On 3/12/24 3:40 PM, olcott wrote:
> 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.
>

You better watch out with that. "Inputs" aren't "Undecidable", Mappings
/ Problems are.

After all the input (H^) (H^) can be correctly decide by a lot of deciders.

If you try to define that the input that a machine gets wrong are just
invalid to give it, the ANY random decider is an ANYTHING decider, as
the inputs it happens to get right it gets right, and the rest just were
incorrect to give to it.

That also means, you don't know if it was valid to ask the decider that
input, and deciding if you can give that input to the decider is
probably an uncomputable problem, so you end up not being able to trust
any deciders answer.

On 3/12/2024 5:30 PM, Richard Damon wrote:
> 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.
>
∀ H ∈ Turing_Machines_Returning_Boolean
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

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

That both of these H/TMD 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 3/12/24 2:44 PM, olcott wrote:
> 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.
>

No, Godel showed (or maybe used a previous proof) that you can use the
Mathematics of Natural Numbers to test if a proof is valid.

You just don't understand it. It really is very related to how Turing
Machines work, which can be converted to a mathematical model.

There is a field that looks at the comparability of Computations to
Logic, so they are all really quite related.

On 12/03/24 23:53, olcott wrote:
> On 3/12/2024 5:30 PM, Richard Damon wrote:
>> 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:
>>>>> 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.
>>
> ∀ H ∈ Turing_Machines_Returning_Boolean
> ∃ TMD ∈ Turing_Machine_Descriptions  |
> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
> Every H/TMD pair (referenced in the above set) has a
> corresponding H/TMD pair that only differs by the return
> value of its Boolean_TM.
>
> That both of these H/TMD 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.
>

Nobody knows what the fuck you are talking about. You have to actually
explain it. The same machine always gives the same return value on the
same input.

On 3/12/24 3:53 PM, olcott wrote:
> On 3/12/2024 5:30 PM, Richard Damon wrote:
>> 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.
>>
> ∀ H ∈ Turing_Machines_Returning_Boolean
> ∃ TMD ∈ Turing_Machine_Descriptions  |
> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
> Every H/TMD pair (referenced in the above set) has a
> corresponding H/TMD pair that only differs by the return
> value of its Boolean_TM.

That isn't in the set above.

>
> That both of these H/TMD 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.
>
>
Nope, since both aren't in the set selected.

You just don't understand what that statement is saying.

I've expalined it, but it seems over you head.

For Every H, we show we can find at least one input (chosen just for
that machine) that it will get wrong.

Thus, no H computes the Halting Property.

Never will two machine that give opposite results be given the exact
same input, and your claiming that just shows you are a stupid
pathological liar.

They CAN'T be as the two will give opposite Predicted Behaviors, and
thus one will be correct, so that input won't be selected for that machine.

In fact, the proven TMD to give to each machine is based on the Decider
it will be given to, and the opposite outputs will generate different
"pathological inputs" so you claim just won't happen.

You are just proving your stupidity.

On 3/12/2024 6:00 PM, Richard Damon wrote:
> On 3/12/24 2:44 PM, olcott wrote:
>> 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.
>>
>
> No, Godel showed (or maybe used a previous proof) that you can use the
> Mathematics of Natural Numbers to test if a proof is valid.
>
> You just don't understand it. It really is very related to how Turing
> Machines work, which can be converted to a mathematical model.
>
> There is a field that looks at the comparability of Computations to
> Logic, so they are all really quite related.

This is refuted.
....We are therefore confronted with a proposition which
asserts its own unprovability. 15 ...(Gödel 1931:43-44)

based on immblis Russell's Paradox reply
¬∃G ∈ F | G ↔ ~(F ⊢ G) // is simply false

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 3/12/2024 6:05 PM, immibis wrote:
> On 12/03/24 23:53, olcott wrote:
>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>> 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:
>>>>>> 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.
>>>
>> ∀ H ∈ Turing_Machines_Returning_Boolean
>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>
>> Every H/TMD pair (referenced in the above set) has a
>> corresponding H/TMD pair that only differs by the return
>> value of its Boolean_TM.
>>
>> That both of these H/TMD 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.
>>
>
> Nobody knows what the fuck you are talking about. You have to actually
> explain it. The same machine always gives the same return value on the
> same input.
>

It has taken me twenty years to translate my intuitions into
words that can possibly understood.

