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computers / comp.ai.philosophy / Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]

SubjectAuthor
* Are my reviewers incompetent or dishonest? [ stupid or liar ? ]olcott
`* Re: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]Richard Damon
 `* Re: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]olcott
  `* Re: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]Richard Damon
   +* Re: Are my reviewers incompetent or dishonest? [ closure on one pointolcott
   |`- Re: Are my reviewers incompetent or dishonest? [ closure on one pointRichard Damon
   `* Re: Are my reviewers incompetent or dishonest? [ closure on one pointolcott
    `* Re: Are my reviewers incompetent or dishonest? [ closure on one pointRichard Damon
     `* Re: Are my reviewers incompetent or dishonest? [ closure on one pointolcott
      +* Re: Are my reviewers incompetent or dishonest? [ closure on one pointRichard Damon
      |`* Re: Are my reviewers incompetent or dishonest? [ closure on one pointolcott
      | `- Re: Are my reviewers incompetent or dishonest? [ closure on one pointRichard Damon
      `* Re: Are my reviewers incompetent or dishonest? [ closure on one pointPython
       +* Re: Are my reviewers incompetent or dishonest? [ closure on one pointRichard Damon
       |`* Re: Are my reviewers incompetent or dishonest? [ closure on one pointolcott
       | `- Re: Are my reviewers incompetent or dishonest? [ closure on one pointRichard Damon
       +* Re: Are my reviewers incompetent or dishonest? [ closure on one pointolcott
       |+* Re: Are my reviewers incompetent or dishonest? [ closure on one pointPython
       ||`* Re: Are my reviewers incompetent or dishonest? [ closure on one pointolcott
       || +* Re: Are my reviewers incompetent or dishonest? [ closure on one pointPython
       || |`- Re: Are my reviewers incompetent or dishonest? [ closure on one pointolcott
       || `- Re: Are my reviewers incompetent or dishonest? [ closure on one pointRichard Damon
       |`- Re: Are my reviewers incompetent or dishonest? [ closure on one pointRichard Damon
       `* Re: Are my reviewers incompetent or dishonest? [ closure on one pointJeff Barnett
        `- Re: Are my reviewers incompetent or dishonest? [ closure on one pointolcott

1
Subject: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.software-eng, sci.logic
Date: Sun, 22 May 2022 21:06 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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Subject: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]
Content-Language: en-US
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.logic
References: <20220514170555.00004550@reddwarf.jmc>
<jPednedJMZKJWhr_nZ2dnUU7_81g4p2d@giganews.com> <87o7zr3od4.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
In-Reply-To: <xwuiK.795$8T.565@fx40.iad>
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On 5/22/2022 12:36 PM, Richard Damon wrote:
On 5/22/22 12:42 PM, olcott wrote:
On 5/22/2022 11:40 AM, Richard Damon wrote:
On 5/22/22 12:31 PM, olcott wrote:
On 5/22/2022 11:21 AM, Richard Damon wrote:
On 5/22/22 10:57 AM, olcott wrote:
On 5/22/2022 6:06 AM, Richard Damon wrote:
On 5/22/22 1:02 AM, olcott wrote:
On 5/21/2022 11:05 PM, Richard Damon wrote:
On 5/21/22 11:59 PM, olcott wrote:
On 5/21/2022 10:54 PM, Richard Damon wrote:
On 5/21/22 11:36 PM, olcott wrote:
On 5/21/2022 10:27 PM, Richard Damon wrote:
On 5/21/22 10:48 PM, olcott wrote:
On 5/21/2022 9:37 PM, Richard Damon wrote:
On 5/21/22 10:28 PM, olcott wrote:
On 5/21/2022 9:20 PM, Richard Damon wrote:

On 5/21/22 9:23 PM, olcott wrote:
On 5/21/2022 8:05 PM, Richard Damon wrote:
On 5/21/22 3:38 PM, olcott wrote:
On 5/20/2022 5:25 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

You have known that the input to H(P,P) is simulated correctly proving
that H(P,P)==0 is correct for the whole six months

If H is intended to be a halt decider (even if only for the one case you
claim to care about) then H(P,P) == 0 is wrong, because P(P) halts.


When we correctly reverse-engineer what the execution trace of the input to H(P,P) would be for one emulation and one nested emulation we can see that the correctly emulated input to H(P,P) would never reach its final state at machine address [0000136c].

A nonsense trace, as it is mixing the execution path of two independent execution units.

In other words you acknowledge that you are technically incompetent to provide the execution trace of one simulation and one nested simulation of the input to H(P,P).

No, I am saying that you are asking for the equivalent of a of a square circle.


So an execution trace of the input to H(P,P) is easy to show when H simulates its input, yet another execution trace of the input to H(P,P) that was invoked by P is "like a square circle" can't possibly exist?

The problem is that your second trace is NOT a piece of the first.

The fact you don't understand that says you just don't know how computers or programs actually work.


When a UTM simulates a TM description that calls a UTM that simulates a
TM description all of this is simply data on the first UTM's tape and the only actual executable is the first UTM.


Yes, and a trace made by that outer UTM will show the states that the second UTM is going through, but NOT the states that second UTM simulates in its own processing.

That second UTM might produce its OWN trace of the states that it has simulated, but that is a SEPERATE trace, and NOT part of the trace from the OUTER UTM.


And this trace is written to the outer UTM's tape as a part of its own data.

Yes, the DATA is there, ENCODED on the tape, but it isn't part of the trace generated by that UTM.

The only actual executable is the outer UTM everything else is a part of the same nested process.

So the only actual valid trace is what that outer simulator actual simulated.


There is a valid trace of every line of code that is emulated. Operating system code has its trace tuned off. This only leaves the user code such as P() and main(). Then we see the 14 lines execution trace of the two level simulation of the input to H(P,P)

No, because the second level emulation is NOT emulated by the top level emulator, its emulator is.

Unless you are lying about what H does, you are just lying that the second level code is emulated by the same emulation process that the first is. (That may well be true, but it means you logic is still built on a lie).


If you are too stupid to understand that H(P,P) derives the same execution trace of its input every time it is called you are far too stupid to evaluate my work.

Ok, then why does the H(P,P) that P calls get stuck in an infinite recursion wneh the top level doesn't?

It is a verified fact that the correct simulation of the input to H(P,P) never reaches its final instruction thus conclusively proving that it never halts.


The only machine that you have shown that does a correct simulation is the version that never aborts. That version fails to answer the question, so fails to be a halt decider.

Any version of H that aborts, and returns a not-halting answer changes P into a Halting Compuation.

The "pathological" use of H by P lets it change as you change H, so if H aborts, it is wrong because THAT P halts, if it doesn't, then it is wrong for not answering.

You seem to miss this fact because you just don't understand the basics of how computations work. Part of your problem is you keep on trying to define H by rules that aren't an actual algorithm, so can't actually be written.




It is an easily verifiable fact that the C function H does correctly determine that the C function named P would never reach its last instruction when correctly emulated by H.

Don't just "Claim" it, so an ACTUAL verification, or you just show yourself to be a liar.


Everyone disagreeing with verified facts is incorrect on the basis of lack of technical competency or lack of honesty.

You haven't verified ANY fact, you have made claims using FAKE data that don't even really support your claim.


Software engineering experts
can reverse-engineer what the correct x86 emulation of the input to H(P,P) would be for one emulation and one nested emulation thus confirming that the provided execution trace is correct. They can do this entirely on the basis of the x86 source-code for P with no need to see the source-code or execution trace of H.

Anyone unable to do this conclusively proves their lack of sufficient technical competence.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

Richard was able to correctly determine that the correct simulation of the input to H(P,P) would emulate the first 7 lines of P.

Richard was utterly baffled beyond all comprehension that the invocation of the same function with the same input would derive the same trace.

Richard even said that this trace is "like a square circle" in that it cannot possibly exist.

Does anyone here believe that Richard is really that stupid? (I don't).


