Re: Olcott's theory

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olcott

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Jul 10, 2021, 1:08:10 PMJul 10
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On 7/10/2021 12:00 PM, Mr Flibble wrote:
> I agree with Olcott that a halt decider can NOT be part of that which
> is being decided (see [Strachey 1965]) which, if Olcott is correct,
> falsifies a collection of proofs (which I don't have the time to
> examine) which rely on that mistake. The mistake Olcott seems to be
> making is inferring that just because those proofs are invalid then that
> somehow means that the halting problem itself is not undecidable.
>
> /Flibble
>

[An impossible program] C. Strachey
The Computer Journal, Volume 7, Issue 4, January 1965, Page 313,
https://doi.org/10.1093/comjnl/7.4.313

https://academic.oup.com/comjnl/article/7/4/313/354243

No I never made that mistake, My goal since 2004 has always been to
simply prove that the proofs are incorrect. Here is my original 2004
claim that everyone has disagreed with in more than 15,000 messages
since 2004:

[Halting Problem Final Conclusion]
comp.theory Peter Olcott Sep 5, 2004, 11:21:57 AM
The Liar Paradox can be shown to be nothing more than
a incorrectly formed statement because of its pathological
self-reference. The Halting Problem can only exist because
of this same sort of pathological self-reference.
https://groups.google.com/g/comp.theory/c/RO9Z9eCabeE/m/Ka8-xS2rdEEJ

--
Copyright 2021 Pete Olcott

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

olcott

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Jul 10, 2021, 2:06:43 PMJul 10
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On 7/10/2021 12:12 PM, Mr Flibble wrote:
> On Sat, 10 Jul 2021 12:08:02 -0500
> olcott <No...@NoWhere.com> wrote:
>
>> The Halting Problem can only exist because
>> of this same sort of pathological self-reference.
>
> ^ that is your mistake: the halting problem still exists even if a
> collection of proofs have a mistake.
>
> Does [Turing 1937] rely on a decider being part of that which is
> being decided? Wikipedia suggests not:
>
> "Turing's proof departs from calculation by recursive functions and
> introduces the notion of computation by machine."
>
> /Flibble
>

*I agree that I have not solved the halting problem*
At most I have only proved that the conventional proofs of the
undecidability of the halting problem that rely on the Strachey form,
are incorrect. This seems to include all textbook proofs.

[An impossible program] C. Strachey
The Computer Journal, Volume 7, Issue 4, January 1965, Page 313,
https://doi.org/10.1093/comjnl/7.4.313

Now seems to be a good time to finally look at the Turing proof.
https://www.cs.virginia.edu/~robins/Turing_Paper_1936.pdf
I am not sure if the above linked copy has the later published correction.

If the Turing proof is isomorphic to the Strachey form, I don't know
what it left to prove that the halting problem is undecidable.

Goldbach's Conjecture is merely undecided and thus not undecidable.
https://en.wikipedia.org/wiki/Goldbach%27s_conjecture

Busy Beaver only seems intractable not undecidable.
https://en.wikipedia.org/wiki/Busy_beaver

olcott

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Jul 10, 2021, 2:32:48 PMJul 10
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On 7/10/2021 1:23 PM, Mr Flibble wrote:
> On Sat, 10 Jul 2021 13:06:35 -0500
> olcott <No...@NoWhere.com> wrote:
>
>> On 7/10/2021 12:12 PM, Mr Flibble wrote:
>>> On Sat, 10 Jul 2021 12:08:02 -0500
>>> olcott <No...@NoWhere.com> wrote:
>>>
>>>> The Halting Problem can only exist because
>>>> of this same sort of pathological self-reference.
>>>
>>> ^ that is your mistake: the halting problem still exists even if a
>>> collection of proofs have a mistake.
>>>
>>> Does [Turing 1937] rely on a decider being part of that which is
>>> being decided? Wikipedia suggests not:
>>>
>>> "Turing's proof departs from calculation by recursive functions and
>>> introduces the notion of computation by machine."
>>>
>>> /Flibble
>>>
>>
>> *I agree that I have not solved the halting problem*
>> At most I have only proved that the conventional proofs of the
>> undecidability of the halting problem that rely on the Strachey form,
>> are incorrect. This seems to include all textbook proofs.
>
> You need to reign in the ego, mate, Strachey deserves the credit not
> you unless you are claiming that you have independently reached the same
> conclusion as Strachey without being aware of Strachey until recently?
>
> /Flibble
>

All of the textbooks cite the Strachey form as proof that the halting
problem is undecidable. Ben, Mike and Kaz agree that the Strachey form
proves that the halting problem is undecidable.

rec routine P
§L:if T[P] go to L
Return §

If T[P] = True the routine P will loop, and it will
only terminate if T[P] = False. In each case T[P] has
exactly the wrong value, and this contradiction shows
that the function T cannot exist.

When Strachey says:
"this contradiction shows that the function T cannot exist."

He is saying that he just proved that a universal halt decider {function
T} does not exist.

olcott

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Jul 10, 2021, 3:09:15 PMJul 10
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On 7/10/2021 1:59 PM, Mr Flibble wrote:
> On Sat, 10 Jul 2021 13:45:38 -0500
> olcott <No...@NoWhere.com> wrote:
>
>> On 7/10/2021 1:38 PM, Mr Flibble wrote:
>>> On Sat, 10 Jul 2021 13:32:40 -0500
>>> No he isn't; he is saying that a decider can not be part of what is
>>> being decided which is quite different.
>>
>> So when Strachey says:
>> "this contradiction shows that the function T cannot exist."
>>
>> Strachey does not mean
>> {this contradiction shows that the function T cannot exist.}
>
> He means T cannot decide on P if T is called from within P; i.e. the
> "pathological self reference" you keep going on about.
>
> /Flibble
>

Yes and he and everyone else here besides you and I believes that this
proves that the halting problem is undecidable.

Mr Flibble

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Jul 10, 2021, 3:14:47 PMJul 10
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On Sat, 10 Jul 2021 14:09:07 -0500
When I read Strachey's letter I didn't get the impression that that was
his conclusion; merely that T cannot decide on P if called from within
P .. i.e. the "Impossible Program".

/Flibble

olcott

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Jul 10, 2021, 3:32:17 PMJul 10
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None-the-less everyone else does get that impression.
All of the textbook halting problem undecidability proofs rely on the
Strachey form as their entire basis.

http://www.liarparadox.org/sipser_165.pdf

Mr Flibble

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Jul 10, 2021, 3:36:02 PMJul 10
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On Sat, 10 Jul 2021 14:32:10 -0500
Then you might be on to something but you need to stop implying the
halting problem itself is not undecidable in your posts as it doesn't
help your case (and is the reason I have been dismissive of your posts
in the past).

/Flibble

olcott

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Jul 10, 2021, 5:04:09 PMJul 10
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On 7/10/2021 2:30 PM, Peter wrote:
> There is more than one proof.  I was taught, not the the halting problem
> for TMs is unsolvable, but that the halting problem for register
> machines is unsolvable.  The reason being, I suppose, that the course
> was taught by one of the inventors of the register machine.  [There are
> a variety of register machines, this one had an unlimited number of
> registers and the instructions

All of the proofs seem to boil down to this one idea:

An impossible program
C. Strachey
The Computer Journal, Volume 7, Issue 4, January 1965, Page 313,
https://doi.org/10.1093/comjnl/7.4.313

> add 1 to register R,
> subtract 1 from register R (so long as it doesn't hold 0),
> if R doesn't hold 0 go to instruction i.
> See Shepherdson & Sturgis, 'Computability of recursive functions',
> /JACM/, vol 10, no 2, 1963, pp217-255.]

olcott

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Jul 10, 2021, 5:12:20 PMJul 10
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Like I already explained in much more words with many more key details,
if none of these conventional (Strachey based) undecidability proofs are
correct then that doesn't seem to leave any other proof of halting
undecidability:

Goldbach's Conjecture is merely undecided and thus not undecidable.
https://en.wikipedia.org/wiki/Goldbach%27s_conjecture

Busy Beaver only seems intractable thus not undecidable.
https://en.wikipedia.org/wiki/Busy_beaver

It is probably a good time for me to take a first look at the actual
Turing proof. I merely need to verify that it is isomorphic to the
Strachey form.

olcott

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Jul 10, 2021, 5:46:36 PMJul 10
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On 7/10/2021 4:39 PM, Mr Flibble wrote:
> On Sat, 10 Jul 2021 16:12:12 -0500
> You are still missing the point: even if [Turing 1937] has the same
> mistake you still cannot prove that the halting problem itself is not
> undecidable just because that particular proof is invalid.
>
> /Flibble
>

None-the-less once the halting problem is no longer provably undecidable
computation loses its definite limits.