A pair of Turing Machines that return Boolean that are identical
besides their return value that cannot decide some property of
the same input are being asked the same YES/NO question having
no correct YES/NO answer.

--
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 6:11 PM, Richard Damon wrote:
> On 3/12/24 3:53 PM, olcott wrote:
>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>> 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.
>>>
>> ∀ H ∈ Turing_Machines_Returning_Boolean
>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>
>> Every H/TMD pair (referenced in the above set) has a
>> corresponding H/TMD pair that only differs by the return
>> value of its Boolean_TM.
>
> That isn't in the set above.
>
>>
>> That both of these H/TMD 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.
>>
>>
> Nope, since both aren't in the set selected.
>

When they are deciders that must get the correct answer both
of them are not in the set.

When they are Turing_Machines_Returning_Boolean the this
set inherently includes identical pairs that only differ
by return value.

> You just don't understand what that statement is saying.
>
> I've expalined it, but it seems over you head.
>
No the problem is that you are not paying attention.

> For Every H, we show we can find at least one input (chosen just for
> that machine) that it will get wrong.
>
When we use machine templates then we can see instances of
the same machine that only differs by return value where both
get the wrong answer on the same input. By same input I mean
the same finite string of numerical values.

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

On 13/03/24 00:24, olcott wrote:
> On 3/12/2024 6:05 PM, immibis wrote:
>> On 12/03/24 23:53, olcott wrote:
>>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>>> 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:
>>>>>>> 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.
>>>>
>>> ∀ H ∈ Turing_Machines_Returning_Boolean
>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>
>>> Every H/TMD pair (referenced in the above set) has a
>>> corresponding H/TMD pair that only differs by the return
>>> value of its Boolean_TM.
>>>
>>> That both of these H/TMD 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.
>>>
>>
>> Nobody knows what the fuck you are talking about. You have to actually
>> explain it. The same machine always gives the same return value on the
>> same input.
>>
>
> It has taken me twenty years to translate my intuitions into
> words that can possibly understood.

You failed.

> A pair of Turing Machines that return Boolean that are identical
> besides their return value that cannot decide some property of
> the same input are being asked the same YES/NO question having
> no correct YES/NO answer.

https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their
return value.
You can't because you are talking nonsense. they don't exist.

On 3/12/2024 5:53 PM, Richard Damon wrote:
> On 3/12/24 3:40 PM, olcott wrote:
>> 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.
>>
>
> You better watch out with that. "Inputs" aren't "Undecidable", Mappings
> / Problems are.
>
> After all the input (H^) (H^) can be correctly decide by a lot of deciders.
>
None-the-less the halting problem proof ceases to function.

> If you try to define that the input that a machine gets wrong are just
> invalid to give it, the ANY random decider is an ANYTHING decider, as
> the inputs it happens to get right it gets right, and the rest just were
> incorrect to give to it.
>

We define a corresponding pair of deciders where
YES means {has the property} and NO mean DONT_KNOW
and the other one
YES {means does not have the property} and NO mean DONT_KNOW.
Two DONT_KNOWs mean semantically invalid input for this decider.

> That also means, you don't know if it was valid to ask the decider that
> input, and deciding if you can give that input to the decider is
> probably an uncomputable problem, so you end up not being able to trust
> any deciders answer.

Already addressed.

--
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 6:38 PM, immibis wrote:
> On 13/03/24 00:24, olcott wrote:
>> On 3/12/2024 6:05 PM, immibis wrote:
>>> On 12/03/24 23:53, olcott wrote:
>>>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>>>> 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:
>>>>>>>> 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.
>>>>>
>>>> ∀ H ∈ Turing_Machines_Returning_Boolean
>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>
>>>> Every H/TMD pair (referenced in the above set) has a
>>>> corresponding H/TMD pair that only differs by the return
>>>> value of its Boolean_TM.
>>>>
>>>> That both of these H/TMD 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.
>>>>
>>>
>>> Nobody knows what the fuck you are talking about. You have to
>>> actually explain it. The same machine always gives the same return
>>> value on the same input.
>>>
>>
>> It has taken me twenty years to translate my intuitions into
>> words that can possibly understood.
>
> You failed.
>
>> A pair of Turing Machines that return Boolean that are identical
>> besides their return value that cannot decide some property of
>> the same input are being asked the same YES/NO question having
>> no correct YES/NO answer.
>
> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
> A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
> Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their
> return value.
> You can't because you are talking nonsense. they don't exist.

Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.

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

On 13/03/24 00:56, olcott wrote:
> On 3/12/2024 6:38 PM, immibis wrote:
>> On 13/03/24 00:24, olcott wrote:
>>> On 3/12/2024 6:05 PM, immibis wrote:
>>>> On 12/03/24 23:53, olcott wrote:
>>>>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>>>>> 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:
>>>>>>>>> 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.
>>>>>>
>>>>> ∀ H ∈ Turing_Machines_Returning_Boolean
>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>
>>>>> Every H/TMD pair (referenced in the above set) has a
>>>>> corresponding H/TMD pair that only differs by the return
>>>>> value of its Boolean_TM.
>>>>>
>>>>> That both of these H/TMD 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.
>>>>>
>>>>
>>>> Nobody knows what the fuck you are talking about. You have to
>>>> actually explain it. The same machine always gives the same return
>>>> value on the same input.
>>>>
>>>
>>> It has taken me twenty years to translate my intuitions into
>>> words that can possibly understood.
>>
>> You failed.
>>
>>> A pair of Turing Machines that return Boolean that are identical
>>> besides their return value that cannot decide some property of
>>> the same input are being asked the same YES/NO question having
>>> no correct YES/NO answer.
>>
>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
>> A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
>> Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their
>> return value.
>> You can't because you are talking nonsense. they don't exist.
>
> Turing machine descriptions that are identical finite strings
> except for the the 1/0 that they write the their exact same
> tape relative location.

So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?

On 3/12/2024 7:10 PM, immibis wrote:
> On 13/03/24 00:56, olcott wrote:
>> On 3/12/2024 6:38 PM, immibis wrote:
>>> On 13/03/24 00:24, olcott wrote:
>>>> On 3/12/2024 6:05 PM, immibis wrote:
>>>>> On 12/03/24 23:53, olcott wrote:
>>>>>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>>>>>> 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:
>>>>>>>>>> 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.
>>>>>>>
>>>>>> ∀ H ∈ Turing_Machines_Returning_Boolean
>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>
>>>>>> Every H/TMD pair (referenced in the above set) has a
>>>>>> corresponding H/TMD pair that only differs by the return
>>>>>> value of its Boolean_TM.
>>>>>>
>>>>>> That both of these H/TMD 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.
>>>>>>
>>>>>
>>>>> Nobody knows what the fuck you are talking about. You have to
>>>>> actually explain it. The same machine always gives the same return
>>>>> value on the same input.
>>>>>
>>>>
>>>> It has taken me twenty years to translate my intuitions into
>>>> words that can possibly understood.
>>>
>>> You failed.
>>>
>>>> A pair of Turing Machines that return Boolean that are identical
>>>> besides their return value that cannot decide some property of
>>>> the same input are being asked the same YES/NO question having
>>>> no correct YES/NO answer.
>>>
>>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
>>> A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
>>> Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their
>>> return value.
>>> You can't because you are talking nonsense. they don't exist.
>>
>> Turing machine descriptions that are identical finite strings
>> except for the the 1/0 that they write the their exact same
>> tape relative location.
>
> So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
Exactly one element of Q differs by writing a 1 instead of a 0.

http://www2.lns.mit.edu/~dsw/turing/doc/tm_manual.txt
One quintuple has a different third element.

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

On 13/03/24 01:18, olcott wrote:
> On 3/12/2024 7:10 PM, immibis wrote:
>> On 13/03/24 00:56, olcott wrote:
>>> On 3/12/2024 6:38 PM, immibis wrote:
>>>> On 13/03/24 00:24, olcott wrote:
>>>>> On 3/12/2024 6:05 PM, immibis wrote:
>>>>>> On 12/03/24 23:53, olcott wrote:
>>>>>>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>>>>>>> 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:
>>>>>>>>>>> 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.
>>>>>>>>
>>>>>>> ∀ H ∈ Turing_Machines_Returning_Boolean
>>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>
>>>>>>> Every H/TMD pair (referenced in the above set) has a
>>>>>>> corresponding H/TMD pair that only differs by the return
>>>>>>> value of its Boolean_TM.
>>>>>>>
>>>>>>> That both of these H/TMD 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.
>>>>>>>
>>>>>>
>>>>>> Nobody knows what the fuck you are talking about. You have to
>>>>>> actually explain it. The same machine always gives the same return
>>>>>> value on the same input.
>>>>>>
>>>>>
>>>>> It has taken me twenty years to translate my intuitions into
>>>>> words that can possibly understood.
>>>>
>>>> You failed.
>>>>
>>>>> A pair of Turing Machines that return Boolean that are identical
>>>>> besides their return value that cannot decide some property of
>>>>> the same input are being asked the same YES/NO question having
>>>>> no correct YES/NO answer.
>>>>
>>>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
>>>> A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
>>>> Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their
>>>> return value.
>>>> You can't because you are talking nonsense. they don't exist.
>>>
>>> Turing machine descriptions that are identical finite strings
>>> except for the the 1/0 that they write the their exact same
>>> tape relative location.
>>
>> So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
> Exactly one element of Q differs by writing a 1 instead of a 0.

