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devel / comp.theory / Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)

SubjectAuthor
* What if a cat barks?olcott
+* What if a cat barks?Chris M. Thomasson
|+* What if a cat barks?Ben Bacarisse
||`* What if a cat barks?Chris M. Thomasson
|| +* What if a cat barks?olcott
|| |`* What if a cat barks?Chris M. Thomasson
|| | `* What if a cat barks?olcott
|| |  `* What if a cat barks?Chris M. Thomasson
|| |   `* What if a cat barks?Richard Damon
|| |    `* What if a cat barks?Chris M. Thomasson
|| |     `* What if a cat barks?Richard Damon
|| |      +- What if a cat barks?Daniel Pehoushek
|| |      `* What if a cat barks?wij
|| |       `* What if a cat barks?Chris M. Thomasson
|| |        `* What if a cat barks?wij
|| |         `- What if a cat barks?Chris M. Thomasson
|| `* What if a cat barks?Ben Bacarisse
||  `* What if a cat barks? [ sound deduction proves that I am correct ]olcott
||   `* What if a cat barks? [ sound deduction proves that I am correct ]Richard Damon
||    +* What if a cat barks? [ sound deduction proves that I am correct ]olcott
||    |`* What if a cat barks? [ sound deduction proves that I am correct ]Richard Damon
||    | `* What if a cat barks? [ sound deduction proves that I am correct ]Malcolm McLean
||    |  +* What if a cat barks? [ sound deduction proves that I am correct ]olcott
||    |  |`- What if a cat barks? [ sound deduction proves that I am correct ]Richard Damon
||    |  +* What if a cat barks? [ sound deduction proves that I am correct ]Ben Bacarisse
||    |  |+- What if a cat barks? [ sound deduction proves that I am correct ]olcott
||    |  |`* What if a cat barks? [ sound deduction proves that I am correct ]Malcolm McLean
||    |  | +* What if a cat barks? [ sound deduction proves that I am correct ]Ben Bacarisse
||    |  | |`* What if a cat barks? [ sound deduction proves that I am correct ]olcott
||    |  | | `- What if a cat barks? [ sound deduction proves that I am correct ]Richard Damon
||    |  | `* What if a cat barks? [ sound deduction proves that I am correct ]olcott
||    |  |  `* What if a cat barks? [ sound deduction proves that I am correct ]Malcolm McLean
||    |  |   `- What if a cat barks? [ sound deduction proves that I am correct ]olcott
||    |  `- What if a cat barks? [ sound deduction proves that I am correct ]Richard Damon
||    `* What if a cat barks? [ sound deduction proves that I am correct ]Ben Bacarisse
||     `- What if a cat barks? [ sound deduction proves that I am correct ]olcott
|`* What if a cat barks?André G. Isaak
| `* What if a cat barks?olcott
|  +* What if a cat barks?olcott
|  |`- What if a cat barks?Daniel Pehoushek
|  +* What if a cat barks?Richard Damon
|  |`* What if a cat barks?olcott
|  | +* What if a cat barks?André G. Isaak
|  | |`* What if a cat barks?olcott
|  | | `- What if a cat barks?Richard Damon
|  | `- What if a cat barks?Richard Damon
|  `* What if a cat barks?André G. Isaak
|   `* What if a cat barks?olcott
|    +* What if a cat barks?Richard Damon
|    |`* What if a cat barks?olcott
|    | `* What if a cat barks?Richard Damon
|    |  `* What if a cat barks?olcott
|    |   `- What if a cat barks?Richard Damon
|    `* What if a cat barks?André G. Isaak
|     `* What if a cat barks?olcott
|      `* What if a cat barks?André G. Isaak
|       `* What if a cat barks?olcott
|        `* What if a cat barks?André G. Isaak
|         `* What if a cat barks?olcott
|          `* What if a cat barks?André G. Isaak
|           `* What if a cat barks?olcott
|            +- What if a cat barks?Daniel Pehoushek
|            `* What if a cat barks?André G. Isaak
|             `* What if a cat barks? [ sound deduction is a proof ]olcott
|              +* What if a cat barks? [ sound deduction is a proof ]André G. Isaak
|              |`* What if a cat barks? [ sound deduction is a proof ]olcott
|              | +* What if a cat barks? [ sound deduction is a proof ]Richard Damon
|              | |`* What if a cat barks? [ sound deduction is a proof ](axiom)olcott
|              | | `* What if a cat barks? [ sound deduction is a proof ](axiom)Richard Damon
|              | |  `* What if a cat barks? [ sound deduction is a proof ](axiom)olcott
|              | |   `* What if a cat barks? [ sound deduction is a proof ](axiom)Richard Damon
|              | |    `* What if a cat barks? [ sound deduction is a proof ](axiom)olcott
|              | |     `* What if a cat barks? [ sound deduction is a proof ](axiom)Richard Damon
|              | |      `* What if a cat barks? [ sound deduction is a proof ](axiom)olcott
|              | |       `* What if a cat barks? [ sound deduction is a proof ](axiom)Richard Damon
|              | |        `* What if a cat barks? [ sound deduction is a proof ](axiom)olcott
|              | |         `* What if a cat barks? [ sound deduction is a proof ](axiom)Richard Damon
|              | |          `* What if a cat barks? [ sound deduction is a proof ](infiniteolcott
|              | |           `* What if a cat barks? [ sound deduction is a proof ](infiniteRichard Damon
|              | |            `* What if a cat barks? [ sound deduction is a proof ](infiniteolcott
|              | |             `* What if a cat barks? [ sound deduction is a proof ](infiniteRichard Damon
|              | |              `* What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)olcott
|              | |               +* What if a cat barks? [ sound deduction is a proof ](infiniteChris M. Thomasson
|              | |               |`* What if a cat barks? [ sound deduction is a proof ](infiniteJeff Barnett
|              | |               | `* What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)Ben Bacarisse
|              | |               |  +* What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)(Bolcott
|              | |               |  |+* What if a cat barks? [ sound deduction is a proof ](infiniteRichard Damon
|              | |               |  ||`* What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)olcott
|              | |               |  || `- What if a cat barks? [ sound deduction is a proof ](infiniteRichard Damon
|              | |               |  |`* What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)(BBen Bacarisse
|              | |               |  | +* What if a cat barks? [ sound deduction is a proof ](infiniteRichard Damon
|              | |               |  | |`- What if a cat barks? [ sound deduction is a proof ](infiniteDaniel Pehoushek
|              | |               |  | `* What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)(kolcott
|              | |               |  |  +* What if a cat barks? [ sound deduction is a proof ](infiniteRichard Damon
|              | |               |  |  |+* What if a cat barks? [ sound deduction is a proof ](infiniteolcott
|              | |               |  |  ||`* What if a cat barks? [ sound deduction is a proof ](infiniteRichard Damon
|              | |               |  |  || `* What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)olcott
|              | |               |  |  ||  `* What if a cat barks? [ sound deduction is a proof ](infiniteRichard Damon
|              | |               |  |  ||   `* What if a cat barks? [ sound deduction is a proof ](infiniteolcott
|              | |               |  |  ||    `* What if a cat barks? [ sound deduction is a proof ](infiniteRichard Damon
|              | |               |  |  ||     `* What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)(dolcott
|              | |               |  |  |`- What if a cat barks? [ sound deduction is a proof ](infiniteDaniel Pehoushek
|              | |               |  |  `* What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)(kBen Bacarisse
|              | |               |  `* What if a cat barks? [ sound deduction is a proof ](infiniteJeff Barnett
|              | |               `* What if a cat barks? [ sound deduction is a proof ](infiniteRichard Damon
|              | `* What if a cat barks? [ sound deduction is a proof ]André G. Isaak
|              `- What if a cat barks? [ sound deduction is a proof ]Richard Damon
+* What if a cat barks?wij
+* What if a cat barks?Malcolm McLean
+- What if a cat barks?Richard Damon
+* What if a cat barks? [ How can a cat bark? ]olcott
`* What if a cat barks?Peter

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

<OlbBI.605613$J_5.348305@fx46.iad>

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](axiom)
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From: Rich...@Damon-Family.org (Richard Damon)
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Date: Thu, 24 Jun 2021 22:39:42 -0400
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 by: Richard Damon - Fri, 25 Jun 2021 02:39 UTC

On 6/24/21 11:42 AM, olcott wrote:
> On 6/24/2021 6:29 AM, Richard Damon wrote:
>> On 6/23/21 11:03 PM, olcott wrote:
>>> If the result of purely hypothetical H that never aborts its
>>> simulation/execution of P would result in the infinite execution of P
>>> then we know axiomatically that H decides that P never halts correctly.
>>
>> Right, The P based on the never aborting H has infinite execution, but
>> that isn't the P we are looking at,
>
> That hypothetical H/P proves beyond all possible doubt that the in the
> actual H(P,P) the input to H never halts on the basis of this axiom:
>
> Premise(1) Every computation that never halts unless its simulation is
> aborted is a computation that never halts. This verified as true on the
> basis of the meaning of its words.
>

Except that that the two P's are different. You are saying that the
Black cat is white because you are looking at the white dog instead.

Remember, EVERY time you change the properties of H, you get a DIFFERENT P.

FAIL.

Maybe you should use a different notion, go back to the ^ notation, and
give every different version of H a different name.

If Hn never aborts its input, then Hn^(Hn^) is a non-halting
computationo as it gets into an infinite loop, but then Hn(Hn^, Hn^)
gets stuck in that same infinite loop, since it doesn't abort its
simulation, and thus fails to answer.

If we switch to Ha that does abort then we don't have the case that
Ha^(Ha^) getting stuck in an infinite loop as Ha^ uses Ha which will
abort, thus all we have is that Ha isn't smart enough to figure out what
Ha^ will do, so it gives the wrong answer, since it is treating Ha^ as
if it was Hn^ which ARE different computations.

Yes, If we did Ha(Hn^, Hn^) it would get the right answer of
non-halting, but that isn't the case that Linz was talking about, but
which you seem to want to try to make it the case you are looking at.

Hn^ is only designed to confound Hn (but Hn doesn't need to be
confounded). To show Ha gets something wrong, Linz proposes we give it
Ha^, not Hn^.

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

<1xbBI.88042$Vh1.61135@fx21.iad>

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Subject: Re: What if a cat barks? [ sound deduction proves that I am correct ]
Newsgroups: comp.theory
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Fri, 25 Jun 2021 02:51 UTC

On 6/24/21 11:39 AM, olcott wrote:

>
> // Simplified Linz Ĥ (Linz:1990:319)
> void P(u32 x)
> {
>   u32 Input_Halts = H(x, x);
>   if (Input_Halts)
>     HERE: goto HERE;
> }
>
>
> int main()
> {
>   u32 Input_Halts = H((u32)P, (u32)P);
>   Output("Input_Halts = ", Input_Halts);
> }
>
> That H(P,P) cannot return a correct halt status to P in the above
> computation is well proven.
>
> That H(P,P) ever needs to return to P at all in the above code is the
> key false assumption that unravels the whole proof.

But H needs to be a computation to be a Halting Decider. Computations
always return the same answer to a given question EVERY time it is
computed, the if H(P,P) answers when directly asked, it must also do so
when asked by P.

I thought we got this cleared up months ago when you admitted you didn't
understand what a computation was.

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

But to be computationally equivalent to a Turing Machine, they need to
be actual Computation, thus your claim that it doesn't need to return
its answer to P is false. Maybe you don't really understad what it means
to be computationally equivalent. Maybe you think that just means looks
something like each other,

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

And How is Halting correctly decided as Non-Halting. You make this claim
with what seems to be a straight face, which is a total absurdity.