--
Copyright 2022 Pete Olcott

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


Subject: Re: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]
From: Richard Damon
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Forte - www.forteinc.com
Date: Sun, 22 May 2022 21:34 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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Subject: Re: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]
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<_tOdnaZ1Xa1sOBf_nZ2dnUU7_8zNnZ2d@giganews.com>
From: Rich...@Damon-Family.org (Richard Damon)
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On 5/22/22 5:06 PM, olcott wrote:
On 5/22/2022 12:36 PM, Richard Damon wrote:
On 5/22/22 12:42 PM, olcott wrote:
On 5/22/2022 11:40 AM, Richard Damon wrote:
On 5/22/22 12:31 PM, olcott wrote:
On 5/22/2022 11:21 AM, Richard Damon wrote:
On 5/22/22 10:57 AM, olcott wrote:
On 5/22/2022 6:06 AM, Richard Damon wrote:
On 5/22/22 1:02 AM, olcott wrote:
On 5/21/2022 11:05 PM, Richard Damon wrote:
On 5/21/22 11:59 PM, olcott wrote:
On 5/21/2022 10:54 PM, Richard Damon wrote:
On 5/21/22 11:36 PM, olcott wrote:
On 5/21/2022 10:27 PM, Richard Damon wrote:
On 5/21/22 10:48 PM, olcott wrote:
On 5/21/2022 9:37 PM, Richard Damon wrote:
On 5/21/22 10:28 PM, olcott wrote:
On 5/21/2022 9:20 PM, Richard Damon wrote:

On 5/21/22 9:23 PM, olcott wrote:
On 5/21/2022 8:05 PM, Richard Damon wrote:
On 5/21/22 3:38 PM, olcott wrote:
On 5/20/2022 5:25 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

You have known that the input to H(P,P) is simulated correctly proving
that H(P,P)==0 is correct for the whole six months

If H is intended to be a halt decider (even if only for the one case you
claim to care about) then H(P,P) == 0 is wrong, because P(P) halts.


When we correctly reverse-engineer what the execution trace of the input to H(P,P) would be for one emulation and one nested emulation we can see that the correctly emulated input to H(P,P) would never reach its final state at machine address [0000136c].

A nonsense trace, as it is mixing the execution path of two independent execution units.

In other words you acknowledge that you are technically incompetent to provide the execution trace of one simulation and one nested simulation of the input to H(P,P).

No, I am saying that you are asking for the equivalent of a of a square circle.


So an execution trace of the input to H(P,P) is easy to show when H simulates its input, yet another execution trace of the input to H(P,P) that was invoked by P is "like a square circle" can't possibly exist?

The problem is that your second trace is NOT a piece of the first.

The fact you don't understand that says you just don't know how computers or programs actually work.


When a UTM simulates a TM description that calls a UTM that simulates a
TM description all of this is simply data on the first UTM's tape and the only actual executable is the first UTM.


Yes, and a trace made by that outer UTM will show the states that the second UTM is going through, but NOT the states that second UTM simulates in its own processing.

That second UTM might produce its OWN trace of the states that it has simulated, but that is a SEPERATE trace, and NOT part of the trace from the OUTER UTM.


And this trace is written to the outer UTM's tape as a part of its own data.

Yes, the DATA is there, ENCODED on the tape, but it isn't part of the trace generated by that UTM.

The only actual executable is the outer UTM everything else is a part of the same nested process.

So the only actual valid trace is what that outer simulator actual simulated.


There is a valid trace of every line of code that is emulated. Operating system code has its trace tuned off. This only leaves the user code such as P() and main(). Then we see the 14 lines execution trace of the two level simulation of the input to H(P,P)

No, because the second level emulation is NOT emulated by the top level emulator, its emulator is.

Unless you are lying about what H does, you are just lying that the second level code is emulated by the same emulation process that the first is. (That may well be true, but it means you logic is still built on a lie).


If you are too stupid to understand that H(P,P) derives the same execution trace of its input every time it is called you are far too stupid to evaluate my work.

Ok, then why does the H(P,P) that P calls get stuck in an infinite recursion wneh the top level doesn't?

It is a verified fact that the correct simulation of the input to H(P,P) never reaches its final instruction thus conclusively proving that it never halts.


The only machine that you have shown that does a correct simulation is the version that never aborts. That version fails to answer the question, so fails to be a halt decider.

Any version of H that aborts, and returns a not-halting answer changes P into a Halting Compuation.

The "pathological" use of H by P lets it change as you change H, so if H aborts, it is wrong because THAT P halts, if it doesn't, then it is wrong for not answering.

You seem to miss this fact because you just don't understand the basics of how computations work. Part of your problem is you keep on trying to define H by rules that aren't an actual algorithm, so can't actually be written.




It is an easily verifiable fact that the C function H does correctly determine that the C function named P would never reach its last instruction when correctly emulated by H.

Don't just "Claim" it, so an ACTUAL verification, or you just show yourself to be a liar.


Everyone disagreeing with verified facts is incorrect on the basis of lack of technical competency or lack of honesty.

You haven't verified ANY fact, you have made claims using FAKE data that don't even really support your claim.


Software engineering experts
can reverse-engineer what the correct x86 emulation of the input to H(P,P) would be for one emulation and one nested emulation thus confirming that the provided execution trace is correct. They can do this entirely on the basis of the x86 source-code for P with no need to see the source-code or execution trace of H.

Interesting point, that you eed to talk about "reverse-engineering" this output implies that you don't actually have a program to generate it.

You don't need to "reverse-engineer" something you have.



Anyone unable to do this conclusively proves their lack of sufficient technical competence.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

Richard was able to correctly determine that the correct simulation of the input to H(P,P) would emulate the first 7 lines of P.

Richard was utterly baffled beyond all comprehension that the invocation of the same function with the same input would derive the same trace.

No, that is NOT what I said, and you repeating that LIE just shows you are not intertested in honest answer. I hope what ever review looks at your final paper (it you ever get to it) and happens to search and come accross these discussions and sees your level of dishonesty.

I comment that it is impossible to give a CORRECT simulation from this point, as that needs to show the instruction at address 000011a2, which has not been defined.

Maybe you can create traces out of stuff you don't know, but I am too honest to make up that sort of crap.



Richard even said that this trace is "like a square circle" in that it cannot possibly exist.

Obviously you just don't have the knowledge to understand what I am saying. A Correct Trace of P will NOT show a second level of simulation as actual just being executed code, as that is NOT what you say is happening.

The fact that you keep repeating that this is the proof of your premise just shows how wrong it is. Your theories depend on the existance for Fairy Dust Powered Unicorn Magic to make things work, which just doesn't exist.

There is no "recursion", and there is only an infinite depth of simulation of NO H aborts its simulation and thus H doesn't answer 0 for H(P,P), so that claim is shown to be just a lie.


Does anyone here believe that Richard is really that stupid? (I don't).



No, but we all know YOU are, and that you LIE that much.


Subject: Re: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Date: Sun, 22 May 2022 21:59 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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Subject: Re: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]
Content-Language: en-US
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic
References: <20220514170555.00004550@reddwarf.jmc>
<SfWdnTcajIIjkxX_nZ2dnUU7_8zNnZ2d@giganews.com> <87bkvr3kqn.fsf@bsb.me.uk>
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From: NoO...@NoWhere.com (olcott)
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On 5/22/2022 4:34 PM, Richard Damon wrote:

On 5/22/22 5:06 PM, olcott wrote:
On 5/22/2022 12:36 PM, Richard Damon wrote:
On 5/22/22 12:42 PM, olcott wrote:
On 5/22/2022 11:40 AM, Richard Damon wrote:
On 5/22/22 12:31 PM, olcott wrote:
On 5/22/2022 11:21 AM, Richard Damon wrote:
On 5/22/22 10:57 AM, olcott wrote:
On 5/22/2022 6:06 AM, Richard Damon wrote:
On 5/22/22 1:02 AM, olcott wrote:
On 5/21/2022 11:05 PM, Richard Damon wrote:
On 5/21/22 11:59 PM, olcott wrote:
On 5/21/2022 10:54 PM, Richard Damon wrote:
On 5/21/22 11:36 PM, olcott wrote:
On 5/21/2022 10:27 PM, Richard Damon wrote:
On 5/21/22 10:48 PM, olcott wrote:
On 5/21/2022 9:37 PM, Richard Damon wrote:
On 5/21/22 10:28 PM, olcott wrote:
On 5/21/2022 9:20 PM, Richard Damon wrote:

On 5/21/22 9:23 PM, olcott wrote:
On 5/21/2022 8:05 PM, Richard Damon wrote:
On 5/21/22 3:38 PM, olcott wrote:
On 5/20/2022 5:25 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

You have known that the input to H(P,P) is simulated correctly proving
that H(P,P)==0 is correct for the whole six months

If H is intended to be a halt decider (even if only for the one case you
claim to care about) then H(P,P) == 0 is wrong, because P(P) halts.


When we correctly reverse-engineer what the execution trace of the input to H(P,P) would be for one emulation and one nested emulation we can see that the correctly emulated input to H(P,P) would never reach its final state at machine address [0000136c].

A nonsense trace, as it is mixing the execution path of two independent execution units.

In other words you acknowledge that you are technically incompetent to provide the execution trace of one simulation and one nested simulation of the input to H(P,P).

No, I am saying that you are asking for the equivalent of a of a square circle.


So an execution trace of the input to H(P,P) is easy to show when H simulates its input, yet another execution trace of the input to H(P,P) that was invoked by P is "like a square circle" can't possibly exist?

