On 5/22/2021 9:56 AM, Kaz Kylheku wrote:
> On 2021-05-22, olcott <NoOne@NoWhere.com> wrote:
>> void H_Hat(u32 P)
>> {
>> u32 Input_Halts = Halts(P, P);
>> if (Input_Halts)
>> HERE: goto HERE;
>> }
>>
>> Halts decides the above case by using the behavior of the above code
>> when a simulator is substituted for the simulating halt decider to
>> correctly answer this question:
>
> For the thousandth time, "substitute" means to produce an entirely
> new S_Hat function /similar/ to the above H_Hat which calls Simulate
> instead of Halts.
>

Yes it does.

> // S_Hat is to Simulate what H_Hat is to Halts
>
> void S_Hat(u32 P)
> {
> u32 Input_Halts = Simulate(P, P);
> if (Input_Halts)
> HERE: goto HERE;
> }
>
> That creates a new test case which shows that the following decision is
> wrong or non-halting:
>
> Simulate(S_Hat, S_Hat)

It would not show that the decision is wrong because it never makes any
decision. It would also not directly show that it is not halting only
waiting an infinite amount of time would directly show this.

Its behavior could be examined and find the the same cycle repeats and
there is nothing in this cycle that could possibly break this cycle.

Simulating halt decider H would decide not halting on H(S_Hat, S_Hat).

> just like this is wrong or non-halting:
>
> Halts(H_Hat, H_Hat)
>

When you fully understand its foundational basis then you will fully
understand that it is not the wrong answer.

>> This question <is> correctly answered by the proxy: Would the input P to
>> a simulator S halt?
>
> You must stick to the diagonal of the test case / decider combination
> matrix.
>

Hell no. Although I never quite understood the diagonal argument it
really seems to me that it only tells you that incorrect questions have
no correct answer.

> Halts(S_Hat, S_Hat); // correct, but off the diagonal
>

I am not sure what off the diagonal means.
It is clear that Halts(S_Hat, S_Hat) == 0 is correct.

> The halting proof is a form of diagonal argument. It says that the
> diagonal entries all fail to decide.
>
> Cases F H G <---- deecciders
>
> F^ [ F(^F,^F) ] G(^F, ^F) H(^F, ^F)
> G^ F(^G,^G) [ G(^G, ^G) ] H(^G, ^G)
> H^ F(^H,^H) G(^H, ^H) [ H(^H, ^H) ]
>
> Only the diagonal [ ] entriess are relevant. When you substitute
> a decider in the test case, to obtain another test case, you
> move vertically through the table, which takes you off the diagonal.
>
> When you replacee the decider being applied to the test case,
> you move horizontally; again off the diagonal.
>
> To stick to the diagonal, you must make both substitutions
> in parallel. E.g. when yu change from the H^ case which "attacks" the H
> decider to the G^ test case which attacks the G decider,
> it is that attacked G decider which cannot decide that case.
> You must go from H(^H, ^H) to G(^G, ^G).
>
> Until you thoroughly internalize the above, you do not
> have a grasp on halting.
>

I understand where you are coming from. I am coming from somewhere else.
If you analyze what I am saying using conventional analysis then what I
am saying is incorrect.

To see that what I am saying <is> correct you must switch to
unconventional analysis.

Because we know that the only difference in the behavior of a simulating
halt decider and a simulator is that the simulating halt decider stops
simulating some of its inputs we can examine the behavior of these
inputs in a simulator to determine whether or not a simulating halt
decider would stop simulating these inputs.

If the simulation of input P to simulator S would never terminate then
we can know that simulating halt decider H must stop simulating input P.

