaPs_ZSXCZX8nZ2dnUU7-YnNnZ2d@giganews.com> <87o8af47y0.fsf@bsb.me.uk> <NsudnY99rthDOJX8nZ2dnUU7-c_NnZ2d@giganews.com> <87zgtx2wxn.fsf@bsb.me.uk> <Rr6dnWKhH4ejIJf8nZ2dnUU7-Q3NnZ2d@giganews.com> <87czqt2ewt.fsf@bsb.me.uk> <8YKdnVBaxrJ_i5b8nZ2dnUU7-aPNnZ2d@giganews.com> <87y29g23sk.fsf@bsb.me.uk> <no-dnYnca_3rxZb8nZ2dnUU7-YPNnZ2d@giganews.com> <20276f21-9bdc-4e64-9b20-1dc04bf8fa8an@googlegroups.com>
From: NoO...@NoWhere.com (olcott)
Date: Thu, 5 Aug 2021 07:07:04 -0500
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On 8/5/2021 4:02 AM, Malcolm McLean wrote:
On Thursday, 5 August 2021 at 04:18:21 UTC+1, olcott wrote:
On 8/4/2021 6:22 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

If the various strings you've chosen to number (badly) are not identical
then your "hat" construction is wrong. Linz's "hat" version makes an
exact copy.

When Bill says that his identical twin brother is not going to go to the
store, and then Bill goes to the store this does not make him a liar.

You are free to write ⟨Ĥ[99]⟩ if you like, but I am also free to change
that back to ⟨Ĥ⟩ because they are identical strings. Anything true of

Ĥ is not a string it is a Turing machine.
Turing machines are not identical to strings.
Ĥ halts the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its final state
whether or not the halt decider at Ĥ.qx stop simulating it.

If you want an actual honest dialogue we must have closure on some of
the points. You must say what things you agree with and not merely
ignore those things that you agree with.

Do you understand that the simulation of ⟨Ĥ⟩ on ⟨Ĥ⟩ never reaches its
final state whether or not the halt decider at Ĥ.qx stop simulating it?

H_Hat <H_Hat> halts. So if it is simulated by a UTM, the simulation
UTM <H_HAT><H_Hat> also halts.

It is very easily proven that ⟨Ĥ⟩ on ⟨Ĥ⟩ cannot possibly ever reach its final state. If it stops running without reaching is final state then this does not count as halting.

But the halt decider embedded in H_Hat is not a UTM. It's a near UTM,
that has abort logic. Somewhere the abort logic must be triggered,
which causes H_Hat<H_Hat> to halt.

If it stops running without reaching is final state then this does not count as halting. It can not possibly reach its final state whether or not its simulation is ever aborted, therefore it never halts.

H<H_Hat><H_Hat> also triggers this abort logic, so H<H_Hat><H_Hat>
reports "false" (non-halting).

If you disagree then to prove that you are not simply being disagreeable
you must proceed with a dialogue on this single point until we reach
mutual agreement. All other points will not be discussed until we reach
mutual agreement on this point.

We agree that H_Hat<H_Hat> halts.
We agree that H <H_Hat> <H_Hat> reports "false" (non-halting).

Can't you see the obvious?


The mistake that you are making is counting stopping without reaching a final state as halting, it is not.

--
Copyright 2021 Pete Olcott

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


Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction._[_Is_Ben_a_Liar_or_simply_woefully_ignorant?_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 11 Aug 2021 14:28 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
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NNTP-Posting-Date: Wed, 11 Aug 2021 09:28:32 -0500
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._[_Is_Ben_a_Liar_or_simply
_woefully_ignorant?_]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
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<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
<5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 09:28:31 -0500
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On 8/10/2021 9:26 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
     
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The question is not: Does Ĥ halt on its input?
Yes it is.

The question is:
Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
The ansswer to this question is provably no!
The question is: does Ĥ applied to ⟨Ĥ⟩ halt.  It does:

Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn  THIS IS NOT A CONTRADICTION
Indeed.  There is no contradiction.  Just an Ĥ that does not meet Linz
spec.

Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
Because it is correct it meets the Linz spec.
I find it startling that you think that, but then it seems you don't yet
know what the key words mean:

if M applied to wM does not halt
means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
No.  Would you like to know "what M applied to wM does not halt" means?
Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
applied to wM halts"?

        the Turing machine halting problem. Simply stated, the problem
        is: given the description of a Turing machine M and an input w,
        does M, when started in the initial configuration q0w, perform a
        computation that eventually halts? (Linz:1990:317).
Yes.  I was offering to help you understand the key words in that text.