The key other aspect of this is that the Tarski undefinability theorem
can be understood to fail for the same reason that the halting theorem
fails.

This can have an explosive effect on AI research. Davidson's truth
conditional semantics can finally be anchored in a formal mathematical
notion of truth.

olcott

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Jul 10, 2021, 5:58:20 PMJul 10
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On 7/10/2021 4:52 PM, Mr Flibble wrote:
> On Sat, 10 Jul 2021 16:46:28 -0500
> No, a lack of a proof showing that the halting problem is undecidable
> does NOT imply that the halting problem is not undecidable; that still
> needs to be proven.
>
> [snip]
>
> /Flibble
>

It transforms what was previously thought to be known as a definite
limit to all computation into no known limit to computation. This is huge.

olcott

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Jul 10, 2021, 10:25:40 PMJul 10
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On 7/10/2021 8:56 PM, Ben Bacarisse wrote:
> olcott <No...@NoWhere.com> writes:
>
>> All of the textbooks have this same proof.
>
> You have not even read the proper proof given in the textbook that you
> appear to have: "An Introduction to Formal Languages and Automata" by
> Peter Linz.
>
>> As long as the Turing proof is based on the same idea I am set.
>
> Turing never published a proof of the halting theorem, though he
> certainly knew it. His 1936 paper was not concerned with halting, but
> with defining "commutable numbers".
>

Yes numbers that live in New Jersey and go to work in New York city.
They commute to work.

> I think it was Martin Davis who coined the term "halting problem", but
> his use of the terms is not what you would recognise, and the proof is
> entirely different.
>
> I don't know who first published a proof of the form you are so obsessed
> with. If anyone knows, I'd like to hear about it. Of course it comes
> in many guises, for Turing machines and other models like register
> machines (as I first encountered it), so maybe it was not originally
> given for TMs at all?
>

This one is the the apparent forerunner to all of the pseudo-code proofs
and was written by the creator of the CPL programming language, the
ancestor to BCPL, B and C.

An impossible program C. Strachey
The Computer Journal, Volume 7, Issue 4, January 1965, Page 313,
https://doi.org/10.1093/comjnl/7.4.313

*Here are Strachey's (verbatim) own words*
Suppose T[R] is a Boolean function taking a routine
(or program) R with no formal or free variables as its
argument and that for all R, T[R] — True if R terminates
if run and that T[R] = False if R does not terminate.
Consider the routine P defined as follows

rec routine P
§L:if T[P] go to L
Return §

If T[P] = True the routine P will loop, and it will
only terminate if T[P] = False. In each case T[P] has
exactly the wrong value, and this contradiction shows
that the function T cannot exist.

Strachey is the creator of CPL ancestor to BCPL then B then C
His code above is written in his CPL programming language.

He was criticized in that his use of terminology is not precisely
correct by modern standards, yet apparently the essence of what he
conveyed is exactly correct.

I am happy to say that my C version works just fine:

// Simplified Linz Ĥ (Linz:1990:319)
void P(u32 x)
{
if (H(x, x))
HERE: goto HERE;
}

I will model my proof after Strachey instead of Linz, yet still
reference Linz.

_P()
[00000b66](01) 55 push ebp
[00000b67](02) 8bec mov ebp,esp
[00000b69](03) 8b4508 mov eax,[ebp+08]
[00000b6c](01) 50 push eax
[00000b6d](03) 8b4d08 mov ecx,[ebp+08]
[00000b70](01) 51 push ecx
[00000b71](05) e8f0fdffff call 00000966 // call H
[00000b76](03) 83c408 add esp,+08
[00000b79](02) 85c0 test eax,eax
[00000b7b](02) 7402 jz 00000b7f
[00000b7d](02) ebfe jmp 00000b7d
[00000b7f](01) 5d pop ebp
[00000b80](01) c3 ret
Size in bytes:(0027) [00000b80]

olcott

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Jul 11, 2021, 11:09:56 AMJul 11
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On 7/11/2021 7:19 AM, Peter wrote:
> olcott wrote:
>>> And you know that how?  Have you read _all_ of the proofs?
>>>
>>
>> All of the textbooks have this same proof.
>
> You haven't read all the textbooks, and there are proofs in papers and
> lectures.  The mention of Martin Davis downstream led me to his book
> /Computability and unsolvability/, McGraw-Hill 1958 (also recent Dover
> republication).  He proves the unsolvability of the halting problem by
> noting that (exists y)T(x,x,y) is uncomputable (T being Kleene's T
> predicate).  You seem not to know what you are talking about.
>

https://en.wikipedia.org/wiki/Kleene%27s_T_predicate

That Gödel numbering crap simply hides the pathological
self-reference(Olcott 2004) error behind so much complexity that it
can't be seen.

When we ask what Boolean value can a halt decider return to an input
that changes its behavior to contradict this value we cannot answer this
question because it is an incorrect type mismatch error question. The
answer is restricted to {true, false} thus excluding the correct answer
of “neither” making the question itself incorrect. The TM / input pairs
that “prove” the halting problem is undecidable have the same
pathological self-reference(Olcott 2004) error as the self-contradictory
liar paradox.

To eliminate this pathological feedback loop error we examine the
behavior of the input with a pure simulator that has no effect
what-so-ever on the behavior of the input. As correct science requires
the dependent variable (the halt status decision) must only have the
independent variable (the behavior of the input) and be isolated from
all other influences. Only when we do it this way do we get the correct
halt status decision for the input.

Until the behavior of its input proves that it will never halt every H
remains a pure simulator of this input. This single fact by itself
proves that the behavior of H has no effect what-so-ever on its halt
status decision. When H stops simulating its input the execution of the
input has been suspended, this does not count as halting.


>> As long as the Turing proof is based on the same idea I am set.
>>

olcott

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Jul 11, 2021, 1:45:14 PMJul 11
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> Why are you ignoring the Davis proof that I referred to?
>

I have been saying that the halting problem counter-examples have the
same erroneous form as the liar paradox since 2004.

comp.theory Peter Olcott Sep 5, 2004, 11:21:57 AM
[Halting Problem Final Conclusion]
The Liar Paradox can be shown to be nothing more than
a incorrectly formed statement because of its pathological
self-reference. The Halting Problem can only exist because
of this same sort of pathological self-reference.
https://groups.google.com/g/comp.theory/c/RO9Z9eCabeE/m/Ka8-xS2rdEEJ

Flibble agrees that all of the proofs that depend on this pathological
self-reference are incorrect.

[Olcott's theory] Mr Flibble Jul 10, 2021, 12:00:56 PM
I agree with Olcott that a halt decider can NOT be part of that which
is being decided (see [Strachey 1965]) which, if Olcott is correct,
falsifies a collection of proofs (which I don't have the time to
examine) which rely on that mistake.
https://groups.google.com/g/comp.theory/c/6cEnndkkrKA/m/gRj0x9KOBgAJ

The incompleteness Theorem merely hides its pathological self-reference
behind the enormous complexity of Gödel numbering. When we recreate this
same proof in HOL without Gödel numbering, then we can see its
pathological self-reference error.