>> The reason that the proxy can be used to answer the original question is
>> that the only difference between a simulating halt decider and a
>> simulator is that the simulating halt decider stops simulating some of
>> its inputs.
>
> Any difference whatsoever makes them different entities, which means
> that there cannot be a single H_Hat based on both of them.
> They are different columns in the table, corresponding to different
> rows via the diagonal.
>

--
Copyright 2021 Pete Olcott

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

On 5/23/2021 9:59 AM, Kaz Kylheku wrote:
> On 2021-05-22, olcott <NoOne@NoWhere.com> wrote:
>> On 5/22/2021 9:56 AM, Kaz Kylheku wrote:
>>> The halting proof is a form of diagonal argument. It says that the
>>> diagonal entries all fail to decide.
>>>
>>> Cases F H G <---- deecciders
>>>
>>> F^ [ F(^F,^F) ] G(^F, ^F) H(^F, ^F)
>>> G^ F(^G,^G) [ G(^G, ^G) ] H(^G, ^G)
>>> H^ F(^H,^H) G(^H, ^H) [ H(^H, ^H) ]
>>>
>>> Only the diagonal [ ] entriess are relevant. When you substitute
>>> a decider in the test case, to obtain another test case, you
>>> move vertically through the table, which takes you off the diagonal.
>>>
>>> When you replacee the decider being applied to the test case,
>>> you move horizontally; again off the diagonal.
>>>
>>> To stick to the diagonal, you must make both substitutions
>>> in parallel. E.g. when yu change from the H^ case which "attacks" the H
>>> decider to the G^ test case which attacks the G decider,
>>> it is that attacked G decider which cannot decide that case.
>>> You must go from H(^H, ^H) to G(^G, ^G).
>>>
>>> Until you thoroughly internalize the above, you do not
>>> have a grasp on halting.
>>>
>>
>> I understand where you are coming from. I am coming from somewhere else.
>> If you analyze what I am saying using conventional analysis then what I
>> am saying is incorrect.
>>
>> To see that what I am saying <is> correct you must switch to
>> unconventional analysis.
>
> The Turing model of computation, and the halting question and its
> proofs are entirely contained within "conventional analysis".
>
> You're using the language of conventional analysis to refer to
> its concepts and constructs, and to dispute its results.
>

As I recently explained to Ben My proof assumes the entire Peter Linz
proof on pages 317-319 and is inserted at the end of page 319, skipping
page 320.

> If you want to use "unconventional analysis", you must rigorously pin
> down what that is, before you even approach the halting problem,
> and present that framework upfront.
>
> Within that framework, you must construct that framework's own model of
> computation. Then pursue that framework's version of the halting
> problem within that framework, and not make any claims with regard to
> conventional Turing halting (which will automatically make you look
> wrong).
>
> You can use similar language like "machine", "halts", and so on, if you
> make it clear up front that these words are in your own framework and do
> not have their conventional meaning.
>
> You also face this problem: what good is your alternative analysis?
> Conventional analysis is reliable and powerful. Can we apply your
> alternative analysis to a broad area of problems and get good results?
>
>> Because we know that the only difference in the behavior of a simulating
>> halt decider and a simulator is that the simulating halt decider stops
>> simulating some of its inputs we can examine the behavior of these
>> inputs in a simulator to determine whether or not a simulating halt
>> decider would stop simulating these inputs.
>
> Sure, but the difference in behavior means we are jumping to a different
> column of the input-decider matrix and are no longer on the diagonal.
> To return to the diagonal while staying in the same column, we must
> switch to a different row: the row for the test case which is based on
> that column's decider. Where row and column match, we have an incorrect
> or nonterminating halting decision, according to this conventional analysis.
>

The whole diagonalization thing is gibberish to me unless it only shows
that incorrect questions do not have correct answers.

> Any reasoning that leads you, through substitutions, to discovering a
> good halting decision, but which is off the diagonal trace, has led you
> to an irrelevant configuration. The proof says only that the
> configurations on the diagonal are incorrect or non-halting. The
> proof does not make any assertions about non-diagonal configuations,
> therefore those configurations are not counterexamples to the proof's
> claims.
>
> The undecidability of halting is precisely the claim that the entire
> diagonal consists of incorrect answers or non-halting. Since for every
> decider there is a diagonal entry, every decider is associated with a
> test case it does not handle.
>
> If every decider has at least one test case it does not handle, then
> there does not exist a universal (= handles all cases) decider:
> there is no algorithm that decides all cases of halting.
>

Generic halt deciding principle for inputs having the pathological
self-reference error

(a) Halt deciding simulator H only affects the behavior of input P in
one way and only under one condition. In all other cases a halt deciding
simulator H has no effect what-so-ever on the behavior of input P and
acts exactly like pure simulator S on input P.

