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devel / comp.theory / Are the halting problem proofs refuted by software engineering? (simple overview)

SubjectAuthor
* Are the halting problem proofs refuted by software engineering?olcott
+- Are the halting problem proofs refuted by software engineering?Richard Damon
+- Are the halting problem proofs refuted by software engineering?Dennis Bush
+- Are the halting problem proofs refuted by software engineering?Mr Flibble
+- Are the halting problem proofs refuted by software engineering?Skep Dick
+- Are the halting problem proofs refuted by software engineering?Mr Flibble
+* Are the halting problem proofs refuted by software engineering? (simple overviewAlan Mackenzie
|+- Are the halting problem proofs refuted by software engineering?Richard Damon
|+* Are the halting problem proofs refuted by software engineering?olcott
||+- Are the halting problem proofs refuted by software engineering?Dennis Bush
||`- Are the halting problem proofs refuted by software engineering?Richard Damon
|`* Are the halting problem proofs refuted by software engineering? (simple overviewBen Bacarisse
| `- Are the halting problem proofs refuted by software engineering?Richard Damon
`- Are the halting problem proofs refuted by software engineering?Horatio Cornholer

1
Are the halting problem proofs refuted by software engineering? (simple overview)

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From: NoO...@NoWhere.com (olcott)
Subject: Are the halting problem proofs refuted by software engineering?
(simple overview)
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 by: olcott - Sat, 30 Jul 2022 14:35 UTC

*Everyone here should be able to confirm this in less than a minute*
It is very easy to see that the input to H(P,P) correctly simulated by H
would remain stuck in infinitely recursive simulation. This prevents H
from ever returning any value to P. If H never returns any value to P
then P cannot possibly do the opposite of what H returns.

H is a simulating halt detector that applies an x86 emulator to simulate
the x86 machine-code of its input until the behavior of this input
matches a non-halting behavior pattern: {Infinite loop, Infinite
recursion, Infinitely recursive emulation}.

Because infinitely recursive emulation has the same control flow as
infinite recursion H(P,P) would never return any value to its emulated
input. This makes it impossible for its emulated input to do the
opposite of whatever H returns.

It is well know in software engineering that no function called in
infinite recursion ever returns to its caller, thus it is incorrect for
H to return any value to P. There are no instructions in P that could
possibly break this cycle of repeated simulations.

typedef void (*ptr)();
int H(ptr p, ptr i); // simulating halt decider

void P(ptr x)
{ int Halt_Status = H(x, x);
if (Halt_Status)
HERE: goto HERE;
return;
}

int main()
{ Output("Input_Halts = ", H(P, P));
}

(1) H(P,P) simulates its input with an x86 emulator.
(2) P calls H(P,P) to do this again.
(3) H(P,P) simulates its input with an x86 emulator.
(4) P calls H(P,P) to do this again.
(5) H(P,P) simulates its input with an x86 emulator.
(6) P calls H(P,P) to do this again...

If a simulating halt decider continues to correctly simulate its input
until it correctly matches a non-halting behavior pattern then this SHD
is necessarily correct when it aborts its simulation and reports
non-halting.

--
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: Are the halting problem proofs refuted by software engineering? (simple overview)

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Subject: Re: Are the halting problem proofs refuted by software engineering?
(simple overview)
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 by: Richard Damon - Sat, 30 Jul 2022 14:54 UTC

On 7/30/22 10:35 AM, olcott wrote:
> *Everyone here should be able to confirm this in less than a minute*
> It is very easy to see that the input to H(P,P) correctly simulated by H
> would remain stuck in infinitely recursive simulation. This prevents H
> from ever returning any value to P. If H never returns any value to P
> then P cannot possibly do the opposite of what H returns.
>
> H is a simulating halt detector that applies an x86 emulator to simulate
> the x86 machine-code of its input until the behavior of this input
> matches a non-halting behavior pattern: {Infinite loop, Infinite
> recursion, Infinitely recursive emulation}.
>
> Because infinitely recursive emulation has the same control flow as
> infinite recursion H(P,P) would never return any value to its emulated
> input. This makes it impossible for its emulated input to do the
> opposite of whatever H returns.
>
> It is well know in software engineering that no function called in
> infinite recursion ever returns to its caller, thus it is incorrect for
> H to return any value to P. There are no instructions in P that could
> possibly break this cycle of repeated simulations.

