∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
(the same way the ZFC disallowed self-referential sets) then
pathological inputs are not allowed to come into existence.

Does the barber that shaves everyone that does not shave
themselves shave himself? is rejected as an incorrect question.
https://en.wikipedia.org/wiki/Barber_paradox#

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 12/03/24 19:12, olcott wrote:
> ∀ H ∈ Turing_Machine_Deciders
> ∃ TMD ∈ Turing_Machine_Descriptions  |
> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
> There is some input TMD to every H such that
> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

And it can be a different TMD to each H.

> When we disallow decider/input pairs that are incorrect
> questions where both YES and NO are the wrong answer

Once we understand that either YES or NO is the right answer, the whole
rebuttal is tossed out as invalid and incorrect.

> Does the barber that shaves everyone that does not shave
> themselves shave himself? is rejected as an incorrect question.

The barber does not exist. The following is true statement:

∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
¬Shaves(Person, Person))

The following is a true statement:

¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
¬Shaves(Person, Person))

On 3/12/24 11:12 AM, olcott wrote:
> ∀ H ∈ Turing_Machine_Deciders
> ∃ TMD ∈ Turing_Machine_Descriptions  |
> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
> There is some input TMD to every H such that
> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
> When we disallow decider/input pairs that are incorrect
> questions where both YES and NO are the wrong answer
> (the same way the ZFC disallowed self-referential sets) then
> pathological inputs are not allowed to come into existence.

But each of the questions in your set, that for each H^ built on each
SPECIFIC H, there is a correct answer, just not the one that H gives, so
the ALL the questions are correct.

>
> Does the barber that shaves everyone that does not shave
> themselves shave himself? is rejected as an incorrect question.
> https://en.wikipedia.org/wiki/Barber_paradox#
>

RED HERRING.

H^ does not "reference" the decider H that you paired it with, it has a
copy of it, so when we examine the question, and modulate the decider to
see if a correct answer is possible, we see that it is, since the input
still keeps its original behavior.

You are just proving your utter stupidity and lack of understanding of
how any of this works.

Likely cause by your using Zeroth order principle and thus not actually
knowing what you are talking about, but are just parroting the words you
have seen.

On 3/12/2024 1:31 PM, immibis wrote:
> On 12/03/24 19:12, olcott wrote:
>> ∀ H ∈ Turing_Machine_Deciders
>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>
>> There is some input TMD to every H such that
>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
> And it can be a different TMD to each H.
>
>> When we disallow decider/input pairs that are incorrect
>> questions where both YES and NO are the wrong answer
>
> Once we understand that either YES or NO is the right answer, the whole
> rebuttal is tossed out as invalid and incorrect.
>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩

>> Does the barber that shaves everyone that does not shave
>> themselves shave himself? is rejected as an incorrect question.
>
> The barber does not exist.

Russell's paradox did not allow this answer within Naive set theory.

> The following is true statement:
>
> ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
> ¬Shaves(Person, Person))
>
> The following is a true statement:
>
> ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
> ¬Shaves(Person, Person))
>

That might be correct I did not check it over and over
again and again to make sure.

The same reasoning seems to rebut Gödel Incompleteness:
....We are therefore confronted with a proposition which
asserts its own unprovability. 15 ... (Gödel 1931:43-44)
¬∃G ∈ F | G := ~(F ⊢ G)

Any G in F that asserts its own unprovability in F is
asserting that there is no sequence of inference steps
in F that prove that they themselves do not exist in F.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/12/24 12:02 PM, olcott wrote:
> On 3/12/2024 1:31 PM, immibis wrote:
>> On 12/03/24 19:12, olcott wrote:
>>> ∀ H ∈ Turing_Machine_Deciders
>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>
>>> There is some input TMD to every H such that
>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>
>> And it can be a different TMD to each H.
>>
>>> When we disallow decider/input pairs that are incorrect
>>> questions where both YES and NO are the wrong answer
>>
>> Once we understand that either YES or NO is the right answer, the
>> whole rebuttal is tossed out as invalid and incorrect.
>>
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩

No, because a given H will only go to one of the answers. THAT will be
wrong, and the other one right.

Remember, you above statement was built on the ASSUMPTION that a correct
H existed, and thus the contradiction you see just says that no such H
exists, not that the original question was incorrect.

