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computers / comp.ai.philosophy / Re: Concise refutation of halting problem proofs V42 [honest dialogue]

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* Concise refutation of halting problem proofs V42 [where people getolcott
+* Re: Concise refutation of halting problem proofs V42 [compute theolcott
|+* Re: Concise refutation of halting problem proofs V42 [compute theolcott
||`* Re: Concise refutation of halting problem proofs V42 [ultimateolcott
|| `* Re: Concise refutation of halting problem proofs V42 [ultimateolcott
||  `- Re: Concise refutation of halting problem proofs V42 [honestolcott
|`- Re: Concise refutation of halting problem proofs V42 [compute theolcott
`- Re: Concise refutation of halting problem proofs V42 [computerolcott

1
Subject: Concise refutation of halting problem proofs V42 [where people get stuck]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 29 Dec 2021 16:49 UTC
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From: NoO...@NoWhere.com (olcott)
Subject: Concise refutation of halting problem proofs V42 [where people get
stuck]
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Proving that embedded_H at Ĥ.qx correctly maps its inputs ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn on the basis of the behavior of the UTM simulation of these inputs.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢*  ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

*My criterion measure with Ben's notational conventions*
H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt

We know that H would correctly decide the halt status of its input on the basis of correctly deciding the halt status of the UTM simulation of its input.

We know this because a UTM simulation of the Turing machine description is computationally equivalent to the direct execution of this same Turing machine.

HERE IS WHERE PEOPLE GET STUCK
HERE IS WHERE PEOPLE GET STUCK
HERE IS WHERE PEOPLE GET STUCK

We know that the copy of H is at Ĥ.qx (AKA embedded_H) applies this same criterion measure to every instance of embedded_H to any recursive depth.

If the UTM simulation of the input to any embedded_H would never stop running without being aborted by this embedded_H then this embedded_H has met its non-halting criteria.


https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf

Linz, Peter 1990. An Introduction to Formal Languages and Automata. Lexington/Toronto: D. C. Heath and Company. (315-320)

--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V42 [compute the mapping]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 30 Dec 2021 04:02 UTC
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Subject: Re: Concise refutation of halting problem proofs V42 [compute the
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On 12/29/2021 1:16 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

*My criterion measure with Ben's notational conventions*
H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt

It's not your criterion, it's mine.  It's also Linz's.  If yours is not
the same as his (and mine), it should be different in some way.  (And we
can safely ignore it, since no one cares about your alternative
criteria.)


It has always been my criterion measure that halt deciders base their halt status decision on the pure simulation of their inputs.

That your notational conventions very cleanly express this idea as applied to the Linz H is a good breakthrough.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

As soon as people understand that this same criterion measure applies recursively to every embedded_H at Ĥ.qx, then they will see how this criterion measure provides the means for embedded_H to correctly compute the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn.



--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V42 [compute the mapping]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 30 Dec 2021 15:15 UTC
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Subject: Re: Concise refutation of halting problem proofs V42 [compute the
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On 12/29/2021 11:18 PM, André G. Isaak wrote:
On 2021-12-29 21:51, olcott wrote:
On 12/29/2021 10:44 PM, Richard Damon wrote:
On 12/29/21 11:02 PM, olcott wrote:
On 12/29/2021 1:16 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

*My criterion measure with Ben's notational conventions*
H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt

It's not your criterion, it's mine.  It's also Linz's.  If yours is not
the same as his (and mine), it should be different in some way. (And we
can safely ignore it, since no one cares about your alternative
criteria.)


It has always been my criterion measure that halt deciders base their halt status decision on the pure simulation of their inputs.

That your notational conventions very cleanly express this idea as applied to the Linz H is a good breakthrough.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

As soon as people understand that this same criterion measure applies recursively to every embedded_H at Ĥ.qx, then they will see how this criterion measure provides the means for embedded_H to correctly compute the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn.


Yes each level can use the test to determine if the halt decider at that level got the right answer or not.