It is also a well know software engineerig principle that the a pure
function will ALWAYS return the same answer (or never return) for ALL
calls to it. Thus, the H(P,P) called via mian and the one called by P(P)
must have the same result, so if H(P,P) called by main returns 0, then H
does NOT get into infinite recursion with P, and thus it can (and MUST)
return that answer to P(P), and thus P(P) returns.

This means that the answer of 0 that H returned was INCORRECT.

Your error is you think that a SINGLE definition of H, that is defined
to be a pure function, can behave differently from two different
identical calls.

This shows you don't understand how computers work.

>
> typedef void (*ptr)();
> int H(ptr p, ptr i); // simulating halt decider
>
> void P(ptr x)
> {
>   int Halt_Status = H(x, x);
>   if (Halt_Status)
>     HERE: goto HERE;
>   return;
> }
>
> int main()
> {
>   Output("Input_Halts = ", H(P, P));
> }
>
> (1) H(P,P) simulates its input with an x86 emulator.
> (2) P calls H(P,P) to do this again.
> (3) H(P,P) simulates its input with an x86 emulator.
> (4) P calls H(P,P) to do this again.
> (5) H(P,P) simulates its input with an x86 emulator.
> (6) P calls H(P,P) to do this again...
>
> If a simulating halt decider continues to correctly simulate its input
> until it correctly matches a non-halting behavior pattern then this SHD
> is necessarily correct when it aborts its simulation and reports
> non-halting.
>

Right, *IF* H doesn't abort, then it gets into infinite recursion, and
both P(P) and H(P,P) are non-hating.

If H DOES abort, then P(P) will Halt, so H(P,P) saying it doesn't is
just WRONG, and is based on P calling an H that doesn't actually exist.

You can't build logical arguements about a program by looking at a
program that doesn't actually exist.

Yes, the aborting H can correctly decide on a P that calls the
non-aborting H and show that it is non-halting, but that isn't the case
that the Halting Problem proof shows. H is shown to be unable to decide
on a P that uses IT, not some variant of itself that behaves
differently), and H does't get that case right.

When you say that H(P,P) isn't asking about P(P), then you are just
asserting that you aren't working on the Halting Problem, as your
example program P is doing exactly that, so if that is wrong, you have
failed to meet the requrement.

FAIL.

Re: Are the halting problem proofs refuted by software engineering? (simple overview)

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Subject: Re: Are the halting problem proofs refuted by software engineering?
(simple overview)
From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Sat, 30 Jul 2022 14:56 UTC

On Saturday, July 30, 2022 at 10:35:47 AM UTC-4, olcott wrote:
> *Everyone here should be able to confirm this in less than a minute*
> It is very easy to see that the input to H(P,P) correctly simulated by H
> would remain stuck in infinitely recursive simulation. This prevents H
> from ever returning any value to P. If H never returns any value to P
> then P cannot possibly do the opposite of what H returns.

So you're saying that there's infinite simulation because H never aborts. Then we'll call it Hn, and P we'll call Pn to make that point clear.

You are correct that the input to Hn(Pn,Pn) correctly simulated by Hn is stuck in infinitely recursive simulation, which means that Hn can't return a value to Pn.

>
> H is a simulating halt detector that applies an x86 emulator to simulate
> the x86 machine-code of its input until the behavior of this input
> matches a non-halting behavior pattern: {Infinite loop, Infinite
> recursion, Infinitely recursive emulation}.
>
> Because infinitely recursive emulation has the same control flow as
> infinite recursion H(P,P) would never return any value to its emulated
> input. This makes it impossible for its emulated input to do the
> opposite of whatever H returns.
>
> It is well know in software engineering that no function called in
> infinite recursion ever returns to its caller, thus it is incorrect for
> H to return any value to P. There are no instructions in P that could
> possibly break this cycle of repeated simulations.

Correct, because the function P which is part of Pn nor the function H which is part of both Hn and Pn have any instructions to break the cycle.

