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devel / comp.theory / My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

SubjectAuthor
* My_Dishonest_reviewers:_André,_Ben,_Mike,_Deolcott
+* _My_Dishonest_reviewers:_André,_Ben,_Mike,_DenniDennis Bush
|`* _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
| +- _My_Dishonest_reviewers:_André,_Ben,_MikeRichard Damon
| `* _My_Dishonest_reviewers:_André,_Ben,_Mike,_DenniDennis Bush
|  `* _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
|   +* _My_Dishonest_reviewers:_André,_Ben,_Mike,_DenniDennis Bush
|   |`* _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
|   | +* _My_Dishonest_reviewers:_André,_Ben,_Mike,_DenniDennis Bush
|   | |`* _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
|   | | +* _My_Dishonest_reviewers:_André,_Ben,_Mike,_DenniDennis Bush
|   | | |`* _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
|   | | | +- _My_Dishonest_reviewers:_André,_Ben,_MikeRichard Damon
|   | | | `* _My_Dishonest_reviewers:_André,_Ben,_Mike,_DenniDennis Bush
|   | | |  `* _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
|   | | |   `* _My_Dishonest_reviewers:_André,_Ben,_Mike,_DenniDennis Bush
|   | | |    `* _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
|   | | |     `* _My_Dishonest_reviewers:_André,_Ben,_Mike,_DenniDennis Bush
|   | | |      `* _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
|   | | |       `* _My_Dishonest_reviewers:_André,_Ben,_Mike,_DenniDennis Bush
|   | | |        +- _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
|   | | |        `* _My_Dishonest_reviewers:_André,_Ben,_Mike,_DenniDennis Bush
|   | | |         `* _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
|   | | |          `- _My_Dishonest_reviewers:_André,_Ben,_MikeRichard Damon
|   | | `- _My_Dishonest_reviewers:_André,_Ben,_MikeRichard Damon
|   | `- _My_Dishonest_reviewers:_André,_Ben,_MikeRichard Damon
|   `- _My_Dishonest_reviewers:_André,_Ben,_MikeRichard Damon
`- _My_Dishonest_reviewers:_André,_Ben,_MikeRichard Damon

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My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: olcott - Sun, 17 Apr 2022 03:40 UTC

If the input to H(P,P) is non-halting then H(P,P)==false is necessary
correct.

Does anyone have a rebuttal to that one, or do you acknowledge that it
is correct?

You must find an example where {an X is a Y} and Z is incorrect for
reporting {an X is a Y}.

If no one can show that the correctly simulated input to H(P,P) reaches
its final state and halts that proves that it is non-halting.

If I was wrong then the correct simulation of the 27 bytes of machine
code at machine address [000009d6] by H would show some correct
execution trace from machine address [000009d6] ending at machine
address [000009f0].

_P()
[000009d6](01) 55 push ebp
[000009d7](02) 8bec mov ebp,esp
[000009d9](03) 8b4508 mov eax,[ebp+08]
[000009dc](01) 50 push eax // push P
[000009dd](03) 8b4d08 mov ecx,[ebp+08]
[000009e0](01) 51 push ecx // push P
[000009e1](05) e840feffff call 00000826 // call H
[000009e6](03) 83c408 add esp,+08
[000009e9](02) 85c0 test eax,eax
[000009eb](02) 7402 jz 000009ef
[000009ed](02) ebfe jmp 000009ed
[000009ef](01) 5d pop ebp
[000009f0](01) c3 ret // Final state
Size in bytes:(0027) [000009f0]

That everyone refuses this challenge proves that it is beyond their
technical capacity or that they are liars.

Halting problem undecidability and infinitely nested simulation (V5)

https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5

--
Copyright 2022 Pete Olcott

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

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Dennis Bush - Sun, 17 Apr 2022 04:14 UTC

On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
> correct.

True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.

>
> Does anyone have a rebuttal to that one, or do you acknowledge that it
> is correct?
>
> You must find an example where {an X is a Y} and Z is incorrect for
> reporting {an X is a Y}.
>
>
> If no one can show that the correctly simulated input to H(P,P) reaches
> its final state and halts that proves that it is non-halting.
>
> If I was wrong then the correct simulation of the 27 bytes of machine
> code at machine address [000009d6] by H would show some correct
> execution trace from machine address [000009d6] ending at machine
> address [000009f0].
>
> _P()
> [000009d6](01) 55 push ebp
> [000009d7](02) 8bec mov ebp,esp
> [000009d9](03) 8b4508 mov eax,[ebp+08]
> [000009dc](01) 50 push eax // push P
> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> [000009e0](01) 51 push ecx // push P
> [000009e1](05) e840feffff call 00000826 // call H
> [000009e6](03) 83c408 add esp,+08
> [000009e9](02) 85c0 test eax,eax
> [000009eb](02) 7402 jz 000009ef
> [000009ed](02) ebfe jmp 000009ed
> [000009ef](01) 5d pop ebp
> [000009f0](01) c3 ret // Final state
> Size in bytes:(0027) [000009f0]

Since an H that never aborts is an uninteresting case, then H must abort. We'll then rename it Ha to make it clear that we're talking about a specific H that aborts, and we'll rename P to Pa to make it explicit that it calls Ha.

When Ha(Pa,Pa) runs, it is unable to simulate its input to a final state. This however is not proof of non-halting as it could be possible that Ha did not simulate for long enough. We can verify this by passing the same input to Hb, which has the same halt criteria as Ha but defers aborting for some k number of steps. We then see that the input to Hb(Pa,Pa), which is the same as the input to Ha(Pa,Pa) does in fact reach a final state and returns true. The trace of the input to Hb(Pa,Pa) is as follows:

_Pa()
[000009d6](01) 55 push ebp
[000009d7](02) 8bec mov ebp,esp
[000009d9](03) 8b4508 mov eax,[ebp+08]
[000009dc](01) 50 push eax // push Pa
[000009dd](03) 8b4d08 mov ecx,[ebp+08]
[000009e0](01) 51 push ecx // push Pa
[000009e1](05) e840feffff call 00000826 // call Ha
Here Ha returns false as per its fixed algorithm
[000009e6](03) 83c408 add esp,+08
[000009e9](02) 85c0 test eax,eax
[000009eb](02) 7402 jz 000009ef
[000009ed](02) ebfe jmp 000009ed
[000009ef](01) 5d pop ebp
[000009f0](01) c3 ret // Final state
Here the input reaches a final state

Because Ha and Hb are both simulating halt deciders and and both are given the same input but return two different values, one must be incorrect.

When a simulating halt decider simulates its input to a final state and returns true, this is necessarily correct and conclusively proves that the input is halting. It also means that any other simulating halt decider that aborts and returns false was incorrect to do so because it was unable to simulate for long enough.

Because Hb(Pa,Pa) simulates its input to a final state, Hb(Pa,Pa) == true is correct which necessarily makes Ha(Pa,Pa) == false incorrect.

If you claim that only Ha can correctly report the halt status of its own input, then you must also accept that any halt decider is the sole source of truth for its input, and that any simulating halt decider that aborts and reports non-halting is necessarily correct. Which means that Ha3(N,5) == false is correct for the same reason that Ha(Pa,Pa) == false. So Ha3(N,5) == false and Ha(Pa,Pa) == false are necessarily either both correct or both incorrect.

>
> That everyone refuses this challenge proves that it is beyond their
> technical capacity or that they are liars.

LIE. Several have shown why this is wrong. You just refuse to acknowledge it.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: olcott - Sun, 17 Apr 2022 04:26 UTC

On 4/16/2022 11:14 PM, Dennis Bush wrote:
> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>> correct.
>
> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
>

I count this as excellent progress.

>>
>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>> is correct?
>>
>> You must find an example where {an X is a Y} and Z is incorrect for
>> reporting {an X is a Y}.
>>
>>
>> If no one can show that the correctly simulated input to H(P,P) reaches
>> its final state and halts that proves that it is non-halting.
>>
>> If I was wrong then the correct simulation of the 27 bytes of machine
>> code at machine address [000009d6] by H would show some correct
>> execution trace from machine address [000009d6] ending at machine
>> address [000009f0].
>>
>> _P()
>> [000009d6](01) 55 push ebp
>> [000009d7](02) 8bec mov ebp,esp
>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>> [000009dc](01) 50 push eax // push P
>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>> [000009e0](01) 51 push ecx // push P
>> [000009e1](05) e840feffff call 00000826 // call H
>> [000009e6](03) 83c408 add esp,+08
>> [000009e9](02) 85c0 test eax,eax
>> [000009eb](02) 7402 jz 000009ef
>> [000009ed](02) ebfe jmp 000009ed
>> [000009ef](01) 5d pop ebp
>> [000009f0](01) c3 ret // Final state
>> Size in bytes:(0027) [000009f0]
>
> Since an H that never aborts is an uninteresting case,
We simply hypothesize BOTH cases:
(a) aborted and
(b) never aborted
and then see where their execution traces would be.

The execution trace of both the aborted and not aborted paths is from
from machine address [000009d6] ending at machine address [0000009e1].

Therefore the simulated input to H(P,P) never makes it to [000009f0] and
halts.

--
Copyright 2022 Pete Olcott

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

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Richard Damon - Sun, 17 Apr 2022 10:58 UTC

On 4/17/22 12:26 AM, olcott wrote:
> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>>> correct.
>>
>> True.  *if* the input to H(P,P) is non-halting then H(P,P)==false is
>> correct.  But first we must establish whether the input to H(P,P) is
>> non-halting.
>>
>
> I count this as excellent progress.
>
>>>
>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>>> is correct?
>>>
>>> You must find an example where {an X is a Y} and Z is incorrect for
>>> reporting {an X is a Y}.
>>>
>>>
>>> If no one can show that the correctly simulated input to H(P,P) reaches
>>> its final state and halts that proves that it is non-halting.
>>>
>>> If I was wrong then the correct simulation of the 27 bytes of machine
>>> code at machine address [000009d6] by H would show some correct
>>> execution trace from machine address [000009d6] ending at machine
>>> address [000009f0].
>>>
>>> _P()
>>> [000009d6](01) 55 push ebp
>>> [000009d7](02) 8bec mov ebp,esp
>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>> [000009dc](01) 50 push eax // push P
>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>> [000009e0](01) 51 push ecx // push P
>>> [000009e1](05) e840feffff call 00000826 // call H
>>> [000009e6](03) 83c408 add esp,+08
>>> [000009e9](02) 85c0 test eax,eax
>>> [000009eb](02) 7402 jz 000009ef
>>> [000009ed](02) ebfe jmp 000009ed
>>> [000009ef](01) 5d pop ebp
>>> [000009f0](01) c3 ret // Final state
>>> Size in bytes:(0027) [000009f0]
>>
>> Since an H that never aborts is an uninteresting case,
> We simply hypothesize BOTH cases:
> (a) aborted and
> (b) never aborted
> and then see where their execution traces would be.
>
> The execution trace of both the aborted and not aborted paths is from
> from machine address [000009d6] ending at machine address [0000009e1].
>
> Therefore the simulated input to H(P,P) never makes it to [000009f0] and
> halts.
>

Nope.

Resultes of case (a) is a non-halting input, but H never answers so it
fails to be a decider.

The results of (b) are shown above, note, the condition is the behavior
of the input being given to H, not the behavior of H on that input.

The behavior of the input is DEFINED as the behavior of the machine the
input describes, which is exactly what the above trace shows.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Richard Damon - Sun, 17 Apr 2022 10:59 UTC

On 4/16/22 11:40 PM, olcott wrote:
> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
> correct.

And if the input to H)P,P) is Halting, then H(P,P) == false is
necessarly incorrect.

>
> Does anyone have a rebuttal to that one, or do you acknowledge that it
> is correct?

Your problem is that you don't prove the premise, at least not under the
right definitions and conditions.

>
> You must find an example where {an X is a Y} and Z is incorrect for
> reporting {an X is a Y}.

As must you, you need to look at the Halting Behavior of the input,
which is DEFINED as the behavior of the machine the input describes
(despite you claims that it can't be that).

>
>
> If no one can show that the correctly simulated input to H(P,P) reaches
> its final state and halts that proves that it is non-halting.

It has been shown several times, and you are just lying that no one can
do this, as all your arguments are filled with lies.

>
> If I was wrong then the correct simulation of the 27 bytes of machine
> code at machine address [000009d6] by H would show some correct
> execution trace from machine address [000009d6] ending at machine
> address [000009f0].
>
> _P()
> [000009d6](01) 55         push ebp
> [000009d7](02) 8bec       mov ebp,esp
> [000009d9](03) 8b4508     mov eax,[ebp+08]
> [000009dc](01) 50         push eax         // push P
> [000009dd](03) 8b4d08     mov ecx,[ebp+08]
> [000009e0](01) 51         push ecx         // push P
> [000009e1](05) e840feffff call 00000826    // call H
> [000009e6](03) 83c408     add esp,+08
> [000009e9](02) 85c0       test eax,eax
> [000009eb](02) 7402       jz 000009ef
> [000009ed](02) ebfe       jmp 000009ed
> [000009ef](01) 5d         pop ebp
> [000009f0](01) c3         ret              // Final state
> Size in bytes:(0027) [000009f0]
>
> That everyone refuses this challenge proves that it is beyond their
> technical capacity or that they are liars.