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
You've missed off the key lines yet again.  Is that deliberate?  They
are the lines that show you are wrong so I am suspicious that you keep
omitting them.

When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
Ungrammatical.

When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
input it is a final state of the executed Ĥ.
Yes.  You don't seem to know why that's wrong.

What is your basis for believing that is wrong?
Ah, a question about what I'm saying.  I can help there.  The basis is
what Linz says about Ĥ.  He says that (translating to your notation)
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt".  But, as you can
see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
state).  Your Ĥ is not doing what it should in this one crucial case.


    the Turing machine halting problem. Simply stated, the problem
    is: given the description of a Turing machine M and an input w,
    does M, when started in the initial configuration q0w, perform a
    computation that eventually halts? (Linz:1990:317).

and so on.  Same old stuff.


When the challenge to support one's assertion with reasoning is simply ignored as you are ignoring it right now one can reasonably construe a deceptive intent.

-- the Turing machine halting problem. Simply stated, the problem
-- is: given the description of a Turing machine M and an input w,
-- does M, when started in the initial configuration q0w, perform a
-- computation that eventually halts? (Linz:1990:317).

PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H

The input to H will be the description (encoded in some form) of M, say
WM, as well as the input w. (Linz:1990:318)

H.q0 WM w ⊢* H.qn
if M applied to W does not halt.

   becomes

H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.


Pages of the Linz text to verify the above quotes in their full context:
http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx

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

When we know that M refers to the Turing machine specified by the first wM then when Ĥ transitions to its final state of Ĥ.qn there is no direct contradiction formed.

Can you admit when you are wrong when you really are wrong?
Can you admit when you are wrong when you really are wrong?
Can you admit when you are wrong when you really are wrong?
Can you admit when you are wrong when you really are wrong?

if M applied to wM does not halt (see above for definition of M)
means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.

Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.

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

I'm sorry my explanation did not help at all.  I'm happy to answer any
other questions you might have if you think it might help you understand
what I (and Linz) are saying.



--
Copyright 2021 Pete Olcott

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


Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction._[_Is_Ben_a_Liar_or_simply_woefully_ignorant?_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 11 Aug 2021 14:58 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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NNTP-Posting-Date: Wed, 11 Aug 2021 09:58:26 -0500
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._[_Is_Ben_a_Liar_or_simply
_woefully_ignorant?_]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc> <87czqxa0zk.fsf@bsb.me.uk>
<woudnXWBxPba95r8nZ2dnUU78ffNnZ2d@giganews.com> <87mtpz64sq.fsf@bsb.me.uk>
<w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk>
<1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <87tuk52h0e.fsf@bsb.me.uk>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
<5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk>
<346dnYhkWPUNQ478nZ2dnUU7-UPNnZ2d@giganews.com>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 09:58:25 -0500
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On 8/11/2021 9:28 AM, olcott wrote:
On 8/10/2021 9:26 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The question is not: Does Ĥ halt on its input?
Yes it is.

The question is:
Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
The ansswer to this question is provably no!
The question is: does Ĥ applied to ⟨Ĥ⟩ halt.  It does:

Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn  THIS IS NOT A CONTRADICTION
Indeed.  There is no contradiction.  Just an Ĥ that does not meet Linz
spec.

Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
Because it is correct it meets the Linz spec.
I find it startling that you think that, but then it seems you don't yet
know what the key words mean:

if M applied to wM does not halt
means if the execution of the machine of the first ⟨Ĥ⟩ on its input of
the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
No.  Would you like to know "what M applied to wM does not halt" means?
Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a case of "M
applied to wM halts"?

        the Turing machine halting problem. Simply stated, the problem
        is: given the description of a Turing machine M and an input w,
        does M, when started in the initial configuration q0w, perform a
        computation that eventually halts? (Linz:1990:317).
Yes.  I was offering to help you understand the key words in that text.

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
You've missed off the key lines yet again.  Is that deliberate?  They
are the lines that show you are wrong so I am suspicious that you keep
omitting them.

When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine and its
input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at Ĥ.qx.
Ungrammatical.

When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
input it is a final state of the executed Ĥ.
Yes.  You don't seem to know why that's wrong.