On 3/12/2024 1:41 PM, Richard Damon wrote:
> On 3/12/24 11:12 AM, olcott wrote:
>> ∀ H ∈ Turing_Machine_Deciders
>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>
>> There is some input TMD to every H such that
>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>
>> When we disallow decider/input pairs that are incorrect
>> questions where both YES and NO are the wrong answer
>> (the same way the ZFC disallowed self-referential sets) then
>> pathological inputs are not allowed to come into existence.
>
> But each of the questions in your set, that for each H^ built on each
> SPECIFIC H, there is a correct answer, just not the one that H gives, so
> the ALL the questions are correct.
>

*Halts(TMD) means true if TMD actually halts and false otherwise*
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
(H TMD ⊢* H.qy != Halts(TMD)) ∧ (H TMD ⊢* H.qn != Halts(TMD))

The questions in my set are only the H/TMD pairs specified above.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/12/24 12:57 PM, olcott wrote:
> On 3/12/2024 1:41 PM, Richard Damon wrote:
>> On 3/12/24 11:12 AM, olcott wrote:
>>> ∀ H ∈ Turing_Machine_Deciders
>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>
>>> There is some input TMD to every H such that
>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>
>>> When we disallow decider/input pairs that are incorrect
>>> questions where both YES and NO are the wrong answer
>>> (the same way the ZFC disallowed self-referential sets) then
>>> pathological inputs are not allowed to come into existence.
>>
>> But each of the questions in your set, that for each H^ built on each
>> SPECIFIC H, there is a correct answer, just not the one that H gives,
>> so the ALL the questions are correct.
>>
>
> *Halts(TMD) means true if TMD actually halts and false otherwise*
> ∀ H ∈ Turing_Machine_Deciders
> ∃ TMD ∈ Turing_Machine_Descriptions  |
> (H TMD ⊢* H.qy != Halts(TMD)) ∧ (H TMD ⊢* H.qn != Halts(TMD))
>
> The questions in my set are only the H/TMD pairs specified above.
>

Right, so for Each H, we can find a TMD, which could be the H^ built on
THAT H.

then if H goes to qy, then HALTS(TMD) is false, so that H was wrong, but
there was a correct answer.

and if H goes to qn, then HALTD(TMD) is true, so that H was wrong, but
there was a correct answer.

So, for EACH INDIVIDUAL H, which is what the conditions look at, there
IS a correct answer for that input we foud, just not the one that THAT H
gave.

No one claims that no Halt Decider can get any correct decisions, which
seems to be what you are trying to claim.

You don't seem to understand logical categorical operations.

On 3/12/2024 2:40 PM, Richard Damon wrote:
> On 3/12/24 12:02 PM, olcott wrote:
>> On 3/12/2024 1:31 PM, immibis wrote:
>>> On 12/03/24 19:12, olcott wrote:
>>>> ∀ H ∈ Turing_Machine_Deciders
>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>
>>>> There is some input TMD to every H such that
>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>
>>> And it can be a different TMD to each H.
>>>
>>>> When we disallow decider/input pairs that are incorrect
>>>> questions where both YES and NO are the wrong answer
>>>
>>> Once we understand that either YES or NO is the right answer, the
>>> whole rebuttal is tossed out as invalid and incorrect.
>>>
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>
> No, because a given H will only go to one of the answers. THAT will be
> wrong, and the other one right.
>

∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.

When you say that the opposite answer is correct you
are sneaking outside of the above specified set.

> Remember, you above statement was built on the ASSUMPTION that a correct
> H existed, and thus the contradiction you see just says that no such H
> exists, not that the original question was incorrect.
>

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 12/03/24 20:02, olcott wrote:
> On 3/12/2024 1:31 PM, immibis wrote:
>> On 12/03/24 19:12, olcott wrote:
>>> ∀ H ∈ Turing_Machine_Deciders
>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>
>>> There is some input TMD to every H such that
>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>
>> And it can be a different TMD to each H.
>>
>>> When we disallow decider/input pairs that are incorrect
>>> questions where both YES and NO are the wrong answer
>>
>> Once we understand that either YES or NO is the right answer, the
>> whole rebuttal is tossed out as invalid and incorrect.
>>
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>

Once we understand that either YES or NO is the right answer, the whole
rebuttal is tossed out as invalid and incorrect.

>>> Does the barber that shaves everyone that does not shave
>>> themselves shave himself? is rejected as an incorrect question.
>>
>> The barber does not exist.
>
> Russell's paradox did not allow this answer within Naive set theory.

Naive set theory says that for every predicate P, the set {x | P(x)}
exists. This axiom was a mistake. This axiom is not in ZFC.

In Turing machines, for every non-empty finite set of alphabet symbols
Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states Q, every
q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R}, ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a
Turing machine. Do you think this is a mistake? Would you remove this
axiom from your version of Turing machines?

(Following the definition used on Wikipedia:
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)

>> The following is true statement:
>>
>> ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
>> ¬Shaves(Person, Person))
>>
>> The following is a true statement:
>>
>> ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
>> ¬Shaves(Person, Person))
>>
>
> That might be correct I did not check it over and over
> again and again to make sure.
>
> The same reasoning seems to rebut Gödel Incompleteness:
> ...We are therefore confronted with a proposition which
> asserts its own unprovability. 15 ... (Gödel 1931:43-44)
> ¬∃G ∈ F | G := ~(F ⊢ G)
>
> Any G in F that asserts its own unprovability in F is
> asserting that there is no sequence of inference steps
> in F that prove that they themselves do not exist in F.

The barber does not exist and the proposition does not exist.