It has been posted and here it is again.

Your refusal to even attempt to point out an error says your position is
just incorrect.

>
>
>
> Halting problem undecidability and infinitely nested simulation (V5)
>
> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
>
>
>
>

So what is at 826, with out that we can't run the machine.

But the trace WOULD be

[000009d6](01) 55 push ebp
[000009d7](02) 8bec mov ebp,esp
[000009d9](03) 8b4508 mov eax,[ebp+08]
[000009dc](01) 50 push eax // push P
[000009dd](03) 8b4d08 mov ecx,[ebp+08]
[000009e0](01) 51 push ecx // push P
[000009e1](05) e840feffff call 00000826 // call H
// Since you have defined that H returns Non-Halting, the above call
// WILL return the value 0.
[000009e6](03) 83c408 add esp,+08
[000009e9](02) 85c0 test eax,eax
[000009eb](02) 7402 jz 000009ef, eax is zero, so will jump

[000009ef](01) 5d pop ebp
[000009f0](01) c3 ret // Final state

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Dennis Bush - Sun, 17 Apr 2022 15:11 UTC

On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
> On 4/16/2022 11:14 PM, Dennis Bush wrote:
> > On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
> >> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
> >> correct.
> >
> > True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
> >
> I count this as excellent progress.

This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.

> >>
> >> Does anyone have a rebuttal to that one, or do you acknowledge that it
> >> is correct?
> >>
> >> You must find an example where {an X is a Y} and Z is incorrect for
> >> reporting {an X is a Y}.
> >>
> >>
> >> If no one can show that the correctly simulated input to H(P,P) reaches
> >> its final state and halts that proves that it is non-halting.
> >>
> >> If I was wrong then the correct simulation of the 27 bytes of machine
> >> code at machine address [000009d6] by H would show some correct
> >> execution trace from machine address [000009d6] ending at machine
> >> address [000009f0].
> >>
> >> _P()
> >> [000009d6](01) 55 push ebp
> >> [000009d7](02) 8bec mov ebp,esp
> >> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >> [000009dc](01) 50 push eax // push P
> >> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >> [000009e0](01) 51 push ecx // push P
> >> [000009e1](05) e840feffff call 00000826 // call H
> >> [000009e6](03) 83c408 add esp,+08
> >> [000009e9](02) 85c0 test eax,eax
> >> [000009eb](02) 7402 jz 000009ef
> >> [000009ed](02) ebfe jmp 000009ed
> >> [000009ef](01) 5d pop ebp
> >> [000009f0](01) c3 ret // Final state
> >> Size in bytes:(0027) [000009f0]
> >
> > Since an H that never aborts is an uninteresting case,
> We simply hypothesize BOTH cases:
> (a) aborted and
> (b) never aborted
> and then see where their execution traces would be.

Those two cases means you're changing the code of H, which means that P becomes a completely different computation.

There are actually multiple different H's that abort. So first some notations:

Hn: a simulating halt decider which never aborts
Pn: a P that calls Hn
Hi, where i is an integer: a set of simulating halt deciders which aborts after some number of steps.
-- These may simulate for a different number of steps from each other, or for the same number but using a completely different algorithm
Pi, a P which calls Hi

So that gives us this:

Hn(Pn,Pn) does not return an answer
H1(P1,P1) is unable to simulate its input to a final state, aborts, and returns false
H2(P2,P2) is unable to simulate its input to a final state, aborts, and returns false
H3(P3,P3) is unable to simulate its input to a final state, aborts, and returns false
....

Then what you seem to be saying is that because all of these halt deciders either don't return or return false for their respective inputs, that means that all of them are correct to return false for their respective inputs.

This is a nonsense claim. Each of these inputs is completely different because the H that each P is built on is completely different, so whether or not one halts has *no* bearing on whether another halts.

For any Hi, we can construct Hk which simulates for some k additional steps then Hi. This will result in Hk(Pi,Pi) simulating its input to a final state and returning true. This is necessarily a correct decision because Hk never leaves UTM mode while simulating its input to a final state. That makes Hi(Pi,Pi) == false incorrect for any i.

The only case where non-halting is correct is the case of Hn(Pn,Pn), and in that case Hn is unable to gives that answer because it never aborts.

The execution trace of a specific Pi is as follows

_P1()
[010009d6](01) 55 push ebp
[010009d7](02) 8bec mov ebp,esp
[010009d9](03) 8b4508 mov eax,[ebp+08]
[010009dc](01) 50 push eax // push P1
[010009dd](03) 8b4d08 mov ecx,[ebp+08]
[010009e0](01) 51 push ecx // push P1
[010009e1](05) e840feffff call 00000826 // call H1
Here H1 returns false as per its fixed algorithm

[010009e6](03) 83c408 add esp,+08
[010009e9](02) 85c0 test eax,eax
[010009eb](02) 7402 jz 000009ef
[010009ed](02) ebfe jmp 000009ed
[010009ef](01) 5d pop ebp
[010009f0](01) c3 ret // Final state
Here the input to H(1+k) reaches a final state

You'll note that the addresses are different. That's to make it explicit that each Pi is distinct from each other to prevent any dishonest equivocation of multiple different computations.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: olcott - Sun, 17 Apr 2022 15:25 UTC

On 4/17/2022 10:11 AM, Dennis Bush wrote:
> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>>>> correct.
>>>
>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
>>>
>> I count this as excellent progress.
>
> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.

None-the-less I count this as significant progress. It may have been
nearly the first time that anyone has ever agreed with anything that I
have ever said in this forum.

>
>>>>
>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>>>> is correct?
>>>>
>>>> You must find an example where {an X is a Y} and Z is incorrect for
>>>> reporting {an X is a Y}.
>>>>
>>>>
>>>> If no one can show that the correctly simulated input to H(P,P) reaches
>>>> its final state and halts that proves that it is non-halting.
>>>>
>>>> If I was wrong then the correct simulation of the 27 bytes of machine
>>>> code at machine address [000009d6] by H would show some correct
>>>> execution trace from machine address [000009d6] ending at machine
>>>> address [000009f0].
>>>>
>>>> _P()
>>>> [000009d6](01) 55 push ebp
>>>> [000009d7](02) 8bec mov ebp,esp
>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>> [000009dc](01) 50 push eax // push P
>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>> [000009e0](01) 51 push ecx // push P
>>>> [000009e1](05) e840feffff call 00000826 // call H
>>>> [000009e6](03) 83c408 add esp,+08
>>>> [000009e9](02) 85c0 test eax,eax
>>>> [000009eb](02) 7402 jz 000009ef
>>>> [000009ed](02) ebfe jmp 000009ed
>>>> [000009ef](01) 5d pop ebp
>>>> [000009f0](01) c3 ret // Final state
>>>> Size in bytes:(0027) [000009f0]
>>>
>>> Since an H that never aborts is an uninteresting case,
>> We simply hypothesize BOTH cases:
>> (a) aborted and
>> (b) never aborted
>> and then see where their execution traces would be.
>
> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.

We really only need to answer this single question:
Is there any case where the correctly simulated input to H(P,P) reaches
its own machine address of [000009f0]?

H either aborts the simulation of its input or does not abort the
simulation of its input. In either case the execution trace never
extends beyond machine address [000009e1].

Thus we can see that whether or not H aborts the simulation of its input
this simulated input never reaches machine address [000009f0].

--
Copyright 2022 Pete Olcott

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

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Dennis Bush - Sun, 17 Apr 2022 15:41 UTC

On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
> On 4/17/2022 10:11 AM, Dennis Bush wrote:
> > On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
> >> On 4/16/2022 11:14 PM, Dennis Bush wrote:
> >>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
> >>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
> >>>> correct.
> >>>
> >>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
> >>>
> >> I count this as excellent progress.
> >
> > This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
> None-the-less I count this as significant progress. It may have been
> nearly the first time that anyone has ever agreed with anything that I
> have ever said in this forum.
> >
> >>>>
> >>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
> >>>> is correct?
> >>>>
> >>>> You must find an example where {an X is a Y} and Z is incorrect for
> >>>> reporting {an X is a Y}.
> >>>>
> >>>>
> >>>> If no one can show that the correctly simulated input to H(P,P) reaches
> >>>> its final state and halts that proves that it is non-halting.
> >>>>
> >>>> If I was wrong then the correct simulation of the 27 bytes of machine
> >>>> code at machine address [000009d6] by H would show some correct
> >>>> execution trace from machine address [000009d6] ending at machine
> >>>> address [000009f0].
> >>>>
> >>>> _P()
> >>>> [000009d6](01) 55 push ebp
> >>>> [000009d7](02) 8bec mov ebp,esp
> >>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >>>> [000009dc](01) 50 push eax // push P
> >>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >>>> [000009e0](01) 51 push ecx // push P
> >>>> [000009e1](05) e840feffff call 00000826 // call H
> >>>> [000009e6](03) 83c408 add esp,+08
> >>>> [000009e9](02) 85c0 test eax,eax
> >>>> [000009eb](02) 7402 jz 000009ef
> >>>> [000009ed](02) ebfe jmp 000009ed
> >>>> [000009ef](01) 5d pop ebp
> >>>> [000009f0](01) c3 ret // Final state
> >>>> Size in bytes:(0027) [000009f0]
> >>>
> >>> Since an H that never aborts is an uninteresting case,
> >> We simply hypothesize BOTH cases:
> >> (a) aborted and
> >> (b) never aborted
> >> and then see where their execution traces would be.
> >
> > Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
> We really only need to answer this single question:
> Is there any case where the correctly simulated input to H(P,P) reaches
> its own machine address of [000009f0]?
>
> H either aborts the simulation of its input or does not abort the
> simulation of its input. In either case the execution trace never
> extends beyond machine address [000009e1].
>
> Thus we can see that whether or not H aborts the simulation of its input
> this simulated input never reaches machine address [000009f0].

If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.

So we'll rename H to Ha and P to Pa to make it clear you're only talking about ONE SPECIFIC H and the SPECIFIC P that calls it.

Your claim is that Ha(Pa,Pa) == false is correct. To test that result, we can construct Hb which simulates for some k additional steps then Ha. This will result in Hb(Pa,Pa) simulating its input to a final state and returning true. This is necessarily a correct decision because Hb never leaves UTM mode while simulating its input to a final state.

That makes Hb(Pa,Pa) == true necessarily correct. And because Hb and Ha are both simulating halt deciders being given the same input but getting different results, that makes Ha(Pa,Pa) == false incorrect.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: olcott - Sun, 17 Apr 2022 15:49 UTC

On 4/17/2022 10:41 AM, Dennis Bush wrote:
> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>>>>>> correct.
>>>>>
>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
>>>>>
>>>> I count this as excellent progress.
>>>
>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
>> None-the-less I count this as significant progress. It may have been
>> nearly the first time that anyone has ever agreed with anything that I
>> have ever said in this forum.
>>>
>>>>>>
>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>>>>>> is correct?
>>>>>>
>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
>>>>>> reporting {an X is a Y}.
>>>>>>
>>>>>>
>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
>>>>>> its final state and halts that proves that it is non-halting.
>>>>>>
>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
>>>>>> code at machine address [000009d6] by H would show some correct
>>>>>> execution trace from machine address [000009d6] ending at machine
>>>>>> address [000009f0].
>>>>>>
>>>>>> _P()
>>>>>> [000009d6](01) 55 push ebp
>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>> [000009dc](01) 50 push eax // push P
>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>> [000009e1](05) e840feffff call // call H
>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>> [000009ef](01) 5d pop ebp
>>>>>> [000009f0](01) c3 ret // Final state
>>>>>> Size in bytes:(0027) [000009f0]
>>>>>
>>>>> Since an H that never aborts is an uninteresting case,
>>>> We simply hypothesize BOTH cases:
>>>> (a) aborted and
>>>> (b) never aborted
>>>> and then see where their execution traces would be.
>>>
>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
>> We really only need to answer this single question:
>> Is there any case where the correctly simulated input to H(P,P) reaches
>> its own machine address of [000009f0]?
>>
>> H either aborts the simulation of its input or does not abort the
>> simulation of its input. In either case the execution trace never
>> extends beyond machine address [000009e1].
>>
>> Thus we can see that whether or not H aborts the simulation of its input
>> this simulated input never reaches machine address [000009f0].
>
> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.

No stupid I am not. I am talking about the above fixed P, and two
different hypothetical scenarios of H at machine address [00000826].
(1) H aborts the simulation of its input.
(2) H does not abort the simulation of its input.

In neither one of these hypothetical scenarios does the simulated P ever
reach its own machine address [000009f0].