That's part of δ but this mistake doesn't matter.

It wasn't clear whether you were talking about a Turing machine that was
somehow identical but gave a different return value, or one that was not
identical. Now you have explained it is not identical.

Why does the fact that Bob says 1+1=2 and Bill says 1+1=3 mean that
neither 2 nor 3 is the correct answer?

On 3/12/2024 8:05 PM, immibis wrote:
> On 13/03/24 01:18, olcott wrote:
>> On 3/12/2024 7:10 PM, immibis wrote:
>>> On 13/03/24 00:56, olcott wrote:
>>>> On 3/12/2024 6:38 PM, immibis wrote:
>>>>> On 13/03/24 00:24, olcott wrote:
>>>>>> On 3/12/2024 6:05 PM, immibis wrote:
>>>>>>> On 12/03/24 23:53, olcott wrote:
>>>>>>>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>>>>>>>> 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:
>>>>>>>>>>>> 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.
>>>>>>>>>
>>>>>>>> ∀ H ∈ Turing_Machines_Returning_Boolean
>>>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>>
>>>>>>>> Every H/TMD pair (referenced in the above set) has a
>>>>>>>> corresponding H/TMD pair that only differs by the return
>>>>>>>> value of its Boolean_TM.
>>>>>>>>
>>>>>>>> That both of these H/TMD 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.
>>>>>>>>
>>>>>>>
>>>>>>> Nobody knows what the fuck you are talking about. You have to
>>>>>>> actually explain it. The same machine always gives the same
>>>>>>> return value on the same input.
>>>>>>>
>>>>>>
>>>>>> It has taken me twenty years to translate my intuitions into
>>>>>> words that can possibly understood.
>>>>>
>>>>> You failed.
>>>>>
>>>>>> A pair of Turing Machines that return Boolean that are identical
>>>>>> besides their return value that cannot decide some property of
>>>>>> the same input are being asked the same YES/NO question having
>>>>>> no correct YES/NO answer.
>>>>>
>>>>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
>>>>> A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
>>>>> Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their
>>>>> return value.
>>>>> You can't because you are talking nonsense. they don't exist.
>>>>
>>>> Turing machine descriptions that are identical finite strings
>>>> except for the the 1/0 that they write the their exact same
>>>> tape relative location.
>>>
>>> So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
>> Exactly one element of Q differs by writing a 1 instead of a 0.
>
> That's part of δ but this mistake doesn't matter.
>
> It wasn't clear whether you were talking about a Turing machine that was
> somehow identical but gave a different return value, or one that was not
> identical. Now you have explained it is not identical.
>
They are identical except for their return value that is specified
in a single state that is different.

*This means that they implement the exact same algorithm*

A protocol can be defined so that Turing machine descriptions always
implement their return value in their state with the largest Natural
Number value. This allows other Turing machines to determine identical
algorithms except for return value.

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

On 13/03/24 02:47, olcott wrote:
> On 3/12/2024 8:05 PM, immibis wrote:
>> On 13/03/24 01:18, olcott wrote:
>>> On 3/12/2024 7:10 PM, immibis wrote:
>>>> So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
>>> Exactly one element of Q differs by writing a 1 instead of a 0.
>>
>> That's part of δ but this mistake doesn't matter.
>>
>> It wasn't clear whether you were talking about a Turing machine that
>> was somehow identical but gave a different return value, or one that
>> was not identical. Now you have explained it is not identical.
>>
> They are identical except for their return value that is specified
> in a single state that is different.
>
> *This means that they implement the exact same algorithm*

OK. Well, one of them gets the right answer and one of them gets the
wrong answer. What is the confusion?