On 3/12/2024 3:31 PM, immibis wrote:
> On 12/03/24 20:02, olcott wrote:
>> On 3/12/2024 1:31 PM, immibis wrote:
>>> On 12/03/24 19:12, olcott wrote:
>>>> ∀ H ∈ Turing_Machine_Deciders
>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>
>>>> There is some input TMD to every H such that
>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>
>>> And it can be a different TMD to each H.
>>>
>>>> When we disallow decider/input pairs that are incorrect
>>>> questions where both YES and NO are the wrong answer
>>>
>>> Once we understand that either YES or NO is the right answer, the
>>> whole rebuttal is tossed out as invalid and incorrect.
>>>
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>
>
> Once we understand that either YES or NO is the right answer, the whole
> rebuttal is tossed out as invalid and incorrect.
>
>>>> Does the barber that shaves everyone that does not shave
>>>> themselves shave himself? is rejected as an incorrect question.
>>>
>>> The barber does not exist.
>>
>> Russell's paradox did not allow this answer within Naive set theory.
>
> Naive set theory says that for every predicate P, the set {x | P(x)}
> exists. This axiom was a mistake. This axiom is not in ZFC.
>
> In Turing machines, for every non-empty finite set of alphabet symbols
> Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states Q, every
> q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R}, ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a
> Turing machine. Do you think this is a mistake? Would you remove this
> axiom from your version of Turing machines?
>
> (Following the definition used on Wikipedia:
> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)
>
>>> The following is true statement:
>>>
>>> ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>> ¬Shaves(Person, Person))
>>>
>>> The following is a true statement:
>>>
>>> ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>> ¬Shaves(Person, Person))
>>>
>>
>> That might be correct I did not check it over and over
>> again and again to make sure.
>>
>> The same reasoning seems to rebut Gödel Incompleteness:
>> ...We are therefore confronted with a proposition which
>> asserts its own unprovability. 15 ... (Gödel 1931:43-44)
>> ¬∃G ∈ F | G := ~(F ⊢ G)
>>
>> Any G in F that asserts its own unprovability in F is
>> asserting that there is no sequence of inference steps
>> in F that prove that they themselves do not exist in F.
>
> The barber does not exist and the proposition does not exist.
>

When we do this exact same thing that ZFC did for self-referential
sets then self-referential (thus pathological) inputs to halt deciders
cannot exist.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 12/03/24 21:11, olcott wrote:
> On 3/12/2024 2:40 PM, Richard Damon wrote:
>> On 3/12/24 12:02 PM, olcott wrote:
>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>> On 12/03/24 19:12, olcott wrote:
>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>
>>>>> There is some input TMD to every H such that
>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>
>>>> And it can be a different TMD to each H.
>>>>
>>>>> When we disallow decider/input pairs that are incorrect
>>>>> questions where both YES and NO are the wrong answer
>>>>
>>>> Once we understand that either YES or NO is the right answer, the
>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>
>> No, because a given H will only go to one of the answers. THAT will be
>> wrong, and the other one right.
>>
>
> ∀ H ∈ Turing_Machine_Deciders
> ∃ TMD ∈ Turing_Machine_Descriptions  |
> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
> Not exactly. A pair of otherwise identical machines that
> (that are contained within the above specified set)
> only differ by return value will both be wrong on the
> same pathological input.

You know this is not true so why lie? One of them will be correct on
that input. If Ĥ is the function D from x86utm, then Ĥ applied to ⟨Ĥ⟩
does halt, YES is the correct answer, and any decider which returns YES
on this input is correct on this input.

On 3/12/2024 3:31 PM, immibis wrote:
> On 12/03/24 20:02, olcott wrote:
>> On 3/12/2024 1:31 PM, immibis wrote:
>>> On 12/03/24 19:12, olcott wrote:
>>>> ∀ H ∈ Turing_Machine_Deciders
>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>
>>>> There is some input TMD to every H such that
>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>
>>> And it can be a different TMD to each H.
>>>
>>>> When we disallow decider/input pairs that are incorrect
>>>> questions where both YES and NO are the wrong answer
>>>
>>> Once we understand that either YES or NO is the right answer, the
>>> whole rebuttal is tossed out as invalid and incorrect.
>>>
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>
>
> Once we understand that either YES or NO is the right answer, the whole
> rebuttal is tossed out as invalid and incorrect.
>
>>>> Does the barber that shaves everyone that does not shave
>>>> themselves shave himself? is rejected as an incorrect question.
>>>
>>> The barber does not exist.
>>
>> Russell's paradox did not allow this answer within Naive set theory.
>
> Naive set theory says that for every predicate P, the set {x | P(x)}
> exists. This axiom was a mistake. This axiom is not in ZFC.
>
> In Turing machines, for every non-empty finite set of alphabet symbols
> Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states Q, every
> q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R}, ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a
> Turing machine. Do you think this is a mistake? Would you remove this
> axiom from your version of Turing machines?
>
> (Following the definition used on Wikipedia:
> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)
>
>>> The following is true statement:
>>>
>>> ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>> ¬Shaves(Person, Person))
>>>
>>> The following is a true statement:
>>>
>>> ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>> ¬Shaves(Person, Person))
>>>
>>
>> That might be correct I did not check it over and over
>> again and again to make sure.
>>
>> The same reasoning seems to rebut Gödel Incompleteness:
>> ...We are therefore confronted with a proposition which
>> asserts its own unprovability. 15 ... (Gödel 1931:43-44)
>> ¬∃G ∈ F | G := ~(F ⊢ G)
>>
>> Any G in F that asserts its own unprovability in F is
>> asserting that there is no sequence of inference steps
>> in F that prove that they themselves do not exist in F.
>
> The barber does not exist and the proposition does not exist.
>