That is like saying 1 is the same as infinity. It is just crazy talk.

Maybe you just don't understand what it means for two things to be
equivalent, and everything can just be equivalent if you really want to.

This IS the natural result of your logic going inconsistent.

Re: What if a cat barks?

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Subject: Re: What if a cat barks?
From: wyni...@gmail.com (wij)
Injection-Date: Fri, 25 Jun 2021 07:39:45 +0000
Content-Type: text/plain; charset="UTF-8"
 by: wij - Fri, 25 Jun 2021 07:39 UTC

On Friday, 25 June 2021 at 07:26:09 UTC+8, Chris M. Thomasson wrote:
> On 6/22/2021 5:45 PM, wij wrote:
> > On Wednesday, 23 June 2021 at 06:57:58 UTC+8, Richard Damon wrote:
> >> On 6/22/21 3:22 PM, Chris M. Thomasson wrote:
> >>> On 6/21/2021 6:54 PM, Richard Damon wrote:
> >>>>
> >>>> Well, the problem is that Turing Machines CAN'T be black boxes. And the
> >>>> definition of the Halting Problem is that the decider is given a full
> >>>> description of the Turing Machine, which is basically like a full
> >>>> listing of the program.
> >>>
> >>> Oh, sorry. I thought his x86 simulator would run an x86 program and
> >>> determine if it would halt or not. My bad. The assembled program would
> >>> have all the information he needs right? Or, would I have to give him
> >>> source code... Humm, I don't know. Does he have an assembler?
> >>>
> >>> When I say "black box", I was basically referring to a program that
> >>> somebody else assembled into an executable.
> >>>
> >>> His simulator, as-is, should be able to simulate any x86 program and
> >>> determine if it halts or not... Sound Kosher?
> >>>
> >>>
> >> The problem is that the term 'black box' tend to mean something that you
> >> can't look inside of, which means that his simulator couldn't get at the
> >> object code of it.
> >>
> >> A program that was a real black box would have its internals encrypted
> >> and have test to make sure that it wasn't being 'spied' on before
> >> decrypting itself. Of course, without support from the OS, or some
> >> controlling program, things can't be perfect black boxes, just enough to
> >> be a pain to look into.
> >
> > Black box program can be executed, like any app., by simulator or OS.
> > The same as oracle can be used in TM, P can be a black box to H, black body
> > in physics.
> >
>
> A valid x86 program should just run on his simulator. The simulator
> would start getting instructions and executing them. The question is,
> will the instruction stream halt, or not? I am guessing that his system
> would spawn a simulator in a sandbox or something, with a target
> program. The simulator fires up, initializes itself, and starts
> executing the target program.

// program p.cpp
//-------------------
//extern bool H(Func p);
void P();

inline static bool H2(Func p) {
// Rewrite H to simulate H (or not).
// The rewrite can be in very different but functionally equivalent ways.
}

static const bool r=H2(P); // equivalent to r=H(P)

void P() {
if(r) {
for(;;) {}; // If H2(P)==true, P is in infinite loop.
}
};

int main() {
P();
};

// Note: P is written this way to demonstrate a way that H cannot possibly detect
// being traced by P.
//-----------

If H is a pure simulator, H(P) is doomed to be in infinite recursive call.
If H(f) is specified to return true iff f() returns, and not allowed to halt,
then the question is equivalent to asking for the value of r.

If r==true, H spec. says the execution of P() will return, but the compiled
executable of p.cpp will disagree (run in infinite loop).
If r==false, H spec. says the execution of P() will not return, but the compiled
executable of p.cpp will return.

I see no way function H can be done even using quantum computers.

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ]
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References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com> <sap6l7$130c$5@gioia.aioe.org> <saqlhk$860$1@dont-email.me> <tZOdneTmHpHNnUz9nZ2dnUU7-V3NnZ2d@giganews.com> <sarcip$jgd$1@dont-email.me> <zZOdnUG2JPuxokz9nZ2dnUU7-UPNnZ2d@giganews.com> <sarhu7$f83$1@dont-email.me> <E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com> <sat46g$afu$1@dont-email.me> <4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com> <satao3$gjk$1@dont-email.me> <YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me> <48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me> <6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com> <sb0kra$ert$1@dont-email.me> <kaudnZS1vPJiaE79nZ2dnUU7-QHNnZ2d@giganews.com> <sb2si0$v6j$1@dont-email.me> <0dWdnYvF0fc0Zkn9nZ2dnUU7-f_NnZ2d@giganews.com> <sb399l$l2e$1@dont-email.me>
From: NoO...@NoWhere.com (olcott)
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 by: olcott - Fri, 25 Jun 2021 13:53 UTC

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

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

hypothetical case
hypothetical case
hypothetical case
hypothetical case

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

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

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

--
Copyright 2021 Pete Olcott

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

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ]
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References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
<sap6l7$130c$5@gioia.aioe.org> <saqlhk$860$1@dont-email.me>
<tZOdneTmHpHNnUz9nZ2dnUU7-V3NnZ2d@giganews.com> <sarcip$jgd$1@dont-email.me>
<zZOdnUG2JPuxokz9nZ2dnUU7-UPNnZ2d@giganews.com> <sarhu7$f83$1@dont-email.me>
<E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com> <sat46g$afu$1@dont-email.me>
<4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com> <satao3$gjk$1@dont-email.me>
<YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me>
<48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Fri, 25 Jun 2021 15:39 UTC

On 6/25/21 9:53 AM, olcott wrote:
> On 6/24/2021 7:51 PM, André G. Isaak wrote:
>> On 2021-06-24 15:40, olcott wrote:
>>
>> <snip>
>>
>>> The question can be rephrased this way in the hypothetical case where
>>> the H/P is replaced with an x86 emulator and the hypothetical case
>>> where the halt decider embedded at Ĥ.qx is replaced with a UTM
>>
>> If you want to test your claim that the halt decider *MUST* terminate
>> its input, you change the behaviour of the halt decider so it can't
>> terminate its input. (Or, better yet, you just run P(P) directly
>> without making any changes at all -- your aversion to this option is
>> truly mystifying).
>>
>
> It is silly that I have to repeat my words like this so that you pay
> attention, you should just pay attention.
>
> hypothetical case
> hypothetical case
> hypothetical case
> hypothetical case
>
>> What you *CAN'T* do is change the behaviour of the input to the
>> decider which is what you are doing if you also change the H which is
>> contained inside P. By doing this, you change P(P) into an entirely
>> different computation.
>>
>> You claim that H is an independent variable and P is a dependent
>> variable. If you change both P and H, then P is not an independent
>> variable. P and H are interdependent variables, which means any test
>> you perform is worthless.
>>
>>> we can know that the original P(P) and Ĥ(⟨Ĥ⟩) are correctly decided
>>> as computations that never halt by a simulating halt decider if the
>>> above computations never halt because a simulating halt acts only as
>>> a x86 emulator / UTM until after its input demonstrates an infinitely
>>> repeating behavior pattern.
>>>
>>> That people have difficulty comprehending the necessary truth of this
>>> does not count as any rebuttal what-so-ever.
>>
>> People comprehend things just fine. They comprehend that to test you
>> claim we must change *only* the topmost H, not the copy of H embedded
>> in P.
>>
>> André
>>
>>
>
> In the computation P(P) if any invocation of the infinite chain of
> invocations must be terminated to prevent the infinite execution of this
> chain then we know that P(P) does specify an infinite chain of invocations.
>
> I have finally gotten my words clear enough that any disagreement would
> look quite foolish. On the above key point this took six months and
> thousands of reviews.
>

The problem with this argument is that P(P) only creates an infinite
chain of invocations if H never aborts any of the simulations it performs.

If any H aborts any of the simulations, then the computation P(P) is Not
an infinite chain of invocations.

Making the argument that P(P) is an infinite chain of invocations is the
equivalent of saying that H will NEVER abort its simulation of the chain.

Once you allow for H to abort the simulation, you no longer can make
your initial claim that P(P) is in infinite invocation.

Yes, SOME H needs to terminate its simulation, but that does not match
the rule that THIS H needs to terminate its simulation, thus you do not
have the grounds to call the computation infinite.

This is the fatal flaw in your logic, you have lost precision in what
you are talking about.

You logic confuses the various copies of H (and P) in the system. They
may be the same algorithm, but they are NOT the same instance of
computation. You ignore context, which seems strange as that is one of
the most important thing to remember when doing software engineering.

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](axiom)
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<sap6l7$130c$5@gioia.aioe.org> <saqlhk$860$1@dont-email.me>
<tZOdneTmHpHNnUz9nZ2dnUU7-V3NnZ2d@giganews.com> <sarcip$jgd$1@dont-email.me>
<zZOdnUG2JPuxokz9nZ2dnUU7-UPNnZ2d@giganews.com> <sarhu7$f83$1@dont-email.me>
<E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com> <sat46g$afu$1@dont-email.me>
<4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com> <satao3$gjk$1@dont-email.me>
<YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me>
<48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me>
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From: NoO...@NoWhere.com (olcott)
Date: Fri, 25 Jun 2021 11:33:53 -0500
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 by: olcott - Fri, 25 Jun 2021 16:33 UTC

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

Not when the H and the P are both hypothetical.

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

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

(1) P halts
(2) P never halts

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

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

--
Copyright 2021 Pete Olcott

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

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](axiom)
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References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
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<zZOdnUG2JPuxokz9nZ2dnUU7-UPNnZ2d@giganews.com> <sarhu7$f83$1@dont-email.me>
<E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com> <sat46g$afu$1@dont-email.me>
<4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com> <satao3$gjk$1@dont-email.me>
<YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me>
<48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Fri, 25 Jun 2021 17:14 UTC

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

WRONG. P is DEFINED based on H. If you Hypothetically create a P that
doesn't follow that form, then you are hypothetically creating nonsense
and can't use it to for anything logical.

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

Right, So we get the relations ship that H1 -> P1, and H2 -> P2

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

WRONG. You have just used the logic that Hypothetically, if cats were
dogs, they would bark, thus cats bark, even in the real world where they
aren't dogs.

Yes, if H doesn't abort its simulation, then if it could give an answer
(which it can't) it would be right to say the P is non-Halting. Thus we
can say if we call the H Hn, and its P Pn, that we can be correct in
calling Pn as non-Halting, it just happens that Hm can't give this
answer because it is non-Halting itself.

If we now consider Ha, which does abort, it can't consider that Pa to be
non-Halting, becasue Pa isn't Pn, but is could correctly decide that Pn
is non-Halting.

But this is fine, as Pn doesn't make a claim to confound Ha, just Hn.

It is Pa that confounds Ha because Ha treats Pa as if it was Pn, which
it is different from.

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

But your logic interprets this statement wrong. Given Pa from above,
UTM(Pa(Pa)) is a halting computation thus THAT invocation of Ha doesn't
need to abort its simulation of Pa, thus it is INCORRECT decision.

A natural consequence of YOUR interpretation of this axiom is that H
itself is non-halting and thus fails to meet the definition of a
decider, and we also get inconsistencies in that we can show some
computations to be both halting and non-halting.

I will also point out that with the proper interpreation of the words,
this doesn't need to be taken as an Axiom, but can acutally be proven
for Turing Machines based on basic definitions, but can be a simple
Theorem. With YOUR meaning it can be proven false.

This seems to be a common problem with your logic. Simple statements
which are very provable with a proper interpretation of the terms become
'obviously true' statements for you with a different interpretation of
the terms (that actually make the statement not true).

You are forced to make there absurd claims, because if you looked at the
proofs of the statements, it becomes clear that you are misusing some of
the terms.

This shows the problem with trying to do logic with natural langauge as
opposed to formal language, which you seem to resist learning.