>
> typedef void (*ptr)();
> int H(ptr p, ptr i); // simulating halt decider
>
> void P(ptr x)
> {
> int Halt_Status = H(x, x);
> if (Halt_Status)
> HERE: goto HERE;
> return;
> }
>
> int main()
> {
> Output("Input_Halts = ", H(P, P));
> }
>
> (1) H(P,P) simulates its input with an x86 emulator.
> (2) P calls H(P,P) to do this again.
> (3) H(P,P) simulates its input with an x86 emulator.
> (4) P calls H(P,P) to do this again.
> (5) H(P,P) simulates its input with an x86 emulator.
> (6) P calls H(P,P) to do this again...
>
> If a simulating halt decider continues to correctly simulate its input
> until it correctly matches a non-halting behavior pattern then this SHD
> is necessarily correct when it aborts its simulation and reports
> non-halting.

But we've already established that Hn can't abort because it's stuck infinitely nested simulation.

Re: Are the halting problem proofs refuted by software engineering? (simple overview)

<20220730161941.0000009d@reddwarf.jmc.corp>

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From: flib...@reddwarf.jmc.corp (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: Are the halting problem proofs refuted by software engineering?
(simple overview)
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 by: Mr Flibble - Sat, 30 Jul 2022 15:19 UTC

On Sat, 30 Jul 2022 09:35:40 -0500
olcott <NoOne@NoWhere.com> wrote:

> *Everyone here should be able to confirm this in less than a minute*
> It is very easy to see that the input to H(P,P) correctly simulated
> by H would remain stuck in infinitely recursive simulation. This
> prevents H from ever returning any value to P. If H never returns any
> value to P then P cannot possibly do the opposite of what H returns.
>
> H is a simulating halt detector that applies an x86 emulator to
> simulate the x86 machine-code of its input until the behavior of this
> input matches a non-halting behavior pattern: {Infinite loop,
> Infinite recursion, Infinitely recursive emulation}.
>
> Because infinitely recursive emulation has the same control flow as
> infinite recursion H(P,P) would never return any value to its
> emulated input. This makes it impossible for its emulated input to do
> the opposite of whatever H returns.
>
> It is well know in software engineering that no function called in
> infinite recursion ever returns to its caller, thus it is incorrect
> for H to return any value to P. There are no instructions in P that
> could possibly break this cycle of repeated simulations.
>
> typedef void (*ptr)();
> int H(ptr p, ptr i); // simulating halt decider
>
> void P(ptr x)
> {
> int Halt_Status = H(x, x);
> if (Halt_Status)
> HERE: goto HERE;
> return;
> }
>
> int main()
> {
> Output("Input_Halts = ", H(P, P));
> }
>
> (1) H(P,P) simulates its input with an x86 emulator.
> (2) P calls H(P,P) to do this again.
> (3) H(P,P) simulates its input with an x86 emulator.
> (4) P calls H(P,P) to do this again.
> (5) H(P,P) simulates its input with an x86 emulator.
> (6) P calls H(P,P) to do this again...
>
> If a simulating halt decider continues to correctly simulate its
> input until it correctly matches a non-halting behavior pattern then
> this SHD is necessarily correct when it aborts its simulation and
> reports non-halting.
>

You are incorrect to map your simulation having to abort its simulation
because your H is recursive with a halting decision of non-halting.

My SHD doesn't suffer from this problem as it isn't recursive in nature:

https://github.com/i42output/halting-problem#readme

/Flibble

Re: Are the halting problem proofs refuted by software engineering? (simple overview)

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Subject: Re: Are the halting problem proofs refuted by software engineering?
(simple overview)
From: skepdic...@gmail.com (Skep Dick)
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 by: Skep Dick - Sat, 30 Jul 2022 18:21 UTC

On Saturday, 30 July 2022 at 16:35:47 UTC+2, olcott wrote:
> *Everyone here should be able to confirm this in less than a minute*

You've been telling that lie for 5+ years to other people; and 15+ years to yourself.