Because Kleene's T predicate uses the same Gödel numbering it is
reasonable to assume that it may have the same hidden pathological
self-reference error as the Gödel proof.
https://en.wikipedia.org/wiki/Kleene%27s_T_predicate

olcott

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Jul 12, 2021, 6:21:19 PMJul 12
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On 7/12/2021 4:57 PM, Andy Walker wrote:
> On 12/07/2021 21:08, Mike Terry wrote:
>> On 12/07/2021 20:27, Jeff Barnett wrote:
>>> On 7/12/2021 7:19 AM, Malcolm McLean wrote:
> [I wrote:]
>>>>> It's undecidable whether an arbitrary TM is a BB; and the
>>>>> BB function [ie, the "score" of a BB of given size] is uncomputable.
>>>>> That's a stronger result than intractability. The proof that the
>>>>> BB function is uncomputable is [reasonably] elementary, and does not
>>>>> rely on the HP [or recursion]. If you had a "halting decider", then
>>>>> you could find BBs by exhaustive [albeit intractable!] search and
>>>>> thus compute the BB function; so there's yet another HP proof.
>>>> If you have a busy beaver function and a simulating decider, the busy
>>>> beaver function tells you how many times you need to step the decider
>>>> before you can be certain it will never halt. So halting is
>>>> decideable if
>>>> busy beaver is decideable.
>
>     Yes [-ish, see below], but more interesting is "HP /un/decidable
> if BB /un/computable".
>
>>> I'm pretty sure that isn't correct.
>
>     Well, it should be "if the BB function is /computable/", otherwise
> it's more-or-less correct.  The "less" part is because you need a slightly
> different version of BB, but there's no important distinction.
>
>> It looks right, but you might be misunderstanding the point being made.
>> The claim is:
>>    IF we had a BB function [meaning if BB function were computable]
>>    THEN that would imply Halting is decidable, by virtue of constructing
>>    a simple simulating based decider as outlined.
>> In more detail, [...].
>
>     Yes.
>
>> There was also a prior claim concerning BB "decideability", which was
>> taken as meaning "whether a given TM is in CLASS_BB" is decideable.
>> A TM in CLASS_BB is one that halts in the maximum number of steps
>> possible for a TM with that number of states. [...]
>> So "CLASS_BB decideable"  ==> "BB function is computable" ==>
>> "Halting is decideable".
>
>     Yes, but more importantly "BB function uncomputable" is known
> separately and quite independently of the HP [there are simple proofs,
> inc the original one by Rado and one given in the Wiki article on BBs
> (which /also/ gives a proof depending on the blank tape HP, equivalent
> to the standard HP)], so we deduce that "CLASS BB" is undecidable.
>
>> But Halting is NOT decideable, therefore we conclude:
>>    a)  BB function is NOT computable
>>    b)  CLASS_BB is NOT decideable
>
>     Again, this is more important the other way round.  A halt-
> decider would be able to to be tweaked [as above] to compute the
> BB function.  But the BB function is uncomputable;  so Halting is
> not decidable.  This proof is elementary, different from either
> Linz proof, and non-recursive.
>

The key question is whether or not BB proves halting undecidability by
itself within the assumption that the conventional HP proofs have been
correctly refuted.

olcott

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Jul 12, 2021, 6:23:07 PMJul 12
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On 7/12/2021 4:59 PM, Mike Terry wrote:
> On 12/07/2021 21:24, Mr Flibble wrote:
>> On Mon, 12 Jul 2021 21:08:02 +0100
>> Mike Terry <news.dead.p...@darjeeling.plus.com> wrote:
>>
>>> But Halting is NOT decideable, therefore we conclude:
>>
>> Assertions made without evidence can be dismissed without evidence
>> [Hitchens].  Any proof predicated on a misunderstanding of [Strachey
>> 1965] is in error.  Which proof are you basing this assertion on?
>> [Turing 1937]?
>>
>> /Flibble
>>
>
> There are several proofs that halting is not decideable.  Take your
> pick.  The obvious choice given PO's claims over many years would be the
> Linz proof, or similar.  (The Linz proof is slightly sloppy in its
> attention to detail, as it's not aimed at mathematicians, but that's
> nothing that can't be routinely patched up.)
>
> Mike.

I refuted Linz as well. When the whole class of all conventional HP
proofs have been refuted, then does the halting problem become
decidability unknown?

olcott

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Jul 12, 2021, 7:23:08 PMJul 12
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On 7/12/2021 5:40 PM, Ben Bacarisse wrote:
> Mr Flibble <fli...@reddwarf.jmc> writes:
>
>> On Mon, 12 Jul 2021 21:08:02 +0100
>> Mike Terry <news.dead.p...@darjeeling.plus.com> wrote:
>>
>>> But Halting is NOT decideable, therefore we conclude:
>>
>> Assertions made without evidence can be dismissed without evidence
>> [Hitchens].
>
> I suspect he knew (and appreciated) the irony of his making such an
> assertion without any evidence.
>
>> Any proof predicated on a misunderstanding of [Strachey
>> 1965] is in error.
>
> I don't think any proper proof is based on such code sketches. PO is
> reduced to clumsy sentences and borrowed code sketches, but that's
> because he has no access to any other models of computation. And he's
> not proving anything either, despite his claims to the contrary.
>
>> Which proof are you basing this assertion on?
>
> I doubt anyone but a novice could answer that. Do I rely on the
> argument using register machines, shown to me when I was an
> undergraduate? I did when I was a novice. Do I rely on Davis? Linz's
> actual proof (rather than the one PO has been confused by) as a
> corollary to the proof that there exists a recursively enumerable
> languaga that is not recursive is pretty simple. There are least three
> others in book at arms length from me right now. I really could not say
> I rely on any one of these.
>

At first glance its seems that Linz theorem 11.3 might be the same sort
of nonsense as the Tarski undefinability theorem that essentially
"proves" that some expressions of language are true and unprovable.

The Tarski proof is fundamentally anchored in the liar paradox:
http://www.liarparadox.org/Tarski_247_248.pdf

Even after thousands of years people still do not understand that self
contradictory expressions of language do not map to a Boolean value only
because they are erroneous.

the symbol 'Pr' which denotes the class of all provable sentences of the
theory under consideration ...

we can construct a sentence x of the science in question which satisfies
the following condition: it is not true that x ∉ Pr if and only if p
or in equivalent formulation: (1) x ∉ Pr if and only if p
where the symbol 'p' represents the whole sentence x

In other words this: ¬Pr(p) ↔ p was derived from this ¬Tr(p) ↔ p
where Pr means Provable() and Tr means True()

http://www.liarparadox.org/Tarski_275_276.pdf


> Which do you rely on?
>
>> [Turing 1937]?
>
> As far as I know, Turing never published a proof of the halting theorem.
> You can derive one from the results in that seminal paper, but that's
> not quite the same thing.

olcott

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Jul 12, 2021, 10:11:07 PMJul 12
to
On 7/12/2021 6:45 PM, Mike Terry wrote:
>>> But Halting is NOT decideable, therefore we conclude:
>>>    a)  BB function is NOT computable
>>>    b)  CLASS_BB is NOT decideable
>>
>>      Again, this is more important the other way round.  A halt-
>> decider would be able to to be tweaked [as above] to compute the
>> BB function.  But the BB function is uncomputable;  so Halting is
>> not decidable.  This proof is elementary, different from either
>> Linz proof, and non-recursive.
>>
>
> Yeah, the halting problem and computing the BB function are equivalent
> problems as each reduces easily to the other.

This would seem to mean that as long as the class of conventional HP
proofs have been refuted that this same refutation could be transformed
to refute BB.

> I've seen direct proofs
> of BB being uncomputable, but don't instantly recall them, but I recall
> them being quite straight forward.  But somehow I think getting PO to
> understand this and getting him to accept the consequences is not going
> to happen.  Even understanding that refuting one proof doesn't mean he
> has refuted the HP result seems too subtle for PO.

When every "input does the opposite of whatever the halt decider
decides" proof has been refuted this includes all of the conventional
proofs.

Now we construct a new Turing machine D with H as a subroutine. This new
TM calls H to determine what M does when the input to M is its own
description (M). Once D has determined this information, it does the
opposite. http://www.liarparadox.org/sipser_165.pdf

>
>
> Mike.

olcott

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Jul 13, 2021, 9:49:36 AMJul 13
to
On 7/13/2021 5:49 AM, Andy Walker wrote:
> On 12/07/2021 23:21, olcott wrote:
>> The key question is whether or not BB proves halting undecidability
>> by itself within the assumption that the conventional HP proofs have
>> been correctly refuted.
>
>     That much is trivial.  It's just the contrapositive of the
> result [given in more detail by MikeT a few posts back] that if we
> have a halt decider then the BB function is computable.
>

Great! This might seem to indicate that at the point in time when it is
accepted that I have correctly refuted the conventional Halting Problem
proofs then the halting problem is transformed from undecidable to
decidability unknown.