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

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

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

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

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

--
Copyright 2021 Pete Olcott

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

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](axiom)
Newsgroups: comp.theory
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
<sap6l7$130c$5@gioia.aioe.org> <saqlhk$860$1@dont-email.me>
<tZOdneTmHpHNnUz9nZ2dnUU7-V3NnZ2d@giganews.com> <sarcip$jgd$1@dont-email.me>
<zZOdnUG2JPuxokz9nZ2dnUU7-UPNnZ2d@giganews.com> <sarhu7$f83$1@dont-email.me>
<E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com> <sat46g$afu$1@dont-email.me>
<4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com> <satao3$gjk$1@dont-email.me>
<YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me>
<48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me>
<6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com> <sb0kra$ert$1@dont-email.me>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Fri, 25 Jun 2021 20:11 UTC

On 6/25/21 2:50 PM, olcott wrote:
> On 6/25/2021 12:14 PM, Richard Damon wrote:
>
>> WRONG. P is DEFINED based on H. If you Hypothetically create a P that
>> doesn't follow that form, then you are hypothetically creating nonsense
>> and can't use it to for anything logical.
>>
>
> Of every possible encoding of simulating partial halt decider H that can
> possibly exist  H*, if H* never aborts the simulation of its input
> results in the infinite execution of the invocation of of P(P) then a
> simulating halt decider H that does abort its simulation of this input
> does correctly decide that this input does specify the never halting
> behavior of an infinite chain of invocations.

Yes, if H* is an element of the set of non-aborting deciders (Hn), P
will result in infinite recursion, but H* will not give an answer to the
question H(P,P), so isn't a decider that needs to be shown wrong by P,
as it shows itself wrong.

That doesn't say what P will do for an H that isn't a member of that
set, since that is now a different P. Your logic that says otherwise is
flawed. Remember, the definition of P is derived from the definition of
H, so change your H, and you change your P, so ANYTHING you determined
from that other P need to be re-verified.

Basically, you have two major classes of possible H, those that will
abort the simulation for the pattern P (cats), and those that don't (dogs).

You can't looks at dogs to answer a question about cats if the attribute
is different between cats and dogs (like do they bark). The Halting
Behavior of P can clearly be different between these two classes, so we
need to look at the right species of H/P to make the decision.

If you can find a H a member of Hn that does give the answer even
without aborting it simulation, then you can have your answer, but that
seems also impossible.

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](axiom)
Newsgroups: comp.theory
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com> <sap6l7$130c$5@gioia.aioe.org> <saqlhk$860$1@dont-email.me> <tZOdneTmHpHNnUz9nZ2dnUU7-V3NnZ2d@giganews.com> <sarcip$jgd$1@dont-email.me> <zZOdnUG2JPuxokz9nZ2dnUU7-UPNnZ2d@giganews.com> <sarhu7$f83$1@dont-email.me> <E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com> <sat46g$afu$1@dont-email.me> <4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com> <satao3$gjk$1@dont-email.me> <YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me> <48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me> <6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com> <sb0kra$ert$1@dont-email.me> <kaudnZS1vPJiaE79nZ2dnUU7-QHNnZ2d@giganews.com> <C0_AI.793830$2A5.649020@fx45.iad> <udKdnabaTsZvOkn9nZ2dnUU7-dvNnZ2d@giganews.com> <OlbBI.605613$J_5.348305@fx46.iad> <fO6dnQEYd73PmEv9nZ2dnUU7-TmdnZ2d@giganews.com> <u9oBI.267517$lyv9.157656@fx35.iad> <eKednajHd_LtuEv9nZ2dnUU7-TvNnZ2d@giganews.com> <ALqBI.113709$od.33914@fx15.iad>
From: NoO...@NoWhere.com (olcott)
Date: Fri, 25 Jun 2021 15:40:04 -0500
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 by: olcott - Fri, 25 Jun 2021 20:40 UTC

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

Which logically entails beyond all possible doubt that the set of
encodings of simulating partial halt deciders H2* that do abort the
simulation of the (P,P) input would correctly report that this input
never halts.

> but H* will not give an answer to the
> question H(P,P), so isn't a decider that needs to be shown wrong by P,
> as it shows itself wrong.
>
> That doesn't say what P will do for an H that isn't a member of that
> set, since that is now a different P. Your logic that says otherwise is
> flawed. Remember, the definition of P is derived from the definition of
> H, so change your H, and you change your P, so ANYTHING you determined
> from that other P need to be re-verified.
>
> Basically, you have two major classes of possible H, those that will
> abort the simulation for the pattern P (cats), and those that don't (dogs).
>
> You can't looks at dogs to answer a question about cats if the attribute
> is different between cats and dogs (like do they bark). The Halting
> Behavior of P can clearly be different between these two classes, so we
> need to look at the right species of H/P to make the decision.
>
> If you can find a H a member of Hn that does give the answer even
> without aborting it simulation, then you can have your answer, but that
> seems also impossible.
>

--
Copyright 2021 Pete Olcott

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

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](axiom)
Newsgroups: comp.theory
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
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<zZOdnUG2JPuxokz9nZ2dnUU7-UPNnZ2d@giganews.com> <sarhu7$f83$1@dont-email.me>
<E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com> <sat46g$afu$1@dont-email.me>
<4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com> <satao3$gjk$1@dont-email.me>
<YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me>
<48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me>
<6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com> <sb0kra$ert$1@dont-email.me>
<kaudnZS1vPJiaE79nZ2dnUU7-QHNnZ2d@giganews.com>
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<udKdnabaTsZvOkn9nZ2dnUU7-dvNnZ2d@giganews.com>
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<fO6dnQEYd73PmEv9nZ2dnUU7-TmdnZ2d@giganews.com>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Fri, 25 Jun 2021 21:59 UTC

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

WHY?

In what universe does the properties of a Dog define the properties of a
Cat.

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

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

<BuOdncUXaL2swkv9nZ2dnUU7-W-dnZ2d@giganews.com>

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](axiom)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
<sarcip$jgd$1@dont-email.me> <zZOdnUG2JPuxokz9nZ2dnUU7-UPNnZ2d@giganews.com>
<sarhu7$f83$1@dont-email.me> <E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com>
<sat46g$afu$1@dont-email.me> <4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com>
<satao3$gjk$1@dont-email.me> <YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com>
<savr2l$b1i$1@dont-email.me> <48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com>
<sb0f72$hf4$1@dont-email.me> <6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com>
<sb0kra$ert$1@dont-email.me> <kaudnZS1vPJiaE79nZ2dnUU7-QHNnZ2d@giganews.com>
<C0_AI.793830$2A5.649020@fx45.iad>
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<u9oBI.267517$lyv9.157656@fx35.iad>
<eKednajHd_LtuEv9nZ2dnUU7-TvNnZ2d@giganews.com>
<ALqBI.113709$od.33914@fx15.iad>
<cvednUP16NqYokv9nZ2dnUU7-bHNnZ2d@giganews.com> <HksBI.267$al1.209@fx26.iad>
From: NoO...@NoWhere.com (olcott)
Date: Fri, 25 Jun 2021 17:56:50 -0500
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 by: olcott - Fri, 25 Jun 2021 22:56 UTC

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

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

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

--
Copyright 2021 Pete Olcott

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

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

<6IuBI.115687$431.109356@fx39.iad>

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https://www.novabbs.com/devel/article-flat.php?id=17206&group=comp.theory#17206

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](axiom)
Newsgroups: comp.theory
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
<zZOdnUG2JPuxokz9nZ2dnUU7-UPNnZ2d@giganews.com> <sarhu7$f83$1@dont-email.me>
<E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com> <sat46g$afu$1@dont-email.me>
<4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com> <satao3$gjk$1@dont-email.me>
<YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me>
<48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me>
<6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com> <sb0kra$ert$1@dont-email.me>
<kaudnZS1vPJiaE79nZ2dnUU7-QHNnZ2d@giganews.com>
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<udKdnabaTsZvOkn9nZ2dnUU7-dvNnZ2d@giganews.com>
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<BuOdncUXaL2swkv9nZ2dnUU7-W-dnZ2d@giganews.com>
From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Sat, 26 Jun 2021 00:40 UTC

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

WRONG.

Your test does not match the plain meaning of the words, as has been
explained many times.

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

You FAULTY language is thus shown to be incorrect and leads to
inconsistent logic, as you whole argument has shown.

I will again repeat, that statement does NOT need to be made an Axiom,
it can actually be formally proven with the right interpretation of the
words, you only need to try to make is an (incorrect) axiom because you
can't show your false version to be true any other way.

Face it, your logic generates so many contradictions that it gets hard
to find anything that actually has a real factual basis.

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

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
invocation chain)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
<sarhu7$f83$1@dont-email.me> <E8Wdnezrd-inyUz9nZ2dnUU7-cHNnZ2d@giganews.com>
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<satao3$gjk$1@dont-email.me> <YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com>
<savr2l$b1i$1@dont-email.me> <48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com>
<sb0f72$hf4$1@dont-email.me> <6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com>
<sb0kra$ert$1@dont-email.me> <kaudnZS1vPJiaE79nZ2dnUU7-QHNnZ2d@giganews.com>
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<BuOdncUXaL2swkv9nZ2dnUU7-W-dnZ2d@giganews.com>
<6IuBI.115687$431.109356@fx39.iad>
From: NoO...@NoWhere.com (olcott)
Date: Fri, 25 Jun 2021 20:01:25 -0500
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 by: olcott - Sat, 26 Jun 2021 01:01 UTC

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

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

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

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

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

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

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

--
Copyright 2021 Pete Olcott

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

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
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Newsgroups: comp.theory
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
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<YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Sat, 26 Jun 2021 01:37 UTC

On 6/25/21 9:01 PM, olcott wrote:
> On 6/25/2021 7:40 PM, Richard Damon wrote:
>> On 6/25/21 6:56 PM, olcott wrote:
>>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>>> On 6/25/21 4:40 PM, olcott wrote:
>>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>>
>>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create a P
>>>>>>>> that
>>>>>>>> doesn't follow that form, then you are hypothetically creating
>>>>>>>> nonsense
>>>>>>>> and can't use it to for anything logical.
>>>>>>>>
>>>>>>>
>>>>>>> Of every possible encoding of simulating partial halt decider H
>>>>>>> that can
>>>>>>> possibly exist  H*, if H* never aborts the simulation of its input
>>>>>>> results in the infinite execution of the invocation of of P(P)
>>>>>>> then a
>>>>>>> simulating halt decider H that does abort its simulation of this
>>>>>>> input
>>>>>>> does correctly decide that this input does specify the never halting
>>>>>>> behavior of an infinite chain of invocations.
>>>>>>
>>>>>> Yes, if H* is an element of the set of non-aborting deciders (Hn), P
>>>>>> will result in infinite recursion,
>>>>>
>>>>> Which logically entails beyond all possible doubt that the set of
>>>>> encodings of simulating partial halt deciders H2* that do abort the
>>>>> simulation of the (P,P) input would correctly report that this input
>>>>> never halts.
>>>>
>>>> WHY?
>>>>
>>>
>>> Axiom(1) Every computation that never halts unless its simulation is
>>> aborted is a computation that never halts. This verified as true on the
>>> basis of the meaning of its words.
>>
>> WRONG.
>>
>> Your test does not match the plain meaning of the words, as has been
>> explained many times.
>>
>
> Those words may be over your head, yet several others understand that
> they are necessarily correct.

I have seen NO ONE agree to your interpretation of it. The plain meaning
is that if it can be shown that if the given instance of the simulator
simulating a given input doesn't stop its simulation that this
simulation will run forevr, then the machine that is being simulated can
be corrected decided as non-Halting.