The issue is that when we are doing this check, you check JUST that level, and not lower levels.

No you are wrong.

No, he isn't.

The specification you give above (where YET AGAIN you omit the conditions)

The conditions are provided by H, because H is copied to Ĥ.qx we need not state them again.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

To make things simple we name the copy of H at Ĥ.qx embedded_H.

Because the criterion measure is applied as if the embedded_H at Ĥ.qx was replaced by a UTM and the TM description of Ĥ would include a TM description of this UTM the execution of Ĥ applied to ⟨Ĥ⟩ would

endlessly repeat:
copy its input ⟨Ĥ⟩ to ⟨Ĥ⟩ and then simulate this input with the UTM...

In order for embedded_H to see what this machine would do it must actually perform a pure simulation of N steps of its input ⟨Ĥ⟩ ⟨Ĥ⟩. This may or may not include simulating embedded_H.

As soon as embedded_H sees that these steps would otherwise endlessly repeat it aborts its simulation and transitions to Ĥ.qn.

Because it is applying a criterion measure that we know is correct:
A halt decider is always correct when it bases its halt status decision on the behavior of the UTM simulation of its inputs.
We know that this transition to Ĥ.qn is correct.

is for a case where the input to Ĥ is a description of itself, but the actual specification of the machine can't make that assumption. So really we should be using:


The criterion measure is merely provided to humans so that humans can understand that the basis of the halt status decision of embedded_H is sound.

Because it is obvious that the basis is sound when applied to H then this same sound basis carries over to embedded_H.

Ĥ.q0 ⟨wM⟩ ⊢* Ĥ.qx wM wM⟩ ⊢* Ĥ.qy ∞ iff UTM(wM, wM) halts
Ĥ.q0 ⟨wM⟩ ⊢* Ĥ.qx wM wM⟩ ⊢* Ĥ.qn   iff UTM(wM, wM) does not halt

Ĥ has absolutely no knowledge of what it will be given as an input. In

For purposes of discussion we are simply ignoring every other input and only focusing on the one input.

It is simpler for the human mind to see all the relevant details in one place and not have to imagine all of the changes to see the case-at-hand.

The way that you specify the criterion measure is vague. When I define the criterion measure for H this applies the criterion measure at the point where H is copied Ĥ.qx.

the case we considered above wM was ⟨Ĥ⟩, but wM could have been a description of any machine. Ĥ has no way of knowing if wM is a description of itself, or a description of some TM which simply prints its input string to the tape in reverse order (which wouldn't involve simulating its input at all).

H thus embedded_H only makes its halt status decision on the basis of the simulation of its input. H can only see what its input does by simulating its input.

So the conditions given above cannot possibly apply to any 'lower level' since Ĥ knows nothing about any of those 'lower levels'.

André


It only applies to lower levels when there is actual pathological self-reference(Olcott 2004) as there is with the definition of Ĥ. Ĥ is asking itself whether or not it stops running and then looping if it says yes.


--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V42 [compute the mapping]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 30 Dec 2021 16:11 UTC
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On 12/30/2021 9:57 AM, Richard Damon wrote:
On 12/30/21 10:20 AM, olcott wrote:
On 12/30/2021 8:19 AM, Richard Damon wrote:
On 12/29/21 11:51 PM, olcott wrote:
On 12/29/2021 10:44 PM, Richard Damon wrote:
On 12/29/21 11:02 PM, olcott wrote:
On 12/29/2021 1:16 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

*My criterion measure with Ben's notational conventions*
H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt

It's not your criterion, it's mine.  It's also Linz's.  If yours is not
the same as his (and mine), it should be different in some way. (And we
can safely ignore it, since no one cares about your alternative
criteria.)


It has always been my criterion measure that halt deciders base their halt status decision on the pure simulation of their inputs.

That your notational conventions very cleanly express this idea as applied to the Linz H is a good breakthrough.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

As soon as people understand that this same criterion measure applies recursively to every embedded_H at Ĥ.qx, then they will see how this criterion measure provides the means for embedded_H to correctly compute the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn.