--
Copyright 2022 Pete Olcott

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

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Sun, 17 Apr 2022 15:59 UTC

On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
> On 4/17/2022 10:41 AM, Dennis Bush wrote:
> > On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
> >> On 4/17/2022 10:11 AM, Dennis Bush wrote:
> >>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
> >>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
> >>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
> >>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
> >>>>>> correct.
> >>>>>
> >>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
> >>>>>
> >>>> I count this as excellent progress.
> >>>
> >>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
> >> None-the-less I count this as significant progress. It may have been
> >> nearly the first time that anyone has ever agreed with anything that I
> >> have ever said in this forum.
> >>>
> >>>>>>
> >>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
> >>>>>> is correct?
> >>>>>>
> >>>>>> You must find an example where {an X is a Y} and Z is incorrect for
> >>>>>> reporting {an X is a Y}.
> >>>>>>
> >>>>>>
> >>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
> >>>>>> its final state and halts that proves that it is non-halting.
> >>>>>>
> >>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
> >>>>>> code at machine address [000009d6] by H would show some correct
> >>>>>> execution trace from machine address [000009d6] ending at machine
> >>>>>> address [000009f0].
> >>>>>>
> >>>>>> _P()
> >>>>>> [000009d6](01) 55 push ebp
> >>>>>> [000009d7](02) 8bec mov ebp,esp
> >>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >>>>>> [000009dc](01) 50 push eax // push P
> >>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>> [000009e0](01) 51 push ecx // push P
> >>>>>> [000009e1](05) e840feffff call // call H
> >>>>>> [000009e6](03) 83c408 add esp,+08
> >>>>>> [000009e9](02) 85c0 test eax,eax
> >>>>>> [000009eb](02) 7402 jz 000009ef
> >>>>>> [000009ed](02) ebfe jmp 000009ed
> >>>>>> [000009ef](01) 5d pop ebp
> >>>>>> [000009f0](01) c3 ret // Final state
> >>>>>> Size in bytes:(0027) [000009f0]
> >>>>>
> >>>>> Since an H that never aborts is an uninteresting case,
> >>>> We simply hypothesize BOTH cases:
> >>>> (a) aborted and
> >>>> (b) never aborted
> >>>> and then see where their execution traces would be.
> >>>
> >>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
> >> We really only need to answer this single question:
> >> Is there any case where the correctly simulated input to H(P,P) reaches
> >> its own machine address of [000009f0]?
> >>
> >> H either aborts the simulation of its input or does not abort the
> >> simulation of its input. In either case the execution trace never
> >> extends beyond machine address [000009e1].
> >>
> >> Thus we can see that whether or not H aborts the simulation of its input
> >> this simulated input never reaches machine address [000009f0].
> >
> > If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
> No stupid I am not. I am talking about the above fixed P, and two
> different hypothetical scenarios of H at machine address [00000826].

So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.

You have two *different* H's (Hn and Ha) and therefore two *different P's (Pn and Pa). Whether or not Pn halts has *no* bearing on whether or not Pa halts. So to be clear about that they must be put at different addresses. This goes back to the H2 and H4 discussion we had on SE chat that you pushed away from because it showed the flaws in your logic.

Hn(Pn,Pn) never returns a value so it is not a halt decider. We can give this input to Ha as Ha(Pn,Pn) to correctly determine that this input does not halt, but that is not the case Ha needs to get right.

Your claim is that Ha(Pa,Pa) == false is correct. To test that result, we can construct Hb which simulates for some k additional steps then Ha. This will result in Hb(Pa,Pa) simulating its input to a final state and returning true. This is necessarily a correct decision because Hb never leaves UTM mode while simulating its input to a final state.

That makes Hb(Pa,Pa) == true necessarily correct. And because Hb and Ha are both simulating halt deciders being given the same input but getting different results, that makes Ha(Pa,Pa) == false incorrect.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: olcott - Sun, 17 Apr 2022 16:06 UTC

On 4/17/2022 10:59 AM, Dennis Bush wrote:
> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>>>>>>>> correct.
>>>>>>>
>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
>>>>>>>
>>>>>> I count this as excellent progress.
>>>>>
>>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
>>>> None-the-less I count this as significant progress. It may have been
>>>> nearly the first time that anyone has ever agreed with anything that I
>>>> have ever said in this forum.
>>>>>
>>>>>>>>
>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>>>>>>>> is correct?
>>>>>>>>
>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
>>>>>>>> reporting {an X is a Y}.
>>>>>>>>
>>>>>>>>
>>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
>>>>>>>> its final state and halts that proves that it is non-halting.
>>>>>>>>
>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
>>>>>>>> code at machine address [000009d6] by H would show some correct
>>>>>>>> execution trace from machine address [000009d6] ending at machine
>>>>>>>> address [000009f0].
>>>>>>>>
>>>>>>>> _P()
>>>>>>>> [000009d6](01) 55 push ebp
>>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>>> [000009e1](05) e840feffff call // call H
>>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>>
>>>>>>> Since an H that never aborts is an uninteresting case,
>>>>>> We simply hypothesize BOTH cases:
>>>>>> (a) aborted and
>>>>>> (b) never aborted
>>>>>> and then see where their execution traces would be.
>>>>>
>>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
>>>> We really only need to answer this single question:
>>>> Is there any case where the correctly simulated input to H(P,P) reaches
>>>> its own machine address of [000009f0]?
>>>>
>>>> H either aborts the simulation of its input or does not abort the
>>>> simulation of its input. In either case the execution trace never
>>>> extends beyond machine address [000009e1].
>>>>
>>>> Thus we can see that whether or not H aborts the simulation of its input
>>>> this simulated input never reaches machine address [000009f0].
>>>
>>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
>> No stupid I am not. I am talking about the above fixed P, and two
>> different hypothetical scenarios of H at machine address [00000826].
>
> So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
_P()
[000009d6](01) 55 push ebp
[000009d7](02) 8bec mov ebp,esp
[000009d9](03) 8b4508 mov eax,[ebp+08]
[000009dc](01) 50 push eax // push P
[000009dd](03) 8b4d08 mov ecx,[ebp+08]
[000009e0](01) 51 push ecx // push P
[000009e1](05) e840feffff call 00000826 // call H
[000009e6](03) 83c408 add esp,+08
[000009e9](02) 85c0 test eax,eax
[000009eb](02) 7402 jz 000009ef
[000009ed](02) ebfe jmp 000009ed
[000009ef](01) 5d pop ebp
[000009f0](01) c3 ret // Final state
Size in bytes:(0027) [000009f0]

I am considering the behavior of the simulated P under two hypothetical
scenarios:
(1) H at machine address [00000826] aborts its simulation.
(2) H at machine address [00000826] does not abort its simulation.

In neither case does the simulated P ever reach its own machine address
of [000009f0].

--
Copyright 2022 Pete Olcott

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

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Dennis Bush - Sun, 17 Apr 2022 16:12 UTC

On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
> On 4/17/2022 10:59 AM, Dennis Bush wrote:
> > On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
> >> On 4/17/2022 10:41 AM, Dennis Bush wrote:
> >>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
> >>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
> >>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
> >>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
> >>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
> >>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
> >>>>>>>> correct.
> >>>>>>>
> >>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
> >>>>>>>
> >>>>>> I count this as excellent progress.
> >>>>>
> >>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
> >>>> None-the-less I count this as significant progress. It may have been
> >>>> nearly the first time that anyone has ever agreed with anything that I
> >>>> have ever said in this forum.
> >>>>>
> >>>>>>>>
> >>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
> >>>>>>>> is correct?
> >>>>>>>>
> >>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
> >>>>>>>> reporting {an X is a Y}.
> >>>>>>>>
> >>>>>>>>
> >>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
> >>>>>>>> its final state and halts that proves that it is non-halting.
> >>>>>>>>
> >>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
> >>>>>>>> code at machine address [000009d6] by H would show some correct
> >>>>>>>> execution trace from machine address [000009d6] ending at machine
> >>>>>>>> address [000009f0].
> >>>>>>>>
> >>>>>>>> _P()
> >>>>>>>> [000009d6](01) 55 push ebp
> >>>>>>>> [000009d7](02) 8bec mov ebp,esp
> >>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >>>>>>>> [000009dc](01) 50 push eax // push P
> >>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>>>> [000009e0](01) 51 push ecx // push P
> >>>>>>>> [000009e1](05) e840feffff call // call H
> >>>>>>>> [000009e6](03) 83c408 add esp,+08
> >>>>>>>> [000009e9](02) 85c0 test eax,eax
> >>>>>>>> [000009eb](02) 7402 jz 000009ef
> >>>>>>>> [000009ed](02) ebfe jmp 000009ed
> >>>>>>>> [000009ef](01) 5d pop ebp
> >>>>>>>> [000009f0](01) c3 ret // Final state
> >>>>>>>> Size in bytes:(0027) [000009f0]
> >>>>>>>
> >>>>>>> Since an H that never aborts is an uninteresting case,
> >>>>>> We simply hypothesize BOTH cases:
> >>>>>> (a) aborted and
> >>>>>> (b) never aborted
> >>>>>> and then see where their execution traces would be.
> >>>>>
> >>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
> >>>> We really only need to answer this single question:
> >>>> Is there any case where the correctly simulated input to H(P,P) reaches
> >>>> its own machine address of [000009f0]?
> >>>>
> >>>> H either aborts the simulation of its input or does not abort the
> >>>> simulation of its input. In either case the execution trace never
> >>>> extends beyond machine address [000009e1].
> >>>>
> >>>> Thus we can see that whether or not H aborts the simulation of its input
> >>>> this simulated input never reaches machine address [000009f0].
> >>>
> >>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
> >> No stupid I am not. I am talking about the above fixed P, and two
> >> different hypothetical scenarios of H at machine address [00000826].
> >
> > So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
> _P()
> [000009d6](01) 55 push ebp
> [000009d7](02) 8bec mov ebp,esp
> [000009d9](03) 8b4508 mov eax,[ebp+08]
> [000009dc](01) 50 push eax // push P
> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> [000009e0](01) 51 push ecx // push P
> [000009e1](05) e840feffff call 00000826 // call H
> [000009e6](03) 83c408 add esp,+08
> [000009e9](02) 85c0 test eax,eax
> [000009eb](02) 7402 jz 000009ef
> [000009ed](02) ebfe jmp 000009ed
> [000009ef](01) 5d pop ebp
> [000009f0](01) c3 ret // Final state
> Size in bytes:(0027) [000009f0]
> I am considering the behavior of the simulated P under two hypothetical
> scenarios:
> (1) H at machine address [00000826] aborts its simulation.
> (2) H at machine address [00000826] does not abort its simulation.

So in other words you're considering the behavior of two separate unrelated computations that deceptively happen to share the same set of function names and happen to live at the same set of addresses when built separately.

No matter how you look at it, you're trying to equivocate Pn and Pa. They are not the same. Pn(Pn) does not halt. Pa(Pa) halts. The simulation Hb(Pa,Pa) halts. Therefore Ha(Pa,Pa) == false is incorrect.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: olcott - Sun, 17 Apr 2022 16:23 UTC

On 4/17/2022 11:12 AM, Dennis Bush wrote:
> On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
>> On 4/17/2022 10:59 AM, Dennis Bush wrote:
>>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
>>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
>>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
>>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>>>>>>>>>> correct.
>>>>>>>>>
>>>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
>>>>>>>>>
>>>>>>>> I count this as excellent progress.
>>>>>>>
>>>>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
>>>>>> None-the-less I count this as significant progress. It may have been
>>>>>> nearly the first time that anyone has ever agreed with anything that I
>>>>>> have ever said in this forum.
>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>>>>>>>>>> is correct?
>>>>>>>>>>
>>>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
>>>>>>>>>> reporting {an X is a Y}.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
>>>>>>>>>> its final state and halts that proves that it is non-halting.
>>>>>>>>>>
>>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
>>>>>>>>>> code at machine address [000009d6] by H would show some correct
>>>>>>>>>> execution trace from machine address [000009d6] ending at machine
>>>>>>>>>> address [000009f0].
>>>>>>>>>>
>>>>>>>>>> _P()
>>>>>>>>>> [000009d6](01) 55 push ebp
>>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>>>>> [000009e1](05) e840feffff call // call H
>>>>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>>>>
>>>>>>>>> Since an H that never aborts is an uninteresting case,
>>>>>>>> We simply hypothesize BOTH cases:
>>>>>>>> (a) aborted and
>>>>>>>> (b) never aborted
>>>>>>>> and then see where their execution traces would be.
>>>>>>>
>>>>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
>>>>>> We really only need to answer this single question:
>>>>>> Is there any case where the correctly simulated input to H(P,P) reaches
>>>>>> its own machine address of [000009f0]?
>>>>>>
>>>>>> H either aborts the simulation of its input or does not abort the
>>>>>> simulation of its input. In either case the execution trace never
>>>>>> extends beyond machine address [000009e1].
>>>>>>
>>>>>> Thus we can see that whether or not H aborts the simulation of its input
>>>>>> this simulated input never reaches machine address [000009f0].
>>>>>
>>>>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
>>>> No stupid I am not. I am talking about the above fixed P, and two
>>>> different hypothetical scenarios of H at machine address [00000826].
>>>
>>> So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
>> _P()
>> [000009d6](01) 55 push ebp
>> [000009d7](02) 8bec mov ebp,esp
>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>> [000009dc](01) 50 push eax // push P
>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>> [000009e0](01) 51 push ecx // push P
>> [000009e1](05) e840feffff call 00000826 // call H
>> [000009e6](03) 83c408 add esp,+08
>> [000009e9](02) 85c0 test eax,eax
>> [000009eb](02) 7402 jz 000009ef
>> [000009ed](02) ebfe jmp 000009ed
>> [000009ef](01) 5d pop ebp
>> [000009f0](01) c3 ret // Final state
>> Size in bytes:(0027) [000009f0]
>> I am considering the behavior of the simulated P under two hypothetical
>> scenarios:
>> (1) H at machine address [00000826] aborts its simulation.
>> (2) H at machine address [00000826] does not abort its simulation.
>
> So in other words you're considering the behavior of two separate unrelated computations that deceptively happen to share the same set of function names and happen to live at the same set of addresses when built separately.
>

That would be the way a God damned liar would say it.