The problem is that your second trace is NOT a piece of the first.

The fact you don't understand that says you just don't know how computers or programs actually work.


When a UTM simulates a TM description that calls a UTM that simulates a
TM description all of this is simply data on the first UTM's tape and the only actual executable is the first UTM.


Yes, and a trace made by that outer UTM will show the states that the second UTM is going through, but NOT the states that second UTM simulates in its own processing.

That second UTM might produce its OWN trace of the states that it has simulated, but that is a SEPERATE trace, and NOT part of the trace from the OUTER UTM.


And this trace is written to the outer UTM's tape as a part of its own data.

Yes, the DATA is there, ENCODED on the tape, but it isn't part of the trace generated by that UTM.

The only actual executable is the outer UTM everything else is a part of the same nested process.

So the only actual valid trace is what that outer simulator actual simulated.


There is a valid trace of every line of code that is emulated. Operating system code has its trace tuned off. This only leaves the user code such as P() and main(). Then we see the 14 lines execution trace of the two level simulation of the input to H(P,P)

No, because the second level emulation is NOT emulated by the top level emulator, its emulator is.

Unless you are lying about what H does, you are just lying that the second level code is emulated by the same emulation process that the first is. (That may well be true, but it means you logic is still built on a lie).


If you are too stupid to understand that H(P,P) derives the same execution trace of its input every time it is called you are far too stupid to evaluate my work.

Ok, then why does the H(P,P) that P calls get stuck in an infinite recursion wneh the top level doesn't?

It is a verified fact that the correct simulation of the input to H(P,P) never reaches its final instruction thus conclusively proving that it never halts.


The only machine that you have shown that does a correct simulation is the version that never aborts. That version fails to answer the question, so fails to be a halt decider.

Any version of H that aborts, and returns a not-halting answer changes P into a Halting Compuation.

The "pathological" use of H by P lets it change as you change H, so if H aborts, it is wrong because THAT P halts, if it doesn't, then it is wrong for not answering.

You seem to miss this fact because you just don't understand the basics of how computations work. Part of your problem is you keep on trying to define H by rules that aren't an actual algorithm, so can't actually be written.




It is an easily verifiable fact that the C function H does correctly determine that the C function named P would never reach its last instruction when correctly emulated by H.

Don't just "Claim" it, so an ACTUAL verification, or you just show yourself to be a liar.


Everyone disagreeing with verified facts is incorrect on the basis of lack of technical competency or lack of honesty.

You haven't verified ANY fact, you have made claims using FAKE data that don't even really support your claim.


Software engineering experts
can reverse-engineer what the correct x86 emulation of the input to H(P,P) would be for one emulation and one nested emulation thus confirming that the provided execution trace is correct. They can do this entirely on the basis of the x86 source-code for P with no need to see the source-code or execution trace of H.

Interesting point, that you eed to talk about "reverse-engineering" this output implies that you don't actually have a program to generate it.


Not at all. This program took me a whole man-year.
The most difficult part was to make the nested simulations work correctly.

You don't need to "reverse-engineer" something you have.


No but everyone else does to confirm its correctness because
THEY DON'T HAVE IT.



Anyone unable to do this conclusively proves their lack of sufficient technical competence.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

Richard was able to correctly determine that the correct simulation of the input to H(P,P) would emulate the first 7 lines of P.

Richard was utterly baffled beyond all comprehension that the invocation of the same function with the same input would derive the same trace.

No, that is NOT what I said, and you repeating that LIE just shows you are not intertested in honest answer. I hope what ever review looks at your final paper (it you ever get to it) and happens to search and come accross these discussions and sees your level of dishonesty.


So then you now agree that the nested invocation of H(P,P) would derive the same execution trace of its input that the the outer one did?

In order to contradict me you must contradict computer science.


--
Copyright 2022 Pete Olcott

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


Subject: Re: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]
From: Richard Damon
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Forte - www.forteinc.com
Date: Sun, 22 May 2022 22:34 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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Subject: Re: Are my reviewers incompetent or dishonest? [ stupid or liar ? ]
Content-Language: en-US
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic
References: <20220514170555.00004550@reddwarf.jmc> <87bkvr3kqn.fsf@bsb.me.uk>
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From: Rich...@Damon-Family.org (Richard Damon)
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On 5/22/22 5:59 PM, olcott wrote:
On 5/22/2022 4:34 PM, Richard Damon wrote:

On 5/22/22 5:06 PM, olcott wrote:
On 5/22/2022 12:36 PM, Richard Damon wrote:
On 5/22/22 12:42 PM, olcott wrote:
On 5/22/2022 11:40 AM, Richard Damon wrote:
On 5/22/22 12:31 PM, olcott wrote:
On 5/22/2022 11:21 AM, Richard Damon wrote:
On 5/22/22 10:57 AM, olcott wrote:
On 5/22/2022 6:06 AM, Richard Damon wrote:
On 5/22/22 1:02 AM, olcott wrote:
On 5/21/2022 11:05 PM, Richard Damon wrote:
On 5/21/22 11:59 PM, olcott wrote:
On 5/21/2022 10:54 PM, Richard Damon wrote:
On 5/21/22 11:36 PM, olcott wrote:
On 5/21/2022 10:27 PM, Richard Damon wrote:
On 5/21/22 10:48 PM, olcott wrote:
On 5/21/2022 9:37 PM, Richard Damon wrote:
On 5/21/22 10:28 PM, olcott wrote:
On 5/21/2022 9:20 PM, Richard Damon wrote:

On 5/21/22 9:23 PM, olcott wrote:
On 5/21/2022 8:05 PM, Richard Damon wrote:
On 5/21/22 3:38 PM, olcott wrote:
On 5/20/2022 5:25 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

You have known that the input to H(P,P) is simulated correctly proving
that H(P,P)==0 is correct for the whole six months

If H is intended to be a halt decider (even if only for the one case you
claim to care about) then H(P,P) == 0 is wrong, because P(P) halts.


When we correctly reverse-engineer what the execution trace of the input to H(P,P) would be for one emulation and one nested emulation we can see that the correctly emulated input to H(P,P) would never reach its final state at machine address [0000136c].

A nonsense trace, as it is mixing the execution path of two independent execution units.

In other words you acknowledge that you are technically incompetent to provide the execution trace of one simulation and one nested simulation of the input to H(P,P).

No, I am saying that you are asking for the equivalent of a of a square circle.


So an execution trace of the input to H(P,P) is easy to show when H simulates its input, yet another execution trace of the input to H(P,P) that was invoked by P is "like a square circle" can't possibly exist?

The problem is that your second trace is NOT a piece of the first.

The fact you don't understand that says you just don't know how computers or programs actually work.


When a UTM simulates a TM description that calls a UTM that simulates a
TM description all of this is simply data on the first UTM's tape and the only actual executable is the first UTM.


Yes, and a trace made by that outer UTM will show the states that the second UTM is going through, but NOT the states that second UTM simulates in its own processing.

That second UTM might produce its OWN trace of the states that it has simulated, but that is a SEPERATE trace, and NOT part of the trace from the OUTER UTM.


And this trace is written to the outer UTM's tape as a part of its own data.

Yes, the DATA is there, ENCODED on the tape, but it isn't part of the trace generated by that UTM.

The only actual executable is the outer UTM everything else is a part of the same nested process.

So the only actual valid trace is what that outer simulator actual simulated.


There is a valid trace of every line of code that is emulated. Operating system code has its trace tuned off. This only leaves the user code such as P() and main(). Then we see the 14 lines execution trace of the two level simulation of the input to H(P,P)

No, because the second level emulation is NOT emulated by the top level emulator, its emulator is.

Unless you are lying about what H does, you are just lying that the second level code is emulated by the same emulation process that the first is. (That may well be true, but it means you logic is still built on a lie).


If you are too stupid to understand that H(P,P) derives the same execution trace of its input every time it is called you are far too stupid to evaluate my work.

Ok, then why does the H(P,P) that P calls get stuck in an infinite recursion wneh the top level doesn't?

It is a verified fact that the correct simulation of the input to H(P,P) never reaches its final instruction thus conclusively proving that it never halts.


The only machine that you have shown that does a correct simulation is the version that never aborts. That version fails to answer the question, so fails to be a halt decider.

Any version of H that aborts, and returns a not-halting answer changes P into a Halting Compuation.

The "pathological" use of H by P lets it change as you change H, so if H aborts, it is wrong because THAT P halts, if it doesn't, then it is wrong for not answering.

You seem to miss this fact because you just don't understand the basics of how computations work. Part of your problem is you keep on trying to define H by rules that aren't an actual algorithm, so can't actually be written.