What is your basis for believing that is wrong?
Ah, a question about what I'm saying.  I can help there.  The basis is
what Linz says about Ĥ.  He says that (translating to your notation)
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt".  But, as you can
see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
state).  Your Ĥ is not doing what it should in this one crucial case.


    the Turing machine halting problem. Simply stated, the problem
    is: given the description of a Turing machine M and an input w,
    does M, when started in the initial configuration q0w, perform a
    computation that eventually halts? (Linz:1990:317).

and so on.  Same old stuff.


When the challenge to support one's assertion with reasoning is simply ignored as you are ignoring it right now one can reasonably construe a deceptive intent.

-- the Turing machine halting problem. Simply stated, the problem
-- is: given the description of a Turing machine M and an input w,
-- does M, when started in the initial configuration q0w, perform a
-- computation that eventually halts? (Linz:1990:317).

PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H

The input to H will be the description (encoded in some form) of M, say
WM, as well as the input w. (Linz:1990:318)

H.q0 WM w ⊢* H.qn
if M applied to W does not halt.

   becomes

H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.


Pages of the Linz text to verify the above quotes in their full context:
http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx

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

When we know that M refers to the Turing machine specified by the first wM then when Ĥ transitions to its final state of Ĥ.qn there is no direct contradiction formed.

Can you admit when you are wrong when you really are wrong?
Can you admit when you are wrong when you really are wrong?
Can you admit when you are wrong when you really are wrong?
Can you admit when you are wrong when you really are wrong?

if M applied to wM does not halt (see above for definition of M)
means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.

Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.

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

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
if M applied to wM halts, and    // M refers to the TM of the first wM parameter to Ĥ.qx

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt // M refers to the TM of the first wM parameter to Ĥ.qx



--
Copyright 2021 Pete Olcott

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


Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction._[_Is_Ben_a_Liar_or_simply_woefully_ignorant?_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Wed, 11 Aug 2021 19:53 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
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NNTP-Posting-Date: Wed, 11 Aug 2021 14:53:11 -0500
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._[_Is_Ben_a_Liar_or_simply
_woefully_ignorant?_]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc>
<w5edne8d06OkxpX8nZ2dnUU7-b_NnZ2d@giganews.com> <87bl6f5qvy.fsf@bsb.me.uk>
<1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <87tuk52h0e.fsf@bsb.me.uk>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
<5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk>
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<Ja-dneAl3poPeI78nZ2dnUU7-UvNnZ2d@giganews.com> <87fsvggdxz.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 14:53:11 -0500
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On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/11/2021 9:28 AM, olcott wrote:

Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

This is a scumbag move.  It's also cowardly and disingenuous --
pretending to be coy about your scumbag opinions.  I'll leave it
unedited as it says everything readers need to know about your
character.

When the challenge to support one's assertion with reasoning is simply
ignored as you are ignoring it right now one can reasonably construe
a deceptive intent.

A scumbag could construe it that way.  Reasonable people could come to
all sorts of other conclusions.

H.q0 WM w ⊢* H.qn
if M applied to W does not halt.
if M applied to MW does not halt.

(typo corrected)

    becomes
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

Yes.  After a lot a pressing (and I mean lots, over several years!) you
eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts.  That's why your H
(and its associated Ĥ) are wrong but for some reason you can't see this.

You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.  And you
will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.

This should be the end of the matter, but apparently your stating facts
that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
wrong.  I don't think I know any way to make progress on this.

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt
...
if M applied to wM does not halt (see above for definition of
M) means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.

Of course.  We all know that.  Unless you are pulling a fast one.  "the
Turing machine of ⟨Ĥ⟩" is just Ĥ.  Is there a reason you are not simply
saying "when Ĥ applied to ⟨Ĥ⟩ does not halt"?

Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a
UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.

Yes, we know the ruse: the computation would not halt if it were not the
computation that it is.  You've been trying to pull off this trick ever
since the infamous "it wouldn't halt if line 15 was commented out"
admission.  Your Ĥ, however, not being a UTM, has the property that

   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

when it should not.  Ĥ.q0 ⟨Ĥ⟩ should eventually transition to qn only
"if Ĥ applied to ⟨Ĥ⟩ does not halt" (or, as you rather suspiciously
write "if the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt").  But
you've told us, time and time again, that Ĥ applied to ⟨Ĥ⟩ halts.  You
keep showing us the summary description of it's configuration sequence.
You keep showing us the final state it transitions to.  It's so obvious,
someone would need about 16 years misunderstanding Turing machines to
avoid seeing it.