olcott

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Jul 13, 2021, 10:00:46 AMJul 13
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On 7/13/2021 5:51 AM, wij wrote:
> On Tuesday, 13 July 2021 at 12:31:47 UTC+8, olcott wrote:
>> On 7/12/2021 10:32 PM, Richard Damon wrote:
>>> On 7/12/21 4:23 PM, olcott wrote:
>>>> On 7/12/2021 4:59 PM, Mike Terry wrote:
>>>>> On 12/07/2021 21:24, Mr Flibble wrote:
>>>>>> On Mon, 12 Jul 2021 21:08:02 +0100
>>>>>> Mike Terry <news.dead.p...@darjeeling.plus.com> wrote:
>>>>>>
>>>>>>> But Halting is NOT decideable, therefore we conclude:
>>>>>>
>>>>>> Assertions made without evidence can be dismissed without evidence
>>>>>> [Hitchens]. Any proof predicated on a misunderstanding of [Strachey
>>>>>> 1965] is in error. Which proof are you basing this assertion on?
>>>>>> [Turing 1937]?
>>>>>>
>>>>>> /Flibble
>>>>>>
>>>>>
>>>>> There are several proofs that halting is not decideable. Take your
>>>>> pick. The obvious choice given PO's claims over many years would be
>>>>> the Linz proof, or similar. (The Linz proof is slightly sloppy in its
>>>>> attention to detail, as it's not aimed at mathematicians, but that's
>>>>> nothing that can't be routinely patched up.)
>>>>>
>>>>> Mike.
>>>>
>>>> I refuted Linz as well. When the whole class of all conventional HP
>>>> proofs have been refuted, then does the halting problem become
>>>> decidability unknown?
>>>>
>>>
>>> You have NOT refuted Linz as your claimed correct answer is not correct.
>>> Since it is shown tha tH^(H^) when run as an independent machine does
>>> reach its final HALT state, and thus it IS a Halting Turing Machine by
>>> definition. Even YOU have admitted that H^(H^) will come to this Halt
>>> and have provided traces that prove it.
>>>
>>> You just want to change the definition so you can work on POOP.
>>>
>> When we ask what Boolean value can a halt decider return to an input
>> that changes its behavior to contradict this value we cannot answer this
>> question because it is an incorrect type mismatch error question.
>
> Nop, H is a total function, every input is valid.

If this is true then the question:
What time is it (yes or no)? is valid.
Because it is very obvious that the above question has a type mismatch
error then every input to a TM that has a type mismatch error is also an
incorrect question.

bool Add(int X, int Y); // Is 2 + 3 true or false?

>> The answer is restricted to {true, false} thus excluding the correct
>> answer of “neither” making the question itself incorrect.
>
> The answer is restricted to {true,false} because it is fundamental of All functions.
> So, computation theory needs only address decision functions.
> The answer set can be extended to {true,false,neither}, but the result is the same
> ... "undecidable".
>

Just like the above function has no correct return value because of a
type mismatch error any input having behavior that contradict the
decision of the halt decider is also incorrect.

It is the pathological self-reference(Olcott 2004) error that Flibble
agreed to as erroneous.

On 7/10/2021 12:00 PM, Mr Flibble wrote:
> I agree with Olcott that a halt decider can NOT be part of that which
> is being decided (see [Strachey 1965]) which, if Olcott is correct,
> falsifies a collection of proofs (which I don't have the time to
> examine) which rely on that mistake.

>> The TM / input pairs that “prove” the halting problem is undecidable
>> have the same pathological self-reference(Olcott 2004) error as the
>> self-contradictory liar paradox.
>
> HP is computational problem, liar paradox is not.
> You need to DEFINE your pathological error to computational problem form.
>

// Must return the actual sum as a Boolean
bool Add(int X, int Y)
{
return (bool)(X + Y);
};

int main()
{
Add(2,3);
}


> --
> Copyright 2021 WIJ
> "If I can see further it is by standing on top of the tower of dwarfs."

olcott

unread,
Jul 13, 2021, 10:36:10 AMJul 13
to
On 7/12/2021 10:27 PM, Richard Damon wrote:
> On 7/12/21 3:51 PM, olcott wrote:
>> On 7/12/2021 10:33 AM, David Brown wrote:
>>> On 12/07/2021 16:25, olcott wrote:
>>>> On 7/12/2021 8:19 AM, Malcolm McLean wrote:
>>>>> On Monday, 12 July 2021 at 14:13:15 UTC+1, Andy Walker wrote:
>>>>>> On 10/07/2021 19:06, olcott wrote:
>>>>>>> At most I have only proved that the conventional proofs of the
>>>>>>> undecidability of the halting problem that rely on the Strachey form,
>>>>>>> are incorrect. This seems to include all textbook proofs.
>>>>>> As you know, Linz gives a quite different proof; and there are
>>>>>> several other proofs, inc one via "Busy Beaver" [see below].
>>>>>>
>>>>>> [...]
>>>>>>> Goldbach's Conjecture is merely undecided and thus not undecidable.
>>>>>> "Thus" is a step too far. But note the reference to Goldbach
>>>>>> in the Wiki article on "Busy Beaver".
>>>>>>
>>>>>> [...]
>>>>>>> Busy Beaver only seems intractable not undecidable.
>>>>>> It's undecidable whether an arbitrary TM is a BB; and the
>>>>>> BB function [ie, the "score" of a BB of given size] is uncomputable.
>>>>>> That's a stronger result than intractability. The proof that the
>>>>>> BB function is uncomputable is [reasonably] elementary, and does not
>>>>>> rely on the HP [or recursion]. If you had a "halting decider", then
>>>>>> you could find BBs by exhaustive [albeit intractable!] search and
>>>>>> thus compute the BB function; so there's yet another HP proof.
>>>>>>
>>>>> If you have a busy beaver function and a simulating decider, the busy
>>>>> beaver function tells you how many times you need to step the decider
>>>>> before you can be certain it will never halt. So halting is
>>>>> decideable if
>>>>> busy beaver is decideable.
>>>>>
>>>>>
>>>>
>>>> Flibble understands this:
>>>> What correct Boolean value can a TM return to an input that does the
>>>> opposite of whatever it decides? : neither, is excluded from the
>>>> solution set thus making the question itself incorrect.
>>>>
>>>
>>> You still don't understand this "proof by contradiction" concept, do you?
>>
>> We can say that the following expression is undecidable on the basis
>> that both true and false derive contradictions:
>> "This sentence is not true." That same sentence is used as the basis of
>> the Tarski undefinability theorem.
>>
>> http://www.liarparadox.org/Tarski_247_248.pdf
>> http://www.liarparadox.org/Tarski_275_276.pdf
>>
>>
>>
>> Although we have proven that it is undecidable on the basis of proof by
>> contradiction we are ignoring the key detail that the proof by
>> contradiction only succeeds because the expression of language is
>> self-contradictory, thus not a truth bearer and therefore erroneous.
>
> Does P(I) Halt IS a Truth Bearing Sentence, as P(I) will either Halt or
> it won't. We may not be able to prove that answer, but that is one
> fundamental property of the Higher level Logic that includes mathematics.
>

That is not the actual question, you leave out the key context that
makes the question incorrect. Ben played this same dishonest dodge since
2006. When you strip off the context of the question you change it into
an entirely different question.

Here is the actual question:
When we ask what Boolean value can a halt decider return to an input
that changes its behavior to contradict this value we cannot answer this
question because it is an incorrect type mismatch error question.

The answer is restricted to {true, false} thus excluding the correct
answer of “neither” making the question itself incorrect.

If we ask a man that has never been married:
Have you stopped beating you wife?
This is an incorrect question.
Every polar question lacking a correct yes/no answer is incorrect.

Every TM/input to a decision problem lacking a correct Boolean return
value is an incorrect TM/input for this decision problem.

The TM / input pairs that “prove” the halting problem is undecidable
have the same pathological self-reference(Olcott 2004) error as the
self-contradictory liar paradox.

It has been at least 2000 years since the liar paradox was discovered
and academicians still do not understand the because self-contradictory
sentences do not map to a Boolean value they are not truth bearers.

> If you want to keep to the concept that something is only true if you
> can prove it, then you must restrict yourself to only First Order Logic,
> or you WILL eventually run into an inconsistency in your logic system.
>

Not at all. Not in the least.
When Gödel's 1931 incompleteness theorem is translated into HOL that has
its own provability predicate thus no need for the purely extraneous
complexity of Gödel numbers then it is very obvious that it is modeled
after the liar paradox.

Gödel himself said that the liar paradox could be used as his
incompleteness basis.

there is also a close relationship with the “liar”
antinomy,14

We are therefore confronted with a proposition which
asserts its own unprovability

14 Every epistemological antinomy can likewise be
used for a similar undecidability proof.

This is HOL with its in provability predicate:
https://www.researchgate.net/publication/331859461_Minimal_Type_Theory_YACC_BNF

I have examples that show the propositions that assert their own
unprovability are only undecidable because there is an infinite cycle in
their directed graph.

The problem is that like the liar paradox that have a vacuous truth
object. This sentence is not true. This sentence is not provable. Have
no truth object that they are true or provable about.

"It is true that this sentence is a dump truck."
has a truth object thus enabling the sentence to be evaluated as false.