On 3/12/24 6:47 PM, olcott wrote:
> On 3/12/2024 8:05 PM, immibis wrote:
>> On 13/03/24 01:18, olcott wrote:
>>> On 3/12/2024 7:10 PM, immibis wrote:
>>>> On 13/03/24 00:56, olcott wrote:
>>>>> On 3/12/2024 6:38 PM, immibis wrote:
>>>>>> On 13/03/24 00:24, olcott wrote:
>>>>>>> On 3/12/2024 6:05 PM, immibis wrote:
>>>>>>>> On 12/03/24 23:53, olcott wrote:
>>>>>>>>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>>>>>>>>> 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:
>>>>>>>>>>>>> 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.
>>>>>>>>>>
>>>>>>>>> ∀ H ∈ Turing_Machines_Returning_Boolean
>>>>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>>>
>>>>>>>>> Every H/TMD pair (referenced in the above set) has a
>>>>>>>>> corresponding H/TMD pair that only differs by the return
>>>>>>>>> value of its Boolean_TM.
>>>>>>>>>
>>>>>>>>> That both of these H/TMD 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.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Nobody knows what the fuck you are talking about. You have to
>>>>>>>> actually explain it. The same machine always gives the same
>>>>>>>> return value on the same input.
>>>>>>>>
>>>>>>>
>>>>>>> It has taken me twenty years to translate my intuitions into
>>>>>>> words that can possibly understood.
>>>>>>
>>>>>> You failed.
>>>>>>
>>>>>>> A pair of Turing Machines that return Boolean that are identical
>>>>>>> besides their return value that cannot decide some property of
>>>>>>> the same input are being asked the same YES/NO question having
>>>>>>> no correct YES/NO answer.
>>>>>>
>>>>>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
>>>>>> A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
>>>>>> Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides
>>>>>> their return value.
>>>>>> You can't because you are talking nonsense. they don't exist.
>>>>>
>>>>> Turing machine descriptions that are identical finite strings
>>>>> except for the the 1/0 that they write the their exact same
>>>>> tape relative location.
>>>>
>>>> So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
>>> Exactly one element of Q differs by writing a 1 instead of a 0.
>>
>> That's part of δ but this mistake doesn't matter.
>>
>> It wasn't clear whether you were talking about a Turing machine that
>> was somehow identical but gave a different return value, or one that
>> was not identical. Now you have explained it is not identical.
>>
> They are identical except for their return value that is specified
> in a single state that is different.
>
> *This means that they implement the exact same algorithm*

Nope, because the algorithm include the final transition to the output.

>
> A protocol can be defined so that Turing machine descriptions always
> implement their return value in their state with the largest Natural
> Number value. This allows other Turing machines to determine identical
> algorithms except for return value.
>

State doesn't have defined "numbers".

On 3/12/24 4:31 PM, olcott wrote:
> On 3/12/2024 6:11 PM, Richard Damon wrote:
>> On 3/12/24 3:53 PM, olcott wrote:
>>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>>> 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.
>>>>
>>> ∀ H ∈ Turing_Machines_Returning_Boolean
>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>
>>> Every H/TMD pair (referenced in the above set) has a
>>> corresponding H/TMD pair that only differs by the return
>>> value of its Boolean_TM.
>>
>> That isn't in the set above.
>>
>>>
>>> That both of these H/TMD 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.
>>>
>>>
>> Nope, since both aren't in the set selected.
>>
>
> When they are deciders that must get the correct answer both
> of them are not in the set.

*IF* they are correct decider.

WHen we select from all Turing Machine Deciders, there is no requirement
that any of them get any particular answer right.

So, ALL deciders are in the set that we cycle through and apply the
following logic to ALL of them.

Each is them paired with an input that it will get wrong, and the
existance of the input was what as just proven, the ^ template

>
> When they are Turing_Machines_Returning_Boolean the this
> set inherently includes identical pairs that only differ
> by return value.

But in the step of select and input that they will get wrong, they will
be givne DIFFERENT inputs.

>
>> You just don't understand what that statement is saying.
>>
>> I've expalined it, but it seems over you head.
>>
> No the problem is that you are not paying attention.