I am only considering the actual behavior of the simulated input to
H(P,P) under the two possible scenarios. In each of these two possible
scenarios P never reaches its own 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

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Richard Damon - Sun, 17 Apr 2022 16:52 UTC

On 4/17/22 11:25 AM, olcott wrote:
> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>>>>> correct.
>>>>
>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is
>>>> correct. But first we must establish whether the input to H(P,P) is
>>>> non-halting.
>>>>
>>> I count this as excellent progress.
>>
>> This has never been in dispute, by me or anyone else.  The fact that
>> you think this shows how poor your reading comprehension is.
>
> None-the-less I count this as significant progress. It may have been
> nearly the first time that anyone has ever agreed with anything that I
> have ever said in this forum.
>
>>
>>>>>
>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>>>>> is correct?
>>>>>
>>>>> You must find an example where {an X is a Y} and Z is incorrect for
>>>>> reporting {an X is a Y}.
>>>>>
>>>>>
>>>>> If no one can show that the correctly simulated input to H(P,P)
>>>>> reaches
>>>>> its final state and halts that proves that it is non-halting.
>>>>>
>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
>>>>> code at machine address [000009d6] by H would show some correct
>>>>> execution trace from machine address [000009d6] ending at machine
>>>>> address [000009f0].
>>>>>
>>>>> _P()
>>>>> [000009d6](01) 55 push ebp
>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>> [000009dc](01) 50 push eax // push P
>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>> [000009e0](01) 51 push ecx // push P
>>>>> [000009e1](05) e840feffff call 00000826 // call H
>>>>> [000009e6](03) 83c408 add esp,+08
>>>>> [000009e9](02) 85c0 test eax,eax
>>>>> [000009eb](02) 7402 jz 000009ef
>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>> [000009ef](01) 5d pop ebp
>>>>> [000009f0](01) c3 ret // Final state
>>>>> Size in bytes:(0027) [000009f0]
>>>>
>>>> Since an H that never aborts is an uninteresting case,
>>> We simply hypothesize BOTH cases:
>>> (a) aborted and
>>> (b) never aborted
>>> and then see where their execution traces would be.
>>
>> Those two cases means you're changing the code of H, which means that
>> P becomes a completely different computation.
>
> We really only need to answer this single question:
> Is there any case where the correctly simulated input to H(P,P) reaches
> its own machine address of [000009f0]?

And the answer is yes, for ANY H that aborts it simulation of H(P,P),
and the P built from that H, then the CORRECT simulation of that input
will Halt.

>
> H either aborts the simulation of its input or does not abort the
> simulation of its input. In either case the execution trace never
> extends beyond machine address [000009e1].

Except that it isn't the simulation of the input by H that matters, only
the CORRECT (i.e. non-aborted) simulation that matters.

>
> Thus we can see that whether or not H aborts the simulation of its input
> this simulated input never reaches machine address [000009f0].
>

Nope. All you have shown is that H can never simulate its input that far.

FAIL.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Richard Damon - Sun, 17 Apr 2022 16:54 UTC

On 4/17/22 11:49 AM, olcott wrote:
> On 4/17/2022 10:41 AM, Dennis Bush wrote:
>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is
>>>>>>> necessary
>>>>>>> correct.
>>>>>>
>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false
>>>>>> is correct. But first we must establish whether the input to
>>>>>> H(P,P) is non-halting.
>>>>>>
>>>>> I count this as excellent progress.
>>>>
>>>> This has never been in dispute, by me or anyone else. The fact that
>>>> you think this shows how poor your reading comprehension is.
>>> None-the-less I count this as significant progress. It may have been
>>> nearly the first time that anyone has ever agreed with anything that I
>>> have ever said in this forum.
>>>>
>>>>>>>
>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge
>>>>>>> that it
>>>>>>> is correct?
>>>>>>>
>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
>>>>>>> reporting {an X is a Y}.
>>>>>>>
>>>>>>>
>>>>>>> If no one can show that the correctly simulated input to H(P,P)
>>>>>>> reaches
>>>>>>> its final state and halts that proves that it is non-halting.
>>>>>>>
>>>>>>> If I was wrong then the correct simulation of the 27 bytes of
>>>>>>> machine
>>>>>>> code at machine address [000009d6] by H would show some correct
>>>>>>> execution trace from machine address [000009d6] ending at machine
>>>>>>> address [000009f0].
>>>>>>>
>>>>>>> _P()
>>>>>>> [000009d6](01) 55 push ebp
>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>> [000009e1](05) e840feffff call  // call H
>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>
>>>>>> Since an H that never aborts is an uninteresting case,
>>>>> We simply hypothesize BOTH cases:
>>>>> (a) aborted and
>>>>> (b) never aborted
>>>>> and then see where their execution traces would be.
>>>>
>>>> Those two cases means you're changing the code of H, which means
>>>> that P becomes a completely different computation.
>>> We really only need to answer this single question:
>>> Is there any case where the correctly simulated input to H(P,P) reaches
>>> its own machine address of [000009f0]?
>>>
>>> H either aborts the simulation of its input or does not abort the
>>> simulation of its input. In either case the execution trace never
>>> extends beyond machine address [000009e1].
>>>
>>> Thus we can see that whether or not H aborts the simulation of its input
>>> this simulated input never reaches machine address [000009f0].
>>
>> If you're talking about a fixed P at a fixed address, that means you
>> are also talking about ONE SPECIFIC H at a specific address with a
>> fixed, unchanging algorithm.
>
> No stupid I am not. I am talking about the above fixed P, and two
> different hypothetical scenarios of H at machine address [00000826].
> (1) H aborts the simulation of its input.
> (2) H does not abort the simulation of its input.

But you P is wrong then, as P is suppoed to contain a copy of H in it,
not use some 'external' H. This just shows you don't understand what a
computation is.

>
> In neither one of these hypothetical scenarios does the simulated P ever
> reach its own machine address [000009f0].
>

And they are all the wrong case and using the wrong simulation.

FAIL, LIAR.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Richard Damon - Sun, 17 Apr 2022 16:56 UTC

On 4/17/22 12:06 PM, olcott wrote:
> On 4/17/2022 10:59 AM, Dennis Bush wrote:
>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is
>>>>>>>>> necessary
>>>>>>>>> correct.
>>>>>>>>
>>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false
>>>>>>>> is correct. But first we must establish whether the input to
>>>>>>>> H(P,P) is non-halting.
>>>>>>>>
>>>>>>> I count this as excellent progress.
>>>>>>
>>>>>> This has never been in dispute, by me or anyone else. The fact
>>>>>> that you think this shows how poor your reading comprehension is.
>>>>> None-the-less I count this as significant progress. It may have been
>>>>> nearly the first time that anyone has ever agreed with anything that I
>>>>> have ever said in this forum.
>>>>>>
>>>>>>>>>
>>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge
>>>>>>>>> that it
>>>>>>>>> is correct?
>>>>>>>>>
>>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect
>>>>>>>>> for
>>>>>>>>> reporting {an X is a Y}.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> If no one can show that the correctly simulated input to H(P,P)
>>>>>>>>> reaches
>>>>>>>>> its final state and halts that proves that it is non-halting.
>>>>>>>>>
>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of
>>>>>>>>> machine
>>>>>>>>> code at machine address [000009d6] by H would show some correct
>>>>>>>>> execution trace from machine address [000009d6] ending at machine
>>>>>>>>> address [000009f0].
>>>>>>>>>
>>>>>>>>> _P()
>>>>>>>>> [000009d6](01) 55 push ebp
>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>>>> [000009e1](05) e840feffff call // call H
>>>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>>>
>>>>>>>> Since an H that never aborts is an uninteresting case,
>>>>>>> We simply hypothesize BOTH cases:
>>>>>>> (a) aborted and
>>>>>>> (b) never aborted
>>>>>>> and then see where their execution traces would be.
>>>>>>
>>>>>> Those two cases means you're changing the code of H, which means
>>>>>> that P becomes a completely different computation.
>>>>> We really only need to answer this single question:
>>>>> Is there any case where the correctly simulated input to H(P,P)
>>>>> reaches
>>>>> its own machine address of [000009f0]?
>>>>>
>>>>> H either aborts the simulation of its input or does not abort the
>>>>> simulation of its input. In either case the execution trace never
>>>>> extends beyond machine address [000009e1].
>>>>>
>>>>> Thus we can see that whether or not H aborts the simulation of its
>>>>> input
>>>>> this simulated input never reaches machine address [000009f0].
>>>>
>>>> If you're talking about a fixed P at a fixed address, that means you
>>>> are also talking about ONE SPECIFIC H at a specific address with a
>>>> fixed, unchanging algorithm.
>>> No stupid I am not. I am talking about the above fixed P, and two
>>> different hypothetical scenarios of H at machine address [00000826].
>>
>> So you're dishonestly changing the algorithm of H to make two
>> different P's look the same.  They are not.
> _P()
> [000009d6](01) 55         push ebp
> [000009d7](02) 8bec       mov ebp,esp
> [000009d9](03) 8b4508     mov eax,[ebp+08]
> [000009dc](01) 50         push eax         // push P
> [000009dd](03) 8b4d08     mov ecx,[ebp+08]
> [000009e0](01) 51         push ecx         // push P
> [000009e1](05) e840feffff call 00000826    // call H
> [000009e6](03) 83c408     add esp,+08
> [000009e9](02) 85c0       test eax,eax
> [000009eb](02) 7402       jz 000009ef
> [000009ed](02) ebfe       jmp 000009ed
> [000009ef](01) 5d         pop ebp
> [000009f0](01) c3         ret              // Final state
> Size in bytes:(0027) [000009f0]
>
> I am considering the behavior of the simulated P under two hypothetical
> scenarios:
> (1) H at machine address [00000826] aborts its simulation.
> (2) H at machine address [00000826] does not abort its simulation.
>
> In neither case does the simulated P ever reach its own machine address
> of [000009f0].
>

So NOT The H^ of Linz, which contains a copy of H.

And thus your H also fails to be the requried one of Linz, since it
can't be given an arbitrary machine to decide on.