An more formal way to say that is if UTM(P,I) is non-halting then it is
correct for H(P,I) to return the non-halting result.

This actually follows since UTM(P,I) will be non-halting if and only if
P(I) is non-halting by the definition of a UTM, so that statement is
trivially proven.

Your interpretation, where even if a copy of the algorithm of H is
included in P and that included copy needs to abort the simulation of
the copy of the machine that it was given, can be PROVEN wrong, as even
you have shown that P(P) in this case does Halt, thus your claimed
correct answer is wrong by the definition of the problem.

Only if you define that your answer isn't actually supposed to be the
answer to the halting problem can you justify your answer to be correct,
but then you proof doesn't achieve the goal you claim.

>
>> Note, it is easy to show that your interpretation is wrong since even
>> you admit that Linz H^, now called P by you will come to its end and
>> halt when given it own representation as its input, and thus is BY
>> DEFINITION a Halting Computation, Thus the H deciding it didn't need to
>> abort its execution to get the wrong answer of Non-Halting.
>>
>
> Because at least one invocation of the infinite invocation chain
> specified by P(P) had to be terminated to prevent the infinite execution
> of this infinite invocation chain it is confirmed beyond all possible
> doubt that P(P) specifies an invocation chain.

WRONG. Given that we have an H that can answer H(P,P) because it knows
at least enough to terminate the pattern you describe, then when we run
P(P) then because the H within it also knows to abort this sequence
(since it is built on the same algorithm) this P is NOT part of an
infinite chain of execution, and thus its H can return its (wrong)
answer to it and that P can then Halt.

It is only a P built on an H that doesn't abort this sequence that gets
caught in the infinite invocation, but that H also can't return its
answer when asked H(P,P) so it doesn't need to be shown to be incorrect,
as it fails by not answering.

Your logic fails to take into account that the algorithm of P depends on
the algorithm of H and thus when you argue of various possible Hs you
need to look at the way that P changes when H changes.

This is like saying that since the derivative of a*x is a, then the
derivative of f(x)*x is f(x) ignoring that f(x) changes value as x
changes. Wrong logic, wrong answer.

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

No, the termination of a single invocation of the potentially infinite
invocation chain aborts the chain FROM THE POINT OF THE SIMULATOR that
aborted its simulation.

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

A potentially infinite recursion that has been broken by a piece of it
was not actually an infinite recursion, but was only finite.

THAT seems over your head.

Yes, a function that is ACTUALLY infinite recursive will not return to
its caller. A function that stopped a potentially infinite recursion is
not actually infinitely recursive so it can, and in fact if it doesn't
it hasn't really helped itself by stopping the potentially infinite
recusrion.

TRUTH IS TRUTH.

H^(H^) is shown to Halt, and thus is a HALTING computation.

Thus the right answer to the Halting question for H^(H^) is Halting.

H(H^,H^) == 0 is thus incorrect.

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
invocation chain)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
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<savr2l$b1i$1@dont-email.me> <48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com>
<sb0f72$hf4$1@dont-email.me> <6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com>
<sb0kra$ert$1@dont-email.me> <kaudnZS1vPJiaE79nZ2dnUU7-QHNnZ2d@giganews.com>
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<hxvBI.20803$9q1.10955@fx09.iad>
From: NoO...@NoWhere.com (olcott)
Date: Fri, 25 Jun 2021 20:46:12 -0500
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 by: olcott - Sat, 26 Jun 2021 01:46 UTC

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

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

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

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

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

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

Now I have told this this several hundred times.

--
Copyright 2021 Pete Olcott

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

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
invocation chain)
Newsgroups: comp.theory
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
<4I-dnW6siND5hU_9nZ2dnUU7-V3NnZ2d@giganews.com> <satao3$gjk$1@dont-email.me>
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<48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me>
<6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com> <sb0kra$ert$1@dont-email.me>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Sat, 26 Jun 2021 02:07 UTC

On 6/25/21 9:46 PM, olcott wrote:
> On 6/25/2021 8:37 PM, Richard Damon wrote:
>> On 6/25/21 9:01 PM, olcott wrote:
>>> On 6/25/2021 7:40 PM, Richard Damon wrote:
>>>> On 6/25/21 6:56 PM, olcott wrote:
>>>>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>>>>> On 6/25/21 4:40 PM, olcott wrote:
>>>>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>>>>
>>>>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create a P
>>>>>>>>>> that
>>>>>>>>>> doesn't follow that form, then you are hypothetically creating
>>>>>>>>>> nonsense
>>>>>>>>>> and can't use it to for anything logical.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Of every possible encoding of simulating partial halt decider H
>>>>>>>>> that can
>>>>>>>>> possibly exist  H*, if H* never aborts the simulation of its input
>>>>>>>>> results in the infinite execution of the invocation of of P(P)
>>>>>>>>> then a
>>>>>>>>> simulating halt decider H that does abort its simulation of this
>>>>>>>>> input
>>>>>>>>> does correctly decide that this input does specify the never
>>>>>>>>> halting
>>>>>>>>> behavior of an infinite chain of invocations.
>>>>>>>>
>>>>>>>> Yes, if H* is an element of the set of non-aborting deciders
>>>>>>>> (Hn), P
>>>>>>>> will result in infinite recursion,
>>>>>>>
>>>>>>> Which logically entails beyond all possible doubt that the set of
>>>>>>> encodings of simulating partial halt deciders H2* that do abort the
>>>>>>> simulation of the (P,P) input would correctly report that this input
>>>>>>> never halts.
>>>>>>
>>>>>> WHY?
>>>>>>
>>>>>
>>>>> Axiom(1) Every computation that never halts unless its simulation is
>>>>> aborted is a computation that never halts. This verified as true on
>>>>> the
>>>>> basis of the meaning of its words.
>>>>
>>>> WRONG.
>>>>
>>>> Your test does not match the plain meaning of the words, as has been
>>>> explained many times.
>>>>
>>>
>>> Those words may be over your head, yet several others understand that
>>> they are necessarily correct.
>>
>> I have seen NO ONE agree to your interpretation of it. The plain meaning
>> is that if it can be shown that if the given instance of the simulator
>> simulating a given input doesn't stop its simulation that this
>> simulation will run forevr, then the machine that is being simulated can
>> be corrected decided as non-Halting.
>>
>> An more formal way to say that is if UTM(P,I) is non-halting then it is
>> correct for H(P,I) to return the non-halting result.
>>
>> This actually follows since UTM(P,I) will be non-halting if and only if
>> P(I) is non-halting by the definition of a UTM, so that statement is
>> trivially proven.
>>
>> Your interpretation, where even if a copy of the algorithm of H is
>> included in P and that included copy needs to abort the simulation of
>> the copy of the machine that it was given, can be PROVEN wrong, as even
>> you have shown that P(P) in this case does Halt, thus your claimed
>> correct answer is wrong by the definition of the problem.
>>
>> Only if you define that your answer isn't actually supposed to be the
>> answer to the halting problem can you justify your answer to be correct,
>> but then you proof doesn't achieve the goal you claim.
>>
>>>
>>>> Note, it is easy to show that your interpretation is wrong since even
>>>> you admit that Linz H^, now called P by you will come to its end and
>>>> halt when given it own representation as its input, and thus is BY
>>>> DEFINITION a Halting Computation, Thus the H deciding it didn't need to
>>>> abort its execution to get the wrong answer of Non-Halting.
>>>>
>>>
>>> Because at least one invocation of the infinite invocation chain
>>> specified by P(P) had to be terminated to prevent the infinite execution
>>> of this infinite invocation chain it is confirmed beyond all possible
>>> doubt that P(P) specifies an invocation chain.
>>
>> WRONG. Given that we have an H that can answer H(P,P) because it knows
>> at least enough to terminate the pattern you describe, then when we run
>> P(P) then because the H within it also knows to abort this sequence
>> (since it is built on the same algorithm) this P is NOT part of an
>> infinite chain of execution, and thus its H can return its (wrong)
>> answer to it and that P can then Halt.
>
> P(P) specifies in infinite invocation sequence that is terminated on its
> third invocation of H(P,P).
>
> P(P) specifies in infinite invocation sequence that is terminated on its
> third invocation of H(P,P).
>
> P(P) specifies in infinite invocation sequence that is terminated on its
> third invocation of H(P,P).
>
> P(P) specifies in infinite invocation sequence that is terminated on its
> third invocation of H(P,P).
>
> P(P) specifies in infinite invocation sequence that is terminated on its
> third invocation of H(P,P).
>
> Now I have told this this several hundred times.
>
>

WRONG.

P(P) starts.

Calls H(P,P)

H starts the simulation.

H simulates P starting

H simulates P calling H

H simulates H starting its simulation

H simulates H simulating P starting

H simulates H simulating P calling H

The first H about here detects what it THINKS is an infinite execution

THe first H aborts its simulation

The first H returns its answer (Non-Halting) to its caller

P then Halts

Showing P is a Halting Computation.

This P was NEVER in an infinite invocation sequence because THIS version
of P is built on an H that won't let the infinite invocation occur as
part of *P*s algorithm.

The P that was shown to be in an infinite invocation sequences wasn't
this P but a different one, one that was built on a different H.

That is like complaining about the auto design of your Rolls Royce
because your FIAT keeps breaking down. It is a DIFFERENT Machine.

Your logic fails because you habitually reuse symbols and names and lose
the context they were defined in.

Pn based on the Hn that doesn't abort is a non-halting computaiton.

Pa based on the Ha that does abort (and decides incorrectly) is a
Halting Computation

You logic shows that Ha can be correct in deciding that Pn is
non-halting. That is fine, and doesn't really prove anything.

To claim that it also shows that Pa is non-halting is like saying a dog
is a cat, and the fact that dogs barks says that cats do too.

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com> <satao3$gjk$1@dont-email.me> <YuOdnXMSQcOgqk_9nZ2dnUU7-QPNnZ2d@giganews.com> <savr2l$b1i$1@dont-email.me> <48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me> <6POdnUnCSvCzVk79nZ2dnUU7-a3NnZ2d@giganews.com> <sb0kra$ert$1@dont-email.me> <kaudnZS1vPJiaE79nZ2dnUU7-QHNnZ2d@giganews.com> <C0_AI.793830$2A5.649020@fx45.iad> <udKdnabaTsZvOkn9nZ2dnUU7-dvNnZ2d@giganews.com> <OlbBI.605613$J_5.348305@fx46.iad> <fO6dnQEYd73PmEv9nZ2dnUU7-TmdnZ2d@giganews.com> <u9oBI.267517$lyv9.157656@fx35.iad> <eKednajHd_LtuEv9nZ2dnUU7-TvNnZ2d@giganews.com> <ALqBI.113709$od.33914@fx15.iad> <cvednUP16NqYokv9nZ2dnUU7-bHNnZ2d@giganews.com> <HksBI.267$al1.209@fx26.iad> <BuOdncUXaL2swkv9nZ2dnUU7-W-dnZ2d@giganews.com> <6IuBI.115687$431.109356@fx39.iad> <RKCdnSS4Ifj44Uv9nZ2dnUU7-VvNnZ2d@giganews.com> <hxvBI.20803$9q1.10955@fx09.iad> <I8WdnT2QqrV5G0v9nZ2dnUU7-WvNnZ2d@giganews.com> <EZvBI.43029$7Y.22867@fx03.iad>
From: NoO...@NoWhere.com (olcott)
Date: Fri, 25 Jun 2021 22:07:02 -0500
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 by: olcott - Sat, 26 Jun 2021 03:07 UTC