When we do this exact same thing that ZFC did for self-referential
sets then Gödel's self-referential expressions that assert their
own unprovability in F also cease to exist.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/12/2024 3:38 PM, immibis wrote:
> On 12/03/24 21:11, olcott wrote:
>> On 3/12/2024 2:40 PM, Richard Damon wrote:
>>> On 3/12/24 12:02 PM, olcott wrote:
>>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>>> On 12/03/24 19:12, olcott wrote:
>>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>
>>>>>> There is some input TMD to every H such that
>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>
>>>>> And it can be a different TMD to each H.
>>>>>
>>>>>> When we disallow decider/input pairs that are incorrect
>>>>>> questions where both YES and NO are the wrong answer
>>>>>
>>>>> Once we understand that either YES or NO is the right answer, the
>>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>>
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>
>>> No, because a given H will only go to one of the answers. THAT will
>>> be wrong, and the other one right.
>>>
>>
>> ∀ H ∈ Turing_Machine_Deciders
>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>
>> Not exactly. A pair of otherwise identical machines that
>> (that are contained within the above specified set)
>> only differ by return value will both be wrong on the
>> same pathological input.
>
> You know this is not true so why lie? One of them will be correct on
> that input.

The above set specifies both of them and specifies that
they are both wrong.

> If Ĥ is the function D from x86utm, then Ĥ applied to ⟨Ĥ⟩
> does halt, YES is the correct answer, and any decider which returns YES
> on this input is correct on this input.
>

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/12/24 1:11 PM, olcott wrote:
> On 3/12/2024 2:40 PM, Richard Damon wrote:
>> On 3/12/24 12:02 PM, olcott wrote:
>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>> On 12/03/24 19:12, olcott wrote:
>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>
>>>>> There is some input TMD to every H such that
>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>
>>>> And it can be a different TMD to each H.
>>>>
>>>>> When we disallow decider/input pairs that are incorrect
>>>>> questions where both YES and NO are the wrong answer
>>>>
>>>> Once we understand that either YES or NO is the right answer, the
>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>
>> No, because a given H will only go to one of the answers. THAT will be
>> wrong, and the other one right.
>>
>
> ∀ H ∈ Turing_Machine_Deciders
> ∃ TMD ∈ Turing_Machine_Descriptions  |
> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
> Not exactly. A pair of otherwise identical machines that
> (that are contained within the above specified set)
> only differ by return value will both be wrong on the
> same pathological input.

You mean a pair of DIFFERENT machines. Any difference is different.

And no, they won't be fooled by the SAME pathological input.

Lets call the machine Ha and Hb, and Hb will ALWAYS generate the
opposite answer for A for all inputs.

We build the pathological input Ha^ built on it calling Ha and doing the
opposite.

If Ha (Ha^) (Ha^) goes to qn, then Ha^ (Ha^) goes to qn and Halts, so Ha
was wrong, but since Hb says the opposite, Hb (Ha^) (Ha^) by that
definition goes to qy, and is right,

If Ha (Ha^) (Ha^) goes to qy, then Ha^ (Ha^) goes to qy and loops, so Ha
was wrong, but since Hb says the opposite, Hb (Ha^) (Ha^) goes to qn and
is right again.

Note, you said SAME INPUT, which is (Ha^) (Ha^) in both cases,

Changing the decider for the same input doesn't change the input to the
new decider, because that isn't how the input works.

So, that infact PROVES that some decider can get the answer rights.

Note though, the clause doesn't say that there is one input that makes
ALL deciders wrong, it says that for all decider, each one has a
(potentially different) input that it gets wrong.

So, for Hb, we can make an Hb^ that Hb (Hb^) (Hb^) will get wrong, but
Ha will get right.

>
> When you say that the opposite answer is correct you
> are sneaking outside of the above specified set.

Nope. You don't know what that says.