It is an easily verifiable fact that the C function H does correctly determine that the C function named P would never reach its last instruction when correctly emulated by H.

Don't just "Claim" it, so an ACTUAL verification, or you just show yourself to be a liar.


Everyone disagreeing with verified facts is incorrect on the basis of lack of technical competency or lack of honesty.

You haven't verified ANY fact, you have made claims using FAKE data that don't even really support your claim.


Software engineering experts
can reverse-engineer what the correct x86 emulation of the input to H(P,P) would be for one emulation and one nested emulation thus confirming that the provided execution trace is correct. They can do this entirely on the basis of the x86 source-code for P with no need to see the source-code or execution trace of H.

Interesting point, that you eed to talk about "reverse-engineering" this output implies that you don't actually have a program to generate it.


Not at all. This program took me a whole man-year.

Then why do we need to "reverse-engineer" the trace?

The most difficult part was to make the nested simulations work correctly.

Which it seems you failed at. Maybe that is why ot took so long. It is hard to make automaticaly believable lies.


You don't need to "reverse-engineer" something you have.


No but everyone else does to confirm its correctness because
THEY DON'T HAVE IT.

No, you present the trace and people can verify it.

Note, it has been pointed out that it does NOT match what you claim is happening, which proves it is WRONG.

The trace implies that H just calls its input, and hides its own behavior.

If it does that, then it can't abort its trace, or isn't actually a computatiopn. (and we see it aborts, and you claim it is a computatiopn)

Thus, the trace is incorrect, or you are lying and the trace doesn't matter.




Anyone unable to do this conclusively proves their lack of sufficient technical competence.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

Richard was able to correctly determine that the correct simulation of the input to H(P,P) would emulate the first 7 lines of P.

Richard was utterly baffled beyond all comprehension that the invocation of the same function with the same input would derive the same trace.

No, that is NOT what I said, and you repeating that LIE just shows you are not intertested in honest answer. I hope what ever review looks at your final paper (it you ever get to it) and happens to search and come accross these discussions and sees your level of dishonesty.


So then you now agree that the nested invocation of H(P,P) would derive the same execution trace of its input that the the outer one did?


Yes seem to have a confusion. Yes, the second simulator will produce and identical trace to the first one, AS A SEPERATE TRACE.

The issue is you can not validly "combine" them into a single trace, that is just a lie. The first simulator needs to show its trace of the instructions of the second simulator doing its simulation.

I suppose if you want, you can omit that with a note of omitted detail, but the the trace goes to the call, says details omited, then the notice that the trace was aborted, but then you don't have any grounds to claim that it was correct (since there aren't any).

In order to contradict me you must contradict computer science.



Nope, YOU do that enough.


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Date: Sun, 22 May 2022 22:50 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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References: <20220514170555.00004550@reddwarf.jmc>
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From: NoO...@NoWhere.com (olcott)
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On 5/22/2022 5:34 PM, Richard Damon wrote:

On 5/22/22 5:59 PM, olcott wrote:
On 5/22/2022 4:34 PM, Richard Damon wrote:

On 5/22/22 5:06 PM, olcott wrote:
On 5/22/2022 12:36 PM, Richard Damon wrote:
On 5/22/22 12:42 PM, olcott wrote:
On 5/22/2022 11:40 AM, Richard Damon wrote:
On 5/22/22 12:31 PM, olcott wrote:
On 5/22/2022 11:21 AM, Richard Damon wrote:
On 5/22/22 10:57 AM, olcott wrote:
On 5/22/2022 6:06 AM, Richard Damon wrote:
On 5/22/22 1:02 AM, olcott wrote:
On 5/21/2022 11:05 PM, Richard Damon wrote:
On 5/21/22 11:59 PM, olcott wrote:
On 5/21/2022 10:54 PM, Richard Damon wrote:
On 5/21/22 11:36 PM, olcott wrote:
On 5/21/2022 10:27 PM, Richard Damon wrote:
On 5/21/22 10:48 PM, olcott wrote:
On 5/21/2022 9:37 PM, Richard Damon wrote:
On 5/21/22 10:28 PM, olcott wrote:
On 5/21/2022 9:20 PM, Richard Damon wrote:

On 5/21/22 9:23 PM, olcott wrote:
On 5/21/2022 8:05 PM, Richard Damon wrote:
On 5/21/22 3:38 PM, olcott wrote:
On 5/20/2022 5:25 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

You have known that the input to H(P,P) is simulated correctly proving
that H(P,P)==0 is correct for the whole six months

If H is intended to be a halt decider (even if only for the one case you
claim to care about) then H(P,P) == 0 is wrong, because P(P) halts.


When we correctly reverse-engineer what the execution trace of the input to H(P,P) would be for one emulation and one nested emulation we can see that the correctly emulated input to H(P,P) would never reach its final state at machine address [0000136c].

A nonsense trace, as it is mixing the execution path of two independent execution units.

In other words you acknowledge that you are technically incompetent to provide the execution trace of one simulation and one nested simulation of the input to H(P,P).

No, I am saying that you are asking for the equivalent of a of a square circle.


So an execution trace of the input to H(P,P) is easy to show when H simulates its input, yet another execution trace of the input to H(P,P) that was invoked by P is "like a square circle" can't possibly exist?

The problem is that your second trace is NOT a piece of the first.

The fact you don't understand that says you just don't know how computers or programs actually work.


When a UTM simulates a TM description that calls a UTM that simulates a
TM description all of this is simply data on the first UTM's tape and the only actual executable is the first UTM.


Yes, and a trace made by that outer UTM will show the states that the second UTM is going through, but NOT the states that second UTM simulates in its own processing.

That second UTM might produce its OWN trace of the states that it has simulated, but that is a SEPERATE trace, and NOT part of the trace from the OUTER UTM.


And this trace is written to the outer UTM's tape as a part of its own data.

Yes, the DATA is there, ENCODED on the tape, but it isn't part of the trace generated by that UTM.

The only actual executable is the outer UTM everything else is a part of the same nested process.

So the only actual valid trace is what that outer simulator actual simulated.


There is a valid trace of every line of code that is emulated. Operating system code has its trace tuned off. This only leaves the user code such as P() and main(). Then we see the 14 lines execution trace of the two level simulation of the input to H(P,P)

No, because the second level emulation is NOT emulated by the top level emulator, its emulator is.

Unless you are lying about what H does, you are just lying that the second level code is emulated by the same emulation process that the first is. (That may well be true, but it means you logic is still built on a lie).


If you are too stupid to understand that H(P,P) derives the same execution trace of its input every time it is called you are far too stupid to evaluate my work.

Ok, then why does the H(P,P) that P calls get stuck in an infinite recursion wneh the top level doesn't?

It is a verified fact that the correct simulation of the input to H(P,P) never reaches its final instruction thus conclusively proving that it never halts.


The only machine that you have shown that does a correct simulation is the version that never aborts. That version fails to answer the question, so fails to be a halt decider.

Any version of H that aborts, and returns a not-halting answer changes P into a Halting Compuation.

The "pathological" use of H by P lets it change as you change H, so if H aborts, it is wrong because THAT P halts, if it doesn't, then it is wrong for not answering.

You seem to miss this fact because you just don't understand the basics of how computations work. Part of your problem is you keep on trying to define H by rules that aren't an actual algorithm, so can't actually be written.




It is an easily verifiable fact that the C function H does correctly determine that the C function named P would never reach its last instruction when correctly emulated by H.

Don't just "Claim" it, so an ACTUAL verification, or you just show yourself to be a liar.


Everyone disagreeing with verified facts is incorrect on the basis of lack of technical competency or lack of honesty.

You haven't verified ANY fact, you have made claims using FAKE data that don't even really support your claim.


Software engineering experts
can reverse-engineer what the correct x86 emulation of the input to H(P,P) would be for one emulation and one nested emulation thus confirming that the provided execution trace is correct. They can do this entirely on the basis of the x86 source-code for P with no need to see the source-code or execution trace of H.

Interesting point, that you eed to talk about "reverse-engineering" this output implies that you don't actually have a program to generate it.


Not at all. This program took me a whole man-year.

Then why do we need to "reverse-engineer" the trace?


It is the only way that you can verify that this trace is correct:

Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
     machine   stack     stack     machine    assembly
     address   address   data      code       language
     ========  ========  ========  =========  =============
....[00001352][0021233e][00212342] 55         push ebp      // enter P
....[00001353][0021233e][00212342] 8bec       mov ebp,esp
....[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
....[00001358][0021233a][00001352] 50         push eax      // push P
....[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
....[0000135c][00212336][00001352] 51         push ecx      // push P
....[0000135d][00212332][00001362] e840feffff call 000011a2 // call H
....[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
....[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
....[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
....[00001358][0025cd62][00001352] 50         push eax      // push P
....[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
....[0000135c][0025cd5e][00001352] 51         push ecx      // push P
....[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H


The most difficult part was to make the nested simulations work correctly.