On 6/25/2021 9:07 PM, Richard Damon wrote:
> On 6/25/21 9:46 PM, olcott wrote:
>> On 6/25/2021 8:37 PM, Richard Damon wrote:
>>> On 6/25/21 9:01 PM, olcott wrote:
>>>> On 6/25/2021 7:40 PM, Richard Damon wrote:
>>>>> On 6/25/21 6:56 PM, olcott wrote:
>>>>>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>>>>>> On 6/25/21 4:40 PM, olcott wrote:
>>>>>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>>>>>
>>>>>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create a P
>>>>>>>>>>> that
>>>>>>>>>>> doesn't follow that form, then you are hypothetically creating
>>>>>>>>>>> nonsense
>>>>>>>>>>> and can't use it to for anything logical.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Of every possible encoding of simulating partial halt decider H
>>>>>>>>>> that can
>>>>>>>>>> possibly exist  H*, if H* never aborts the simulation of its input
>>>>>>>>>> results in the infinite execution of the invocation of of P(P)
>>>>>>>>>> then a
>>>>>>>>>> simulating halt decider H that does abort its simulation of this
>>>>>>>>>> input
>>>>>>>>>> does correctly decide that this input does specify the never
>>>>>>>>>> halting
>>>>>>>>>> behavior of an infinite chain of invocations.
>>>>>>>>>
>>>>>>>>> Yes, if H* is an element of the set of non-aborting deciders
>>>>>>>>> (Hn), P
>>>>>>>>> will result in infinite recursion,
>>>>>>>>
>>>>>>>> Which logically entails beyond all possible doubt that the set of
>>>>>>>> encodings of simulating partial halt deciders H2* that do abort the
>>>>>>>> simulation of the (P,P) input would correctly report that this input
>>>>>>>> never halts.
>>>>>>>
>>>>>>> WHY?
>>>>>>>
>>>>>>
>>>>>> Axiom(1) Every computation that never halts unless its simulation is
>>>>>> aborted is a computation that never halts. This verified as true on
>>>>>> the
>>>>>> basis of the meaning of its words.
>>>>>
>>>>> WRONG.
>>>>>
>>>>> Your test does not match the plain meaning of the words, as has been
>>>>> explained many times.
>>>>>
>>>>
>>>> Those words may be over your head, yet several others understand that
>>>> they are necessarily correct.
>>>
>>> I have seen NO ONE agree to your interpretation of it. The plain meaning
>>> is that if it can be shown that if the given instance of the simulator
>>> simulating a given input doesn't stop its simulation that this
>>> simulation will run forevr, then the machine that is being simulated can
>>> be corrected decided as non-Halting.
>>>
>>> An more formal way to say that is if UTM(P,I) is non-halting then it is
>>> correct for H(P,I) to return the non-halting result.
>>>
>>> This actually follows since UTM(P,I) will be non-halting if and only if
>>> P(I) is non-halting by the definition of a UTM, so that statement is
>>> trivially proven.
>>>
>>> Your interpretation, where even if a copy of the algorithm of H is
>>> included in P and that included copy needs to abort the simulation of
>>> the copy of the machine that it was given, can be PROVEN wrong, as even
>>> you have shown that P(P) in this case does Halt, thus your claimed
>>> correct answer is wrong by the definition of the problem.
>>>
>>> Only if you define that your answer isn't actually supposed to be the
>>> answer to the halting problem can you justify your answer to be correct,
>>> but then you proof doesn't achieve the goal you claim.
>>>
>>>>
>>>>> Note, it is easy to show that your interpretation is wrong since even
>>>>> you admit that Linz H^, now called P by you will come to its end and
>>>>> halt when given it own representation as its input, and thus is BY
>>>>> DEFINITION a Halting Computation, Thus the H deciding it didn't need to
>>>>> abort its execution to get the wrong answer of Non-Halting.
>>>>>
>>>>
>>>> Because at least one invocation of the infinite invocation chain
>>>> specified by P(P) had to be terminated to prevent the infinite execution
>>>> of this infinite invocation chain it is confirmed beyond all possible
>>>> doubt that P(P) specifies an invocation chain.
>>>
>>> WRONG. Given that we have an H that can answer H(P,P) because it knows
>>> at least enough to terminate the pattern you describe, then when we run
>>> P(P) then because the H within it also knows to abort this sequence
>>> (since it is built on the same algorithm) this P is NOT part of an
>>> infinite chain of execution, and thus its H can return its (wrong)
>>> answer to it and that P can then Halt.
>>
>> P(P) specifies in infinite invocation sequence that is terminated on its
>> third invocation of H(P,P).
>>
>> P(P) specifies in infinite invocation sequence that is terminated on its
>> third invocation of H(P,P).
>>
>> P(P) specifies in infinite invocation sequence that is terminated on its
>> third invocation of H(P,P).
>>
>> P(P) specifies in infinite invocation sequence that is terminated on its
>> third invocation of H(P,P).
>>
>> P(P) specifies in infinite invocation sequence that is terminated on its
>> third invocation of H(P,P).
>>
>> Now I have told this this several hundred times.
>>
>>
>
> WRONG.
>
> P(P) starts.
>
> Calls H(P,P)
>
> H starts the simulation.
>
> H simulates P starting
>
> H simulates P calling H
>
> H simulates H starting its simulation
>
> H simulates H simulating P starting
>
> H simulates H simulating P calling H
>
> The first H about here detects what it THINKS is an infinite execution
>
> THe first H aborts its simulation
>
> The first H returns its answer (Non-Halting) to its caller
>
> P then Halts
>
> Showing P is a Halting Computation.

As you already admitted P ONLY halts because some H aborts some P
otherwise P never ever halts.

As you already admitted P ONLY halts because some H aborts some P
otherwise P never ever halts.

As you already admitted P ONLY halts because some H aborts some P
otherwise P never ever halts.

As you already admitted P ONLY halts because some H aborts some P
otherwise P never ever halts.

As you already admitted P ONLY halts because some H aborts some P
otherwise P never ever halts.

--
Copyright 2021 Pete Olcott

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

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

<sb6nfj$i1b$1@gioia.aioe.org>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=17212&group=comp.theory#17212

 copy link   Newsgroups: comp.theory comp.ai.philosophy comp.software-eng
Path: i2pn2.org!i2pn.org!aioe.org!NBiuIU74OKL7NpIOsbuNjQ.user.gioia.aioe.org.POSTED!not-for-mail
From: chris.m....@gmail.com (Chris M. Thomasson)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng
Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
invocation chain)
Date: Sat, 26 Jun 2021 01:11:32 -0700
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 by: Chris M. Thomasson - Sat, 26 Jun 2021 08:11 UTC

On 6/25/2021 8:07 PM, olcott wrote:
> On 6/25/2021 9:07 PM, Richard Damon wrote:
>> On 6/25/21 9:46 PM, olcott wrote:
>>> On 6/25/2021 8:37 PM, Richard Damon wrote:
>>>> On 6/25/21 9:01 PM, olcott wrote:
>>>>> On 6/25/2021 7:40 PM, Richard Damon wrote:
>>>>>> On 6/25/21 6:56 PM, olcott wrote:
>>>>>>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>>>>>>> On 6/25/21 4:40 PM, olcott wrote:
>>>>>>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>>>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>>>>>>
>>>>>>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create
>>>>>>>>>>>> a P
>>>>>>>>>>>> that
>>>>>>>>>>>> doesn't follow that form, then you are hypothetically creating
>>>>>>>>>>>> nonsense
>>>>>>>>>>>> and can't use it to for anything logical.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Of every possible encoding of simulating partial halt decider H
>>>>>>>>>>> that can
>>>>>>>>>>> possibly exist  H*, if H* never aborts the simulation of its
>>>>>>>>>>> input
>>>>>>>>>>> results in the infinite execution of the invocation of of P(P)
>>>>>>>>>>> then a
>>>>>>>>>>> simulating halt decider H that does abort its simulation of this
>>>>>>>>>>> input
>>>>>>>>>>> does correctly decide that this input does specify the never
>>>>>>>>>>> halting
>>>>>>>>>>> behavior of an infinite chain of invocations.
>>>>>>>>>>
>>>>>>>>>> Yes, if H* is an element of the set of non-aborting deciders
>>>>>>>>>> (Hn), P
>>>>>>>>>> will result in infinite recursion,
>>>>>>>>>
>>>>>>>>> Which logically entails beyond all possible doubt that the set of
>>>>>>>>> encodings of simulating partial halt deciders H2* that do abort
>>>>>>>>> the
>>>>>>>>> simulation of the (P,P) input would correctly report that this
>>>>>>>>> input
>>>>>>>>> never halts.
>>>>>>>>
>>>>>>>> WHY?
>>>>>>>>
>>>>>>>
>>>>>>> Axiom(1) Every computation that never halts unless its simulation is
>>>>>>> aborted is a computation that never halts. This verified as true on
>>>>>>> the
>>>>>>> basis of the meaning of its words.
>>>>>>
>>>>>> WRONG.
>>>>>>
>>>>>> Your test does not match the plain meaning of the words, as has been
>>>>>> explained many times.
>>>>>>
>>>>>
>>>>> Those words may be over your head, yet several others understand that
>>>>> they are necessarily correct.
>>>>
>>>> I have seen NO ONE agree to your interpretation of it. The plain
>>>> meaning
>>>> is that if it can be shown that if the given instance of the simulator
>>>> simulating a given input doesn't stop its simulation that this
>>>> simulation will run forevr, then the machine that is being simulated
>>>> can
>>>> be corrected decided as non-Halting.
>>>>
>>>> An more formal way to say that is if UTM(P,I) is non-halting then it is
>>>> correct for H(P,I) to return the non-halting result.
>>>>
>>>> This actually follows since UTM(P,I) will be non-halting if and only if
>>>> P(I) is non-halting by the definition of a UTM, so that statement is
>>>> trivially proven.
>>>>
>>>> Your interpretation, where even if a copy of the algorithm of H is
>>>> included in P and that included copy needs to abort the simulation of
>>>> the copy of the machine that it was given, can be PROVEN wrong, as even
>>>> you have shown that P(P) in this case does Halt, thus your claimed
>>>> correct answer is wrong by the definition of the problem.
>>>>
>>>> Only if you define that your answer isn't actually supposed to be the
>>>> answer to the halting problem can you justify your answer to be
>>>> correct,
>>>> but then you proof doesn't achieve the goal you claim.
>>>>
>>>>>
>>>>>> Note, it is easy to show that your interpretation is wrong since even
>>>>>> you admit that Linz H^, now called P by you will come to its end and
>>>>>> halt when given it own representation as its input, and thus is BY
>>>>>> DEFINITION a Halting Computation, Thus the H deciding it didn't
>>>>>> need to
>>>>>> abort its execution to get the wrong answer of Non-Halting.
>>>>>>
>>>>>
>>>>> Because at least one invocation of the infinite invocation chain
>>>>> specified by P(P) had to be terminated to prevent the infinite
>>>>> execution
>>>>> of this infinite invocation chain it is confirmed beyond all possible
>>>>> doubt that P(P) specifies an invocation chain.
>>>>
>>>> WRONG. Given that we have an H that can answer H(P,P) because it knows
>>>> at least enough to terminate the pattern you describe, then when we run
>>>> P(P) then because the H within it also knows to abort this sequence
>>>> (since it is built on the same algorithm) this P is NOT part of an
>>>> infinite chain of execution, and thus its H can return its (wrong)
>>>> answer to it and that P can then Halt.
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> Now I have told this this several hundred times.
>>>
>>>
>>
>> WRONG.
>>
>> P(P) starts.
>>
>> Calls H(P,P)
>>
>> H starts the simulation.
>>
>> H simulates P starting
>>
>> H simulates P calling H
>>
>> H simulates H starting its simulation
>>
>> H simulates H simulating P starting
>>
>> H simulates H simulating P calling H
>>
>> The first H about here detects what it THINKS is an infinite execution
>>
>> THe first H aborts its simulation
>>
>> The first H returns its answer (Non-Halting) to its caller
>>
>> P then Halts
>>
>> Showing P is a Halting Computation.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>

Are you nuts, or are recovering from a recent head injury?