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
if M applied to wM halts, and // M refers to the TM of the first wM
parameter to Ĥ.qx

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt // M refers to the TM of the first wM
parameter to Ĥ.qx

The case you care about has M = Ĥ and wM = ⟨Ĥ⟩ as you've written it out
above.


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

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn  // see highlighted portion of Linz text to confirm:
if M applied to wM does not halt     // M refers to the TM of the first wM parameter to Ĥ.qx

Turing machine Ĥ is applied to its input ⟨Ĥ⟩. It copies this input such that this input and the copy of this input become the first  and second parameters to the simulating halt decider at Ĥ.qx. When  Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ decides that the simulation of its first parameter on the input of its second parameter never halt it correctly transitions to its own final state of  Ĥ.qn.

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

--
Copyright 2021 Pete Olcott

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


Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction._[_Is_Ben_a_Liar_or_simply_woefully_ignorant?_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Thu, 12 Aug 2021 00:26 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
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NNTP-Posting-Date: Wed, 11 Aug 2021 19:26:17 -0500
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._[_Is_Ben_a_Liar_or_simply
_woefully_ignorant?_]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc>
<1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <87tuk52h0e.fsf@bsb.me.uk>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
<5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk>
<346dnYhkWPUNQ478nZ2dnUU7-UPNnZ2d@giganews.com>
<Ja-dneAl3poPeI78nZ2dnUU7-UvNnZ2d@giganews.com> <87fsvggdxz.fsf@bsb.me.uk>
<4sCdnb5sOOQ6t4n8nZ2dnUU7-X3NnZ2d@giganews.com> <87pmujfs4o.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 19:26:14 -0500
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On 8/11/2021 7:01 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/11/2021 9:28 AM, olcott wrote:
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no
contradiction. [ Is Ben a Liar or simply woefully ignorant? ]

This is a scumbag move.  It's also cowardly and disingenuous --
pretending to be coy about your scumbag opinions.  I'll leave it
unedited as it says everything readers need to know about your
character.

This is still a scumbag move.

When the challenge to support one's assertion with reasoning is simply
ignored as you are ignoring it right now one can reasonably construe
a deceptive intent.
A scumbag could construe it that way.  Reasonable people could come to
all sorts of other conclusions.

H.q0 WM w ⊢* H.qn
if M applied to W does not halt.
if M applied to MW does not halt.
(typo corrected)

     becomes
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

Yes.  After a lot a pressing (and I mean lots, over several years!) you
eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts.  That's why your H
(and its associated Ĥ) are wrong but for some reason you can't see this.

You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.  And you
will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.

This should be the end of the matter, but apparently your stating facts
that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
wrong.  I don't think I know any way to make progress on this.

You just read past this explanation of why your H and Ĥ are wrong and
you decided that you understood everything I'd said, did you?  And you
also decided that nothing in these three paragraphs needs to be
challenged?


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

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

// M refers to the TM of the first wM parameter to Ĥ.qx

Now that you accept that the above is true we can move on to the next point. My proof must proceed exactly one point at a time an cannot possibly move to the next point until the current point is fully accepted.

That you believe that the fact that Ĥ applied to ⟨Ĥ⟩ transitions to its final state of Ĥ.qn and halts nullifies the fact that Ĥ.qx wM wM does correctly decide that its input never halts is the next point.

It is easier to discuss this with the H(P,P) model of the exact same thing because with the H(P,P) model we can examine every single detail.

_P()
[00000d02](01)  55          push ebp
[00000d03](02)  8bec        mov ebp,esp
[00000d05](03)  8b4508      mov eax,[ebp+08]
[00000d08](01)  50          push eax       // push 2nd Param
[00000d09](03)  8b4d08      mov ecx,[ebp+08]
[00000d0c](01)  51          push ecx       // push 1st Param
[00000d0d](05)  e870feffff  call 00000b82  // call H
[00000d12](03)  83c408      add esp,+08
[00000d15](02)  85c0        test eax,eax
[00000d17](02)  7402        jz 00000d1b
[00000d19](02)  ebfe        jmp 00000d19
[00000d1b](01)  5d          pop ebp
[00000d1c](01)  c3          ret
Size in bytes:(0027) [00000d1c]