Re: Are the halting problem proofs refuted by software engineering? (simple overview)

<20220730210905.000028c6@reddwarf.jmc.corp>

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From: flib...@reddwarf.jmc.corp (Mr Flibble)
Newsgroups: comp.theory
Subject: Re: Are the halting problem proofs refuted by software engineering?
(simple overview)
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 by: Mr Flibble - Sat, 30 Jul 2022 20:09 UTC

On Sat, 30 Jul 2022 09:35:40 -0500
olcott <NoOne@NoWhere.com> wrote:

> *Everyone here should be able to confirm this in less than a minute*
> It is very easy to see that the input to H(P,P) correctly simulated
> by H would remain stuck in infinitely recursive simulation. This
> prevents H from ever returning any value to P. If H never returns any
> value to P then P cannot possibly do the opposite of what H returns.
>
> H is a simulating halt detector that applies an x86 emulator to
> simulate the x86 machine-code of its input until the behavior of this
> input matches a non-halting behavior pattern: {Infinite loop,
> Infinite recursion, Infinitely recursive emulation}.
>
> Because infinitely recursive emulation has the same control flow as
> infinite recursion H(P,P) would never return any value to its
> emulated input. This makes it impossible for its emulated input to do
> the opposite of whatever H returns.
>
> It is well know in software engineering that no function called in
> infinite recursion ever returns to its caller, thus it is incorrect
> for H to return any value to P. There are no instructions in P that
> could possibly break this cycle of repeated simulations.
>
> typedef void (*ptr)();
> int H(ptr p, ptr i); // simulating halt decider
>
> void P(ptr x)
> {
> int Halt_Status = H(x, x);
> if (Halt_Status)
> HERE: goto HERE;
> return;
> }
>
> int main()
> {
> Output("Input_Halts = ", H(P, P));
> }
>
> (1) H(P,P) simulates its input with an x86 emulator.
> (2) P calls H(P,P) to do this again.
> (3) H(P,P) simulates its input with an x86 emulator.
> (4) P calls H(P,P) to do this again.
> (5) H(P,P) simulates its input with an x86 emulator.
> (6) P calls H(P,P) to do this again...
>
> If a simulating halt decider continues to correctly simulate its
> input until it correctly matches a non-halting behavior pattern then
> this SHD is necessarily correct when it aborts its simulation and
> reports non-halting.

Only my signaling halt decider refutes the HP proofs: yours does not as
it cannot produce correct halting decisions.

/Flibble

Re: Are the halting problem proofs refuted by software engineering? (simple overview)

<tc454j$olk$1@news.muc.de>

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From: acm...@muc.de (Alan Mackenzie)
Newsgroups: comp.theory
Subject: Re: Are the halting problem proofs refuted by software engineering? (simple overview)
Date: Sat, 30 Jul 2022 20:42:59 -0000 (UTC)
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 by: Alan Mackenzie - Sat, 30 Jul 2022 20:42 UTC

olcott <NoOne@nowhere.com> wrote:

[ .... ]

No, they're not. Mathematical proofs, by their nature, are not refuted
by anything.

> --
> Copyright 2022 Pete Olcott

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

--
Alan Mackenzie (Nuremberg, Germany).

Re: Are the halting problem proofs refuted by software engineering? (simple overview)

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Subject: Re: Are the halting problem proofs refuted by software engineering?
(simple overview)
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 by: Richard Damon - Sat, 30 Jul 2022 21:24 UTC

On 7/30/22 4:42 PM, Alan Mackenzie wrote:
> olcott <NoOne@nowhere.com> wrote:
>
> [ .... ]
>
> No, they're not. Mathematical proofs, by their nature, are not refuted
> by anything.
>

They are refuted by finding the error in their logic, or pointing out
that they used a premise that hasn't been proved.

Since NONE of Olcotts "proofs" ever give a source for their premises,
nothing he has write is actually a proof, and everything that has been
chalanged has been refuted.

Yes, CONJECTURES (things that haven't been actually proven) can be
refuted with a simple counter example, but as you say, a proof can not be.

Re: Are the halting problem proofs refuted by software engineering? (simple overview)

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(simple overview)
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 by: olcott - Sat, 30 Jul 2022 21:33 UTC

On 7/30/2022 3:42 PM, Alan Mackenzie wrote:
> olcott <NoOne@nowhere.com> wrote:
>
> [ .... ]
>
> No, they're not. Mathematical proofs, by their nature, are not refuted
> by anything.