Yes each level can use the test to determine if the halt decider at that level got the right answer or not.

The issue is that when we are doing this check, you check JUST that level, and not lower levels.

No you are wrong.



Source? (Some REAL reference)


H.q0 wM w ⊢* H.qy  iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn  iff UTM(wM, w) does not halt

The source is the correct reasoning of how the above criterion measure would be applied when H is copied to Ĥ.qx.

We copy the ALGORITHM of H, the detailed step by step instructions



You keep thinking that a subsequent level UTM magically turns into embedded_H and aborts its simulation.

There IS no UTM in the machine. PERIOD. (unless H IS a UTM)

H always bases its halt status decision on the behavior of the UTM simulation of its input. Thus embedded_H bases its halt status decision on the behavior of the UTM simulation of its input. This means that H and embedded_H continue a pure simulation of N steps of their input until their input halts on its own or this input demonstrates an infinite behavior pattern.

Since it is obvious to humans that replacing embedded_H at Ĥ.qx with a UTM would cause Ĥ applied to ⟨Ĥ⟩ to never stop running humans can see that the pure simulation of N steps of the input to embedded_H would never halt on their own for every N from 0 to ∞.

Thus correctly applying the criterion measure of H to the input to embedded_H results in the "does not halt" criterion measure being met.


--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V42 [ultimate criterion measure]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
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On 12/30/2021 10:11 AM, Richard Damon wrote:
On 12/30/21 10:58 AM, olcott wrote:
On 12/30/2021 9:48 AM, Richard Damon wrote:

We don't care if embedded_H is 'Correct', we care that it give the same answer as H.


If it is correct and gives a different answer than H then it remains correct even if it gives a different answer than H.


But since it IS a copy of H, then if it gives a different answer than 'H' does, it shows that H fails to be a Computation, and thus Can't be a Decider, and thus can't be a Halt Decider.
H and embedded_H are able to give a different halt status determinations on the basis that embedded_H had its copy of H corrupted by the appended infinite loop.

This means that a finite string comparison performed by H will determine that embedded_H is not identical to H.

When embedded_H is about to simulate embedded_H with identical inputs it terminates this simulation as matching the infinitely nested simulation infinite behavior pattern.


--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V42 [ultimate criterion measure]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 30 Dec 2021 19:33 UTC
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On 12/30/2021 12:21 PM, André G. Isaak wrote:
On 2021-12-30 09:54, olcott wrote:
On 12/30/2021 10:11 AM, Richard Damon wrote:
On 12/30/21 10:58 AM, olcott wrote:
On 12/30/2021 9:48 AM, Richard Damon wrote:

We don't care if embedded_H is 'Correct', we care that it give the same answer as H.


If it is correct and gives a different answer than H then it remains correct even if it gives a different answer than H.


But since it IS a copy of H, then if it gives a different answer than 'H' does, it shows that H fails to be a Computation, and thus Can't be a Decider, and thus can't be a Halt Decider.
H and embedded_H are able to give a different halt status determinations on the basis that embedded_H had its copy of H corrupted by the appended infinite loop.

This means that a finite string comparison performed by H will determine that embedded_H is not identical to H.

And which two strings are being compared here? H hasn't been provided a string describing itself nor a string describing embedded_H as part of its input. All it has is a string describing Ĥ. What does it compare this string to?

André


Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

We as humans can see that while embedded_H performs a pure simulation of its input ⟨Ĥ⟩ ⟨Ĥ⟩ that this simulated input:

copies its input ⟨Ĥ⟩ to ⟨Ĥ⟩ and then invokes embedded_H with this input over and over without ever stopping.

Since a pure simulation of the input never stops we know that the UTM simulation of this same input never stops because a UTM performs a pure simulation of its input.

Thus we know that the "does not halt" halt status criteria of embedded_H has been met.

In order for embedded_H to see this it needs to see that an identical sequence of configurations keeps repeating.