....[00000d0d][00101829][00000d12] e870feffff  call 00000b82  // call H

Begin Local Halt Decider Simulation at Machine Address:d02
....[00000d02][002118f1][002118f5] 55          push ebp
....[00000d03][002118f1][002118f5] 8bec        mov ebp,esp
....[00000d05][002118f1][002118f5] 8b4508      mov eax,[ebp+08]
....[00000d08][002118ed][00000d02] 50          push eax       // push P
....[00000d09][002118ed][00000d02] 8b4d08      mov ecx,[ebp+08]
....[00000d0c][002118e9][00000d02] 51          push ecx       // push P
....[00000d0d][002118e5][00000d12] e870feffff  call 00000b82  // call H

....[00000d02][0025c319][0025c31d] 55          push ebp
....[00000d03][0025c319][0025c31d] 8bec        mov ebp,esp
....[00000d05][0025c319][0025c31d] 8b4508      mov eax,[ebp+08]
....[00000d08][0025c315][00000d02] 50          push eax       // push P
....[00000d09][0025c315][00000d02] 8b4d08      mov ecx,[ebp+08]
....[00000d0c][0025c311][00000d02] 51          push ecx       // push P
....[00000d0d][0025c30d][00000d12] e870feffff  call 00000b82  // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

It is very obvious to people that know the x86 language very well that while H acts as a pure simulator of P(P) that P cannot possibly stop running.

The same thing applies to the infinite cycle from Ĥ.qx to Ĥ.q0 while the simulating halt decider at Ĥ.qx acts as a UTM.

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

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

// M refers to the TM of the first wM parameter to Ĥ.qx


There's not much more I can add if you do indeed understand and accept
these remarks of mine.  But, as I keep saying, I'm happy to answer any
questions you may have about my explanation.



--
Copyright 2021 Pete Olcott

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


Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction._[_Halting_Problem_is_solved for Ĥ on ⟨Ĥ⟩ ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Thu, 12 Aug 2021 02:14 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
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NNTP-Posting-Date: Wed, 11 Aug 2021 21:14:01 -0500
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._[_Halting_Problem_is_solv
ed for Ĥ on ⟨Ĥ⟩ ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc>
<1ZednUt-qbrXlZf8nZ2dnUU7-QnNnZ2d@giganews.com> <87tuk52h0e.fsf@bsb.me.uk>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
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<zKudnTxkSrW1xYn8nZ2dnUU7-bnNnZ2d@giganews.com> <87r1eze981.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Wed, 11 Aug 2021 21:14:00 -0500
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On 8/11/2021 8:35 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/11/2021 11:10 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 8/11/2021 9:28 AM, olcott wrote:

H.q0 WM w ⊢* H.qn
if M applied to W does not halt.
if M applied to MW does not halt.
(typo corrected)

     becomes
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.
Yes.  After a lot a pressing (and I mean lots, over several years!) you
eventually admitted that H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ does indeed transition to H.qn.
You also keep telling us that Ĥ applied to ⟨Ĥ⟩ halts.  That's why your H
(and its associated Ĥ) are wrong but for some reason you can't see this.
You will plainly state that H rejects the string "⟨Ĥ⟩ ⟨Ĥ⟩" which it
should do if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.  And you
will, time and time again, show us exactly how Ĥ applied to ⟨Ĥ⟩ halts.
This should be the end of the matter, but apparently your stating facts
that show that H and Ĥ are wrong does not mean you know that H and Ĥ are
wrong.  I don't think I know any way to make progress on this.

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
if M applied to wM does not halt
...
if M applied to wM does not halt (see above for definition of
M) means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
Of course.  We all know that.  Unless you are pulling a fast one.  "the
Turing machine of ⟨Ĥ⟩" is just Ĥ.  Is there a reason you are not simply
saying "when Ĥ applied to ⟨Ĥ⟩ does not halt"?

I didn't expect an answer.  You may be reserving this phrase as a
get-out clause for later, so saying what you mean too early will block
that escape route.

Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a
UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.
Yes, we know the ruse: the computation would not halt if it were not the
computation that it is.  You've been trying to pull off this trick ever
since the infamous "it wouldn't halt if line 15 was commented out"
admission.  Your Ĥ, however, not being a UTM, has the property that
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
when it should not.  Ĥ.q0 ⟨Ĥ⟩ should eventually transition to qn only
"if Ĥ applied to ⟨Ĥ⟩ does not halt" (or, as you rather suspiciously
write "if the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt").