<sarcasm>
Because everyone knows that all mathematicians are inherently
infallible, thus none of their proofs can contain any mistakes or
omissions.
</sarcasm>

*The actual mistake is an error of omission*
I found a loophole that no one every noticed before because no one ever
bothered to examine the effect of a simulating halt decider on the
conventional pathological inputs and *think this all-the-way through*

Most everyone simply rejects simulation out-of-hand and thus never gets
to the point of considering the possibility of a simulating halt decider.

If you paid close attention to what I just said in (the post you are
responding to) you would see that it is correct.

>
>> --
>> Copyright 2022 Pete Olcott
>
>> "Talent hits a target no one else can hit;
>> Genius hits a target no one else can see."
>> Arthur Schopenhauer
>

--
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: Are the halting problem proofs refuted by software engineering? (simple overview)

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Are the halting problem proofs refuted by software engineering? (simple overview)
Date: Sat, 30 Jul 2022 22:46:45 +0100
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 by: Ben Bacarisse - Sat, 30 Jul 2022 21:46 UTC

Alan Mackenzie <acm@muc.de> writes:

> olcott <NoOne@nowhere.com> wrote:
>
> [ .... ]
>
> No, they're not. Mathematical proofs, by their nature, are not refuted
> by anything.

This is key. I am sure now (and I should have been sure years ago) that
PO does not know what a proof is. He thinks of mathematical proofs like
courtroom arguments -- clever stories that can be used to trick naive
juries (the students who "learn by rote") into believing falsehoods.
And just as those sorts of "proofs" can be refuted by some piece of
hitherto unknown evidence, mathematical proofs can be refuted by saying
something no one has previously said!

I once tried to get PO to give an example of proved theorem that he had
rock-solid confidence in -- something he was more sure of (because of
the proof) than, say, the likelihood of the sun rising tomorrow. He
ducked that question for ages.

--
Ben.

Re: Are the halting problem proofs refuted by software engineering? (simple overview)

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Subject: Re: Are the halting problem proofs refuted by software engineering?
(simple overview)
From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Sat, 30 Jul 2022 21:47 UTC

On Saturday, July 30, 2022 at 5:33:27 PM UTC-4, olcott wrote:
> On 7/30/2022 3:42 PM, Alan Mackenzie wrote:
> > olcott <No...@nowhere.com> wrote:
> >
> > [ .... ]
> >
> > No, they're not. Mathematical proofs, by their nature, are not refuted
> > by anything.
> <sarcasm>
> Because everyone knows that all mathematicians are inherently
> infallible, thus none of their proofs can contain any mistakes or
> omissions.
> </sarcasm>
>
> *The actual mistake is an error of omission*
> I found a loophole that no one every noticed before because no one ever
> bothered to examine the effect of a simulating halt decider on the
> conventional pathological inputs and *think this all-the-way through*

The halting problem proofs don't specifically mention simulation because they don't need to. The proof applies to *any* potential halt decider, which includes those that simulate.

Your problem is that you think that Ha(Pa,Pa) is supposed to be deciding on Pn(Pn), or alternately that a halt decider H should determine if there exists an implementation of the function H that can simulate the function P(P) to a final state.

But the halting problem is about algorithms, not C functions. That means it is completely disallowed to consider any alternate behavior of the function H because doing so changes the algorithm that is being decided on. Since the algorithm P to be decided on is fixed, that means the function H is also fixed.

The bottom line: If H(P,P)==0, then P(P) will halt. And by the definition of halting (the *actual* definition, not your alternate one) and the mapping a halt decider is required to compute, H(P,P)==0 is the wrong answer.

>
> Most everyone simply rejects simulation out-of-hand and thus never gets
> to the point of considering the possibility of a simulating halt decider.

Not at all, they just point out that it's already been considered.

Re: Are the halting problem proofs refuted by software engineering? (simple overview)

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 by: Richard Damon - Sat, 30 Jul 2022 22:04 UTC

On 7/30/22 5:33 PM, olcott wrote:
> On 7/30/2022 3:42 PM, Alan Mackenzie wrote:
>> olcott <NoOne@nowhere.com> wrote:
>>
>> [ .... ]
>>
>> No, they're not.  Mathematical proofs, by their nature, are not refuted
>> by anything.
>
> <sarcasm>
> Because everyone knows that all mathematicians are inherently
> infallible, thus none of their proofs can contain any mistakes or
> omissions.
> </sarcasm>

Mistakes are found, and things corrected. But they are few for th things
that are actually accepted.