"This sentence is true"
"This sentence is false"
"This sentence is provable"

Have no truth object, thus specify an infinite cycle in the directed
graph of their resolution.

> Yes, if you make that restriction to your logic, you can use that
> property, but you can not express the fullness of mathematics in your logic.
>
> This is well established.
>

I reestablished it correctly this time.

olcott

unread,
Jul 13, 2021, 11:01:29 AMJul 13
to
On 7/13/2021 9:38 AM, wij wrote:
> Whatever program will be a string of 0's and 1's in the end. A total decision function
> does not care what it means.
>
>>>> The answer is restricted to {true, false} thus excluding the correct
>>>> answer of “neither” making the question itself incorrect.
>>>
>>> The answer is restricted to {true,false} because it is fundamental of All functions.
>>> So, computation theory needs only address decision functions.
>>> The answer set can be extended to {true,false,neither}, but the result is the same
>>> ... "undecidable".
>>>
>> Just like the above function has no correct return value because of a
>> type mismatch error any input having behavior that contradict the
>> decision of the halt decider is also incorrect.
>
> Then, your H is not a halting decider.

When-so-ever any yes/no question lacks a correct yes/no answer this
question is incorrect.

When-so-ever a TM/input pair to a decision problem lacks a correct
Boolean return value the TM/input pair is incorrect.

>> It is the pathological self-reference(Olcott 2004) error that Flibble
>> agreed to as erroneous.
>> On 7/10/2021 12:00 PM, Mr Flibble wrote:
>>> I agree with Olcott that a halt decider can NOT be part of that which
>>> is being decided (see [Strachey 1965]) which, if Olcott is correct,
>>> falsifies a collection of proofs (which I don't have the time to
>>> examine) which rely on that mistake.
>>>> The TM / input pairs that “prove” the halting problem is undecidable
>>>> have the same pathological self-reference(Olcott 2004) error as the
>>>> self-contradictory liar paradox.
>
> We need 'fact'. What does it have anything to do with the halting problem?
>

When-so-ever any yes/no question lacks a correct yes/no answer this
question is incorrect.

When-so-ever a TM/input pair to a decision problem lacks a correct
Boolean return value the TM/input pair is incorrect.

>>> HP is computational problem, liar paradox is not.
>>> You need to DEFINE your pathological error to computational problem form.
>>>
>> // Must return the actual sum as a Boolean
>> bool Add(int X, int Y)
>> {
>> return (bool)(X + Y);
>> };
>>
>> int main()
>> {
>> Add(2,3);
>> }
>>
>
> Yes, Add(int,int) does exactly what it is asked to do, so what?

olcott

unread,
Jul 13, 2021, 11:35:21 AMJul 13
to
On 7/13/2021 10:23 AM, wij wrote:
> Halting problem is a valid question:
>
> // [Syn] Decide whether the function %f, given the argument %a, will halt or not,
> //
> // [Ret] true: f(a) will return
> // false: otherwise (f(a) will not return)
> //
> bool H(Func f, Arg a);
>
> But according to GUA https://groups.google.com/g/comp.theory/c/65ZaXe9Sabk,
> H tries to decide the (dynamic) property of f that f can defy, thus, H is
> undecidable.
>
> Your H is a false teller.


H(P,P) never halts. If H(P,P) ever stops running this is because its
infinitely nested simulation has had its execution suspended. This does
not count as halting.


>
> --
> Copyright 2021 WIJ
> "If I can see further it is by standing on top of the tower of dwarfs."
>


olcott

unread,
Jul 13, 2021, 6:16:33 PMJul 13
to
On 7/13/2021 1:15 PM, Peter wrote:
> olcott wrote:
>> [...]
>> When Gödel's 1931 incompleteness theorem is translated into HOL that
>> has its own provability predicate thus no need for the purely
>> extraneous complexity of Gödel numbers then it is very obvious that it
>> is modeled after the liar paradox.
>
> It's almost certainly of no interest to you, but a proof of the
> incompleteness of second order predicate calculus may be found in
>
> Hans Hermes, (trans Herman & Plassmann) /Enumerability, decidability,
> computability/, 2nd edition, Springer, 1964.
>
> It does not use Gödel numbering.  Nor is it "is modeled after the liar
> paradox."
>

I did not initially add that it must also have its own provability
predicate like the Tarski undefinability theorem.

Even when we get these things boiled down to their simplest possible
essence no academicians can begin to understand what is really going on.

They still don't understand that the actual liar paradox is simply not a
truth bearer because it is self contradictory.

They all seem to think that it is some puzzle that is far too difficult
to ever figure out. The Tarski undefinability theory has the liar
paradox as its only basis.

http://www.liarparadox.org/Tarski_247_248.pdf

http://www.liarparadox.org/Tarski_Proof_275_276.pdf

olcott

unread,
Jul 13, 2021, 6:53:40 PMJul 13
to
On 7/13/2021 5:24 PM, wij wrote:
> On Wednesday, 14 July 2021 at 00:13:39 UTC+8, wij wrote:
>> What are you talking about?
>> It read to me your H is a telepathy program. If so, your H is qualified as a
>> undecidable program. This is the correct answer accepted by GUA, but
>> such a H does not conform to the spec. of a halting decider.
>> Copyright 2021 WIJ
>> "If I can see further it is by standing on top of the tower of dwarfs."
>
> I wrote it too fast, correction to previous reply:
> It read to me that your H is a telepathic program whose answer oscillates and
> can not output a fixed answer. According to GUA, such a H can be call an
> "undecidable" program...
>

All undecidable inputs are incorrect inputs.
The Tarski undefinability theorem only exists on the basis of the
impossibility of proving that a self-contradictory sentence is true.

olcott

unread,
Jul 13, 2021, 8:29:26 PMJul 13
to
On 7/13/2021 7:17 PM, olcott wrote:
> On 7/13/2021 7:06 PM, wij wrote:
>> On Wednesday, 14 July 2021 at 07:48:22 UTC+8, olcott wrote:
>>> On 7/13/2021 6:22 PM, wij wrote:
>>>> How do you compute "All undecidable inputs"?
>>>>
>>>> // [Syn] Decide whether or not $f is a undecidable input
>>>> //
>>>> // [Ret] true: $f is a undecidable input
>>>> // false: otherwise
>>>> //
>>>> D(Prog p);
>>>>
>>>> According to GUA
>>>> https://groups.google.com/g/comp.theory/c/65ZaXe9Sabk, D tries
>>>> to decide the (dynamic) property of p that p can defy, thus, D is
>>>> undecidable.
>>>>
>>>> I am afraid you are saying something you do not really understand.
>>>>
>>>>> The Tarski undefinability theorem only exists on the basis of the
>>>>> impossibility of proving that a self-contradictory sentence is true.
>>>>
>>>> Tarski undefinability theorem or the sort are sub-instances of GUA.
>>>>
>>>> --
>>>> Copyright 2021 WIJ
>>>>
>>> What time is it (yes or no)? is equally undecidable.
>>
>> I can provide a bunch answers to the irrelevant question. But
>
> What is the correct (yes or no) answer to the question What time is it?
>
> What is the correct halt status that a TM can return to an input that
> does the opposite of whatever the decider decides?
>
> What is the correct answer to the question:
> Have you stopped beating your wife?
> posed to a many that has never been married?
>
> All three examples have no correct answer only because they are
> incorrect questions.
>
>> Do you really honestly want to talk about your H(P,P) or the subject
>> of your paper
>> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

This is circular, thus nonsense:
'snow is white' is true if, and only if, snow is white