>
>> Remember, you above statement was built on the ASSUMPTION that a
>> correct H existed, and thus the contradiction you see just says that
>> no such H exists, not that the original question was incorrect.
>>
>
>

On 3/12/24 1:35 PM, olcott wrote:
> On 3/12/2024 3:31 PM, immibis wrote:
>> On 12/03/24 20:02, olcott wrote:
>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>> On 12/03/24 19:12, olcott wrote:
>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>
>>>>> There is some input TMD to every H such that
>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>
>>>> And it can be a different TMD to each H.
>>>>
>>>>> When we disallow decider/input pairs that are incorrect
>>>>> questions where both YES and NO are the wrong answer
>>>>
>>>> Once we understand that either YES or NO is the right answer, the
>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>
>>
>> Once we understand that either YES or NO is the right answer, the
>> whole rebuttal is tossed out as invalid and incorrect.
>>
>>>>> Does the barber that shaves everyone that does not shave
>>>>> themselves shave himself? is rejected as an incorrect question.
>>>>
>>>> The barber does not exist.
>>>
>>> Russell's paradox did not allow this answer within Naive set theory.
>>
>> Naive set theory says that for every predicate P, the set {x | P(x)}
>> exists. This axiom was a mistake. This axiom is not in ZFC.
>>
>> In Turing machines, for every non-empty finite set of alphabet symbols
>> Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states Q, every
>> q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R}, ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a
>> Turing machine. Do you think this is a mistake? Would you remove this
>> axiom from your version of Turing machines?
>>
>> (Following the definition used on Wikipedia:
>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)
>>
>>>> The following is true statement:
>>>>
>>>> ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>>> ¬Shaves(Person, Person))
>>>>
>>>> The following is a true statement:
>>>>
>>>> ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>>> ¬Shaves(Person, Person))
>>>>
>>>
>>> That might be correct I did not check it over and over
>>> again and again to make sure.
>>>
>>> The same reasoning seems to rebut Gödel Incompleteness:
>>> ...We are therefore confronted with a proposition which
>>> asserts its own unprovability. 15 ... (Gödel 1931:43-44)
>>> ¬∃G ∈ F | G := ~(F ⊢ G)
>>>
>>> Any G in F that asserts its own unprovability in F is
>>> asserting that there is no sequence of inference steps
>>> in F that prove that they themselves do not exist in F.
>>
>> The barber does not exist and the proposition does not exist.
>>
>
> When we do this exact same thing that ZFC did for self-referential
> sets then self-referential (thus pathological) inputs to halt deciders
> cannot exist.
>

And your computation system isn't Turing Complete, by definition.

You are saying there are some computations that Turing Machines can do
(like H^) that can not exist in your system.

That option has been rejected as viable for Computation Theory

On 3/12/24 1:38 PM, olcott wrote:
> On 3/12/2024 3:31 PM, immibis wrote:
>> On 12/03/24 20:02, olcott wrote:
>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>> On 12/03/24 19:12, olcott wrote:
>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>
>>>>> There is some input TMD to every H such that
>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>
>>>> And it can be a different TMD to each H.
>>>>
>>>>> When we disallow decider/input pairs that are incorrect
>>>>> questions where both YES and NO are the wrong answer
>>>>
>>>> Once we understand that either YES or NO is the right answer, the
>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>
>>
>> Once we understand that either YES or NO is the right answer, the
>> whole rebuttal is tossed out as invalid and incorrect.
>>
>>>>> Does the barber that shaves everyone that does not shave
>>>>> themselves shave himself? is rejected as an incorrect question.
>>>>
>>>> The barber does not exist.
>>>
>>> Russell's paradox did not allow this answer within Naive set theory.
>>
>> Naive set theory says that for every predicate P, the set {x | P(x)}
>> exists. This axiom was a mistake. This axiom is not in ZFC.
>>
>> In Turing machines, for every non-empty finite set of alphabet symbols
>> Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states Q, every
>> q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R}, ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a
>> Turing machine. Do you think this is a mistake? Would you remove this
>> axiom from your version of Turing machines?
>>
>> (Following the definition used on Wikipedia:
>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)
>>
>>>> The following is true statement:
>>>>
>>>> ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>>> ¬Shaves(Person, Person))
>>>>
>>>> The following is a true statement:
>>>>
>>>> ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>>> ¬Shaves(Person, Person))
>>>>
>>>
>>> That might be correct I did not check it over and over
>>> again and again to make sure.
>>>
>>> The same reasoning seems to rebut Gödel Incompleteness:
>>> ...We are therefore confronted with a proposition which
>>> asserts its own unprovability. 15 ... (Gödel 1931:43-44)
>>> ¬∃G ∈ F | G := ~(F ⊢ G)
>>>
>>> Any G in F that asserts its own unprovability in F is
>>> asserting that there is no sequence of inference steps
>>> in F that prove that they themselves do not exist in F.
>>
>> The barber does not exist and the proposition does not exist.
>>
>
> When we do this exact same thing that ZFC did for self-referential
> sets then Gödel's self-referential expressions that assert their
> own unprovability in F also cease to exist.
>

And you end up with a very weak logic system that can't even have the
full properties of the Natuarl Numbers.