Note, you are showing you lack of understanding of the terms.

You find FLAWS in proofs, where the proof uses an incorrect logical
step, or an incorrect (or unproven) premises was used.

almost ALL your premises are unproved (since you never show an actual
reference to where you got them from).

>
> *The actual mistake is an error of omission*
> I found a loophole that no one every noticed before because no one ever
> bothered to examine the effect of a simulating halt decider on the
> conventional pathological inputs and *think this all-the-way through*

Nope, you THINK there is a loop hole, but that is only because you are
just totally ignorant of ANY of the material you are talking about.

The omission is your omission of actually learning the material you are
talking about.

You are NOT talking about the Halting Problem, but only your POOP, and
no one cares about your POOP.

>
> Most everyone simply rejects simulation out-of-hand and thus never gets
> to the point of considering the possibility of a simulating halt decider.

YOU have admitted to rejecting the fundamental definitions.

OF COURSE we reject your simulation, because you have admitted that you
aren't following the rules.

A simulating Halt decider is a well know method of attempt. Its problems
are also well known. The fact that you don't understand this just shows
how little you know.

The problem with simulating Halt Deciders is they don't really perform
any better that just running the machine itself. Yes, there are a
battery of well know patterns that can be added to a simulator to
quickly check for the major issues that get a machine into the simple
infinte loops, and they are good for that.

They just don't help with the tough problems where you really want a
Halt Decider to give you a QUICK answer about a tough problem.

>
> If you paid close attention to what I just said in (the post you are
> responding to) you would see that it is correct.
>
>>
>>> --
>>> 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: Are the halting problem proofs refuted by software engineering? (simple overview)

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 by: Richard Damon - Sat, 30 Jul 2022 22:09 UTC

On 7/30/22 5:46 PM, Ben Bacarisse wrote:
> Alan Mackenzie <acm@muc.de> writes:
>
>> olcott <NoOne@nowhere.com> wrote:
>>
>> [ .... ]
>>
>> No, they're not. Mathematical proofs, by their nature, are not refuted
>> by anything.
>
> This is key. I am sure now (and I should have been sure years ago) that
> PO does not know what a proof is. He thinks of mathematical proofs like
> courtroom arguments -- clever stories that can be used to trick naive
> juries (the students who "learn by rote") into believing falsehoods.
> And just as those sorts of "proofs" can be refuted by some piece of
> hitherto unknown evidence, mathematical proofs can be refuted by saying
> something no one has previously said!
>
> I once tried to get PO to give an example of proved theorem that he had
> rock-solid confidence in -- something he was more sure of (because of
> the proof) than, say, the likelihood of the sun rising tomorrow. He
> ducked that question for ages.
>

If you listen to some of the corners of Peter's discussions, he has some
familiarity with other branches of Philosophy where things often go not
to "proof", but to "Arguments".

Things like arguing over what should be the definition of Knowledge.

He totally doesn't understand that in Formal Logic, which is the broad
classsificaiton that all this falls under, that is NOT how you work on
things.

Yes, he absolutely doesn't know how to write an actual proof.

He thinks he knows a bit because he can cut and paste the fancy squigles
used in some circles, but he doesn't actually know what they really mean.

Re: Are the halting problem proofs refuted by software engineering? (simple overview)

<tc4dgr$che$1@gioia.aioe.org>

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Newsgroups: comp.theory
Subject: Re: Are the halting problem proofs refuted by software engineering?
(simple overview)
Date: Sat, 30 Jul 2022 16:06:01 -0700
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 by: Horatio Cornholer - Sat, 30 Jul 2022 23:06 UTC

On 7/30/2022 7:35 AM, olcott wrote:
> *Everyone here should be able to confirm this in less than a minute*
> It is very easy to see that the input to H(P,P) correctly simulated by H

Holy shit give it a rest you impotent blithering megatard.

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