(b) When an infinite computation is specified as input P to a simulating
halt decider H the halt decider H must stop the simulation of this input
P. When this same input is simulated by simulator S, its simulation
never stops.

(c) When a finite computation is specified as input P to a simulating
halt decider H the halt decider H simulates its input exactly as if H
was simulator S.

Because of the above simulating halt decider H can decide input P having
the pathological self-reference error on the basis of the proxy input P2
having the pathological self-reference error removed.

When input P invokes embedded simulating halt decider H on itself it
creates proxy input P2 replacing H with simulator S. The embedded halt
decider H examines the execution trace of proxy input P2. The embedded
halt decider correctly decides its input P on the basis of how it
decides input P2.

When we apply the Generic halt deciding principle to the embedded
simulating halt decider at state Ĥ.qx to its input ([Ĥ],[Ĥ]) we find
that the simulation of this input must be aborted.

Because the necessity to abort the simulation of an input is equated
with this input specifying
an infinite computation the simulating halt decider Ĥ.qx correctly
transitions to its final Ĥ.qn state deciding not halting on its input.

--
Copyright 2021 Pete Olcott

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

On 5/23/2021 9:59 AM, Kaz Kylheku wrote:
> On 2021-05-22, olcott <NoOne@NoWhere.com> wrote:
>> On 5/22/2021 9:56 AM, Kaz Kylheku wrote:
>>> The halting proof is a form of diagonal argument. It says that the
>>> diagonal entries all fail to decide.
>>>
>>> Cases F H G <---- deecciders
>>>
>>> F^ [ F(^F,^F) ] G(^F, ^F) H(^F, ^F)
>>> G^ F(^G,^G) [ G(^G, ^G) ] H(^G, ^G)
>>> H^ F(^H,^H) G(^H, ^H) [ H(^H, ^H) ]
>>>
>>> Only the diagonal [ ] entriess are relevant. When you substitute
>>> a decider in the test case, to obtain another test case, you
>>> move vertically through the table, which takes you off the diagonal.
>>>
>>> When you replacee the decider being applied to the test case,
>>> you move horizontally; again off the diagonal.
>>>
>>> To stick to the diagonal, you must make both substitutions
>>> in parallel. E.g. when yu change from the H^ case which "attacks" the H
>>> decider to the G^ test case which attacks the G decider,
>>> it is that attacked G decider which cannot decide that case.
>>> You must go from H(^H, ^H) to G(^G, ^G).
>>>
>>> Until you thoroughly internalize the above, you do not
>>> have a grasp on halting.
>>>
>>
>> I understand where you are coming from. I am coming from somewhere else.
>> If you analyze what I am saying using conventional analysis then what I
>> am saying is incorrect.
>>
>> To see that what I am saying <is> correct you must switch to
>> unconventional analysis.
>
> The Turing model of computation, and the halting question and its
> proofs are entirely contained within "conventional analysis".
>
> You're using the language of conventional analysis to refer to
> its concepts and constructs, and to dispute its results.
>
> If you want to use "unconventional analysis", you must rigorously pin
> down what that is, before you even approach the halting problem,
> and present that framework upfront.
>
> Within that framework, you must construct that framework's own model of
> computation. Then pursue that framework's version of the halting
> problem within that framework, and not make any claims with regard to
> conventional Turing halting (which will automatically make you look
> wrong).
>
> You can use similar language like "machine", "halts", and so on, if you
> make it clear up front that these words are in your own framework and do
> not have their conventional meaning.
>
> You also face this problem: what good is your alternative analysis?
> Conventional analysis is reliable and powerful. Can we apply your
> alternative analysis to a broad area of problems and get good results?
>
>> Because we know that the only difference in the behavior of a simulating
>> halt decider and a simulator is that the simulating halt decider stops
>> simulating some of its inputs we can examine the behavior of these
>> inputs in a simulator to determine whether or not a simulating halt
>> decider would stop simulating these inputs.
>
> Sure, but the difference in behavior means we are jumping to a different
> column of the input-decider matrix and are no longer on the diagonal.
> To return to the diagonal while staying in the same column, we must
> switch to a different row: the row for the test case which is based on
> that column's decider. Where row and column match, we have an incorrect
> or nonterminating halting decision, according to this conventional analysis.
>
> Any reasoning that leads you, through substitutions, to discovering a
> good halting decision, but which is off the diagonal trace, has led you
> to an irrelevant configuration. The proof says only that the
> configurations on the diagonal are incorrect or non-halting. The
> proof does not make any assertions about non-diagonal configuations,
> therefore those configurations are not counterexamples to the proof's
> claims.
>
> The undecidability of halting is precisely the claim that the entire
> diagonal consists of incorrect answers or non-halting. Since for every
> decider there is a diagonal entry, every decider is associated with a
> test case it does not handle.
>
> If every decider has at least one test case it does not handle, then
> there does not exist a universal (= handles all cases) decider:
> there is no algorithm that decides all cases of halting.
>