No, you keep on making STUPID mistakes, like thinking that select a
input that the machine will get wrong needs to be the same for two
differnt machines.

>
>> For Every H, we show we can find at least one input (chosen just for
>> that machine) that it will get wrong.
>>
> When we use machine templates then we can see instances of
> the same machine that only differs by return value where both
> get the wrong answer on the same input. By same input I mean
> the same finite string of numerical values.
>

But if they returned differnt values, they will have different descriptions.

Otherwise, how could a UTM get the right answer, since it only gets the
description.

On 3/12/24 4:47 PM, olcott wrote:
> On 3/12/2024 5:53 PM, Richard Damon wrote:
>> On 3/12/24 3:40 PM, olcott wrote:
>>> 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.
>>>
>>
>> You better watch out with that. "Inputs" aren't "Undecidable",
>> Mappings / Problems are.
>>
>> After all the input (H^) (H^) can be correctly decide by a lot of
>> deciders.
>>
> None-the-less the halting problem proof ceases to function.

Doesn't meed to. You have already defined that you system is weaker than
Turing Complete, so no longer has impact on the Halting Theoren for
Turing Complete systems.

>
>> If you try to define that the input that a machine gets wrong are just
>> invalid to give it, the ANY random decider is an ANYTHING decider, as
>> the inputs it happens to get right it gets right, and the rest just
>> were incorrect to give to it.
>>
>
> We define a corresponding pair of deciders where
> YES means {has the property} and NO mean DONT_KNOW
> and the other one
> YES {means does not have the property} and NO mean DONT_KNOW.
> Two DONT_KNOWs mean semantically invalid input for this decider.

So, a pair that just answers "Don't Know" is a correct anything decider
that just delairs all inputs invalid.

Not very useful.

>
>> That also means, you don't know if it was valid to ask the decider
>> that input, and deciding if you can give that input to the decider is
>> probably an uncomputable problem, so you end up not being able to
>> trust any deciders answer.
>
> Already addressed.
>

Only by letting the decider arbirtarily declare all input invalid.

On 3/12/2024 9:17 PM, immibis wrote:
> On 13/03/24 02:47, olcott wrote:
>> On 3/12/2024 8:05 PM, immibis wrote:
>>> On 13/03/24 01:18, olcott wrote:
>>>> On 3/12/2024 7:10 PM, immibis wrote:
>>>>> So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
>>>> Exactly one element of Q differs by writing a 1 instead of a 0.
>>>
>>> That's part of δ but this mistake doesn't matter.
>>>
>>> It wasn't clear whether you were talking about a Turing machine that
>>> was somehow identical but gave a different return value, or one that
>>> was not identical. Now you have explained it is not identical.
>>>
>> They are identical except for their return value that is specified
>> in a single state that is different.
>>
>> *This means that they implement the exact same algorithm*
>
> OK. Well, one of them gets the right answer and one of them gets the
> wrong answer. What is the confusion?

The Linz Ĥ.H machine gets the wrong answer on its own
machine description no matter how its Linz H is defined.

This means that it gets the wrong answer on YES and the
wrong answer on NO.