TOTAL FAILURE.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Richard Damon - Sun, 17 Apr 2022 16:56 UTC

On 4/17/22 12:23 PM, olcott wrote:
> On 4/17/2022 11:12 AM, Dennis Bush wrote:
>> On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
>>> On 4/17/2022 10:59 AM, Dennis Bush wrote:
>>>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
>>>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
>>>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
>>>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>>>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is
>>>>>>>>>>> necessary
>>>>>>>>>>> correct.
>>>>>>>>>>
>>>>>>>>>> True. *if* the input to H(P,P) is non-halting then
>>>>>>>>>> H(P,P)==false is correct. But first we must establish whether
>>>>>>>>>> the input to H(P,P) is non-halting.
>>>>>>>>>>
>>>>>>>>> I count this as excellent progress.
>>>>>>>>
>>>>>>>> This has never been in dispute, by me or anyone else. The fact
>>>>>>>> that you think this shows how poor your reading comprehension is.
>>>>>>> None-the-less I count this as significant progress. It may have been
>>>>>>> nearly the first time that anyone has ever agreed with anything
>>>>>>> that I
>>>>>>> have ever said in this forum.
>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Does anyone have a rebuttal to that one, or do you
>>>>>>>>>>> acknowledge that it
>>>>>>>>>>> is correct?
>>>>>>>>>>>
>>>>>>>>>>> You must find an example where {an X is a Y} and Z is
>>>>>>>>>>> incorrect for
>>>>>>>>>>> reporting {an X is a Y}.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> If no one can show that the correctly simulated input to
>>>>>>>>>>> H(P,P) reaches
>>>>>>>>>>> its final state and halts that proves that it is non-halting.
>>>>>>>>>>>
>>>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of
>>>>>>>>>>> machine
>>>>>>>>>>> code at machine address [000009d6] by H would show some correct
>>>>>>>>>>> execution trace from machine address [000009d6] ending at
>>>>>>>>>>> machine
>>>>>>>>>>> address [000009f0].
>>>>>>>>>>>
>>>>>>>>>>> _P()
>>>>>>>>>>> [000009d6](01) 55 push ebp
>>>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>>>>>> [000009e1](05) e840feffff call // call H
>>>>>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>>>>>
>>>>>>>>>> Since an H that never aborts is an uninteresting case,
>>>>>>>>> We simply hypothesize BOTH cases:
>>>>>>>>> (a) aborted and
>>>>>>>>> (b) never aborted
>>>>>>>>> and then see where their execution traces would be.
>>>>>>>>
>>>>>>>> Those two cases means you're changing the code of H, which means
>>>>>>>> that P becomes a completely different computation.
>>>>>>> We really only need to answer this single question:
>>>>>>> Is there any case where the correctly simulated input to H(P,P)
>>>>>>> reaches
>>>>>>> its own machine address of [000009f0]?
>>>>>>>
>>>>>>> H either aborts the simulation of its input or does not abort the
>>>>>>> simulation of its input. In either case the execution trace never
>>>>>>> extends beyond machine address [000009e1].
>>>>>>>
>>>>>>> Thus we can see that whether or not H aborts the simulation of
>>>>>>> its input
>>>>>>> this simulated input never reaches machine address [000009f0].
>>>>>>
>>>>>> If you're talking about a fixed P at a fixed address, that means
>>>>>> you are also talking about ONE SPECIFIC H at a specific address
>>>>>> with a fixed, unchanging algorithm.
>>>>> No stupid I am not. I am talking about the above fixed P, and two
>>>>> different hypothetical scenarios of H at machine address [00000826].
>>>>
>>>> So you're dishonestly changing the algorithm of H to make two
>>>> different P's look the same. They are not.
>>> _P()
>>> [000009d6](01) 55 push ebp
>>> [000009d7](02) 8bec mov ebp,esp
>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>> [000009dc](01) 50 push eax // push P
>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>> [000009e0](01) 51 push ecx // push P
>>> [000009e1](05) e840feffff call 00000826 // call H
>>> [000009e6](03) 83c408 add esp,+08
>>> [000009e9](02) 85c0 test eax,eax
>>> [000009eb](02) 7402 jz 000009ef
>>> [000009ed](02) ebfe jmp 000009ed
>>> [000009ef](01) 5d pop ebp
>>> [000009f0](01) c3 ret // Final state
>>> Size in bytes:(0027) [000009f0]
>>> I am considering the behavior of the simulated P under two hypothetical
>>> scenarios:
>>> (1) H at machine address [00000826] aborts its simulation.
>>> (2) H at machine address [00000826] does not abort its simulation.
>>
>> So in other words you're considering the behavior of two separate
>> unrelated computations that deceptively happen to share the same set
>> of function names and happen to live at the same set of addresses when
>> built separately.
>>
>
> That would be the way a God damned liar would say it.
>
> I am only considering the actual behavior of the simulated input to
> H(P,P) under the two possible scenarios. In each of these two possible
> scenarios P never reaches its own final state.
>
>

No, you are not, you are just lying about what you are doing and proving
your incompetence.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Dennis Bush - Sun, 17 Apr 2022 17:35 UTC

On Sunday, April 17, 2022 at 12:23:16 PM UTC-4, olcott wrote:
> On 4/17/2022 11:12 AM, Dennis Bush wrote:
> > On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
> >> On 4/17/2022 10:59 AM, Dennis Bush wrote:
> >>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
> >>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
> >>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
> >>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
> >>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
> >>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
> >>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
> >>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
> >>>>>>>>>> correct.
> >>>>>>>>>
> >>>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
> >>>>>>>>>
> >>>>>>>> I count this as excellent progress.
> >>>>>>>
> >>>>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
> >>>>>> None-the-less I count this as significant progress. It may have been
> >>>>>> nearly the first time that anyone has ever agreed with anything that I
> >>>>>> have ever said in this forum.
> >>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
> >>>>>>>>>> is correct?
> >>>>>>>>>>
> >>>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
> >>>>>>>>>> reporting {an X is a Y}.
> >>>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
> >>>>>>>>>> its final state and halts that proves that it is non-halting.
> >>>>>>>>>>
> >>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
> >>>>>>>>>> code at machine address [000009d6] by H would show some correct
> >>>>>>>>>> execution trace from machine address [000009d6] ending at machine
> >>>>>>>>>> address [000009f0].
> >>>>>>>>>>
> >>>>>>>>>> _P()
> >>>>>>>>>> [000009d6](01) 55 push ebp
> >>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
> >>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >>>>>>>>>> [000009dc](01) 50 push eax // push P
> >>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>>>>>> [000009e0](01) 51 push ecx // push P
> >>>>>>>>>> [000009e1](05) e840feffff call // call H
> >>>>>>>>>> [000009e6](03) 83c408 add esp,+08
> >>>>>>>>>> [000009e9](02) 85c0 test eax,eax
> >>>>>>>>>> [000009eb](02) 7402 jz 000009ef
> >>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
> >>>>>>>>>> [000009ef](01) 5d pop ebp
> >>>>>>>>>> [000009f0](01) c3 ret // Final state
> >>>>>>>>>> Size in bytes:(0027) [000009f0]
> >>>>>>>>>
> >>>>>>>>> Since an H that never aborts is an uninteresting case,
> >>>>>>>> We simply hypothesize BOTH cases:
> >>>>>>>> (a) aborted and
> >>>>>>>> (b) never aborted
> >>>>>>>> and then see where their execution traces would be.
> >>>>>>>
> >>>>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
> >>>>>> We really only need to answer this single question:
> >>>>>> Is there any case where the correctly simulated input to H(P,P) reaches
> >>>>>> its own machine address of [000009f0]?
> >>>>>>
> >>>>>> H either aborts the simulation of its input or does not abort the
> >>>>>> simulation of its input. In either case the execution trace never
> >>>>>> extends beyond machine address [000009e1].
> >>>>>>
> >>>>>> Thus we can see that whether or not H aborts the simulation of its input
> >>>>>> this simulated input never reaches machine address [000009f0].
> >>>>>
> >>>>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
> >>>> No stupid I am not. I am talking about the above fixed P, and two
> >>>> different hypothetical scenarios of H at machine address [00000826].
> >>>
> >>> So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
> >> _P()
> >> [000009d6](01) 55 push ebp
> >> [000009d7](02) 8bec mov ebp,esp
> >> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >> [000009dc](01) 50 push eax // push P
> >> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >> [000009e0](01) 51 push ecx // push P
> >> [000009e1](05) e840feffff call 00000826 // call H
> >> [000009e6](03) 83c408 add esp,+08
> >> [000009e9](02) 85c0 test eax,eax
> >> [000009eb](02) 7402 jz 000009ef
> >> [000009ed](02) ebfe jmp 000009ed
> >> [000009ef](01) 5d pop ebp
> >> [000009f0](01) c3 ret // Final state
> >> Size in bytes:(0027) [000009f0]
> >> I am considering the behavior of the simulated P under two hypothetical
> >> scenarios:
> >> (1) H at machine address [00000826] aborts its simulation.
> >> (2) H at machine address [00000826] does not abort its simulation.
> >
> > So in other words you're considering the behavior of two separate unrelated computations that deceptively happen to share the same set of function names and happen to live at the same set of addresses when built separately.
> >
> That would be the way a God damned liar would say it.
>
> I am only considering the actual behavior of the simulated input to
> H(P,P) under the two possible scenarios. In each of these two possible
> scenarios P never reaches its own final state.

In other words, you're considering whether Hn can simulate Pn to a final state, and *independently* whether Ha can simulate Pa to a final state.

Pn and Pa are not the same. Pn(Pn) does not halt. Pa(Pa) halts. Hn(Pn,Pn) never stops running so it can't answer. The simulation Ha(Pa,Pa) is not simulated to a final state but the simulation Hb(Pa,Pa) *is* simulated to a final state.

Therefore Ha(Pa,Pa) == false is incorrect.

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: olcott - Sun, 17 Apr 2022 18:07 UTC

On 4/17/2022 12:35 PM, Dennis Bush wrote:
> On Sunday, April 17, 2022 at 12:23:16 PM UTC-4, olcott wrote:
>> On 4/17/2022 11:12 AM, Dennis Bush wrote:
>>> On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
>>>> On 4/17/2022 10:59 AM, Dennis Bush wrote:
>>>>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
>>>>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
>>>>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
>>>>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>>>>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>>>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>>>>>>>>>>>> correct.
>>>>>>>>>>>
>>>>>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
>>>>>>>>>>>
>>>>>>>>>> I count this as excellent progress.
>>>>>>>>>
>>>>>>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
>>>>>>>> None-the-less I count this as significant progress. It may have been
>>>>>>>> nearly the first time that anyone has ever agreed with anything that I
>>>>>>>> have ever said in this forum.
>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>>>>>>>>>>>> is correct?
>>>>>>>>>>>>
>>>>>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
>>>>>>>>>>>> reporting {an X is a Y}.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
>>>>>>>>>>>> its final state and halts that proves that it is non-halting.
>>>>>>>>>>>>
>>>>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
>>>>>>>>>>>> code at machine address [000009d6] by H would show some correct
>>>>>>>>>>>> execution trace from machine address [000009d6] ending at machine
>>>>>>>>>>>> address [000009f0].
>>>>>>>>>>>>
>>>>>>>>>>>> _P()
>>>>>>>>>>>> [000009d6](01) 55 push ebp
>>>>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>>>>>>> [000009e1](05) e840feffff call // call H
>>>>>>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>>>>>>
>>>>>>>>>>> Since an H that never aborts is an uninteresting case,
>>>>>>>>>> We simply hypothesize BOTH cases:
>>>>>>>>>> (a) aborted and
>>>>>>>>>> (b) never aborted
>>>>>>>>>> and then see where their execution traces would be.
>>>>>>>>>
>>>>>>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
>>>>>>>> We really only need to answer this single question:
>>>>>>>> Is there any case where the correctly simulated input to H(P,P) reaches
>>>>>>>> its own machine address of [000009f0]?
>>>>>>>>
>>>>>>>> H either aborts the simulation of its input or does not abort the
>>>>>>>> simulation of its input. In either case the execution trace never
>>>>>>>> extends beyond machine address [000009e1].
>>>>>>>>
>>>>>>>> Thus we can see that whether or not H aborts the simulation of its input
>>>>>>>> this simulated input never reaches machine address [000009f0].
>>>>>>>
>>>>>>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
>>>>>> No stupid I am not. I am talking about the above fixed P, and two
>>>>>> different hypothetical scenarios of H at machine address [00000826].
>>>>>
>>>>> So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
>>>> _P()
>>>> [000009d6](01) 55 push ebp
>>>> [000009d7](02) 8bec mov ebp,esp
>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>> [000009dc](01) 50 push eax // push P
>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>> [000009e0](01) 51 push ecx // push P
>>>> [000009e1](05) e840feffff call 00000826 // call H
>>>> [000009e6](03) 83c408 add esp,+08
>>>> [000009e9](02) 85c0 test eax,eax
>>>> [000009eb](02) 7402 jz 000009ef
>>>> [000009ed](02) ebfe jmp 000009ed
>>>> [000009ef](01) 5d pop ebp
>>>> [000009f0](01) c3 ret // Final state
>>>> Size in bytes:(0027) [000009f0]
>>>> I am considering the behavior of the simulated P under two hypothetical
>>>> scenarios:
>>>> (1) H at machine address [00000826] aborts its simulation.
>>>> (2) H at machine address [00000826] does not abort its simulation.
>>>
>>> So in other words you're considering the behavior of two separate unrelated computations that deceptively happen to share the same set of function names and happen to live at the same set of addresses when built separately.
>>>
>> That would be the way a God damned liar would say it.
>>
>> I am only considering the actual behavior of the simulated input to
>> H(P,P) under the two possible scenarios. In each of these two possible
>> scenarios P never reaches its own final state.
>
> In other words, you're considering whether Hn can simulate Pn to a final state, and *independently* whether Ha can simulate Pa to a final state.
>
> Pn and Pa are not the same. Pn(Pn) does not halt. Pa(Pa) halts. Hn(Pn,Pn) never stops running so it can't answer. The simulation Ha(Pa,Pa) is not simulated to a final state but the simulation Hb(Pa,Pa) *is* simulated to a final state.
>
> Therefore Ha(Pa,Pa) == false is incorrect.

I have conclusively proved that the correct simulation of the input to
H(P,P) would never reach its own final state whether or not H aborts the
simulation of its input.