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

<faf6990d-6146-45f1-a2d9-be79919177b4@notatt.com>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=17213&group=comp.theory#17213

 copy link   Newsgroups: comp.theory comp.ai.philosophy comp.software-eng
Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: jbb...@notatt.com (Jeff Barnett)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng
Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
invocation chain)
Date: Sat, 26 Jun 2021 04:27:46 -0600
Organization: A noiseless patient Spider
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 by: Jeff Barnett - Sat, 26 Jun 2021 10:27 UTC

On 6/26/2021 2:11 AM, Chris M. Thomasson wrote:
> On 6/25/2021 8:07 PM, olcott wrote:
>> On 6/25/2021 9:07 PM, Richard Damon wrote:
>>> On 6/25/21 9:46 PM, olcott wrote:
>>>> On 6/25/2021 8:37 PM, Richard Damon wrote:
>>>>> On 6/25/21 9:01 PM, olcott wrote:
>>>>>> On 6/25/2021 7:40 PM, Richard Damon wrote:
>>>>>>> On 6/25/21 6:56 PM, olcott wrote:
>>>>>>>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>>>>>>>> On 6/25/21 4:40 PM, olcott wrote:
>>>>>>>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>>>>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>>>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically
>>>>>>>>>>>>> create a P
>>>>>>>>>>>>> that
>>>>>>>>>>>>> doesn't follow that form, then you are hypothetically creating
>>>>>>>>>>>>> nonsense
>>>>>>>>>>>>> and can't use it to for anything logical.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Of every possible encoding of simulating partial halt decider H
>>>>>>>>>>>> that can
>>>>>>>>>>>> possibly exist  H*, if H* never aborts the simulation of its
>>>>>>>>>>>> input
>>>>>>>>>>>> results in the infinite execution of the invocation of of P(P)
>>>>>>>>>>>> then a
>>>>>>>>>>>> simulating halt decider H that does abort its simulation of
>>>>>>>>>>>> this
>>>>>>>>>>>> input
>>>>>>>>>>>> does correctly decide that this input does specify the never
>>>>>>>>>>>> halting
>>>>>>>>>>>> behavior of an infinite chain of invocations.
>>>>>>>>>>>
>>>>>>>>>>> Yes, if H* is an element of the set of non-aborting deciders
>>>>>>>>>>> (Hn), P
>>>>>>>>>>> will result in infinite recursion,
>>>>>>>>>>
>>>>>>>>>> Which logically entails beyond all possible doubt that the set of
>>>>>>>>>> encodings of simulating partial halt deciders H2* that do
>>>>>>>>>> abort the
>>>>>>>>>> simulation of the (P,P) input would correctly report that this
>>>>>>>>>> input
>>>>>>>>>> never halts.
>>>>>>>>>
>>>>>>>>> WHY?
>>>>>>>>>
>>>>>>>>
>>>>>>>> Axiom(1) Every computation that never halts unless its
>>>>>>>> simulation is
>>>>>>>> aborted is a computation that never halts. This verified as true on
>>>>>>>> the
>>>>>>>> basis of the meaning of its words.
>>>>>>>
>>>>>>> WRONG.
>>>>>>>
>>>>>>> Your test does not match the plain meaning of the words, as has been
>>>>>>> explained many times.
>>>>>>>
>>>>>>
>>>>>> Those words may be over your head, yet several others understand that
>>>>>> they are necessarily correct.
>>>>>
>>>>> I have seen NO ONE agree to your interpretation of it. The plain
>>>>> meaning
>>>>> is that if it can be shown that if the given instance of the simulator
>>>>> simulating a given input doesn't stop its simulation that this
>>>>> simulation will run forevr, then the machine that is being
>>>>> simulated can
>>>>> be corrected decided as non-Halting.
>>>>>
>>>>> An more formal way to say that is if UTM(P,I) is non-halting then
>>>>> it is
>>>>> correct for H(P,I) to return the non-halting result.
>>>>>
>>>>> This actually follows since UTM(P,I) will be non-halting if and
>>>>> only if
>>>>> P(I) is non-halting by the definition of a UTM, so that statement is
>>>>> trivially proven.
>>>>>
>>>>> Your interpretation, where even if a copy of the algorithm of H is
>>>>> included in P and that included copy needs to abort the simulation of
>>>>> the copy of the machine that it was given, can be PROVEN wrong, as
>>>>> even
>>>>> you have shown that P(P) in this case does Halt, thus your claimed
>>>>> correct answer is wrong by the definition of the problem.
>>>>>
>>>>> Only if you define that your answer isn't actually supposed to be the
>>>>> answer to the halting problem can you justify your answer to be
>>>>> correct,
>>>>> but then you proof doesn't achieve the goal you claim.
>>>>>
>>>>>>
>>>>>>> Note, it is easy to show that your interpretation is wrong since
>>>>>>> even
>>>>>>> you admit that Linz H^, now called P by you will come to its end and
>>>>>>> halt when given it own representation as its input, and thus is BY
>>>>>>> DEFINITION a Halting Computation, Thus the H deciding it didn't
>>>>>>> need to
>>>>>>> abort its execution to get the wrong answer of Non-Halting.
>>>>>>>
>>>>>>
>>>>>> Because at least one invocation of the infinite invocation chain
>>>>>> specified by P(P) had to be terminated to prevent the infinite
>>>>>> execution
>>>>>> of this infinite invocation chain it is confirmed beyond all possible
>>>>>> doubt that P(P) specifies an invocation chain.
>>>>>
>>>>> WRONG. Given that we have an H that can answer H(P,P) because it knows
>>>>> at least enough to terminate the pattern you describe, then when we
>>>>> run
>>>>> P(P) then because the H within it also knows to abort this sequence
>>>>> (since it is built on the same algorithm) this P is NOT part of an
>>>>> infinite chain of execution, and thus its H can return its (wrong)
>>>>> answer to it and that P can then Halt.
>>>>
>>>> P(P) specifies in infinite invocation sequence that is terminated on
>>>> its
>>>> third invocation of H(P,P).
>>>>
>>>> P(P) specifies in infinite invocation sequence that is terminated on
>>>> its
>>>> third invocation of H(P,P).
>>>>
>>>> P(P) specifies in infinite invocation sequence that is terminated on
>>>> its
>>>> third invocation of H(P,P).
>>>>
>>>> P(P) specifies in infinite invocation sequence that is terminated on
>>>> its
>>>> third invocation of H(P,P).
>>>>
>>>> P(P) specifies in infinite invocation sequence that is terminated on
>>>> its
>>>> third invocation of H(P,P).
>>>>
>>>> Now I have told this this several hundred times.
>>>>
>>>>
>>>
>>> WRONG.
>>>
>>> P(P) starts.
>>>
>>> Calls H(P,P)
>>>
>>> H starts the simulation.
>>>
>>> H simulates P starting
>>>
>>> H simulates P calling H
>>>
>>> H simulates H starting its simulation
>>>
>>> H simulates H simulating P starting
>>>
>>> H simulates H simulating P calling H
>>>
>>> The first H about here detects what it THINKS is an infinite execution
>>>
>>> THe first H aborts its simulation
>>>
>>> The first H returns its answer (Non-Halting) to its caller
>>>
>>> P then Halts
>>>
>>> Showing P is a Halting Computation.
>>
>> As you already admitted P ONLY halts because some H aborts some P
>> otherwise P never ever halts.
>>
>> As you already admitted P ONLY halts because some H aborts some P
>> otherwise P never ever halts.
>>
>> As you already admitted P ONLY halts because some H aborts some P
>> otherwise P never ever halts.
>>
>> As you already admitted P ONLY halts because some H aborts some P
>> otherwise P never ever halts.
>>
>> As you already admitted P ONLY halts because some H aborts some P
>> otherwise P never ever halts.
>>
>
> Are you nuts, or are recovering from a recent head injury?


Click here to read the complete article
Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
invocation chain)
Newsgroups: comp.theory
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
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From: Rich...@Damon-Family.org (Richard Damon)
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Date: Sat, 26 Jun 2021 06:32:27 -0400
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 by: Richard Damon - Sat, 26 Jun 2021 10:32 UTC

On 6/25/21 11:07 PM, olcott wrote:
> On 6/25/2021 9:07 PM, Richard Damon wrote:
>> On 6/25/21 9:46 PM, olcott wrote:
>>> On 6/25/2021 8:37 PM, Richard Damon wrote:
>>>> On 6/25/21 9:01 PM, olcott wrote:
>>>>> On 6/25/2021 7:40 PM, Richard Damon wrote:
>>>>>> On 6/25/21 6:56 PM, olcott wrote:
>>>>>>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>>>>>>> On 6/25/21 4:40 PM, olcott wrote:
>>>>>>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>>>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>>>>>>
>>>>>>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create
>>>>>>>>>>>> a P
>>>>>>>>>>>> that
>>>>>>>>>>>> doesn't follow that form, then you are hypothetically creating
>>>>>>>>>>>> nonsense
>>>>>>>>>>>> and can't use it to for anything logical.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Of every possible encoding of simulating partial halt decider H
>>>>>>>>>>> that can
>>>>>>>>>>> possibly exist  H*, if H* never aborts the simulation of its
>>>>>>>>>>> input
>>>>>>>>>>> results in the infinite execution of the invocation of of P(P)
>>>>>>>>>>> then a
>>>>>>>>>>> simulating halt decider H that does abort its simulation of this
>>>>>>>>>>> input
>>>>>>>>>>> does correctly decide that this input does specify the never
>>>>>>>>>>> halting
>>>>>>>>>>> behavior of an infinite chain of invocations.
>>>>>>>>>>
>>>>>>>>>> Yes, if H* is an element of the set of non-aborting deciders
>>>>>>>>>> (Hn), P
>>>>>>>>>> will result in infinite recursion,
>>>>>>>>>
>>>>>>>>> Which logically entails beyond all possible doubt that the set of
>>>>>>>>> encodings of simulating partial halt deciders H2* that do abort
>>>>>>>>> the
>>>>>>>>> simulation of the (P,P) input would correctly report that this
>>>>>>>>> input
>>>>>>>>> never halts.
>>>>>>>>
>>>>>>>> WHY?
>>>>>>>>
>>>>>>>
>>>>>>> Axiom(1) Every computation that never halts unless its simulation is
>>>>>>> aborted is a computation that never halts. This verified as true on
>>>>>>> the
>>>>>>> basis of the meaning of its words.
>>>>>>
>>>>>> WRONG.
>>>>>>
>>>>>> Your test does not match the plain meaning of the words, as has been
>>>>>> explained many times.
>>>>>>
>>>>>
>>>>> Those words may be over your head, yet several others understand that
>>>>> they are necessarily correct.
>>>>
>>>> I have seen NO ONE agree to your interpretation of it. The plain
>>>> meaning
>>>> is that if it can be shown that if the given instance of the simulator
>>>> simulating a given input doesn't stop its simulation that this
>>>> simulation will run forevr, then the machine that is being simulated
>>>> can
>>>> be corrected decided as non-Halting.
>>>>
>>>> An more formal way to say that is if UTM(P,I) is non-halting then it is
>>>> correct for H(P,I) to return the non-halting result.
>>>>
>>>> This actually follows since UTM(P,I) will be non-halting if and only if
>>>> P(I) is non-halting by the definition of a UTM, so that statement is
>>>> trivially proven.
>>>>
>>>> Your interpretation, where even if a copy of the algorithm of H is
>>>> included in P and that included copy needs to abort the simulation of
>>>> the copy of the machine that it was given, can be PROVEN wrong, as even
>>>> you have shown that P(P) in this case does Halt, thus your claimed
>>>> correct answer is wrong by the definition of the problem.
>>>>
>>>> Only if you define that your answer isn't actually supposed to be the
>>>> answer to the halting problem can you justify your answer to be
>>>> correct,
>>>> but then you proof doesn't achieve the goal you claim.
>>>>
>>>>>
>>>>>> Note, it is easy to show that your interpretation is wrong since even
>>>>>> you admit that Linz H^, now called P by you will come to its end and
>>>>>> halt when given it own representation as its input, and thus is BY
>>>>>> DEFINITION a Halting Computation, Thus the H deciding it didn't
>>>>>> need to
>>>>>> abort its execution to get the wrong answer of Non-Halting.
>>>>>>
>>>>>
>>>>> Because at least one invocation of the infinite invocation chain
>>>>> specified by P(P) had to be terminated to prevent the infinite
>>>>> execution
>>>>> of this infinite invocation chain it is confirmed beyond all possible
>>>>> doubt that P(P) specifies an invocation chain.
>>>>
>>>> WRONG. Given that we have an H that can answer H(P,P) because it knows
>>>> at least enough to terminate the pattern you describe, then when we run
>>>> P(P) then because the H within it also knows to abort this sequence
>>>> (since it is built on the same algorithm) this P is NOT part of an
>>>> infinite chain of execution, and thus its H can return its (wrong)
>>>> answer to it and that P can then Halt.
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> Now I have told this this several hundred times.
>>>
>>>
>>
>> WRONG.
>>
>> P(P) starts.
>>
>> Calls H(P,P)
>>
>> H starts the simulation.
>>
>> H simulates P starting
>>
>> H simulates P calling H
>>
>> H simulates H starting its simulation
>>
>> H simulates H simulating P starting
>>
>> H simulates H simulating P calling H
>>
>> The first H about here detects what it THINKS is an infinite execution
>>
>> THe first H aborts its simulation
>>
>> The first H returns its answer (Non-Halting) to its caller
>>
>> P then Halts
>>
>> Showing P is a Halting Computation.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>