Which it seems you failed at. Maybe that is why ot took so long. It is hard to make automaticaly believable lies.


You don't need to "reverse-engineer" something you have.


No but everyone else does to confirm its correctness because
THEY DON'T HAVE IT.

No, you present the trace and people can verify it.

Note, it has been pointed out that it does NOT match what you claim is happening, which proves it is WRONG.

THIS HAS NEVER BEEN MORE THAN A BASELESS ASSERTION.


The trace implies that H just calls its input, and hides its own behavior.


I hrrd code the system to never output the trace of operating system code.

If it does that, then it can't abort its trace, or isn't actually a computatiopn. (and we see it aborts, and you claim it is a computatiopn)

Thus, the trace is incorrect, or you are lying and the trace doesn't matter.




Anyone unable to do this conclusively proves their lack of sufficient technical competence.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp

Click here to read the complete article
Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Richard Damon
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Forte - www.forteinc.com
Date: Sun, 22 May 2022 23:04 UTC
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Gecko/20100101 Thunderbird/91.9.1
Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
Content-Language: en-US
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References: <20220514170555.00004550@reddwarf.jmc>
<I6idnSDl8NcUDRT_nZ2dnUU7_83NnZ2d@giganews.com> <s5hiK.45$cq8.28@fx03.iad>
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From: Rich...@Damon-Family.org (Richard Damon)
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On 5/22/22 6:50 PM, olcott wrote:
On 5/22/2022 5:34 PM, Richard Damon wrote:

On 5/22/22 5:59 PM, olcott wrote:
On 5/22/2022 4:34 PM, Richard Damon wrote:

On 5/22/22 5:06 PM, olcott wrote:
On 5/22/2022 12:36 PM, Richard Damon wrote:
On 5/22/22 12:42 PM, olcott wrote:
On 5/22/2022 11:40 AM, Richard Damon wrote:
On 5/22/22 12:31 PM, olcott wrote:
On 5/22/2022 11:21 AM, Richard Damon wrote:
On 5/22/22 10:57 AM, olcott wrote:
On 5/22/2022 6:06 AM, Richard Damon wrote:
On 5/22/22 1:02 AM, olcott wrote:
On 5/21/2022 11:05 PM, Richard Damon wrote:
On 5/21/22 11:59 PM, olcott wrote:
On 5/21/2022 10:54 PM, Richard Damon wrote:
On 5/21/22 11:36 PM, olcott wrote:
On 5/21/2022 10:27 PM, Richard Damon wrote:
On 5/21/22 10:48 PM, olcott wrote:
On 5/21/2022 9:37 PM, Richard Damon wrote:
On 5/21/22 10:28 PM, olcott wrote:
On 5/21/2022 9:20 PM, Richard Damon wrote:

On 5/21/22 9:23 PM, olcott wrote:
On 5/21/2022 8:05 PM, Richard Damon wrote:
On 5/21/22 3:38 PM, olcott wrote:
On 5/20/2022 5:25 PM, Ben wrote:
olcott <NoOne@NoWhere.com> writes:

You have known that the input to H(P,P) is simulated correctly proving
that H(P,P)==0 is correct for the whole six months

If H is intended to be a halt decider (even if only for the one case you
claim to care about) then H(P,P) == 0 is wrong, because P(P) halts.


When we correctly reverse-engineer what the execution trace of the input to H(P,P) would be for one emulation and one nested emulation we can see that the correctly emulated input to H(P,P) would never reach its final state at machine address [0000136c].

A nonsense trace, as it is mixing the execution path of two independent execution units.

In other words you acknowledge that you are technically incompetent to provide the execution trace of one simulation and one nested simulation of the input to H(P,P).

No, I am saying that you are asking for the equivalent of a of a square circle.


So an execution trace of the input to H(P,P) is easy to show when H simulates its input, yet another execution trace of the input to H(P,P) that was invoked by P is "like a square circle" can't possibly exist?

The problem is that your second trace is NOT a piece of the first.

The fact you don't understand that says you just don't know how computers or programs actually work.


When a UTM simulates a TM description that calls a UTM that simulates a
TM description all of this is simply data on the first UTM's tape and the only actual executable is the first UTM.


Yes, and a trace made by that outer UTM will show the states that the second UTM is going through, but NOT the states that second UTM simulates in its own processing.

That second UTM might produce its OWN trace of the states that it has simulated, but that is a SEPERATE trace, and NOT part of the trace from the OUTER UTM.


And this trace is written to the outer UTM's tape as a part of its own data.

Yes, the DATA is there, ENCODED on the tape, but it isn't part of the trace generated by that UTM.

The only actual executable is the outer UTM everything else is a part of the same nested process.

So the only actual valid trace is what that outer simulator actual simulated.


There is a valid trace of every line of code that is emulated. Operating system code has its trace tuned off. This only leaves the user code such as P() and main(). Then we see the 14 lines execution trace of the two level simulation of the input to H(P,P)

No, because the second level emulation is NOT emulated by the top level emulator, its emulator is.

Unless you are lying about what H does, you are just lying that the second level code is emulated by the same emulation process that the first is. (That may well be true, but it means you logic is still built on a lie).


If you are too stupid to understand that H(P,P) derives the same execution trace of its input every time it is called you are far too stupid to evaluate my work.

Ok, then why does the H(P,P) that P calls get stuck in an infinite recursion wneh the top level doesn't?

It is a verified fact that the correct simulation of the input to H(P,P) never reaches its final instruction thus conclusively proving that it never halts.


The only machine that you have shown that does a correct simulation is the version that never aborts. That version fails to answer the question, so fails to be a halt decider.

Any version of H that aborts, and returns a not-halting answer changes P into a Halting Compuation.

The "pathological" use of H by P lets it change as you change H, so if H aborts, it is wrong because THAT P halts, if it doesn't, then it is wrong for not answering.

You seem to miss this fact because you just don't understand the basics of how computations work. Part of your problem is you keep on trying to define H by rules that aren't an actual algorithm, so can't actually be written.




It is an easily verifiable fact that the C function H does correctly determine that the C function named P would never reach its last instruction when correctly emulated by H.

Don't just "Claim" it, so an ACTUAL verification, or you just show yourself to be a liar.


Everyone disagreeing with verified facts is incorrect on the basis of lack of technical competency or lack of honesty.

You haven't verified ANY fact, you have made claims using FAKE data that don't even really support your claim.


Software engineering experts
can reverse-engineer what the correct x86 emulation of the input to H(P,P) would be for one emulation and one nested emulation thus confirming that the provided execution trace is correct. They can do this entirely on the basis of the x86 source-code for P with no need to see the source-code or execution trace of H.

Interesting point, that you eed to talk about "reverse-engineering" this output implies that you don't actually have a program to generate it.


Not at all. This program took me a whole man-year.

Then why do we need to "reverse-engineer" the trace?


It is the only way that you can verify that this trace is correct:

Since it ISN"T correct, that is meaningless.


Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
     machine   stack     stack     machine    assembly
     address   address   data      code       language
     ========  ========  ========  =========  =============
...[00001352][0021233e][00212342] 55         push ebp      // enter P
...[00001353][0021233e][00212342] 8bec       mov ebp,esp
...[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
...[00001358][0021233a][00001352] 50         push eax      // push P
...[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][00212336][00001352] 51         push ecx      // push P
...[0000135d][00212332][00001362] e840feffff call 000011a2 // call H

ERROR HERE. The code below in not executed as part of the same process as the code above.
...[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
...[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
...[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
...[00001358][0025cd62][00001352] 50         push eax      // push P
...[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][0025cd5e][00001352] 51         push ecx      // push P
...[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H


The most difficult part was to make the nested simulations work correctly.

Which it seems you failed at. Maybe that is why ot took so long. It is hard to make automaticaly believable lies.


You don't need to "reverse-engineer" something you have.


No but everyone else does to confirm its correctness because
THEY DON'T HAVE IT.

No, you present the trace and people can verify it.

Note, it has been pointed out that it does NOT match what you claim is happening, which proves it is WRONG.

THIS HAS NEVER BEEN MORE THAN A BASELESS ASSERTION.

You admit your claim is baseless?

You can't call my claim baseless, as I fully explain my reasoning, which you do not even attempt to refute, so you must be accepting.




The trace implies that H just calls its input, and hides its own behavior.


I hrrd code the system to never output the trace of operating system code.

WHAT operationg system code. There is no such thing in a Computation.

The H that P calls is code that is part of the compuation and needs to be treated as such.

That is one of you big lies.