We as humans can see that the input to embedded_H is always identical.
We as humans can see that the copy of embedded_H is always identical.
We as humans can see that these two things can verified by finite string comparison.

That an identical copy of a function is called with identical inputs proves infinite behavior.


When embedded_H is about to simulate embedded_H with identical inputs it terminates this simulation as matching the infinitely nested simulation infinite behavior pattern.






--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V42 [computer scientist]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Sat, 1 Jan 2022 05:08 UTC
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On 12/31/2021 10:54 PM, Richard Damon wrote:

On 12/31/21 11:48 PM, olcott wrote:
On 12/31/2021 10:29 PM, Richard Damon wrote:
On 12/31/21 11:19 PM, olcott wrote:
On 12/31/2021 10:02 PM, Richard Damon wrote:
On 12/31/21 10:57 PM, olcott wrote:
On 12/31/2021 9:50 PM, Richard Damon wrote:
On 12/31/21 10:39 PM, olcott wrote:
On 12/31/2021 9:10 PM, Richard Damon wrote:

On 12/31/21 9:50 PM, olcott wrote:
On 12/31/2021 8:39 PM, Richard Damon wrote:
On 12/31/21 8:50 PM, olcott wrote:
On 12/31/2021 7:37 PM, André G. Isaak wrote:
On 2021-12-31 18:17, olcott wrote:
On 12/31/2021 7:11 PM, André G. Isaak wrote:
On 2021-12-31 17:05, olcott wrote:
On 12/31/2021 5:33 PM, André G. Isaak wrote:
On 2021-12-31 16:02, olcott wrote:
On 12/31/2021 4:06 PM, André G. Isaak wrote:

An actual computer scientist will understand that embedded_H does compute the mapping from these inputs finite strings ⟨Ĥ⟩ ⟨Ĥ⟩ to this final state Ĥ.qn on the basis that the actual input would never halt.

You're not really in a position to state what an actual computer scientist would understand. Only an actual computer scientist can do that.


It is a self-evident truth that:
(a) The pure simulation of the Turing machine description of a machine is computationally equivalent to the direct execution of this machine.

(b) The pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never halt.

(c) If the pure simulation of the input to a halt decider would never halt then the halt decider correctly decides that this input does not halt.

A computer scientist would understand these things.

It would appear that you ignored (and cut) all the actual points in my post.

Why don't we simplify things a bit. When Ĥ ⟨Ĥ⟩ is called, how does Ĥ determine that its input describes itself? You claim this is done by string comparisons, but which strings are being compared? The only string Ĥ has access to its input string. What does it compare this string with?

André


So far I have not gotten to any point of closure on anything that I have said. I must insist on points of closure for continued dialogue.

Do you agree that the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never halt?

Of course I don't, since that claim is simply false.

Now why don't you actually answer the question I asked?

André


We must stay on this point until we have mutual agreement.
Why do you say it is false?


BIG QUESTION.

Is it even a proper question?

Is it a fact that embedded_H just does a pure simulation?

Or. is there an abort condition?

Remember, any time you change any of the properties of the H that you built H^ from, any analysis from previous H^s need to be thrown out, or at least reconfirmed with the new H.

It is the case that when embedded_H simulates 0 to ∞ steps of its input that its input never halts.


Yes, but then it doesn't answer and fails.


It is the case that when embedded_H simulates 0 to ∞ steps of its input that its input never halts

CONCLUSIVELY PROVING THAT THIS INPUT NEVER HALTS EVEN IF IT IS ABORTED



Yes, But H can't do that aborting, because you just said that embedded_H didn't abort it.


Do you think that it is possible to tell that an infinite loop will never stop running without actually having to wait forever?

void Infinite_Loop(int N)
{
   HERE: goto HERE;
}

_Infinite_Loop()
[00000cb5](01)  55              push ebp
[00000cb6](02)  8bec            mov ebp,esp
[00000cb8](02)  ebfe            jmp 00000cb8
[00000cba](01)  5d              pop ebp
[00000cbb](01)  c3              ret
Size in bytes:(0007) [00000cbb]



In some cases, yes.