Yes, P halts because the H it contains terminated the simulation of
another copy of its description.

YOUR problem is that you think that actually means something, it doesn't

Halting is DEFINED as the ultimate operation of the machine. Did it
finish or not. The fact that in its path to finishing it did a
simulation that it terminated doesn't affect that.


Click here to read the complete article
Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)
Date: Sat, 26 Jun 2021 15:51:29 +0100
Organization: A noiseless patient Spider
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 by: Ben Bacarisse - Sat, 26 Jun 2021 14:51 UTC

I've trimmed the newsgroups. I suggest everyone else does so as well.
There is no need to damage more than one group with this junk.

Jeff Barnett <jbb@notatt.com> writes:

> On 6/26/2021 2:11 AM, Chris M. Thomasson wrote:
>> Are you nuts, or are recovering from a recent head injury?
>
> It's not recent. It seems he's been at it for a few decades.

I first cam across PO when he was about to get rich selling some
software based on two junk patents he'd filed. The software was vapour
ware (at the time) but the expectation of riches was real enough. The
situation may be reversed now.

I appear to have first talked about halting with PO in 2012. I had a
look at one of the replies I made back then. See how predictable it all
is:

"You see? No connection to what I said. No discussion. Just a
re-statement of the same false claim. Should I state, yet again, why
it's false? Is there any reason to think you'd address the argument
if I did so?"

That's a shade off 20 years ago. It really is futile.

--
Ben.

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

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
invocation chain)
Newsgroups: comp.theory
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
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From: NoO...@NoWhere.com (olcott)
Date: Sat, 26 Jun 2021 09:55:57 -0500
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 by: olcott - Sat, 26 Jun 2021 14:55 UTC

On 6/26/2021 5:32 AM, Richard Damon wrote:
> On 6/25/21 11:07 PM, olcott wrote:
>> On 6/25/2021 9:07 PM, Richard Damon wrote:
>>> On 6/25/21 9:46 PM, olcott wrote:
>>>> On 6/25/2021 8:37 PM, Richard Damon wrote:
>>>>> On 6/25/21 9:01 PM, olcott wrote:
>>>>>> On 6/25/2021 7:40 PM, Richard Damon wrote:
>>>>>>> On 6/25/21 6:56 PM, olcott wrote:
>>>>>>>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>>>>>>>> On 6/25/21 4:40 PM, olcott wrote:
>>>>>>>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>>>>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>>>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create
>>>>>>>>>>>>> a P
>>>>>>>>>>>>> that
>>>>>>>>>>>>> doesn't follow that form, then you are hypothetically creating
>>>>>>>>>>>>> nonsense
>>>>>>>>>>>>> and can't use it to for anything logical.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Of every possible encoding of simulating partial halt decider H
>>>>>>>>>>>> that can
>>>>>>>>>>>> possibly exist  H*, if H* never aborts the simulation of its
>>>>>>>>>>>> input
>>>>>>>>>>>> results in the infinite execution of the invocation of of P(P)
>>>>>>>>>>>> then a
>>>>>>>>>>>> simulating halt decider H that does abort its simulation of this
>>>>>>>>>>>> input
>>>>>>>>>>>> does correctly decide that this input does specify the never
>>>>>>>>>>>> halting
>>>>>>>>>>>> behavior of an infinite chain of invocations.
>>>>>>>>>>>
>>>>>>>>>>> Yes, if H* is an element of the set of non-aborting deciders
>>>>>>>>>>> (Hn), P
>>>>>>>>>>> will result in infinite recursion,
>>>>>>>>>>
>>>>>>>>>> Which logically entails beyond all possible doubt that the set of
>>>>>>>>>> encodings of simulating partial halt deciders H2* that do abort
>>>>>>>>>> the
>>>>>>>>>> simulation of the (P,P) input would correctly report that this
>>>>>>>>>> input
>>>>>>>>>> never halts.
>>>>>>>>>
>>>>>>>>> WHY?
>>>>>>>>>
>>>>>>>>
>>>>>>>> Axiom(1) Every computation that never halts unless its simulation is
>>>>>>>> aborted is a computation that never halts. This verified as true on
>>>>>>>> the
>>>>>>>> basis of the meaning of its words.
>>>>>>>
>>>>>>> WRONG.
>>>>>>>
>>>>>>> Your test does not match the plain meaning of the words, as has been
>>>>>>> explained many times.
>>>>>>>
>>>>>>
>>>>>> Those words may be over your head, yet several others understand that
>>>>>> they are necessarily correct.
>>>>>
>>>>> I have seen NO ONE agree to your interpretation of it. The plain
>>>>> meaning
>>>>> is that if it can be shown that if the given instance of the simulator
>>>>> simulating a given input doesn't stop its simulation that this
>>>>> simulation will run forevr, then the machine that is being simulated
>>>>> can
>>>>> be corrected decided as non-Halting.
>>>>>
>>>>> An more formal way to say that is if UTM(P,I) is non-halting then it is
>>>>> correct for H(P,I) to return the non-halting result.
>>>>>
>>>>> This actually follows since UTM(P,I) will be non-halting if and only if
>>>>> P(I) is non-halting by the definition of a UTM, so that statement is
>>>>> trivially proven.
>>>>>
>>>>> Your interpretation, where even if a copy of the algorithm of H is
>>>>> included in P and that included copy needs to abort the simulation of
>>>>> the copy of the machine that it was given, can be PROVEN wrong, as even
>>>>> you have shown that P(P) in this case does Halt, thus your claimed
>>>>> correct answer is wrong by the definition of the problem.
>>>>>
>>>>> Only if you define that your answer isn't actually supposed to be the
>>>>> answer to the halting problem can you justify your answer to be
>>>>> correct,
>>>>> but then you proof doesn't achieve the goal you claim.
>>>>>
>>>>>>
>>>>>>> Note, it is easy to show that your interpretation is wrong since even
>>>>>>> you admit that Linz H^, now called P by you will come to its end and
>>>>>>> halt when given it own representation as its input, and thus is BY
>>>>>>> DEFINITION a Halting Computation, Thus the H deciding it didn't
>>>>>>> need to
>>>>>>> abort its execution to get the wrong answer of Non-Halting.
>>>>>>>
>>>>>>
>>>>>> Because at least one invocation of the infinite invocation chain
>>>>>> specified by P(P) had to be terminated to prevent the infinite
>>>>>> execution
>>>>>> of this infinite invocation chain it is confirmed beyond all possible
>>>>>> doubt that P(P) specifies an invocation chain.
>>>>>
>>>>> WRONG. Given that we have an H that can answer H(P,P) because it knows
>>>>> at least enough to terminate the pattern you describe, then when we run
>>>>> P(P) then because the H within it also knows to abort this sequence
>>>>> (since it is built on the same algorithm) this P is NOT part of an
>>>>> infinite chain of execution, and thus its H can return its (wrong)
>>>>> answer to it and that P can then Halt.
>>>>
>>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>>> third invocation of H(P,P).
>>>>
>>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>>> third invocation of H(P,P).
>>>>
>>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>>> third invocation of H(P,P).
>>>>
>>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>>> third invocation of H(P,P).
>>>>
>>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>>> third invocation of H(P,P).
>>>>
>>>> Now I have told this this several hundred times.
>>>>
>>>>
>>>
>>> WRONG.
>>>
>>> P(P) starts.
>>>
>>> Calls H(P,P)
>>>
>>> H starts the simulation.
>>>
>>> H simulates P starting
>>>
>>> H simulates P calling H
>>>
>>> H simulates H starting its simulation
>>>
>>> H simulates H simulating P starting
>>>
>>> H simulates H simulating P calling H
>>>
>>> The first H about here detects what it THINKS is an infinite execution
>>>
>>> THe first H aborts its simulation
>>>
>>> The first H returns its answer (Non-Halting) to its caller
>>>
>>> P then Halts
>>>
>>> Showing P is a Halting Computation.
>>
>> As you already admitted P ONLY halts because some H aborts some P
>> otherwise P never ever halts.
>>
>> As you already admitted P ONLY halts because some H aborts some P
>> otherwise P never ever halts.
>>
>> As you already admitted P ONLY halts because some H aborts some P
>> otherwise P never ever halts.
>>
>> As you already admitted P ONLY halts because some H aborts some P
>> otherwise P never ever halts.
>>
>> As you already admitted P ONLY halts because some H aborts some P
>> otherwise P never ever halts.
>>
>
> Yes, P halts because the H it contains terminated the simulation of
> another copy of its description.
>
> YOUR problem is that you think that actually means something, it doesn't
>


Click here to read the complete article
Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)

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https://www.novabbs.com/devel/article-flat.php?id=17219&group=comp.theory#17219

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite
invocation chain)
Newsgroups: comp.theory
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com>
<48GdnbWRNeocCU79nZ2dnUU7-TXNnZ2d@giganews.com> <sb0f72$hf4$1@dont-email.me>
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From: Rich...@Damon-Family.org (Richard Damon)
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Date: Sat, 26 Jun 2021 11:06:29 -0400
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 by: Richard Damon - Sat, 26 Jun 2021 15:06 UTC