If it does that, then it can't abort its trace, or isn't actually a computatiopn. (and we see it aborts, and you claim it is a computatiopn)

Thus, the trace is incorrect, or you are lying and the trace doesn't matter.




Anyone unable to do this conclusively proves their lack of sufficient technical competence.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08

Click here to read the complete article
Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Date: Sun, 22 May 2022 23:16 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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Newsgroups: comp.theory,comp.ai.philosophy,sci.logic
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From: NoO...@NoWhere.com (olcott)
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On 5/22/2022 5:34 PM, Richard Damon wrote:

On 5/22/22 5:59 PM, olcott wrote:
On 5/22/2022 4:34 PM, Richard Damon wrote:

So then you now agree that the nested invocation of H(P,P) would derive the same execution trace of its input that the the outer one did?


Yes seem to have a confusion. Yes, the second simulator will produce and identical trace to the first one, AS A SEPERATE TRACE.


So the first invocation of H(P,P) derives this trace
Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
....[00001352][0021233e][00212342] 55         push ebp      // enter P
....[00001353][0021233e][00212342] 8bec       mov ebp,esp
....[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
....[00001358][0021233a][00001352] 50         push eax      // push P
....[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
....[0000135c][00212336][00001352] 51         push ecx      // push P
....[0000135d][00212332][00001362] e840feffff call 000011a2 // call H

And the second (nested) invocation of H(P,P) derives this trace
....[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
....[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
....[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
....[00001358][0025cd62][00001352] 50         push eax      // push P
....[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
....[0000135c][0025cd5e][00001352] 51         push ecx      // push P
....[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H

 From this we can see the pattern that the C/x86 function named P will never reach its last instruction, thus conclusively proving that when H(P,P) rejects its input as non-halting it is correct.


The issue is you can not validly "combine" them into a single trace, that is just a lie. The first simulator needs to show its trace of the instructions of the second simulator doing its simulation.


So when I show the trace of main() this trace is a lie? I show the trace of every line of user code and none of these traces are a lie.

I suppose if you want, you can omit that with a note of omitted detail, but the the trace goes to the call, says details omited, then the notice that the trace was aborted, but then you don't have any grounds to claim that it was correct (since there aren't any).

In order to contradict me you must contradict computer science.



Nope, YOU do that enough.


--
Copyright 2022 Pete Olcott

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


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Richard Damon
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Forte - www.forteinc.com
Date: Sun, 22 May 2022 23:52 UTC
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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From: Rich...@Damon-Family.org (Richard Damon)
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On 5/22/22 7:16 PM, olcott wrote:
On 5/22/2022 5:34 PM, Richard Damon wrote:

On 5/22/22 5:59 PM, olcott wrote:
On 5/22/2022 4:34 PM, Richard Damon wrote:

So then you now agree that the nested invocation of H(P,P) would derive the same execution trace of its input that the the outer one did?


Yes seem to have a confusion. Yes, the second simulator will produce and identical trace to the first one, AS A SEPERATE TRACE.


So the first invocation of H(P,P) derives this trace
Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
...[00001352][0021233e][00212342] 55         push ebp      // enter P
...[00001353][0021233e][00212342] 8bec       mov ebp,esp
...[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
...[00001358][0021233a][00001352] 50         push eax      // push P
...[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][00212336][00001352] 51         push ecx      // push P
...[0000135d][00212332][00001362] e840feffff call 000011a2 // call H

And the second (nested) invocation of H(P,P) derives this trace
...[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
...[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
...[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
...[00001358][0025cd62][00001352] 50         push eax      // push P
...[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][0025cd5e][00001352] 51         push ecx      // push P
...[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H

 From this we can see the pattern that the C/x86 function named P will never reach its last instruction, thus conclusively proving that when H(P,P) rejects its input as non-halting it is correct.


No, because the whole second trace is conditioned on the halt deciding of H. So, if the top level H aborts at that point, then the trace become incorrect by being incomplete (it doesn't show the FULL Halting behavior of the input, which refers to the machine it represents, which did not halt there.

When we put this input into an ACTUAL CORRECT emulator that doesn't abort, it will show a third (nested) simulation followed by the simulator that P called deciding to abort, and then P Halting.

The issue is you can not validly "combine" them into a single trace, that is just a lie. The first simulator needs to show its trace of the instructions of the second simulator doing its simulation.


So when I show the trace of main() this trace is a lie? I show the trace of every line of user code and none of these traces are a lie.

The phrase "User Code" is a lie, as H needs to be User Code, as that is all Turing Machines have (at least as far as tracing goes).

I think the problem is that your H doesn't actually exist as a real independent algorithm, but is enmeshed in your UTM x86, and thus isn't actually a Computation.


I suppose if you want, you can omit that with a note of omitted detail, but the the trace goes to the call, says details omited, then the notice that the trace was aborted, but then you don't have any grounds to claim that it was correct (since there aren't any).

In order to contradict me you must contradict computer science.



Nope, YOU do that enough.





Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Date: Sun, 22 May 2022 23:59 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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From: NoO...@NoWhere.com (olcott)
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On 5/22/2022 6:52 PM, Richard Damon wrote:
On 5/22/22 7:16 PM, olcott wrote:
On 5/22/2022 5:34 PM, Richard Damon wrote:

On 5/22/22 5:59 PM, olcott wrote:
On 5/22/2022 4:34 PM, Richard Damon wrote:

So then you now agree that the nested invocation of H(P,P) would derive the same execution trace of its input that the the outer one did?


Yes seem to have a confusion. Yes, the second simulator will produce and identical trace to the first one, AS A SEPERATE TRACE.


So the first invocation of H(P,P) derives this trace
Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
...[00001352][0021233e][00212342] 55         push ebp      // enter P
...[00001353][0021233e][00212342] 8bec       mov ebp,esp
...[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
...[00001358][0021233a][00001352] 50         push eax      // push P
...[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][00212336][00001352] 51         push ecx      // push P
...[0000135d][00212332][00001362] e840feffff call 000011a2 // call H

And the second (nested) invocation of H(P,P) derives this trace
...[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
...[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
...[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
...[00001358][0025cd62][00001352] 50         push eax      // push P
...[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][0025cd5e][00001352] 51         push ecx      // push P
...[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H

 From this we can see the pattern that the C/x86 function named P will never reach its last instruction, thus conclusively proving that when H(P,P) rejects its input as non-halting it is correct.


No, because the whole second trace is conditioned on the halt deciding of H. So, if the top level H aborts at that point, then the trace become incorrect by being incomplete (it doesn't show the FULL Halting behavior of the input, which refers to the machine it represents, which did not halt there.
The fact that I just proved that the simulated input to H(P,P) never reaches its own final state conclusively proves that it is non-halting regardless of whether or not it stops running.

--
Copyright 2022 Pete Olcott

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


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Richard Damon
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Forte - www.forteinc.com
Date: Mon, 23 May 2022 00:21 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Path: i2pn2.org!i2pn.org!usenet.blueworldhosting.com!feed1.usenet.blueworldhosting.com!peer02.iad!feed-me.highwinds-media.com!news.highwinds-media.com!fx33.iad.POSTED!not-for-mail
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Gecko/20100101 Thunderbird/91.9.1
Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
Content-Language: en-US
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References: <20220514170555.00004550@reddwarf.jmc>
<NtGdndBRE5xLAhT_nZ2dnUU7_83NnZ2d@giganews.com>
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From: Rich...@Damon-Family.org (Richard Damon)
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On 5/22/22 7:59 PM, olcott wrote:
On 5/22/2022 6:52 PM, Richard Damon wrote:
On 5/22/22 7:16 PM, olcott wrote:
On 5/22/2022 5:34 PM, Richard Damon wrote:

On 5/22/22 5:59 PM, olcott wrote:
On 5/22/2022 4:34 PM, Richard Damon wrote:

So then you now agree that the nested invocation of H(P,P) would derive the same execution trace of its input that the the outer one did?


Yes seem to have a confusion. Yes, the second simulator will produce and identical trace to the first one, AS A SEPERATE TRACE.


So the first invocation of H(P,P) derives this trace
Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
...[00001352][0021233e][00212342] 55         push ebp      // enter P
...[00001353][0021233e][00212342] 8bec       mov ebp,esp
...[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
...[00001358][0021233a][00001352] 50         push eax      // push P
...[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][00212336][00001352] 51         push ecx      // push P
...[0000135d][00212332][00001362] e840feffff call 000011a2 // call H

And the second (nested) invocation of H(P,P) derives this trace
...[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
...[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
...[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
...[00001358][0025cd62][00001352] 50         push eax      // push P
...[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][0025cd5e][00001352] 51         push ecx      // push P
...[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H

 From this we can see the pattern that the C/x86 function named P will never reach its last instruction, thus conclusively proving that when H(P,P) rejects its input as non-halting it is correct.