This is not circular:
"snow is white" ↔ has_physical_property(snow, color(white))

olcott

unread,
Jul 13, 2021, 9:40:16 PMJul 13
to
On 7/13/2021 8:14 PM, wij wrote:
> On Wednesday, 14 July 2021 at 08:17:22 UTC+8, olcott wrote:
>> On 7/13/2021 7:06 PM, wij wrote:
>>> On Wednesday, 14 July 2021 at 07:48:22 UTC+8, olcott wrote:
>>>> On 7/13/2021 6:22 PM, wij wrote:
>>>>> How do you compute "All undecidable inputs"?
>>>>>
>>>>> // [Syn] Decide whether or not $f is a undecidable input
>>>>> //
>>>>> // [Ret] true: $f is a undecidable input
>>>>> // false: otherwise
>>>>> //
>>>>> D(Prog p);
>>>>>
>>>>> According to GUA https://groups.google.com/g/comp.theory/c/65ZaXe9Sabk, D tries
>>>>> to decide the (dynamic) property of p that p can defy, thus, D is undecidable.
>>>>>
>>>>> I am afraid you are saying something you do not really understand.
>>>>>
>>>>>> The Tarski undefinability theorem only exists on the basis of the
>>>>>> impossibility of proving that a self-contradictory sentence is true.
>>>>>
>>>>> Tarski undefinability theorem or the sort are sub-instances of GUA.
>>>>>
>>>>> --
>>>>> Copyright 2021 WIJ
>>>>>
>>>> What time is it (yes or no)? is equally undecidable.
>>>
>>> I can provide a bunch answers to the irrelevant question. But
>> What is the correct (yes or no) answer to the question What time is it?
>
> Why do you keep asking irrelevant questions? Is it the way to evade your
> failure you dare not to face?
>
>> What is the correct halt status that a TM can return to an input that
>> does the opposite of whatever the decider decides?
>
> GUA https://groups.google.com/g/comp.theory/c/65ZaXe9Sabk
>
>> What is the correct answer to the question:
>> Have you stopped beating your wife?
>> posed to a many that has never been married?
>
> Why do you keep asking irrelevant questions? Is it the way to evade your
> failure you dare not to face?
>
>> All three examples have no correct answer only because they are
>> incorrect questions.
>
> It is YOU making them 'incorrect questions' to yourself.
>
> In computation theory, TM can only halt or not halt.
> Why do you keep asking irrelevant questions? Is it the way to evade your
> failure you dare not to face?
>

Although a TM either halts or does not halt when we do not freaking
ignore the context of this question (As dishonest Ben has done for 15
years)

Then the halt decider cannot "decide" between true and false only
because within the full context of the question {where the input does
the opposite of whatever the halt decider decides} both True and False
are the wrong answer, thus proving that the question itself is incorrect
when posed to this halt decider.

olcott

unread,
Jul 13, 2021, 10:08:55 PMJul 13
to
On 7/13/2021 8:58 PM, wij wrote:
> Halt decider H is intended a total function. All input in the problem domain are valid.
> There is no such thing as 'incorrect input' in the problem domain.

Yes that is the key universal misconception that only Flibble and I
understand to be a misconception.

When we comprehend that every yes/no question having no possible correct
yes/no answer only includes incorrect questions such as asking a man
that has never been married: Have you stopped beating your wife?

Then we understand that this same analytical framework also applies to
decision problems such that a TM/input pair is defined such that both
true/false are the wrong answer.

> Otherwise, you need to define 'incorrect input', but then, this is another
> undecidable problem. I.e. you can not define it.
>
> --
> Copyright 2021 WIJ

olcott

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Jul 13, 2021, 11:39:22 PMJul 13
to
On 7/13/2021 9:52 PM, wij wrote:
> What precisely the misconception are you referring to?

I explain that in the past that you skipped below:

>> When we comprehend that every yes/no question having no possible correct
>> yes/no answer only includes incorrect questions such as asking a man
>> that has never been married: Have you stopped beating your wife?
>>
>> Then we understand that this same analytical framework also applies to
>> decision problems such that a TM/input pair is defined such that both
>> true/false are the wrong answer.

"Undecidable" TM/input pairs are only "undecidable" because they are
incorrect.

olcott

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Jul 14, 2021, 12:18:00 AMJul 14
to
On 7/13/2021 11:00 PM, wij wrote:
> Why "undecidable" is incorrect?

The same reason that a yes/no question having no correct yes/no answer
is incorrect.

> With my GUA https://groups.google.com/g/comp.theory/c/65ZaXe9Sabk, one need
> not to define 'incorrect input'. I think no one can refute GUA.
>

The above link is only Sipser:
http://www.liarparadox.org/Sipser_165_167.pdf

> --
> Copyright 2021 WIJ
> "If I can see further it is by standing on top of the tower of dwarfs."
>


olcott

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Jul 14, 2021, 10:56:26 AMJul 14
to
On 7/13/2021 11:45 PM, Richard Damon wrote:
> But that ISN'T the question of the Halting Problem.

That is the question of this specific TM/input pair applied to the
halting problem.

When we ask someone that has never been married :
Have you stopped beating your wife?

Within the full context that it is someone that has never been married
then this question becomes incorrect. If we simply ignore the context
then the analysis is incomplete.

For the decision problem of the halting problem some TM/input pairs are
incorrect and others are correct, the context of TM/input pair must be
considered or the analysis is incomplete.

Since the above two questions are isomorphic we can know with complete
certainty that the context of the TM/input pair must be considered.

I am the creator of the whole idea of incorrect questions. I created
this concept in this forum for the sole purpose of showing that
pathological self-reference(Olcott 2004) is an error.

> The question of the
> Halting Problem NEVER refers to the decider, only to the machine being
> decided on. PERIOD.
>>
>> The answer is restricted to {true, false} thus excluding the correct
>> answer of “neither” making the question itself incorrect.
>>
>> If we ask a man that has never been married:
>> Have you stopped beating you wife?
>> This is an incorrect question.
>> Every polar question lacking a correct yes/no answer is incorrect.
>
> STRAWMEN
>
>>
>> Every TM/input to a decision problem lacking a correct Boolean return
>> value is an incorrect TM/input for this decision problem.
>
> EVERY TM/Input has a correct value for the Halting Problem, either that
> machine/input combination will halt, or it won't. Absolutely one answer
> is correct.
>
>>
>> The TM / input pairs that “prove” the halting problem is undecidable
>> have the same pathological self-reference(Olcott 2004) error as the
>> self-contradictory liar paradox.
>
> H^(H^) is either a Halting Computation or it isn't. There is no neither.
>
> The key point to remember is that H^ is DEFINED to be based on GIVEN
> machine H, as ^ is actually a machine transformation. Thus to even ask
> about H^, we need to have first fixed what H is. Once we have fixed that
> H, then H^(H^) has a defininte halting behavior, and it will always be
> the opposite of what H(H^,H^) predicts (as long as H does predict this
> question). Thus H is always shown to not be a proper Halting decider.
>
> There is no 'pathological' self-reference, as there is absolutely no
> requrement that H can get the answer right, so the only 'pathology' in
> the situation is to the existance of H.
>
>
>>
>> It has been at least 2000 years since the liar paradox was discovered
>> and academicians still do not understand the because self-contradictory
>> sentences do not map to a Boolean value they are not truth bearers.
>
> But the Halting Problem ISN'T your 'pathological' question.
>>
>>> If you want to keep to the concept that something is only true if you
>>> can prove it, then you must restrict yourself to only First Order Logic,
>>> or you WILL eventually run into an inconsistency in your logic system.
>>>
>>
>> Not at all. Not in the least.
>> When Gödel's 1931 incompleteness theorem is translated into HOL that has
>> its own provability predicate thus no need for the purely extraneous
>> complexity of Gödel numbers then it is very obvious that it is modeled
>> after the liar paradox>
> No, you haven't. PERIOD. You seem incapable of handling any complex logic.

olcott

unread,
Jul 14, 2021, 11:01:34 AMJul 14
to
On 7/13/2021 11:56 PM, Richard Damon wrote:
> YOU do not understand the TRUE meaning of a Truth Bearer.
>

A "Truth bearer" is an expression of language capable of having a
Boolean property.

In all cases where it is impossible to resolve an expression of language
to a Boolean value this expression is not a truth bearer.

The key case of this is the liar Paradox. Something must be wrong with
the world if no academician has figured out that self-contradictory
sentences are not truth bearers in more than 2000 years.