On 6/26/21 10:55 AM, olcott wrote:
> On 6/26/2021 5:32 AM, Richard Damon wrote:
>> On 6/25/21 11:07 PM, olcott wrote:
>>> On 6/25/2021 9:07 PM, Richard Damon wrote:
>>>> On 6/25/21 9:46 PM, olcott wrote:
>>>>> On 6/25/2021 8:37 PM, Richard Damon wrote:
>>>>>> On 6/25/21 9:01 PM, olcott wrote:
>>>>>>> On 6/25/2021 7:40 PM, Richard Damon wrote:
>>>>>>>> On 6/25/21 6:56 PM, olcott wrote:
>>>>>>>>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>>>>>>>>> On 6/25/21 4:40 PM, olcott wrote:
>>>>>>>>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>>>>>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>>>>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create
>>>>>>>>>>>>>> a P
>>>>>>>>>>>>>> that
>>>>>>>>>>>>>> doesn't follow that form, then you are hypothetically
>>>>>>>>>>>>>> creating
>>>>>>>>>>>>>> nonsense
>>>>>>>>>>>>>> and can't use it to for anything logical.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Of every possible encoding of simulating partial halt
>>>>>>>>>>>>> decider H
>>>>>>>>>>>>> that can
>>>>>>>>>>>>> possibly exist  H*, if H* never aborts the simulation of its
>>>>>>>>>>>>> input
>>>>>>>>>>>>> results in the infinite execution of the invocation of of P(P)
>>>>>>>>>>>>> then a
>>>>>>>>>>>>> simulating halt decider H that does abort its simulation of
>>>>>>>>>>>>> this
>>>>>>>>>>>>> input
>>>>>>>>>>>>> does correctly decide that this input does specify the never
>>>>>>>>>>>>> halting
>>>>>>>>>>>>> behavior of an infinite chain of invocations.
>>>>>>>>>>>>
>>>>>>>>>>>> Yes, if H* is an element of the set of non-aborting deciders
>>>>>>>>>>>> (Hn), P
>>>>>>>>>>>> will result in infinite recursion,
>>>>>>>>>>>
>>>>>>>>>>> Which logically entails beyond all possible doubt that the
>>>>>>>>>>> set of
>>>>>>>>>>> encodings of simulating partial halt deciders H2* that do abort
>>>>>>>>>>> the
>>>>>>>>>>> simulation of the (P,P) input would correctly report that this
>>>>>>>>>>> input
>>>>>>>>>>> never halts.
>>>>>>>>>>
>>>>>>>>>> WHY?
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Axiom(1) Every computation that never halts unless its
>>>>>>>>> simulation is
>>>>>>>>> aborted is a computation that never halts. This verified as
>>>>>>>>> true on
>>>>>>>>> the
>>>>>>>>> basis of the meaning of its words.
>>>>>>>>
>>>>>>>> WRONG.
>>>>>>>>
>>>>>>>> Your test does not match the plain meaning of the words, as has
>>>>>>>> been
>>>>>>>> explained many times.
>>>>>>>>
>>>>>>>
>>>>>>> Those words may be over your head, yet several others understand
>>>>>>> that
>>>>>>> they are necessarily correct.
>>>>>>
>>>>>> I have seen NO ONE agree to your interpretation of it. The plain
>>>>>> meaning
>>>>>> is that if it can be shown that if the given instance of the
>>>>>> simulator
>>>>>> simulating a given input doesn't stop its simulation that this
>>>>>> simulation will run forevr, then the machine that is being simulated
>>>>>> can
>>>>>> be corrected decided as non-Halting.
>>>>>>
>>>>>> An more formal way to say that is if UTM(P,I) is non-halting then
>>>>>> it is
>>>>>> correct for H(P,I) to return the non-halting result.
>>>>>>
>>>>>> This actually follows since UTM(P,I) will be non-halting if and
>>>>>> only if
>>>>>> P(I) is non-halting by the definition of a UTM, so that statement is
>>>>>> trivially proven.
>>>>>>
>>>>>> Your interpretation, where even if a copy of the algorithm of H is
>>>>>> included in P and that included copy needs to abort the simulation of
>>>>>> the copy of the machine that it was given, can be PROVEN wrong, as
>>>>>> even
>>>>>> you have shown that P(P) in this case does Halt, thus your claimed
>>>>>> correct answer is wrong by the definition of the problem.
>>>>>>
>>>>>> Only if you define that your answer isn't actually supposed to be the
>>>>>> answer to the halting problem can you justify your answer to be
>>>>>> correct,
>>>>>> but then you proof doesn't achieve the goal you claim.
>>>>>>
>>>>>>>
>>>>>>>> Note, it is easy to show that your interpretation is wrong since
>>>>>>>> even
>>>>>>>> you admit that Linz H^, now called P by you will come to its end
>>>>>>>> and
>>>>>>>> halt when given it own representation as its input, and thus is BY
>>>>>>>> DEFINITION a Halting Computation, Thus the H deciding it didn't
>>>>>>>> need to
>>>>>>>> abort its execution to get the wrong answer of Non-Halting.
>>>>>>>>
>>>>>>>
>>>>>>> Because at least one invocation of the infinite invocation chain
>>>>>>> specified by P(P) had to be terminated to prevent the infinite
>>>>>>> execution
>>>>>>> of this infinite invocation chain it is confirmed beyond all
>>>>>>> possible
>>>>>>> doubt that P(P) specifies an invocation chain.
>>>>>>
>>>>>> WRONG. Given that we have an H that can answer H(P,P) because it
>>>>>> knows
>>>>>> at least enough to terminate the pattern you describe, then when
>>>>>> we run
>>>>>> P(P) then because the H within it also knows to abort this sequence
>>>>>> (since it is built on the same algorithm) this P is NOT part of an
>>>>>> infinite chain of execution, and thus its H can return its (wrong)
>>>>>> answer to it and that P can then Halt.
>>>>>
>>>>> P(P) specifies in infinite invocation sequence that is terminated
>>>>> on its
>>>>> third invocation of H(P,P).
>>>>>
>>>>> P(P) specifies in infinite invocation sequence that is terminated
>>>>> on its
>>>>> third invocation of H(P,P).
>>>>>
>>>>> P(P) specifies in infinite invocation sequence that is terminated
>>>>> on its
>>>>> third invocation of H(P,P).
>>>>>
>>>>> P(P) specifies in infinite invocation sequence that is terminated
>>>>> on its
>>>>> third invocation of H(P,P).
>>>>>
>>>>> P(P) specifies in infinite invocation sequence that is terminated
>>>>> on its
>>>>> third invocation of H(P,P).
>>>>>
>>>>> Now I have told this this several hundred times.
>>>>>
>>>>>
>>>>
>>>> WRONG.
>>>>
>>>> P(P) starts.
>>>>
>>>> Calls H(P,P)
>>>>
>>>> H starts the simulation.
>>>>
>>>> H simulates P starting
>>>>
>>>> H simulates P calling H
>>>>
>>>> H simulates H starting its simulation
>>>>
>>>> H simulates H simulating P starting
>>>>
>>>> H simulates H simulating P calling H
>>>>
>>>> The first H about here detects what it THINKS is an infinite execution
>>>>
>>>> THe first H aborts its simulation
>>>>
>>>> The first H returns its answer (Non-Halting) to its caller
>>>>
>>>> P then Halts
>>>>
>>>> Showing P is a Halting Computation.
>>>
>>> As you already admitted P ONLY halts because some H aborts some P
>>> otherwise P never ever halts.
>>>
>>> As you already admitted P ONLY halts because some H aborts some P
>>> otherwise P never ever halts.
>>>
>>> As you already admitted P ONLY halts because some H aborts some P
>>> otherwise P never ever halts.
>>>
>>> As you already admitted P ONLY halts because some H aborts some P
>>> otherwise P never ever halts.
>>>
>>> As you already admitted P ONLY halts because some H aborts some P
>>> otherwise P never ever halts.
>>>
>>
>> Yes, P halts because the H it contains terminated the simulation of
>> another copy of its description.
>>
>> YOUR problem is that you think that actually means something, it doesn't
>>
>
> Whenever one invocation of the infinite invocation chain of infinite
> recursion is aborted the whole chain terminates.


Click here to read the complete article
Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)(Ben lies)

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Subject: Re: What if a cat barks? [ sound deduction is a proof ](infinite invocation chain)(Ben lies)
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng
References: <BpqdnWBR5LTFj039nZ2dnUU7-XPNnZ2d@giganews.com> <sb0kra$ert$1@dont-email.me> <kaudnZS1vPJiaE79nZ2dnUU7-QHNnZ2d@giganews.com> <C0_AI.793830$2A5.649020@fx45.iad> <udKdnabaTsZvOkn9nZ2dnUU7-dvNnZ2d@giganews.com> <OlbBI.605613$J_5.348305@fx46.iad> <fO6dnQEYd73PmEv9nZ2dnUU7-TmdnZ2d@giganews.com> <u9oBI.267517$lyv9.157656@fx35.iad> <eKednajHd_LtuEv9nZ2dnUU7-TvNnZ2d@giganews.com> <ALqBI.113709$od.33914@fx15.iad> <cvednUP16NqYokv9nZ2dnUU7-bHNnZ2d@giganews.com> <HksBI.267$al1.209@fx26.iad> <BuOdncUXaL2swkv9nZ2dnUU7-W-dnZ2d@giganews.com> <6IuBI.115687$431.109356@fx39.iad> <RKCdnSS4Ifj44Uv9nZ2dnUU7-VvNnZ2d@giganews.com> <hxvBI.20803$9q1.10955@fx09.iad> <I8WdnT2QqrV5G0v9nZ2dnUU7-WvNnZ2d@giganews.com> <EZvBI.43029$7Y.22867@fx03.iad> <yr2dnWJMMpJLBEv9nZ2dnUU7-dvNnZ2d@giganews.com> <sb6nfj$i1b$1@gioia.aioe.org> <faf6990d-6146-45f1-a2d9-be79919177b4@notatt.com> <87fsx4n0ge.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Sat, 26 Jun 2021 10:52:20 -0500
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 by: olcott - Sat, 26 Jun 2021 15:52 UTC

On 6/26/2021 9:51 AM, Ben Bacarisse wrote:
> I've trimmed the newsgroups. I suggest everyone else does so as well.
> There is no need to damage more than one group with this junk.
>
> Jeff Barnett <jbb@notatt.com> writes:
>
>> On 6/26/2021 2:11 AM, Chris M. Thomasson wrote:
>>> Are you nuts, or are recovering from a recent head injury?
>>
>> It's not recent. It seems he's been at it for a few decades.
>
> I first cam across PO when he was about to get rich selling some
> software based on two junk patents he'd filed. The software was vapour
> ware (at the time) but the expectation of riches was real enough. The
> situation may be reversed now.
>
> I appear to have first talked about halting with PO in 2012. I had a
> look at one of the replies I made back then. See how predictable it all
> is:
>
> "You see? No connection to what I said. No discussion. Just a
> re-statement of the same false claim. Should I state, yet again, why
> it's false? Is there any reason to think you'd address the argument
> if I did so?"
>
> That's a shade off 20 years ago. It really is futile.
>

On 10/17/2006 7:03 PM, Ben Bacarisse wrote:
> "Peter Olcott" <NoSpam@SeeScreen.com> writes:

It is the case that P(P) has infinite execution unless one its infinite
chain of invocations is aborted.

You know there is no rebuttal to this because it is an easily verifiable
fact, even Richard acknowledged that P(P) never halts

On 6/25/2021 3:11 PM, Richard Damon wrote:
>
> Yes, if H* is an element of the set of non-aborting deciders (Hn), P
> will result in infinite recursion,

So you will misdirect with ad hominem and rhetoric because you know that
there is no plausible rebuttal using logic.

--
Copyright 2021 Pete Olcott

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

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