On 3/12/2024 4:23 PM, Richard Damon wrote:
> On 3/12/24 1:11 PM, olcott wrote:
>> On 3/12/2024 2:40 PM, Richard Damon wrote:
>>> On 3/12/24 12:02 PM, olcott wrote:
>>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>>> On 12/03/24 19:12, olcott wrote:
>>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>
>>>>>> There is some input TMD to every H such that
>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>
>>>>> And it can be a different TMD to each H.
>>>>>
>>>>>> When we disallow decider/input pairs that are incorrect
>>>>>> questions where both YES and NO are the wrong answer
>>>>>
>>>>> Once we understand that either YES or NO is the right answer, the
>>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>>
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>
>>> No, because a given H will only go to one of the answers. THAT will
>>> be wrong, and the other one right.
>>>
>>
>> ∀ H ∈ Turing_Machine_Deciders
>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>
>> Not exactly. A pair of otherwise identical machines that
>> (that are contained within the above specified set)
>> only differ by return value will both be wrong on the
>> same pathological input.
>
> You mean a pair of DIFFERENT machines. Any difference is different.

Every decider/input pair (referenced in the above set) has a
corresponding decider/input pair that only differs by the return
value of its decider.

That both of these decider/input pairs get the wrong answer proves
that their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/12/2024 4:29 PM, Richard Damon wrote:
> On 3/12/24 1:35 PM, olcott wrote:
>> On 3/12/2024 3:31 PM, immibis wrote:
>>> On 12/03/24 20:02, olcott wrote:
>>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>>> On 12/03/24 19:12, olcott wrote:
>>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>
>>>>>> There is some input TMD to every H such that
>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>
>>>>> And it can be a different TMD to each H.
>>>>>
>>>>>> When we disallow decider/input pairs that are incorrect
>>>>>> questions where both YES and NO are the wrong answer
>>>>>
>>>>> Once we understand that either YES or NO is the right answer, the
>>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>>
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>
>>>
>>> Once we understand that either YES or NO is the right answer, the
>>> whole rebuttal is tossed out as invalid and incorrect.
>>>
>>>>>> Does the barber that shaves everyone that does not shave
>>>>>> themselves shave himself? is rejected as an incorrect question.
>>>>>
>>>>> The barber does not exist.
>>>>
>>>> Russell's paradox did not allow this answer within Naive set theory.
>>>
>>> Naive set theory says that for every predicate P, the set {x | P(x)}
>>> exists. This axiom was a mistake. This axiom is not in ZFC.
>>>
>>> In Turing machines, for every non-empty finite set of alphabet
>>> symbols Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states
>>> Q, every q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R},
>>> ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a Turing machine. Do you think this is a mistake?
>>> Would you remove this axiom from your version of Turing machines?
>>>
>>> (Following the definition used on Wikipedia:
>>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)
>>>
>>>>> The following is true statement:
>>>>>
>>>>> ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>>>> ¬Shaves(Person, Person))
>>>>>
>>>>> The following is a true statement:
>>>>>
>>>>> ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>>>> ¬Shaves(Person, Person))
>>>>>
>>>>
>>>> That might be correct I did not check it over and over
>>>> again and again to make sure.
>>>>
>>>> The same reasoning seems to rebut Gödel Incompleteness:
>>>> ...We are therefore confronted with a proposition which
>>>> asserts its own unprovability. 15 ... (Gödel 1931:43-44)
>>>> ¬∃G ∈ F | G := ~(F ⊢ G)
>>>>
>>>> Any G in F that asserts its own unprovability in F is
>>>> asserting that there is no sequence of inference steps
>>>> in F that prove that they themselves do not exist in F.
>>>
>>> The barber does not exist and the proposition does not exist.
>>>
>>
>> When we do this exact same thing that ZFC did for self-referential
>> sets then self-referential (thus pathological) inputs to halt deciders
>> cannot exist.
>>
>
> And your computation system isn't Turing Complete, by definition.
>

I don't think that this has anything to do with Turing completeness.
It is more a matter of reestablishing the notion of computation on
a new foundation the same way that ZFC did for Naive set theory.

ZFC removed logically impossible decision problem instances. My new
foundation for computation (NFFC) only removes logically impossible
decision problem instances. Turing machines remain the same.

> You are saying there are some computations that Turing Machines can do
> (like H^) that can not exist in your system.
>
> That option has been rejected as viable for Computation Theory