https://www.youtube.com/watch?v=jM6osxSX9GA

It is not at all the horribly convoluted mess of the diagonal lemma:
https://en.wikipedia.org/wiki/Diagonal_lemma

--
Copyright 2021 Pete Olcott

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

On 5/23/2021 9:59 AM, Kaz Kylheku wrote:
> On 2021-05-22, olcott <NoOne@NoWhere.com> wrote:
>> On 5/22/2021 9:56 AM, Kaz Kylheku wrote:

>> Because we know that the only difference in the behavior of a simulating
>> halt decider and a simulator is that the simulating halt decider stops
>> simulating some of its inputs we can examine the behavior of these
>> inputs in a simulator to determine whether or not a simulating halt
>> decider would stop simulating these inputs.
>
> Sure, but the difference in behavior means we are jumping to a different
> column of the input-decider matrix and are no longer on the diagonal.
> To return to the diagonal while staying in the same column, we must
> switch to a different row: the row for the test case which is based on
> that column's decider. Where row and column match, we have an incorrect
> or nonterminating halting decision, according to this conventional analysis.
>
> Any reasoning that leads you, through substitutions, to discovering a
> good halting decision, but which is off the diagonal trace, has led you
> to an irrelevant configuration.

The substitutions are not actually made they are a teaching tool.

> The proof says only that the
> configurations on the diagonal are incorrect or non-halting. The
> proof does not make any assertions about non-diagonal configuations,
> therefore those configurations are not counterexamples to the proof's
> claims.

Now that I actually studied the very simple Sipser diagonalization proof
https://www.youtube.com/watch?v=jM6osxSX9GA

I can address this objection on the basis of adapting my H_Hat to become
a little closer to the Sipser D:

bool D(u32 P)
{ if ( H(P, P) )
HERE: goto HERE;
return false;
}

D correctly decides not halting on itself when H is based on simulating
its input.

On 5/27/2021 7:45 PM, Ben Bacarisse wrote:
> A halt decider is a TM that answers yes/true/1 for all instances
> representing halting computations and no/false/0 otherwise.

D(<D>) is decided to be a non-halting computation on the basis that an
aspect of this computation was aborted.

Its embedded halt decider H(<D>,<D>) aborts the simulation of its input,
which is an aspect of the complete D(<D>) computation.

> The undecidability of halting is precisely the claim that the entire
> diagonal consists of incorrect answers or non-halting. Since for every
> decider there is a diagonal entry, every decider is associated with a
> test case it does not handle.
>
> If every decider has at least one test case it does not handle, then
> there does not exist a universal (= handles all cases) decider:
> there is no algorithm that decides all cases of halting.
>

--
Copyright 2021 Pete Olcott

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