> Truth Bearer does NOT mean provable, unless you have decided to restrict
> your logic to only provable things. If you do this, then you can not use
> higher than First Order Logic or you become inconsistent. And even if
> you use only First Order Logic, unless your filed of knowledge is so
> limited as to be finitely enumerated, it can not actually prove that it
> has stayed consistant.
>
> If you are going to try to live with the proposition that All Truth must
> be Provable, live within its limits, and get out of the fields that are
> beyond it.
>
> This was settled a Century ago. Your problem is that you aren't
> accepting that, going beyond that allowable limits of your logic, and
> you are failing to recognize that your logic system is inconsistent. The
> mere fact that we can prove a machine to be both Halting and Non-Halting
> is proof of that.
>
> You seem to fail to understand that you can't 'DISPROVE' something that
> has been proved by proving its opposite. THAT doesn't work. That method
> only works if the theorem is unproven.
>
> When these contradictions pop up, that real answer is this shows that
> either one of the proofs has an error (that should be able to be found)
> or the logic system has broken and gone inconsistent, and the whole
> theory needs to be rewound to find the axiom that broke it.
>
> In your case, this is fairly easy, as you keep asserting as postulates
> things that are provably false, so it is easy to see where the errors
> came in.

olcott

unread,
Jul 15, 2021, 10:24:34 AMJul 15
to
On 7/15/2021 5:43 AM, Malcolm McLean wrote:
> On Thursday, 15 July 2021 at 04:48:23 UTC+1, Richard Damon wrote:
>> On 7/14/21 8:56 AM, olcott wrote:
>>>
>>>
>>> When we ask someone that has never been married :
>>> Have you stopped beating your wife?
>> Will you stop making dumb comments?
>>
> It spoils them when you have to explain them.
> But the point of the joke is not that the the person of whom the question
> is being asked is a bachelor. It's assumed that he is married.
>
> It's that the question presumes that he habitually beat his wife in the
> past. So if he answers "no" that means he is still beating his wife, and
> if he answers "yes" then that confirms that he beat his wife in the past.
> There's no answer he can give which implies "I never beat my wife".
>
> Technically the correct answer is "no". But that's even more damning than
> "yes".
>

Technically both answers imply an incorrect supposition thus making both
answers incorrect.

My original incorrect question was: What time is it (yes or no)?
I needed a new one that is correct in some contexts and incorrect in
others.

olcott

unread,
Jul 15, 2021, 10:35:47 AMJul 15
to
On 7/15/2021 8:54 AM, Richard Damon wrote:
> On 7/15/21 4:43 AM, Malcolm McLean wrote:
>> On Thursday, 15 July 2021 at 04:48:23 UTC+1, Richard Damon wrote:
>>> On 7/14/21 8:56 AM, olcott wrote:
>>>>
>>>>
>>>> When we ask someone that has never been married :
>>>> Have you stopped beating your wife?
>>> Will you stop making dumb comments?
>>>
>> It spoils them when you have to explain them.
>> But the point of the joke is not that the the person of whom the question
>> is being asked is a bachelor. It's assumed that he is married.
>>
>> It's that the question presumes that he habitually beat his wife in the
>> past. So if he answers "no" that means he is still beating his wife, and
>> if he answers "yes" then that confirms that he beat his wife in the past.
>> There's no answer he can give which implies "I never beat my wife".
>>
>> Technically the correct answer is "no". But that's even more damning than
>> "yes".
>>
>
> You missed the Irony, for HIM there is a clear answer to the question,
> even if the question would be poorly defined for most people.
>
>

When-so-ever any yes/no question has no correct yes/no answer then
question itself is incorrect.

When-so-ever any decision problem TM/input pair has no correct Boolean
final state that the TM of this pair can transition to then this
TM/input pair is incorrect for this decision problem.


This is the original best example of the Pathological self-reference
(Olcott 2004) error:

Daryl McCullough Jun 25, 2004, 6:30:39 PM

Both Godel's proof and Turing's proof have the flavor of using
self-reference to force someone to make a mistake. Both cases
seem a little like the following paradox (call it the "Gotcha"
paradox).

You ask someone (we'll call him "Jack") to give a truthful
yes/no answer to the following question:

Will Jack's answer to this question be no?

Jack can't possibly give a correct yes/no answer to the question.
https://groups.google.com/g/sci.logic/c/4kIXI1kxmsI/m/hRroMoQZx2IJ

The answer posed to Jack is an incorrect question because Pathological
self-reference (Olcott 2004) immediately contradicts any answer that
Jack can provide, thus making the question itself incorrect in the exact
same way that the TM/input pair of the HP make the question incorrect.

olcott

unread,
Jul 15, 2021, 11:05:25 AMJul 15
to
On 7/14/2021 11:03 PM, Richard Damon wrote:
> Right, and the answer to the Halting Problem, "Does P(I) come to a
> Halting state in a finite number of steps?" always, yes always. has a
> Yes or No anser.

Yes continue to dishonestly change the question by ignoring the
mandatory context of the question. Does being a liar make you feel good?

What Boolean value can H return to an input that does the opposite of
whatever H decides making sure to contradict the halt status that H
returns such that H returns the correct Boolean value halt status to P?

>>
>> In all cases where it is impossible to resolve an expression of language
>> to a Boolean value this expression is not a truth bearer.
>
> And the Halting Problem never doesn't have an answer.

Only because you dishonestly remove the context of the question.

I claim there are no grocery stores that are bigger than the universe
and your reply is that I am wrong Walmart is a big store.

>>
>> The key case of this is the liar Paradox. Something must be wrong with
>> the world if no academician has figured out that self-contradictory
>> sentences are not truth bearers in more than 2000 years.
>
> Right, you get the Liar paradox when you assume that H exists and ask
> what it should produce in this case, the lack of answer shows that H
> doesn't exist, not that the Halting Problem has a problem.
>

When-so-ever any decision problem TM/input pair has no correct Boolean
value this is not an undecidable instance where the TM cannot decide
which Boolean value is correct. It is an incorrect instance where both
Boolean values are the wrong answer.

> The WRONG QUESTIOH isn't a Truth Bearer.

olcott

unread,
Jul 20, 2021, 9:41:52 AMJul 20
to
> In other words too much complexity for you to understand.  It is a
> common crank's error:
>
>     I don't understand it = it is erroneous.
>

Unless we boil these ideas down to their barest essence and expressly
state every detail key insights are lost.

// Strachey(1965) "An impossible program"
// CPL translated to C
// https://doi.org/10.1093/comjnl/7.4.313
void P(u32 x)
{
if (H(x, x))
HERE: goto HERE;
}

int main()
{
Output("Input_Halts = ", H((u32)P, (u32)P));
}

The HP diagonalization proof does correctly prove that no H can possibly
return a correct halts status to P. The diagonalization proof glosses
over all of the details of why this is impossible.

None-of-the HP proofs consider the possibility that a simulating halt
decider would derive the correct halts status for P(P) by aborting P(P).
They simply didn't bother to think this step through.

CONVENTION T. A formally correct definition of the symbol ‘Tr’,
formulated in
the metalanguage, will be called an adequate definition of truth if it
has the following
consequences:
(α) all sentences which are obtained from the expression ‘x ∈ Tr if and
only if p’
by substituting for the symbol ‘x’ a structural-descriptive name of any
sentence of the
language in question and for the symbol ‘p’ the expression which forms
the translation of
this sentence into the metalanguage;
(β) the sentence ‘for any x, if x ∈ Tr then x ∈ S’ (in other words ‘Tr S’).

‘x ∈ Tr if and only if p’ becomes:
x ∈ Tr ↔ p // p ↔ True(x)

translated into the Liar Paradox:
x ∉ Tr ↔ p // p ↔ ¬True(x)

translated into the 1931 incompleteness:
x ∉ Pr ↔ p // p ↔ ¬Provable(x)

Tarski derives his whole undefinability theorem on the basis of the liar
paradox which he converts into the nearly the simplest possibly form of
Gödel's 1931 incompleteness theorem.

http://www.liarparadox.org/Tarski_275_276.pdf

Tarski has "proved" the some "true" sentences cannot be properly
mathematically formalized on the basis that the liar paradox cannot be
proved to be true. He tried to proved that a lie is true.

olcott

unread,
Jul 20, 2021, 9:44:32 AMJul 20
to
> In other words too much complexity for you to understand.  It is a
> common crank's error:
>
>     I don't understand it = it is erroneous.
>

No, What I am saying is the trying to prove that a self contradictory
statement is true is a nitwit idea.