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/12/2024 4:31 PM, Richard Damon wrote:
> On 3/12/24 1:38 PM, olcott wrote:
>> On 3/12/2024 3:31 PM, immibis wrote:
>>> On 12/03/24 20:02, olcott wrote:
>>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>>> On 12/03/24 19:12, olcott wrote:
>>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>
>>>>>> There is some input TMD to every H such that
>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>
>>>>> And it can be a different TMD to each H.
>>>>>
>>>>>> When we disallow decider/input pairs that are incorrect
>>>>>> questions where both YES and NO are the wrong answer
>>>>>
>>>>> Once we understand that either YES or NO is the right answer, the
>>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>>
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>
>>>
>>> Once we understand that either YES or NO is the right answer, the
>>> whole rebuttal is tossed out as invalid and incorrect.
>>>
>>>>>> Does the barber that shaves everyone that does not shave
>>>>>> themselves shave himself? is rejected as an incorrect question.
>>>>>
>>>>> The barber does not exist.
>>>>
>>>> Russell's paradox did not allow this answer within Naive set theory.
>>>
>>> Naive set theory says that for every predicate P, the set {x | P(x)}
>>> exists. This axiom was a mistake. This axiom is not in ZFC.
>>>
>>> In Turing machines, for every non-empty finite set of alphabet
>>> symbols Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states
>>> Q, every q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R},
>>> ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a Turing machine. Do you think this is a mistake?
>>> Would you remove this axiom from your version of Turing machines?
>>>
>>> (Following the definition used on Wikipedia:
>>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)
>>>
>>>>> The following is true statement:
>>>>>
>>>>> ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>>>> ¬Shaves(Person, Person))
>>>>>
>>>>> The following is a true statement:
>>>>>
>>>>> ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>>>> ¬Shaves(Person, Person))
>>>>>
>>>>
>>>> That might be correct I did not check it over and over
>>>> again and again to make sure.
>>>>
>>>> The same reasoning seems to rebut Gödel Incompleteness:
>>>> ...We are therefore confronted with a proposition which
>>>> asserts its own unprovability. 15 ... (Gödel 1931:43-44)
>>>> ¬∃G ∈ F | G := ~(F ⊢ G)
>>>>
>>>> Any G in F that asserts its own unprovability in F is
>>>> asserting that there is no sequence of inference steps
>>>> in F that prove that they themselves do not exist in F.
>>>
>>> The barber does not exist and the proposition does not exist.
>>>
>>
>> When we do this exact same thing that ZFC did for self-referential
>> sets then Gödel's self-referential expressions that assert their
>> own unprovability in F also cease to exist.
>>
>
> And you end up with a very weak logic system that can't even have the
> full properties of the Natuarl Numbers.

Natural numbers never really did have the property of provability.
This was something artificially contrived that never really belonged
to them.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 12/03/24 22:34, olcott wrote:
> Every decider/input pair (referenced in the above set) has a
> corresponding decider/input pair that only differs by the return
> value of its decider.
>
> That both of these decider/input pairs get the wrong answer proves
> that their question was incorrect because the opposite answer to the
> same question is also proven to be incorrect.

It's become obvious that you don't understand what "correct answer" means.

On 12/03/24 22:42, olcott wrote:
> ZFC removed logically impossible decision problem instances. My new
> foundation for computation (NFFC) only removes logically impossible
> decision problem instances. Turing machines remain the same.

If Turing machines remain the same, then every halt decider still has a
Turing machine which it gets wrong.

On 3/12/24 2:34 PM, olcott wrote:
> On 3/12/2024 4:23 PM, Richard Damon wrote:
>> On 3/12/24 1:11 PM, olcott wrote:
>>> On 3/12/2024 2:40 PM, Richard Damon wrote:
>>>> On 3/12/24 12:02 PM, olcott wrote:
>>>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>>>> On 12/03/24 19:12, olcott wrote:
>>>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>
>>>>>>> There is some input TMD to every H such that
>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>
>>>>>> And it can be a different TMD to each H.
>>>>>>
>>>>>>> When we disallow decider/input pairs that are incorrect
>>>>>>> questions where both YES and NO are the wrong answer
>>>>>>
>>>>>> Once we understand that either YES or NO is the right answer, the
>>>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>>>
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>
>>>> No, because a given H will only go to one of the answers. THAT will
>>>> be wrong, and the other one right.
>>>>
>>>
>>> ∀ H ∈ Turing_Machine_Deciders
>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>
>>> Not exactly. A pair of otherwise identical machines that
>>> (that are contained within the above specified set)
>>> only differ by return value will both be wrong on the
>>> same pathological input.
>>
>> You mean a pair of DIFFERENT machines. Any difference is different.
>
> Every decider/input pair (referenced in the above set) has a
> corresponding decider/input pair that only differs by the return
> value of its decider.

Nope.

Lets break it down.

FIRST, we go through each of the H's due to:

∀ H ∈ Turing_Machine_Deciders

Then for Each of them we find that we can find an input, that might be a
different input of each decider:

∃ TMD ∈ Turing_Machine_Descriptions

The Results of which, that decider give a wrong answer for that input:

Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

So, in no case do we have an assertioig that it is the SAME input for
anytwo deciders that we chose.

Thus, you have NO basis for you claim that there is ANY pairing of two
deciders, that answer differently of getting the same input.

>
> That both of these decider/input pairs get the wrong answer proves
> that their question was incorrect because the opposite answer to the
> same question is also proven to be incorrect.
>

But that isn't claimed, so your logic is baseless.

On 3/12/2024 5:11 PM, immibis wrote:
> On 12/03/24 22:34, olcott wrote:
>> Every decider/input pair (referenced in the above set) has a
>> corresponding decider/input pair that only differs by the return
>> value of its decider.
>>
>> That both of these decider/input pairs get the wrong answer proves
>> that their question was incorrect because the opposite answer to the
>> same question is also proven to be incorrect.
>
> It's become obvious that you don't understand what "correct answer" means.
>

∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

Every decider/input pair (referenced in the above set) has a
corresponding decider/input pair that only differs by the
return value of its decider.