Honest people knowing the x86 language will see right through your
bullshit dishonest double-talk.

--
Copyright 2022 Pete Olcott

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

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: Dennis Bush - Sun, 17 Apr 2022 18:26 UTC

On Sunday, April 17, 2022 at 2:07:15 PM UTC-4, olcott wrote:
> On 4/17/2022 12:35 PM, Dennis Bush wrote:
> > On Sunday, April 17, 2022 at 12:23:16 PM UTC-4, olcott wrote:
> >> On 4/17/2022 11:12 AM, Dennis Bush wrote:
> >>> On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
> >>>> On 4/17/2022 10:59 AM, Dennis Bush wrote:
> >>>>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
> >>>>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
> >>>>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
> >>>>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
> >>>>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
> >>>>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
> >>>>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
> >>>>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
> >>>>>>>>>>>> correct.
> >>>>>>>>>>>
> >>>>>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
> >>>>>>>>>>>
> >>>>>>>>>> I count this as excellent progress.
> >>>>>>>>>
> >>>>>>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
> >>>>>>>> None-the-less I count this as significant progress. It may have been
> >>>>>>>> nearly the first time that anyone has ever agreed with anything that I
> >>>>>>>> have ever said in this forum.
> >>>>>>>>>
> >>>>>>>>>>>>
> >>>>>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
> >>>>>>>>>>>> is correct?
> >>>>>>>>>>>>
> >>>>>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
> >>>>>>>>>>>> reporting {an X is a Y}.
> >>>>>>>>>>>>
> >>>>>>>>>>>>
> >>>>>>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
> >>>>>>>>>>>> its final state and halts that proves that it is non-halting.
> >>>>>>>>>>>>
> >>>>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
> >>>>>>>>>>>> code at machine address [000009d6] by H would show some correct
> >>>>>>>>>>>> execution trace from machine address [000009d6] ending at machine
> >>>>>>>>>>>> address [000009f0].
> >>>>>>>>>>>>
> >>>>>>>>>>>> _P()
> >>>>>>>>>>>> [000009d6](01) 55 push ebp
> >>>>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
> >>>>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >>>>>>>>>>>> [000009dc](01) 50 push eax // push P
> >>>>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>>>>>>>> [000009e0](01) 51 push ecx // push P
> >>>>>>>>>>>> [000009e1](05) e840feffff call // call H
> >>>>>>>>>>>> [000009e6](03) 83c408 add esp,+08
> >>>>>>>>>>>> [000009e9](02) 85c0 test eax,eax
> >>>>>>>>>>>> [000009eb](02) 7402 jz 000009ef
> >>>>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
> >>>>>>>>>>>> [000009ef](01) 5d pop ebp
> >>>>>>>>>>>> [000009f0](01) c3 ret // Final state
> >>>>>>>>>>>> Size in bytes:(0027) [000009f0]
> >>>>>>>>>>>
> >>>>>>>>>>> Since an H that never aborts is an uninteresting case,
> >>>>>>>>>> We simply hypothesize BOTH cases:
> >>>>>>>>>> (a) aborted and
> >>>>>>>>>> (b) never aborted
> >>>>>>>>>> and then see where their execution traces would be.
> >>>>>>>>>
> >>>>>>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
> >>>>>>>> We really only need to answer this single question:
> >>>>>>>> Is there any case where the correctly simulated input to H(P,P) reaches
> >>>>>>>> its own machine address of [000009f0]?
> >>>>>>>>
> >>>>>>>> H either aborts the simulation of its input or does not abort the
> >>>>>>>> simulation of its input. In either case the execution trace never
> >>>>>>>> extends beyond machine address [000009e1].
> >>>>>>>>
> >>>>>>>> Thus we can see that whether or not H aborts the simulation of its input
> >>>>>>>> this simulated input never reaches machine address [000009f0].
> >>>>>>>
> >>>>>>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
> >>>>>> No stupid I am not. I am talking about the above fixed P, and two
> >>>>>> different hypothetical scenarios of H at machine address [00000826].
> >>>>>
> >>>>> So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
> >>>> _P()
> >>>> [000009d6](01) 55 push ebp
> >>>> [000009d7](02) 8bec mov ebp,esp
> >>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >>>> [000009dc](01) 50 push eax // push P
> >>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >>>> [000009e0](01) 51 push ecx // push P
> >>>> [000009e1](05) e840feffff call 00000826 // call H
> >>>> [000009e6](03) 83c408 add esp,+08
> >>>> [000009e9](02) 85c0 test eax,eax
> >>>> [000009eb](02) 7402 jz 000009ef
> >>>> [000009ed](02) ebfe jmp 000009ed
> >>>> [000009ef](01) 5d pop ebp
> >>>> [000009f0](01) c3 ret // Final state
> >>>> Size in bytes:(0027) [000009f0]
> >>>> I am considering the behavior of the simulated P under two hypothetical
> >>>> scenarios:
> >>>> (1) H at machine address [00000826] aborts its simulation.
> >>>> (2) H at machine address [00000826] does not abort its simulation.
> >>>
> >>> So in other words you're considering the behavior of two separate unrelated computations that deceptively happen to share the same set of function names and happen to live at the same set of addresses when built separately.
> >>>
> >> That would be the way a God damned liar would say it.
> >>
> >> I am only considering the actual behavior of the simulated input to
> >> H(P,P) under the two possible scenarios. In each of these two possible
> >> scenarios P never reaches its own final state.
> >
> > In other words, you're considering whether Hn can simulate Pn to a final state, and *independently* whether Ha can simulate Pa to a final state.
> >
> > Pn and Pa are not the same. Pn(Pn) does not halt. Pa(Pa) halts. Hn(Pn,Pn) never stops running so it can't answer. The simulation Ha(Pa,Pa) is not simulated to a final state but the simulation Hb(Pa,Pa) *is* simulated to a final state.
> >
> > Therefore Ha(Pa,Pa) == false is incorrect.
> I have conclusively proved that the correct simulation of the input to
> H(P,P) would never reach its own final state whether or not H aborts the
> simulation of its input.


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 by: olcott - Sun, 17 Apr 2022 18:32 UTC

On 4/17/2022 1:26 PM, Dennis Bush wrote:
> On Sunday, April 17, 2022 at 2:07:15 PM UTC-4, olcott wrote:
>> On 4/17/2022 12:35 PM, Dennis Bush wrote:
>>> On Sunday, April 17, 2022 at 12:23:16 PM UTC-4, olcott wrote:
>>>> On 4/17/2022 11:12 AM, Dennis Bush wrote:
>>>>> On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
>>>>>> On 4/17/2022 10:59 AM, Dennis Bush wrote:
>>>>>>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
>>>>>>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
>>>>>>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
>>>>>>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>>>>>>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>>>>>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>>>>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>>>>>>>>>>>>>> correct.
>>>>>>>>>>>>>
>>>>>>>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
>>>>>>>>>>>>>
>>>>>>>>>>>> I count this as excellent progress.
>>>>>>>>>>>
>>>>>>>>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
>>>>>>>>>> None-the-less I count this as significant progress. It may have been
>>>>>>>>>> nearly the first time that anyone has ever agreed with anything that I
>>>>>>>>>> have ever said in this forum.
>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>>>>>>>>>>>>>> is correct?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
>>>>>>>>>>>>>> reporting {an X is a Y}.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
>>>>>>>>>>>>>> its final state and halts that proves that it is non-halting.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
>>>>>>>>>>>>>> code at machine address [000009d6] by H would show some correct
>>>>>>>>>>>>>> execution trace from machine address [000009d6] ending at machine
>>>>>>>>>>>>>> address [000009f0].
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> _P()
>>>>>>>>>>>>>> [000009d6](01) 55 push ebp
>>>>>>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>>>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>>>>>>>>> [000009e1](05) e840feffff call // call H
>>>>>>>>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>>>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>>>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>>>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>>>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>>>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>>>>>>>>
>>>>>>>>>>>>> Since an H that never aborts is an uninteresting case,
>>>>>>>>>>>> We simply hypothesize BOTH cases:
>>>>>>>>>>>> (a) aborted and
>>>>>>>>>>>> (b) never aborted
>>>>>>>>>>>> and then see where their execution traces would be.
>>>>>>>>>>>
>>>>>>>>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
>>>>>>>>>> We really only need to answer this single question:
>>>>>>>>>> Is there any case where the correctly simulated input to H(P,P) reaches
>>>>>>>>>> its own machine address of [000009f0]?
>>>>>>>>>>
>>>>>>>>>> H either aborts the simulation of its input or does not abort the
>>>>>>>>>> simulation of its input. In either case the execution trace never
>>>>>>>>>> extends beyond machine address [000009e1].
>>>>>>>>>>
>>>>>>>>>> Thus we can see that whether or not H aborts the simulation of its input
>>>>>>>>>> this simulated input never reaches machine address [000009f0].
>>>>>>>>>
>>>>>>>>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
>>>>>>>> No stupid I am not. I am talking about the above fixed P, and two
>>>>>>>> different hypothetical scenarios of H at machine address [00000826].
>>>>>>>
>>>>>>> So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
>>>>>> _P()
>>>>>> [000009d6](01) 55 push ebp
>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>> [000009dc](01) 50 push eax // push P
>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>> [000009e1](05) e840feffff call 00000826 // call H
>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>> [000009ef](01) 5d pop ebp
>>>>>> [000009f0](01) c3 ret // Final state
>>>>>> Size in bytes:(0027) [000009f0]
>>>>>> I am considering the behavior of the simulated P under two hypothetical
>>>>>> scenarios:
>>>>>> (1) H at machine address [00000826] aborts its simulation.
>>>>>> (2) H at machine address [00000826] does not abort its simulation.
>>>>>
>>>>> So in other words you're considering the behavior of two separate unrelated computations that deceptively happen to share the same set of function names and happen to live at the same set of addresses when built separately.
>>>>>
>>>> That would be the way a God damned liar would say it.
>>>>
>>>> I am only considering the actual behavior of the simulated input to
>>>> H(P,P) under the two possible scenarios. In each of these two possible
>>>> scenarios P never reaches its own final state.
>>>
>>> In other words, you're considering whether Hn can simulate Pn to a final state, and *independently* whether Ha can simulate Pa to a final state.
>>>
>>> Pn and Pa are not the same. Pn(Pn) does not halt. Pa(Pa) halts. Hn(Pn,Pn) never stops running so it can't answer. The simulation Ha(Pa,Pa) is not simulated to a final state but the simulation Hb(Pa,Pa) *is* simulated to a final state.
>>>
>>> Therefore Ha(Pa,Pa) == false is incorrect.
>> I have conclusively proved that the correct simulation of the input to
>> H(P,P) would never reach its own final state whether or not H aborts the
>> simulation of its input.
>
> FALSE. At most, you have shown that no H can simulate the P built from it to a final state. This is not the same as non-halting. Just because H1(P1,P1)==false and H2(P2,P2)==false and so forth doesn't mean that they all give the correct answer.


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s,_Richard_V3
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 by: Dennis Bush - Sun, 17 Apr 2022 18:38 UTC