No, because the whole second trace is conditioned on the halt deciding of H. So, if the top level H aborts at that point, then the trace become incorrect by being incomplete (it doesn't show the FULL Halting behavior of the input, which refers to the machine it represents, which did not halt there.
The fact that I just proved that the simulated input to H(P,P) never reaches its own final state conclusively proves that it is non-halting regardless of whether or not it stops running.


You proved that the INCOMPLETE simulation of the input by H doesn't reach a final state.

NOT that the CORRECT simulation of the input by a CORRECT simulator doesn't.

You just are proving that you don't know the meaning of CORRECT. (or are just lying).


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Date: Mon, 23 May 2022 01:01 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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Newsgroups: comp.theory,comp.ai.philosophy,sci.logic
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From: NoO...@NoWhere.com (olcott)
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On 5/22/2022 7:21 PM, Richard Damon wrote:
On 5/22/22 7:59 PM, olcott wrote:
On 5/22/2022 6:52 PM, Richard Damon wrote:
On 5/22/22 7:16 PM, olcott wrote:
On 5/22/2022 5:34 PM, Richard Damon wrote:

On 5/22/22 5:59 PM, olcott wrote:
On 5/22/2022 4:34 PM, Richard Damon wrote:

So then you now agree that the nested invocation of H(P,P) would derive the same execution trace of its input that the the outer one did?


Yes seem to have a confusion. Yes, the second simulator will produce and identical trace to the first one, AS A SEPERATE TRACE.


So the first invocation of H(P,P) derives this trace
Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
...[00001352][0021233e][00212342] 55         push ebp      // enter P
...[00001353][0021233e][00212342] 8bec       mov ebp,esp
...[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
...[00001358][0021233a][00001352] 50         push eax      // push P
...[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][00212336][00001352] 51         push ecx      // push P
...[0000135d][00212332][00001362] e840feffff call 000011a2 // call H

And the second (nested) invocation of H(P,P) derives this trace
...[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
...[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
...[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
...[00001358][0025cd62][00001352] 50         push eax      // push P
...[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][0025cd5e][00001352] 51         push ecx      // push P
...[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H

 From this we can see the pattern that the C/x86 function named P will never reach its last instruction, thus conclusively proving that when H(P,P) rejects its input as non-halting it is correct.


No, because the whole second trace is conditioned on the halt deciding of H. So, if the top level H aborts at that point, then the trace become incorrect by being incomplete (it doesn't show the FULL Halting behavior of the input, which refers to the machine it represents, which did not halt there.
The fact that I just proved that the simulated input to H(P,P) never reaches its own final state conclusively proves that it is non-halting regardless of whether or not it stops running.


You proved that the INCOMPLETE simulation of the input by H doesn't reach a final state.


So in other words you simply are not bright enough to understand that the emulated P can't possibly ever reach its own last instruction.
I honestly can't believe that you are that stupid.

NOT that the CORRECT simulation of the input by a CORRECT simulator doesn't.

You just are proving that you don't know the meaning of CORRECT. (or are just lying).


--
Copyright 2022 Pete Olcott

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


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Richard Damon
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Forte - www.forteinc.com
Date: Mon, 23 May 2022 01:11 UTC
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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From: Rich...@Damon-Family.org (Richard Damon)
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On 5/22/22 9:01 PM, olcott wrote:
On 5/22/2022 7:21 PM, Richard Damon wrote:
On 5/22/22 7:59 PM, olcott wrote:
On 5/22/2022 6:52 PM, Richard Damon wrote:
On 5/22/22 7:16 PM, olcott wrote:
On 5/22/2022 5:34 PM, Richard Damon wrote:

On 5/22/22 5:59 PM, olcott wrote:
On 5/22/2022 4:34 PM, Richard Damon wrote:

So then you now agree that the nested invocation of H(P,P) would derive the same execution trace of its input that the the outer one did?


Yes seem to have a confusion. Yes, the second simulator will produce and identical trace to the first one, AS A SEPERATE TRACE.


So the first invocation of H(P,P) derives this trace
Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
...[00001352][0021233e][00212342] 55         push ebp      // enter P
...[00001353][0021233e][00212342] 8bec       mov ebp,esp
...[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
...[00001358][0021233a][00001352] 50         push eax      // push P
...[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][00212336][00001352] 51         push ecx      // push P
...[0000135d][00212332][00001362] e840feffff call 000011a2 // call H

And the second (nested) invocation of H(P,P) derives this trace
...[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
...[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
...[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
...[00001358][0025cd62][00001352] 50         push eax      // push P
...[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][0025cd5e][00001352] 51         push ecx      // push P
...[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H

 From this we can see the pattern that the C/x86 function named P will never reach its last instruction, thus conclusively proving that when H(P,P) rejects its input as non-halting it is correct.


No, because the whole second trace is conditioned on the halt deciding of H. So, if the top level H aborts at that point, then the trace become incorrect by being incomplete (it doesn't show the FULL Halting behavior of the input, which refers to the machine it represents, which did not halt there.
The fact that I just proved that the simulated input to H(P,P) never reaches its own final state conclusively proves that it is non-halting regardless of whether or not it stops running.


You proved that the INCOMPLETE simulation of the input by H doesn't reach a final state.


So in other words you simply are not bright enough to understand that the emulated P can't possibly ever reach its own last instruction.
I honestly can't believe that you are that stupid.

H can not emulate the input P to the final state, yes.

But that isn't the definition of Halting.

Halting is defined by the ACTUAL MACHINE P reaching the final state, or a CORRECT (unaborted) simulation reaching the final state.

P DOES do this if H(P,P) returns 0, so that can't be the correct answer.

YOu are just confirming your ignorance of the topic, as you keep thinking that an aborted simulation not reaching the final state shows the machine is non-halting.


NOT that the CORRECT simulation of the input by a CORRECT simulator doesn't.

You just are proving that you don't know the meaning of CORRECT. (or are just lying).





Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Python
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Aioe.org NNTP Server
Date: Mon, 23 May 2022 11:55 UTC
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From: pyt...@example.invalid (Python)
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic
Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
Date: Mon, 23 May 2022 13:55:35 +0200
Organization: Aioe.org NNTP Server
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Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.




Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Richard Damon
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Forte - www.forteinc.com
Date: Mon, 23 May 2022 12:22 UTC
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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From: Rich...@Damon-Family.org (Richard Damon)
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On 5/23/22 7:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.



As near as I can figure it, Peter thinks that because H stops simulating its P input, that P never halts, because that copy never gets to the final state that a fully run P will get to.

Yes, he confuses all sorts of things, forcing distinctions between things that are supposed to be identical and merges things that are supposed to be distinct.

It seems he as confinced himself of his own false ideas, and twists reality as much as needed to try to get it to conform to what he thinks he needs to make it fit.

He can't afford to look at rebuttals, because they show him more areas that he needs to twist to make things conform and he is running out of things to distort to make it happen.

This is the typical result when you takes as a core foundation something that isn't true.


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Date: Mon, 23 May 2022 13:29 UTC
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From: NoO...@NoWhere.com (olcott)
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On 5/23/2022 6:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.



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

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


--
Copyright 2022 Pete Olcott

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


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Date: Mon, 23 May 2022 13:33 UTC
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
Content-Language: en-US
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On 5/23/2022 7:22 AM, Richard Damon wrote:
On 5/23/22 7:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.



As near as I can figure it, Peter thinks that because H stops simulating its P input, that P never halts, because that copy never gets to the final state that a fully run P will get to.


Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
....[00001352][0021233e][00212342] 55         push ebp      // enter P
....[00001353][0021233e][00212342] 8bec       mov ebp,esp
....[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
....[00001358][0021233a][00001352] 50         push eax      // push P
....[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
....[0000135c][00212336][00001352] 51         push ecx      // push P
....[0000135d][00212332][00001362] e840feffff call 000011a2 // call H
....[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
....[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
....[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
....[00001358][0025cd62][00001352] 50         push eax      // push P
....[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
....[0000135c][0025cd5e][00001352] 51         push ecx      // push P
....[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

In other words you are simply too freaking stupid to understand that the correctly simulated P never reaches its last instruction whether or not this simulation is ever aborted.

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]


--
Copyright 2022 Pete Olcott

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


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Python
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Aioe.org NNTP Server
Date: Mon, 23 May 2022 13:58 UTC
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From: pyt...@example.invalid (Python)
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic
Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
Date: Mon, 23 May 2022 15:58:00 +0200
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Peter Olcott, demented kook, wrote:
On 5/23/2022 6:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.



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

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

This is not even remotely related to your absurd claims, come on Peter!




Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Date: Mon, 23 May 2022 14:33 UTC
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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From: NoO...@NoWhere.com (olcott)
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On 5/23/2022 8:58 AM, Python wrote:
Peter Olcott, demented kook, wrote:
On 5/23/2022 6:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.



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

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

This is not even remotely related to your absurd claims, come on Peter!



Halting DOES NOT MEAN STOPS RUNNING,
Halting means reaches its final state.

--
Copyright 2022 Pete Olcott

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


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Python
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Aioe.org NNTP Server
Date: Mon, 23 May 2022 14:43 UTC
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Path: i2pn2.org!i2pn.org!aioe.org!7a25jG6pUKCqa0zKnKnvdg.user.46.165.242.75.POSTED!not-for-mail
From: pyt...@example.invalid (Python)
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic
Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
Date: Mon, 23 May 2022 16:43:47 +0200
Organization: Aioe.org NNTP Server
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View all headers
Peter Olcott, old fart, wrote:
On 5/23/2022 8:58 AM, Python wrote:
Peter Olcott, demented kook, wrote:
On 5/23/2022 6:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.



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

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

This is not even remotely related to your absurd claims, come on Peter!



Halting DOES NOT MEAN STOPS RUNNING,
Halting means reaches its final state.


*facepalm*




Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Date: Mon, 23 May 2022 14:47 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
Content-Language: en-US
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic
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From: NoO...@NoWhere.com (olcott)
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On 5/23/2022 9:43 AM, Python wrote:
Peter Olcott, old fart, wrote:
On 5/23/2022 8:58 AM, Python wrote:
Peter Olcott, demented kook, wrote:
On 5/23/2022 6:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.



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

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

This is not even remotely related to your absurd claims, come on Peter!



Halting DOES NOT MEAN STOPS RUNNING,
Halting means reaches its final state.


*facepalm*

I know that you are ignorant about this, yet at least you are not a liar. The definition of halting does mean reaches final state, this is a term of the art of computer science thus overrides and supersedes its common meaning.


--
Copyright 2022 Pete Olcott

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


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Jeff Barnett
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: A noiseless patient Spider
Date: Mon, 23 May 2022 18:39 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
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,sci.logic
Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
Date: Mon, 23 May 2022 12:39:00 -0600
Organization: A noiseless patient Spider
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On 5/23/2022 5:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.
Insightful observation. I have been curious, since he is now a pathological liar, whether he lies about cancer too. I wouldn't be surprised. As to cognitive disorders, I believe he his learning disabilities compounded by the attention span of a roach. I don't believe that any thing said here will cure any of his observed issues.

As a test, I speculate he will respond to this message, if at all, with either many identical lines of capitol letters or yet another copy of his faked up trace. Gentlemen, Place Your Bets!
--
Jeff Barnett


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Date: Mon, 23 May 2022 18:51 UTC
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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From: NoO...@NoWhere.com (olcott)
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On 5/23/2022 1:39 PM, Jeff Barnett wrote:
On 5/23/2022 5:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.
Insightful observation. I have been curious, since he is now a pathological liar, whether he lies about cancer too. I wouldn't be surprised. As to cognitive disorders, I believe he his learning disabilities compounded by the attention span of a roach. I don't believe that any thing said here will cure any of his observed issues.

As a test, I speculate he will respond to this message, if at all, with either many identical lines of capitol letters or yet another copy of his faked up trace. Gentlemen, Place Your Bets!

_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]

People with sufficient technical competence that are not God damned liars can easily verify that the following trace provided for the x86 emulated input to H(P,P) for one emulation and one nested emulation is correct entirely on the basis of the above x86 source code for P.

Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
....[00001352][0021233e][00212342] 55         push ebp      // enter P
....[00001353][0021233e][00212342] 8bec       mov ebp,esp
....[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
....[00001358][0021233a][00001352] 50         push eax      // push P
....[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
....[0000135c][00212336][00001352] 51         push ecx      // push P
....[0000135d][00212332][00001362] e840feffff call 000011a2 // call H
....[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
....[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
....[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
....[00001358][0025cd62][00001352] 50         push eax      // push P
....[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
....[0000135c][0025cd5e][00001352] 51         push ecx      // push P
....[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H


--
Copyright 2022 Pete Olcott

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


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Richard Damon
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Forte - www.forteinc.com
Date: Mon, 23 May 2022 23:00 UTC
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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On 5/23/22 9:33 AM, olcott wrote:
On 5/23/2022 7:22 AM, Richard Damon wrote:
On 5/23/22 7:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.



As near as I can figure it, Peter thinks that because H stops simulating its P input, that P never halts, because that copy never gets to the final state that a fully run P will get to.


Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
...[00001352][0021233e][00212342] 55         push ebp      // enter P
...[00001353][0021233e][00212342] 8bec       mov ebp,esp
...[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
...[00001358][0021233a][00001352] 50         push eax      // push P
...[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][00212336][00001352] 51         push ecx      // push P
...[0000135d][00212332][00001362] e840feffff call 000011a2 // call H
...[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
...[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
...[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
...[00001358][0025cd62][00001352] 50         push eax      // push P
...[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
...[0000135c][0025cd5e][00001352] 51         push ecx      // push P
...[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

In other words you are simply too freaking stupid to understand that the correctly simulated P never reaches its last instruction whether or not this simulation is ever aborted.

No, you confuse the concept of a CORRECT simulation with what H deos.

H generates an ABORTED simulation. An ABORTED SIMULATION are NEVER correct, because the machine they are simulating continue to run after that point.

You just don't understand that the behavior of a PROGRAM is not dependent on the behavior of an individual execution which isn't allowed to complete.

P(P) Halts, the CORRECT simulation of the input to H(P,P), i.e the equivalent to UTM(P,P) does too.

THe fact that H can't do that just shows that it can't do it, and it is just plain wrong if it claims to be a Halt Decider.

If it gives up the claim to being a Halt Decider, then it could possibly redefine what the "behavior of its input" is, but then is can't be used to be a counter example for the Halting Problem.


Your failure to understand just proves that you do not understand the very basics of computations, and apparently a substantial part of your connection with reality.


_P()
[00001352](01)  55              push ebp
[00001353](02)  8bec            mov ebp,esp
[00001355](03)  8b4508          mov eax,[ebp+08]
[00001358](01)  50              push eax      // push P
[00001359](03)  8b4d08          mov ecx,[ebp+08]
[0000135c](01)  51              push ecx      // push P
[0000135d](05)  e840feffff      call 000011a2 // call H
[00001362](03)  83c408          add esp,+08
[00001365](02)  85c0            test eax,eax
[00001367](02)  7402            jz 0000136b
[00001369](02)  ebfe            jmp 00001369
[0000136b](01)  5d              pop ebp
[0000136c](01)  c3              ret
Size in bytes:(0027) [0000136c]





Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Richard Damon
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Forte - www.forteinc.com
Date: Mon, 23 May 2022 23:05 UTC
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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From: Rich...@Damon-Family.org (Richard Damon)
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On 5/23/22 9:29 AM, olcott wrote:
On 5/23/2022 6:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.



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

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



Right, and nothing you have done shows that the TURING MACHINE didn't halt when the Turing Machine that represent your decider said it wouldnt't

All you have done with all your arguments is prove that you just don't understand what that is saying.

A partial simulation is NOT a Turing Machine, or shows what a Turing Machine would do.

To make your arguments, you FIRST need to define the EXACT algorithm your H uses, and once that is fixed, it becomes clear that for whatever algorithm you give it, it is wrong about H^/P not halting.

All your proof over "All H" does is prove that NO H, built by your template can ever prove that this H^/P will Halt, which does NOT prove that it is non-Halting.

You logic is FULL of inconsistencies like the above that have been pointed out to you. Your inability to see them shows a lack of logical processing in your brain.


Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point ? ]
From: Richard Damon
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic
Organization: Forte - www.forteinc.com
Date: Mon, 23 May 2022 23:06 UTC
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Subject: Re: Are my reviewers incompetent or dishonest? [ closure on one point
? ]
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From: Rich...@Damon-Family.org (Richard Damon)
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On 5/23/22 10:33 AM, olcott wrote:
On 5/23/2022 8:58 AM, Python wrote:
Peter Olcott, demented kook, wrote:
On 5/23/2022 6:55 AM, Python wrote:
Peter Olcott wrote:
... it is non-halting regardless of whether or not it stops running.

This is quite a serious cognitive dissonance to be able to
write down such a sentence and believe it.

Cancer is not the worse illness you have, Peter.



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

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

This is not even remotely related to your absurd claims, come on Peter!



Halting DOES NOT MEAN STOPS RUNNING,
Halting means reaches its final state.

And the fact that H stopped running P doesn't mean that it is Non-Halting.


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