That both of these decider/input pairs get the wrong answer proves
that their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.

E C R Hehner. Objective and Subjective Specifications
WST Workshop on Termination, Oxford. 2018 July 18.
See https://www.cs.toronto.edu/~hehner/OSS.pdf

Saying that it doesn't matter who answers the question simply
ignores that Hehner's Carol can't correctly answer her question:
Can Carol correctly answer “no” to this [yes/no] question?

*Richard found a loophole in Hehner's original wording*
"Can Carol correctly answer “no” to this question?"
The loophole is that answers equivalent to "no" would be correct.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/12/2024 5:11 PM, immibis wrote:
> On 12/03/24 22:42, olcott wrote:
>> ZFC removed logically impossible decision problem instances. My new
>> foundation for computation (NFFC) only removes logically impossible
>> decision problem instances. Turing machines remain the same.
>
> If Turing machines remain the same, then every halt decider still has a
> Turing machine which it gets wrong.

Turing machines can remain the same yet the notion of computation
would change. Undecidable inputs simply become construed as semantically
invalid inputs.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/12/24 2:42 PM, olcott wrote:
> On 3/12/2024 4:29 PM, Richard Damon wrote:
>> On 3/12/24 1:35 PM, olcott wrote:
>>> On 3/12/2024 3:31 PM, immibis wrote:
>>>> On 12/03/24 20:02, olcott wrote:
>>>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>>>> On 12/03/24 19:12, olcott wrote:
>>>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>
>>>>>>> There is some input TMD to every H such that
>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>
>>>>>> And it can be a different TMD to each H.
>>>>>>
>>>>>>> When we disallow decider/input pairs that are incorrect
>>>>>>> questions where both YES and NO are the wrong answer
>>>>>>
>>>>>> Once we understand that either YES or NO is the right answer, the
>>>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>>>
>>>>>
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>>
>>>>
>>>> Once we understand that either YES or NO is the right answer, the
>>>> whole rebuttal is tossed out as invalid and incorrect.
>>>>
>>>>>>> Does the barber that shaves everyone that does not shave
>>>>>>> themselves shave himself? is rejected as an incorrect question.
>>>>>>
>>>>>> The barber does not exist.
>>>>>
>>>>> Russell's paradox did not allow this answer within Naive set theory.
>>>>
>>>> Naive set theory says that for every predicate P, the set {x | P(x)}
>>>> exists. This axiom was a mistake. This axiom is not in ZFC.
>>>>
>>>> In Turing machines, for every non-empty finite set of alphabet
>>>> symbols Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of
>>>> states Q, every q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R},
>>>> ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a Turing machine. Do you think this is a
>>>> mistake? Would you remove this axiom from your version of Turing
>>>> machines?
>>>>
>>>> (Following the definition used on Wikipedia:
>>>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)
>>>>
>>>>>> The following is true statement:
>>>>>>
>>>>>> ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>>>>> ¬Shaves(Person, Person))
>>>>>>
>>>>>> The following is a true statement:
>>>>>>
>>>>>> ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔
>>>>>> ¬Shaves(Person, Person))
>>>>>>
>>>>>
>>>>> That might be correct I did not check it over and over
>>>>> again and again to make sure.
>>>>>
>>>>> The same reasoning seems to rebut Gödel Incompleteness:
>>>>> ...We are therefore confronted with a proposition which
>>>>> asserts its own unprovability. 15 ... (Gödel 1931:43-44)
>>>>> ¬∃G ∈ F | G := ~(F ⊢ G)
>>>>>
>>>>> Any G in F that asserts its own unprovability in F is
>>>>> asserting that there is no sequence of inference steps
>>>>> in F that prove that they themselves do not exist in F.
>>>>
>>>> The barber does not exist and the proposition does not exist.
>>>>
>>>
>>> When we do this exact same thing that ZFC did for self-referential
>>> sets then self-referential (thus pathological) inputs to halt deciders
>>> cannot exist.
>>>
>>
>> And your computation system isn't Turing Complete, by definition.
>>
>
> I don't think that this has anything to do with Turing completeness.
> It is more a matter of reestablishing the notion of computation on
> a new foundation the same way that ZFC did for Naive set theory.

Just means you don't know the MEANING of Turing Completness, just like
most things in the field.

>
> ZFC removed logically impossible decision problem instances. My new
> foundation for computation (NFFC) only removes logically impossible
> decision problem instances. Turing machines remain the same.

Nope. ZFC was not about "decision problems" it was about the methodology
to build sets.

You just don't understand what any of that means,

If you want to build this NFFC, you better get cracking on it, Sounds
like you don't have much time. You will need to first learn enough about
what people actualy expect in a definition of a system so you can make
sure you define what is needed, and then show that it is actually useful
for something beside you just arguing about it. It sounds like it will
NOT be a "Turing Complete" system, so showing its usefullness will be an
uphill battle.

>
>> You are saying there are some computations that Turing Machines can do
>> (like H^) that can not exist in your system.
>>
>> That option has been rejected as viable for Computation Theory
>