We can see that the above never halts.

We can see that it halts.  It's right there in a fact that has been
undisputed for months:

  Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn


The above Ĥ halts only because Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input never halts.

   the Turing machine halting problem. Simply stated, the problem
   is: given the description of a Turing machine M and an input w,
   does M, when started in the initial configuration q0w, perform a
   computation that eventually halts? (Linz:1990:317).

Thus the halting problem is solved for this input because the halting problem only applies to inputs. It does not apply to computations that are not inputs.


--
Copyright 2021 Pete Olcott

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


Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct_and_forms_no_contradiction._[_Ben_accepts_one_point? ] (typo fixed )
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.math.symbolic
Date: Thu, 12 Aug 2021 20:36 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
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NNTP-Posting-Date: Thu, 12 Aug 2021 15:36:02 -0500
Subject: Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is
correct_and_forms_no_contradiction._[_Ben_accepts_one_point
? ] (typo fixed )
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc>
<zcadnTSOD5rtZ5f8nZ2dnUU7-T3NnZ2d@giganews.com> <877dh03l3c.fsf@bsb.me.uk>
<Z5adnd038KGXwJb8nZ2dnUU7-I_NnZ2d@giganews.com> <8735rn1qvj.fsf@bsb.me.uk>
<goydnfCCIYUWE5H8nZ2dnUU7-e_NnZ2d@giganews.com> <87eeb7z4d1.fsf@bsb.me.uk>
<0_Sdnb6Qe8XGOZH8nZ2dnUU7-U3NnZ2d@giganews.com> <87zgtslqpv.fsf@bsb.me.uk>
<4JOdnRS2SLR7MYz8nZ2dnUU7-YnNnZ2d@giganews.com> <87zgtoizgp.fsf@bsb.me.uk>
<Z6ednWt7SpIIv478nZ2dnUU7-WHNnZ2d@giganews.com> <87pmukiwr5.fsf@bsb.me.uk>
<5c6dnbn2gMMIsI78nZ2dnUU7-dnNnZ2d@giganews.com> <87eeb0iuo1.fsf@bsb.me.uk>
<346dnYhkWPUNQ478nZ2dnUU7-UPNnZ2d@giganews.com>
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<yuGdneaGkeN-84n8nZ2dnUU7-cnNnZ2d@giganews.com> <87bl62cu25.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Thu, 12 Aug 2021 15:36:00 -0500
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On 8/12/2021 3:00 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

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

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

// M refers to the TM of the first wM parameter to Ĥ.qx

Now that you accept that the above is true...

I don't.  It's a garbled formula arising from a silly error on your
part.  I'd like to know where (you think) I said I accept this nonsense
so I can correct any such impression.

... we can move on to the next point. My proof must proceed exactly
one point at a time an cannot possibly move to the next point until
the current point is fully accepted.

It would be simpler if we worked though the reasons you are wrong
because there are fewer steps.

You've stated that

   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

and you accept (at least you keep quoting) that Linz requires that this
should be the case only if Ĥ applied to ⟨Ĥ⟩ does not halt.  QED.

That you believe that the fact that Ĥ applied to ⟨Ĥ⟩ transitions to
its final state of Ĥ.qn and halts nullifies the fact that Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
does correctly decide that its input never halts is the next point.

⟨Ĥ⟩ ⟨Ĥ⟩ is a string that encodes a halting computation as shown a few
lines above.  That Ĥ applied to ⟨Ĥ⟩ halts does not "nullify" anything,
it's just wrong as clearly stated by Linz.


Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
⟨Ĥ⟩ ⟨Ĥ⟩ is not a string that encodes a halting computation.

Ĥ is a TM that halts only because
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input never halts.

When you examine this in its x86utm equivalent of H(P,P) there are no loopholes that can slip through the cracks because every detail is explicitly encoded in the x86 language.

When we examine this as Ĥ applied to ⟨Ĥ⟩ there are millions of pages of Turing machine code that cannot be explicitly specified.

None-the-less the key element of all this is the fact that if we assume that the simulating halt decider at Ĥ.qx is simply a UTM then it becomes quite obvious that we have an infinite cycle from Ĥ.qx to Ĥ.q0.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

If we know that we have an infinite cycle then this knowledge all by itself proves that the simulating halt decider at Ĥ.qx must abort the simulation of its input which proves that this input never halts.


--
Copyright 2021 Pete Olcott

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


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