On Sunday, April 17, 2022 at 2:33:01 PM UTC-4, olcott wrote:
> On 4/17/2022 1:26 PM, Dennis Bush wrote:
> > On Sunday, April 17, 2022 at 2:07:15 PM UTC-4, olcott wrote:
> >> On 4/17/2022 12:35 PM, Dennis Bush wrote:
> >>> On Sunday, April 17, 2022 at 12:23:16 PM UTC-4, olcott wrote:
> >>>> On 4/17/2022 11:12 AM, Dennis Bush wrote:
> >>>>> On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
> >>>>>> On 4/17/2022 10:59 AM, Dennis Bush wrote:
> >>>>>>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
> >>>>>>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
> >>>>>>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
> >>>>>>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
> >>>>>>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
> >>>>>>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
> >>>>>>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
> >>>>>>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
> >>>>>>>>>>>>>> correct.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
> >>>>>>>>>>>>>
> >>>>>>>>>>>> I count this as excellent progress.
> >>>>>>>>>>>
> >>>>>>>>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
> >>>>>>>>>> None-the-less I count this as significant progress. It may have been
> >>>>>>>>>> nearly the first time that anyone has ever agreed with anything that I
> >>>>>>>>>> have ever said in this forum.
> >>>>>>>>>>>
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
> >>>>>>>>>>>>>> is correct?
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
> >>>>>>>>>>>>>> reporting {an X is a Y}.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
> >>>>>>>>>>>>>> its final state and halts that proves that it is non-halting.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
> >>>>>>>>>>>>>> code at machine address [000009d6] by H would show some correct
> >>>>>>>>>>>>>> execution trace from machine address [000009d6] ending at machine
> >>>>>>>>>>>>>> address [000009f0].
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> _P()
> >>>>>>>>>>>>>> [000009d6](01) 55 push ebp
> >>>>>>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
> >>>>>>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >>>>>>>>>>>>>> [000009dc](01) 50 push eax // push P
> >>>>>>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>>>>>>>>>> [000009e0](01) 51 push ecx // push P
> >>>>>>>>>>>>>> [000009e1](05) e840feffff call // call H
> >>>>>>>>>>>>>> [000009e6](03) 83c408 add esp,+08
> >>>>>>>>>>>>>> [000009e9](02) 85c0 test eax,eax
> >>>>>>>>>>>>>> [000009eb](02) 7402 jz 000009ef
> >>>>>>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
> >>>>>>>>>>>>>> [000009ef](01) 5d pop ebp
> >>>>>>>>>>>>>> [000009f0](01) c3 ret // Final state
> >>>>>>>>>>>>>> Size in bytes:(0027) [000009f0]
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> Since an H that never aborts is an uninteresting case,
> >>>>>>>>>>>> We simply hypothesize BOTH cases:
> >>>>>>>>>>>> (a) aborted and
> >>>>>>>>>>>> (b) never aborted
> >>>>>>>>>>>> and then see where their execution traces would be.
> >>>>>>>>>>>
> >>>>>>>>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
> >>>>>>>>>> We really only need to answer this single question:
> >>>>>>>>>> Is there any case where the correctly simulated input to H(P,P) reaches
> >>>>>>>>>> its own machine address of [000009f0]?
> >>>>>>>>>>
> >>>>>>>>>> H either aborts the simulation of its input or does not abort the
> >>>>>>>>>> simulation of its input. In either case the execution trace never
> >>>>>>>>>> extends beyond machine address [000009e1].
> >>>>>>>>>>
> >>>>>>>>>> Thus we can see that whether or not H aborts the simulation of its input
> >>>>>>>>>> this simulated input never reaches machine address [000009f0].
> >>>>>>>>>
> >>>>>>>>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
> >>>>>>>> No stupid I am not. I am talking about the above fixed P, and two
> >>>>>>>> different hypothetical scenarios of H at machine address [00000826].
> >>>>>>>
> >>>>>>> So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
> >>>>>> _P()
> >>>>>> [000009d6](01) 55 push ebp
> >>>>>> [000009d7](02) 8bec mov ebp,esp
> >>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >>>>>> [000009dc](01) 50 push eax // push P
> >>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>> [000009e0](01) 51 push ecx // push P
> >>>>>> [000009e1](05) e840feffff call 00000826 // call H
> >>>>>> [000009e6](03) 83c408 add esp,+08
> >>>>>> [000009e9](02) 85c0 test eax,eax
> >>>>>> [000009eb](02) 7402 jz 000009ef
> >>>>>> [000009ed](02) ebfe jmp 000009ed
> >>>>>> [000009ef](01) 5d pop ebp
> >>>>>> [000009f0](01) c3 ret // Final state
> >>>>>> Size in bytes:(0027) [000009f0]
> >>>>>> I am considering the behavior of the simulated P under two hypothetical
> >>>>>> scenarios:
> >>>>>> (1) H at machine address [00000826] aborts its simulation.
> >>>>>> (2) H at machine address [00000826] does not abort its simulation.
> >>>>>
> >>>>> So in other words you're considering the behavior of two separate unrelated computations that deceptively happen to share the same set of function names and happen to live at the same set of addresses when built separately.
> >>>>>
> >>>> That would be the way a God damned liar would say it.
> >>>>
> >>>> I am only considering the actual behavior of the simulated input to
> >>>> H(P,P) under the two possible scenarios. In each of these two possible
> >>>> scenarios P never reaches its own final state.
> >>>
> >>> In other words, you're considering whether Hn can simulate Pn to a final state, and *independently* whether Ha can simulate Pa to a final state.
> >>>
> >>> Pn and Pa are not the same. Pn(Pn) does not halt. Pa(Pa) halts. Hn(Pn,Pn) never stops running so it can't answer. The simulation Ha(Pa,Pa) is not simulated to a final state but the simulation Hb(Pa,Pa) *is* simulated to a final state.
> >>>
> >>> Therefore Ha(Pa,Pa) == false is incorrect.
> >> I have conclusively proved that the correct simulation of the input to
> >> H(P,P) would never reach its own final state whether or not H aborts the
> >> simulation of its input.
> >
> > FALSE. At most, you have shown that no H can simulate the P built from it to a final state. This is not the same as non-halting. Just because H1(P1,P1)==false and H2(P2,P2)==false and so forth doesn't mean that they all give the correct answer.
> If the correctly simulated input to any simulating halt decider would
> never reach its own final state then this SHD would always be correct to
> reject this input.


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 by: olcott - Sun, 17 Apr 2022 18:42 UTC

On 4/17/2022 1:38 PM, Dennis Bush wrote:
> On Sunday, April 17, 2022 at 2:33:01 PM UTC-4, olcott wrote:
>> On 4/17/2022 1:26 PM, Dennis Bush wrote:
>>> On Sunday, April 17, 2022 at 2:07:15 PM UTC-4, olcott wrote:
>>>> On 4/17/2022 12:35 PM, Dennis Bush wrote:
>>>>> On Sunday, April 17, 2022 at 12:23:16 PM UTC-4, olcott wrote:
>>>>>> On 4/17/2022 11:12 AM, Dennis Bush wrote:
>>>>>>> On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
>>>>>>>> On 4/17/2022 10:59 AM, Dennis Bush wrote:
>>>>>>>>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
>>>>>>>>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
>>>>>>>>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
>>>>>>>>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>>>>>>>>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>>>>>>>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>>>>>>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>>>>>>>>>>>>>>>> correct.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>> I count this as excellent progress.
>>>>>>>>>>>>>
>>>>>>>>>>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
>>>>>>>>>>>> None-the-less I count this as significant progress. It may have been
>>>>>>>>>>>> nearly the first time that anyone has ever agreed with anything that I
>>>>>>>>>>>> have ever said in this forum.
>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>>>>>>>>>>>>>>>> is correct?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
>>>>>>>>>>>>>>>> reporting {an X is a Y}.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
>>>>>>>>>>>>>>>> its final state and halts that proves that it is non-halting.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
>>>>>>>>>>>>>>>> code at machine address [000009d6] by H would show some correct
>>>>>>>>>>>>>>>> execution trace from machine address [000009d6] ending at machine
>>>>>>>>>>>>>>>> address [000009f0].
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> _P()
>>>>>>>>>>>>>>>> [000009d6](01) 55 push ebp
>>>>>>>>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>>>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>>>>>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>>>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>>>>>>>>>>> [000009e1](05) e840feffff call // call H
>>>>>>>>>>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>>>>>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>>>>>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>>>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>>>>>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>>>>>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>>>>>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Since an H that never aborts is an uninteresting case,
>>>>>>>>>>>>>> We simply hypothesize BOTH cases:
>>>>>>>>>>>>>> (a) aborted and
>>>>>>>>>>>>>> (b) never aborted
>>>>>>>>>>>>>> and then see where their execution traces would be.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
>>>>>>>>>>>> We really only need to answer this single question:
>>>>>>>>>>>> Is there any case where the correctly simulated input to H(P,P) reaches
>>>>>>>>>>>> its own machine address of [000009f0]?
>>>>>>>>>>>>
>>>>>>>>>>>> H either aborts the simulation of its input or does not abort the
>>>>>>>>>>>> simulation of its input. In either case the execution trace never
>>>>>>>>>>>> extends beyond machine address [000009e1].
>>>>>>>>>>>>
>>>>>>>>>>>> Thus we can see that whether or not H aborts the simulation of its input
>>>>>>>>>>>> this simulated input never reaches machine address [000009f0].
>>>>>>>>>>>
>>>>>>>>>>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
>>>>>>>>>> No stupid I am not. I am talking about the above fixed P, and two
>>>>>>>>>> different hypothetical scenarios of H at machine address [00000826].
>>>>>>>>>
>>>>>>>>> So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
>>>>>>>> _P()
>>>>>>>> [000009d6](01) 55 push ebp
>>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>>> [000009e1](05) e840feffff call 00000826 // call H
>>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>>> I am considering the behavior of the simulated P under two hypothetical
>>>>>>>> scenarios:
>>>>>>>> (1) H at machine address [00000826] aborts its simulation.
>>>>>>>> (2) H at machine address [00000826] does not abort its simulation.
>>>>>>>
>>>>>>> So in other words you're considering the behavior of two separate unrelated computations that deceptively happen to share the same set of function names and happen to live at the same set of addresses when built separately.
>>>>>>>
>>>>>> That would be the way a God damned liar would say it.
>>>>>>
>>>>>> I am only considering the actual behavior of the simulated input to
>>>>>> H(P,P) under the two possible scenarios. In each of these two possible
>>>>>> scenarios P never reaches its own final state.
>>>>>
>>>>> In other words, you're considering whether Hn can simulate Pn to a final state, and *independently* whether Ha can simulate Pa to a final state.
>>>>>
>>>>> Pn and Pa are not the same. Pn(Pn) does not halt. Pa(Pa) halts. Hn(Pn,Pn) never stops running so it can't answer. The simulation Ha(Pa,Pa) is not simulated to a final state but the simulation Hb(Pa,Pa) *is* simulated to a final state.
>>>>>
>>>>> Therefore Ha(Pa,Pa) == false is incorrect.
>>>> I have conclusively proved that the correct simulation of the input to
>>>> H(P,P) would never reach its own final state whether or not H aborts the
>>>> simulation of its input.
>>>
>>> FALSE. At most, you have shown that no H can simulate the P built from it to a final state. This is not the same as non-halting. Just because H1(P1,P1)==false and H2(P2,P2)==false and so forth doesn't mean that they all give the correct answer.
>> If the correctly simulated input to any simulating halt decider would
>> never reach its own final state then this SHD would always be correct to
>> reject this input.
>
> So now you're back to "if X then Y" without establishing whether X is true. I.E. you made a non-statement that makes no attempt at refuting what I said.


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Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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Subject: Re:_My_Dishonest_reviewers:_André,_Ben,_Mike,_Denni
s,_Richard_V3
From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Sun, 17 Apr 2022 18:47 UTC