// Strachey(1965) "An impossible program"
// CPL translated to C
// https://doi.org/10.1093/comjnl/7.4.313
void P(u32 x)
{
if (H(x, x))
HERE: goto HERE;
}

int main()
{
Output("Input_Halts = ", H((u32)P, (u32)P));
}

_P()
[00000c36](01) 55 push ebp
[00000c37](02) 8bec mov ebp,esp
[00000c39](03) 8b4508 mov eax,[ebp+08] // 2nd Param
[00000c3c](01) 50 push eax
[00000c3d](03) 8b4d08 mov ecx,[ebp+08] // 1st Param
[00000c40](01) 51 push ecx
[00000c41](05) e820fdffff call 00000966 // call H
[00000c46](03) 83c408 add esp,+08
[00000c49](02) 85c0 test eax,eax
[00000c4b](02) 7402 jz 00000c4f
[00000c4d](02) ebfe jmp 00000c4d
[00000c4f](01) 5d pop ebp
[00000c50](01) c3 ret
Size in bytes:(0027) [00000c50]

_main()
[00000c56](01) 55 push ebp
[00000c57](02) 8bec mov ebp,esp
[00000c59](05) 68360c0000 push 00000c36
[00000c5e](05) 68360c0000 push 00000c36
[00000c63](05) e8fefcffff call 00000966
[00000c68](03) 83c408 add esp,+08
[00000c6b](01) 50 push eax
[00000c6c](05) 6857030000 push 00000357
[00000c71](05) e810f7ffff call 00000386
[00000c76](03) 83c408 add esp,+08
[00000c79](02) 33c0 xor eax,eax
[00000c7b](01) 5d pop ebp
[00000c7c](01) c3 ret
Size in bytes:(0039) [00000c7c]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[00000c56][0010172a][00000000] 55 push ebp
[00000c57][0010172a][00000000] 8bec mov ebp,esp
[00000c59][00101726][00000c36] 68360c0000 push 00000c36
[00000c5e][00101722][00000c36] 68360c0000 push 00000c36
[00000c63][0010171e][00000c68] e8fefcffff call 00000966 // call H(P,P)

Begin Local Halt Decider Simulation at Machine Address:c36
[00000c36][002117ca][002117ce] 55 push ebp
[00000c37][002117ca][002117ce] 8bec mov ebp,esp
[00000c39][002117ca][002117ce] 8b4508 mov eax,[ebp+08]
[00000c3c][002117c6][00000c36] 50 push eax
[00000c3d][002117c6][00000c36] 8b4d08 mov ecx,[ebp+08]
[00000c40][002117c2][00000c36] 51 push ecx
[00000c41][002117be][00000c46] e820fdffff call 00000966 // call H(P,P)
[00000c36][0025c1f2][0025c1f6] 55 push ebp
[00000c37][0025c1f2][0025c1f6] 8bec mov ebp,esp
[00000c39][0025c1f2][0025c1f6] 8b4508 mov eax,[ebp+08]
[00000c3c][0025c1ee][00000c36] 50 push eax
[00000c3d][0025c1ee][00000c36] 8b4d08 mov ecx,[ebp+08]
[00000c40][0025c1ea][00000c36] 51 push ecx
[00000c41][0025c1e6][00000c46] e820fdffff call 00000966 // call H(P,P)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

[00000c68][0010172a][00000000] 83c408 add esp,+08
[00000c6b][00101726][00000000] 50 push eax
[00000c6c][00101722][00000357] 6857030000 push 00000357
[00000c71][00101722][00000357] e810f7ffff call 00000386
Input_Halts = 0
[00000c76][0010172a][00000000] 83c408 add esp,+08
[00000c79][0010172a][00000000] 33c0 xor eax,eax
[00000c7b][0010172e][00100000] 5d pop ebp
[00000c7c][00101732][00000068] c3 ret
Number_of_User_Instructions(27)
Number of Instructions Executed(23721)



> I'm like you in that I'm not very bright.  When I was at school I
> couldn't understand German (the modern foreign language we were all
> expected to learn).  Did I excuse myself by claiming that German was
> erroneous?  No.  I had no excuse beyond "I don't get it".  (German was
> chosen pretty much at random - I might have mentioned a lot of other
> things.)
>
>> . When we recreate this same proof in HOL without Gödel numbering
>
> See Hans Hermes, /Enumerability, decidability, computability/, 2nd
> translated ed., Springer

olcott

unread,
Jul 20, 2021, 3:41:58 PMJul 20
to
Tarski says that the above is the liar paradox.
He then arbitrarily swaps "Tr" meaning "true" for "Pr" meaning provable
and ends up with x ∉ Pr ↔ x ∈ Tr as the basis of his whole proof.

¬Provable(x) ↔ True(x)

In other words he has a sentence x that says of itself that it is true
that it is not provable.

The two pages on this link are his whole proof:
http://www.liarparadox.org/Tarski_275_276.pdf

My unique work in this field breaks new ground.
Everyone has always assumed that declarative sentences are either true
or false. It never occurred to them that declarative sentences have two
Boolean properties and not just one.

https://www.researchgate.net/publication/323866366_The_Notion_of_Truth_in_Natural_and_Formal_Languages

Apparently in all of the years prior to my original (2016) paper no one
ever separated the analysis of propositions into their atomic units of
semantic compositionality:
(a) Assertion (Boolean) // What the expression of language claims to
be True
(b) Satisfied (Boolean) // Whether or not the expression of language
is True

"This sentence is not true." is indeed not true, its assertion is true.
Because it is self-contradictory its assertion is not satisfied.

>> translated into the 1931 incompleteness:
>> x ∉ Pr ↔ p      // p ↔ ¬Provable(x)
>
> Two strange uses of the word 'translated'.  Translation, if it is of any
> use, is supposed to preserve meaning.

olcott

unread,
Jul 21, 2021, 10:41:04 AMJul 21
to
> Where?
>

When you understand that this convention T is his starting point:
x ∈ Tr ↔ p // p ↔ True(x)

then applying Theorem I
http://www.liarparadox.org/Tarski_247_248.pdf we get:
x ∉ Tr ↔ p // p ↔ ¬True(x)

The above link explains how Theorem I translates convention T
into the Liar Paradox

In his proof he does two steps are once
He does not show this intermediate step:

In accordance with the first
part of Th. I we can obtain the negation of one of the sentences
in condition (α) of convention T

Convention T is translated into the Liar Paradox
x ∉ Tr ↔ p // p ↔ ¬True(x)

He shows two steps combined:
In accordance with the first
part of Th. I we can obtain the negation of one of the sentences
in condition (α) of convention T of § 3 as a consequence of the
definition of the symbol 'Pr' (provided we replace 'Tr' in this
convention by 'Pr').

x ∈ Tr ↔ p // p ↔ True(x)
becomes
x ∉ Pr ↔ p // p ↔ ¬Provable(x)

So it is proven that Tarski's whole proof is anchored in the Liar Paradox.

Because Gödel himself said that the Liar Paradox can be used for a
similar proof, refuting Tarski refutes Gödel:

ON FORMALLY UNDECIDABLE PROPOSITIONS OF PRINCIPIA
MATHEMATICA AND RELATED SYSTEMS I by Kurt Godel Vienna
The analogy between this result and Richard’s antinomy
leaps to the eye; there is also a close relationship
with the “liar” antinomy,14

14 Every epistemological antinomy can likewise be used
for a similar undecidability proof.


>> and ends up with x ∉ Pr ↔ x ∈ Tr as the basis of his whole proof.
>>
>> ¬Provable(x) ↔ True(x)
>>
>> In other words he has a sentence x that says of itself that it is true
>> that it is not provable.
>>
>> The two pages on this link are his whole proof:
>> http://www.liarparadox.org/Tarski_275_276.pdf
>>
>> My unique work in this field breaks new ground.
>> Everyone has always assumed that declarative sentences are either true
>> or false
>
> Everyone?  There were ancient Greeks who did not believe it.

olcott

unread,
Jul 21, 2021, 12:00:22 PMJul 21
to
> And _what_ justifies such a replacement?  My 'Where?' (which I confess
> was brief enough to be misunderstood) meant that as an alternative would
> be 'on what page of what work fdoes Tarski do such a thing?'
>
>> ).
>>
>> x ∈ Tr ↔ p       // p ↔ True(x)
>>    becomes
>> x ∉ Pr ↔ p       // p ↔ ¬Provable(x)
>>
>> So it is proven that Tarski's whole proof is anchored in the Liar
>> Paradox.
>>
>> Because Gödel himself said that the Liar Paradox can be used for a
>> similar proof, refuting Tarski refutes Gödel:
>>
>>     ON FORMALLY UNDECIDABLE PROPOSITIONS OF PRINCIPIA
>>     MATHEMATICA AND RELATED SYSTEMS I by Kurt Godel Vienna
>>     The analogy between this result and Richard’s antinomy
>>     leaps to the eye; there is also a close relationship
>>     with the “liar” antinomy,14
>
> And what did Gödel mean by 'a close relationship'?  I don't think he
> ever used it for what you call 'a similar proof', so can you supply the
> details?

14 Every epistemological antinomy can likewise be used
for a similar undecidability proof.

This means that the epistemological antinomy of the Liar Paradox can
likewise be used for a similar undecidability proof.

That he did not use it for a similar proof does not matter. That Tarski
used it for a similar proof matters in that refuting Tarski refutes
Gödel. Refuting Tarski is 100,000-fold simpler than refuting Gödel.
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