You seem to like your Herring Red.

The question is not is it possible to answer some other halting questions, but if H can answer a particular one.

If it is the case that embedded_H does correctly determine its input would never halt if it simulates 0 to ∞ steps of this input

and it makes this determination in a finite number of steps
and it aborts the simulation of this input


Which it can't as proven at:

Mesage ID  <FOnzJ.162569$np6.119786@fx46.iad>
Date: 2021-12-30 19:31:49 GMT
Subject: Re: Concise refutation of halting problem proofs V42 [compute the mapping]

FAIL.

That is irrelevant at this point in the dialogue.


WHY?

The fact that you just claim, with NO proof to have the Unicorn Halt Decider to do your bidding.

You logic is UNSOUND, as are you.



this does not cause the input to halt because an input must reach its final state to halt

i.e that ACTUAL Machine must not reach a halting state, not an aborted simulation, but H& does reach that state.


Your words are incoherent.

When we hypothesize that embedded_H can determine that input to embedded_H never reaches its final state in 0 to ∞ steps of simulation and it can do this in a finite number of steps

You Hypothesize the impossible.


This is the only point that I am willing to discuss with anyone until mutual agreement is achieved.

If embedded_H does correctly determine that its input cannot possibly reach the final state of this input then it is correct for embedded_H to aborts its simulation of this input and transition to Ĥ.qn.


--
Copyright 2021 Pete Olcott

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


Subject: Re: Concise refutation of halting problem proofs V42 [honest dialogue]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Organization: A noiseless patient Spider
Date: Sun, 2 Jan 2022 16:07 UTC
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From: polco...@gmail.com (olcott)
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math
Subject: Re: Concise refutation of halting problem proofs V42 [honest
dialogue]
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On 12/31/2021 7:37 PM, André G. Isaak wrote:
On 2021-12-31 18:17, olcott wrote:
On 12/31/2021 7:11 PM, André G. Isaak wrote:
On 2021-12-31 17:05, olcott wrote:
On 12/31/2021 5:33 PM, André G. Isaak wrote:
On 2021-12-31 16:02, olcott wrote:
On 12/31/2021 4:06 PM, André G. Isaak wrote:

An actual computer scientist will understand that embedded_H does compute the mapping from these inputs finite strings ⟨Ĥ⟩ ⟨Ĥ⟩ to this final state Ĥ.qn on the basis that the actual input would never halt.

You're not really in a position to state what an actual computer scientist would understand. Only an actual computer scientist can do that.


It is a self-evident truth that:
(a) The pure simulation of the Turing machine description of a machine is computationally equivalent to the direct execution of this machine.

(b) The pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never halt.

(c) If the pure simulation of the input to a halt decider would never halt then the halt decider correctly decides that this input does not halt.

A computer scientist would understand these things.

It would appear that you ignored (and cut) all the actual points in my post.

Why don't we simplify things a bit. When Ĥ ⟨Ĥ⟩ is called, how does Ĥ determine that its input describes itself? You claim this is done by string comparisons, but which strings are being compared? The only string Ĥ has access to its input string. What does it compare this string with?

André


So far I have not gotten to any point of closure on anything that I have said. I must insist on points of closure for continued dialogue.

Do you agree that the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never halt?

Of course I don't, since that claim is simply false.

Now why don't you actually answer the question I asked?

André


For simplicity we will refer to the copy of Linz H at Ĥ.qx embedded_H.

Simplified syntax adapted from bottom of page 319:
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

If embedded_H would never stop simulating its input ⟨Ĥ⟩ ⟨Ĥ⟩
this input would never reach a final state and stop running.

These steps would keep repeating:
Ĥ copies its input ⟨Ĥ⟩ to ⟨Ĥ⟩ then embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩...

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

That you disagree with self-evident truth proves that you do not want any honest dialogue.



--
Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer


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