On Sunday, April 17, 2022 at 2:43:00 PM UTC-4, olcott wrote:
> On 4/17/2022 1:38 PM, Dennis Bush wrote:
> > On Sunday, April 17, 2022 at 2:33:01 PM UTC-4, olcott wrote:
> >> On 4/17/2022 1:26 PM, Dennis Bush wrote:
> >>> On Sunday, April 17, 2022 at 2:07:15 PM UTC-4, olcott wrote:
> >>>> On 4/17/2022 12:35 PM, Dennis Bush wrote:
> >>>>> On Sunday, April 17, 2022 at 12:23:16 PM UTC-4, olcott wrote:
> >>>>>> On 4/17/2022 11:12 AM, Dennis Bush wrote:
> >>>>>>> On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
> >>>>>>>> On 4/17/2022 10:59 AM, Dennis Bush wrote:
> >>>>>>>>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
> >>>>>>>>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
> >>>>>>>>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
> >>>>>>>>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
> >>>>>>>>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
> >>>>>>>>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
> >>>>>>>>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
> >>>>>>>>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
> >>>>>>>>>>>>>>>> correct.
> >>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
> >>>>>>>>>>>>>>>
> >>>>>>>>>>>>>> I count this as excellent progress.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
> >>>>>>>>>>>> None-the-less I count this as significant progress. It may have been
> >>>>>>>>>>>> nearly the first time that anyone has ever agreed with anything that I
> >>>>>>>>>>>> have ever said in this forum.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
> >>>>>>>>>>>>>>>> is correct?
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
> >>>>>>>>>>>>>>>> reporting {an X is a Y}.
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
> >>>>>>>>>>>>>>>> its final state and halts that proves that it is non-halting.
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
> >>>>>>>>>>>>>>>> code at machine address [000009d6] by H would show some correct
> >>>>>>>>>>>>>>>> execution trace from machine address [000009d6] ending at machine
> >>>>>>>>>>>>>>>> address [000009f0].
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> _P()
> >>>>>>>>>>>>>>>> [000009d6](01) 55 push ebp
> >>>>>>>>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
> >>>>>>>>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >>>>>>>>>>>>>>>> [000009dc](01) 50 push eax // push P
> >>>>>>>>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>>>>>>>>>>>> [000009e0](01) 51 push ecx // push P
> >>>>>>>>>>>>>>>> [000009e1](05) e840feffff call // call H
> >>>>>>>>>>>>>>>> [000009e6](03) 83c408 add esp,+08
> >>>>>>>>>>>>>>>> [000009e9](02) 85c0 test eax,eax
> >>>>>>>>>>>>>>>> [000009eb](02) 7402 jz 000009ef
> >>>>>>>>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
> >>>>>>>>>>>>>>>> [000009ef](01) 5d pop ebp
> >>>>>>>>>>>>>>>> [000009f0](01) c3 ret // Final state
> >>>>>>>>>>>>>>>> Size in bytes:(0027) [000009f0]
> >>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>> Since an H that never aborts is an uninteresting case,
> >>>>>>>>>>>>>> We simply hypothesize BOTH cases:
> >>>>>>>>>>>>>> (a) aborted and
> >>>>>>>>>>>>>> (b) never aborted
> >>>>>>>>>>>>>> and then see where their execution traces would be.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
> >>>>>>>>>>>> We really only need to answer this single question:
> >>>>>>>>>>>> Is there any case where the correctly simulated input to H(P,P) reaches
> >>>>>>>>>>>> its own machine address of [000009f0]?
> >>>>>>>>>>>>
> >>>>>>>>>>>> H either aborts the simulation of its input or does not abort the
> >>>>>>>>>>>> simulation of its input. In either case the execution trace never
> >>>>>>>>>>>> extends beyond machine address [000009e1].
> >>>>>>>>>>>>
> >>>>>>>>>>>> Thus we can see that whether or not H aborts the simulation of its input
> >>>>>>>>>>>> this simulated input never reaches machine address [000009f0].
> >>>>>>>>>>>
> >>>>>>>>>>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
> >>>>>>>>>> No stupid I am not. I am talking about the above fixed P, and two
> >>>>>>>>>> different hypothetical scenarios of H at machine address [00000826].
> >>>>>>>>>
> >>>>>>>>> So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
> >>>>>>>> _P()
> >>>>>>>> [000009d6](01) 55 push ebp
> >>>>>>>> [000009d7](02) 8bec mov ebp,esp
> >>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
> >>>>>>>> [000009dc](01) 50 push eax // push P
> >>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
> >>>>>>>> [000009e0](01) 51 push ecx // push P
> >>>>>>>> [000009e1](05) e840feffff call 00000826 // call H
> >>>>>>>> [000009e6](03) 83c408 add esp,+08
> >>>>>>>> [000009e9](02) 85c0 test eax,eax
> >>>>>>>> [000009eb](02) 7402 jz 000009ef
> >>>>>>>> [000009ed](02) ebfe jmp 000009ed
> >>>>>>>> [000009ef](01) 5d pop ebp
> >>>>>>>> [000009f0](01) c3 ret // Final state
> >>>>>>>> Size in bytes:(0027) [000009f0]
> >>>>>>>> I am considering the behavior of the simulated P under two hypothetical
> >>>>>>>> scenarios:
> >>>>>>>> (1) H at machine address [00000826] aborts its simulation.
> >>>>>>>> (2) H at machine address [00000826] does not abort its simulation.
> >>>>>>>
> >>>>>>> So in other words you're considering the behavior of two separate unrelated computations that deceptively happen to share the same set of function names and happen to live at the same set of addresses when built separately.
> >>>>>>>
> >>>>>> That would be the way a God damned liar would say it.
> >>>>>>
> >>>>>> I am only considering the actual behavior of the simulated input to
> >>>>>> H(P,P) under the two possible scenarios. In each of these two possible
> >>>>>> scenarios P never reaches its own final state.
> >>>>>
> >>>>> In other words, you're considering whether Hn can simulate Pn to a final state, and *independently* whether Ha can simulate Pa to a final state.
> >>>>>
> >>>>> Pn and Pa are not the same. Pn(Pn) does not halt. Pa(Pa) halts. Hn(Pn,Pn) never stops running so it can't answer. The simulation Ha(Pa,Pa) is not simulated to a final state but the simulation Hb(Pa,Pa) *is* simulated to a final state.
> >>>>>
> >>>>> Therefore Ha(Pa,Pa) == false is incorrect.
> >>>> I have conclusively proved that the correct simulation of the input to
> >>>> H(P,P) would never reach its own final state whether or not H aborts the
> >>>> simulation of its input.
> >>>
> >>> FALSE. At most, you have shown that no H can simulate the P built from it to a final state. This is not the same as non-halting. Just because H1(P1,P1)==false and H2(P2,P2)==false and so forth doesn't mean that they all give the correct answer.
> >> If the correctly simulated input to any simulating halt decider would
> >> never reach its own final state then this SHD would always be correct to
> >> reject this input.
> >
> > So now you're back to "if X then Y" without establishing whether X is true. I.E. you made a non-statement that makes no attempt at refuting what I said.
> On 4/17/2022 1:32 PM, olcott wrote:
> > computation that halts … the Turing machine will halt
> > whenever it enters a final state. (Linz:1990:234)
> Sure that is what a God damned liar would say when they simply erase the
> part that proved my point.


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Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard V3

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 by: olcott - Sun, 17 Apr 2022 18:55 UTC

On 4/17/2022 1:47 PM, Dennis Bush wrote:
> On Sunday, April 17, 2022 at 2:43:00 PM UTC-4, olcott wrote:
>> On 4/17/2022 1:38 PM, Dennis Bush wrote:
>>> On Sunday, April 17, 2022 at 2:33:01 PM UTC-4, olcott wrote:
>>>> On 4/17/2022 1:26 PM, Dennis Bush wrote:
>>>>> On Sunday, April 17, 2022 at 2:07:15 PM UTC-4, olcott wrote:
>>>>>> On 4/17/2022 12:35 PM, Dennis Bush wrote:
>>>>>>> On Sunday, April 17, 2022 at 12:23:16 PM UTC-4, olcott wrote:
>>>>>>>> On 4/17/2022 11:12 AM, Dennis Bush wrote:
>>>>>>>>> On Sunday, April 17, 2022 at 12:06:31 PM UTC-4, olcott wrote:
>>>>>>>>>> On 4/17/2022 10:59 AM, Dennis Bush wrote:
>>>>>>>>>>> On Sunday, April 17, 2022 at 11:49:56 AM UTC-4, olcott wrote:
>>>>>>>>>>>> On 4/17/2022 10:41 AM, Dennis Bush wrote:
>>>>>>>>>>>>> On Sunday, April 17, 2022 at 11:25:37 AM UTC-4, olcott wrote:
>>>>>>>>>>>>>> On 4/17/2022 10:11 AM, Dennis Bush wrote:
>>>>>>>>>>>>>>> On Sunday, April 17, 2022 at 12:26:51 AM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>> On 4/16/2022 11:14 PM, Dennis Bush wrote:
>>>>>>>>>>>>>>>>> On Saturday, April 16, 2022 at 11:40:20 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>>>> If the input to H(P,P) is non-halting then H(P,P)==false is necessary
>>>>>>>>>>>>>>>>>> correct.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> True. *if* the input to H(P,P) is non-halting then H(P,P)==false is correct. But first we must establish whether the input to H(P,P) is non-halting.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I count this as excellent progress.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> This has never been in dispute, by me or anyone else. The fact that you think this shows how poor your reading comprehension is.
>>>>>>>>>>>>>> None-the-less I count this as significant progress. It may have been
>>>>>>>>>>>>>> nearly the first time that anyone has ever agreed with anything that I
>>>>>>>>>>>>>> have ever said in this forum.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Does anyone have a rebuttal to that one, or do you acknowledge that it
>>>>>>>>>>>>>>>>>> is correct?
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> You must find an example where {an X is a Y} and Z is incorrect for
>>>>>>>>>>>>>>>>>> reporting {an X is a Y}.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> If no one can show that the correctly simulated input to H(P,P) reaches
>>>>>>>>>>>>>>>>>> its final state and halts that proves that it is non-halting.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes of machine
>>>>>>>>>>>>>>>>>> code at machine address [000009d6] by H would show some correct
>>>>>>>>>>>>>>>>>> execution trace from machine address [000009d6] ending at machine
>>>>>>>>>>>>>>>>>> address [000009f0].
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> _P()
>>>>>>>>>>>>>>>>>> [000009d6](01) 55 push ebp
>>>>>>>>>>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>>>>>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>>>>>>>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>>>>>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>>>>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>>>>>>>>>>>>> [000009e1](05) e840feffff call // call H
>>>>>>>>>>>>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>>>>>>>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>>>>>>>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>>>>>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>>>>>>>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>>>>>>>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>>>>>>>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Since an H that never aborts is an uninteresting case,
>>>>>>>>>>>>>>>> We simply hypothesize BOTH cases:
>>>>>>>>>>>>>>>> (a) aborted and
>>>>>>>>>>>>>>>> (b) never aborted
>>>>>>>>>>>>>>>> and then see where their execution traces would be.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Those two cases means you're changing the code of H, which means that P becomes a completely different computation.
>>>>>>>>>>>>>> We really only need to answer this single question:
>>>>>>>>>>>>>> Is there any case where the correctly simulated input to H(P,P) reaches
>>>>>>>>>>>>>> its own machine address of [000009f0]?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> H either aborts the simulation of its input or does not abort the
>>>>>>>>>>>>>> simulation of its input. In either case the execution trace never
>>>>>>>>>>>>>> extends beyond machine address [000009e1].
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Thus we can see that whether or not H aborts the simulation of its input
>>>>>>>>>>>>>> this simulated input never reaches machine address [000009f0].
>>>>>>>>>>>>>
>>>>>>>>>>>>> If you're talking about a fixed P at a fixed address, that means you are also talking about ONE SPECIFIC H at a specific address with a fixed, unchanging algorithm.
>>>>>>>>>>>> No stupid I am not. I am talking about the above fixed P, and two
>>>>>>>>>>>> different hypothetical scenarios of H at machine address [00000826].
>>>>>>>>>>>
>>>>>>>>>>> So you're dishonestly changing the algorithm of H to make two different P's look the same. They are not.
>>>>>>>>>> _P()
>>>>>>>>>> [000009d6](01) 55 push ebp
>>>>>>>>>> [000009d7](02) 8bec mov ebp,esp
>>>>>>>>>> [000009d9](03) 8b4508 mov eax,[ebp+08]
>>>>>>>>>> [000009dc](01) 50 push eax // push P
>>>>>>>>>> [000009dd](03) 8b4d08 mov ecx,[ebp+08]
>>>>>>>>>> [000009e0](01) 51 push ecx // push P
>>>>>>>>>> [000009e1](05) e840feffff call 00000826 // call H
>>>>>>>>>> [000009e6](03) 83c408 add esp,+08
>>>>>>>>>> [000009e9](02) 85c0 test eax,eax
>>>>>>>>>> [000009eb](02) 7402 jz 000009ef
>>>>>>>>>> [000009ed](02) ebfe jmp 000009ed
>>>>>>>>>> [000009ef](01) 5d pop ebp
>>>>>>>>>> [000009f0](01) c3 ret // Final state
>>>>>>>>>> Size in bytes:(0027) [000009f0]
>>>>>>>>>> I am considering the behavior of the simulated P under two hypothetical
>>>>>>>>>> scenarios:
>>>>>>>>>> (1) H at machine address [00000826] aborts its simulation.
>>>>>>>>>> (2) H at machine address [00000826] does not abort its simulation.
>>>>>>>>>
>>>>>>>>> So in other words you're considering the behavior of two separate unrelated computations that deceptively happen to share the same set of function names and happen to live at the same set of addresses when built separately.
>>>>>>>>>
>>>>>>>> That would be the way a God damned liar would say it.
>>>>>>>>
>>>>>>>> I am only considering the actual behavior of the simulated input to
>>>>>>>> H(P,P) under the two possible scenarios. In each of these two possible
>>>>>>>> scenarios P never reaches its own final state.
>>>>>>>
>>>>>>> In other words, you're considering whether Hn can simulate Pn to a final state, and *independently* whether Ha can simulate Pa to a final state.
>>>>>>>
>>>>>>> Pn and Pa are not the same. Pn(Pn) does not halt. Pa(Pa) halts. Hn(Pn,Pn) never stops running so it can't answer. The simulation Ha(Pa,Pa) is not simulated to a final state but the simulation Hb(Pa,Pa) *is* simulated to a final state.
>>>>>>>
>>>>>>> Therefore Ha(Pa,Pa) == false is incorrect.
>>>>>> I have conclusively proved that the correct simulation of the input to
>>>>>> H(P,P) would never reach its own final state whether or not H aborts the
>>>>>> simulation of its input.
>>>>>
>>>>> FALSE. At most, you have shown that no H can simulate the P built from it to a final state. This is not the same as non-halting. Just because H1(P1,P1)==false and H2(P2,P2)==false and so forth doesn't mean that they all give the correct answer.
>>>> If the correctly simulated input to any simulating halt decider would
>>>> never reach its own final state then this SHD would always be correct to
>>>> reject this input.
>>>
>>> So now you're back to "if X then Y" without establishing whether X is true. I.E. you made a non-statement that makes no attempt at refuting what I said.
>> On 4/17/2022 1:32 PM, olcott wrote:
>>> computation that halts … the Turing machine will halt
>>> whenever it enters a final state. (Linz:1990:234)
>> Sure that is what a God damned liar would say when they simply erase the
>> part that proved my point.
>
> It doesn't prove your point. My point still stands.
>
> If a simulating halt decider simulates its input to a final state, this is conclusive proof that the input is halting.


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