--
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 9:21 PM, Richard Damon wrote:
> On 3/12/24 6:47 PM, olcott wrote:
>> On 3/12/2024 8:05 PM, immibis wrote:
>>> On 13/03/24 01:18, olcott wrote:
>>>> On 3/12/2024 7:10 PM, immibis wrote:
>>>>> On 13/03/24 00:56, olcott wrote:
>>>>>> On 3/12/2024 6:38 PM, immibis wrote:
>>>>>>> On 13/03/24 00:24, olcott wrote:
>>>>>>>> On 3/12/2024 6:05 PM, immibis wrote:
>>>>>>>>> On 12/03/24 23:53, olcott wrote:
>>>>>>>>>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>>>>>>>>>> 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:
>>>>>>>>>>>>>> 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.
>>>>>>>>>>>
>>>>>>>>>> ∀ H ∈ Turing_Machines_Returning_Boolean
>>>>>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>>>>
>>>>>>>>>> Every H/TMD pair (referenced in the above set) has a
>>>>>>>>>> corresponding H/TMD pair that only differs by the return
>>>>>>>>>> value of its Boolean_TM.
>>>>>>>>>>
>>>>>>>>>> That both of these H/TMD 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.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Nobody knows what the fuck you are talking about. You have to
>>>>>>>>> actually explain it. The same machine always gives the same
>>>>>>>>> return value on the same input.
>>>>>>>>>
>>>>>>>>
>>>>>>>> It has taken me twenty years to translate my intuitions into
>>>>>>>> words that can possibly understood.
>>>>>>>
>>>>>>> You failed.
>>>>>>>
>>>>>>>> A pair of Turing Machines that return Boolean that are identical
>>>>>>>> besides their return value that cannot decide some property of
>>>>>>>> the same input are being asked the same YES/NO question having
>>>>>>>> no correct YES/NO answer.
>>>>>>>
>>>>>>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
>>>>>>> A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
>>>>>>> Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides
>>>>>>> their return value.
>>>>>>> You can't because you are talking nonsense. they don't exist.
>>>>>>
>>>>>> Turing machine descriptions that are identical finite strings
>>>>>> except for the the 1/0 that they write the their exact same
>>>>>> tape relative location.
>>>>>
>>>>> So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
>>>> Exactly one element of Q differs by writing a 1 instead of a 0.
>>>
>>> That's part of δ but this mistake doesn't matter.
>>>
>>> It wasn't clear whether you were talking about a Turing machine that
>>> was somehow identical but gave a different return value, or one that
>>> was not identical. Now you have explained it is not identical.
>>>
>> They are identical except for their return value that is specified
>> in a single state that is different.
>>
>> *This means that they implement the exact same algorithm*
>
> Nope, because the algorithm include the final transition to the output.

The decision criteria is identical

>>
>> A protocol can be defined so that Turing machine descriptions always
>> implement their return value in their state with the largest Natural
>> Number value. This allows other Turing machines to determine identical
>> algorithms except for return value.
>>
>
> State doesn't have defined "numbers".

It must have something like this.

--
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 4:14 PM, olcott wrote:
> On 3/12/2024 6:00 PM, Richard Damon wrote:
>> On 3/12/24 2:44 PM, olcott wrote:
>>> 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.
>>>
>>
>> No, Godel showed (or maybe used a previous proof) that you can use the
>> Mathematics of Natural Numbers to test if a proof is valid.
>>
>> You just don't understand it. It really is very related to how Turing
>> Machines work, which can be converted to a mathematical model.
>>
>> There is a field that looks at the comparability of Computations to
>> Logic, so they are all really quite related.
>
> This is refuted.
> ...We are therefore confronted with a proposition which
> asserts its own unprovability. 15 ...(Gödel 1931:43-44)

Right, seen in the meta-Theory from F.

>
> based on immblis Russell's Paradox reply
> ¬∃G ∈ F | G ↔ ~(F ⊢ G)  // is simply false

Then a proof must exist in F that G is True, So G can't be false.

Note, The statement G is NOT a statement about itself being provable,
that is only a semantic revealed in the RIGHT meta-theory.

>
> 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.
>

No FINITE sequence of inference steps.

Note, the key is that G doesn't assert that, G is a statement about
math, that only when interpreted in a meta-theory sees that.

G is actually a statment about the existance of a number that matches a
complected and carefully constructed relationship, that is fully
computable. The Existance or non-existance of such a number is a pure
binary thing, either it WILL or it WON'T.

The complicated relationship deals with a way to encode as a number ANY
analyitic statement in F (since all statements are just strings, and
strings can be encoded into a number), and a calculation to see if that
statement actually is a proof starting with the enumerated truth makers
of F, through the valid and allowed logical operatons to the statement
of G. A number that satisfies this relation, encodes a proof of G, and
any such proof, can always be encoded in to a number.

G is the statement that no such number exist. So, if G is false, then
such a number exists, and then in the Meta-Theory, we can decode that
number into a proof in F that G must be true.

If this is the case, then F must be inconsistant, and the proof starts
with the presumption that we are dealing with a consistant logic system.

But, if G is true, then there can be no such number, and thus from our
knowledge in the mete-theory, we know that no proof CAN exist for that fact.

Thus, the only possiblity for this statement, is to be True, but unprovable.

The big part of the theory is showing that such a relationship can be
made, and that shows that proof checkers are a computable function.

Given a system, and an enumeration of its basic truths, we can check if
a proof present actually proves the result it claims to.

On 3/12/2024 9:28 PM, Richard Damon wrote:
> On 3/12/24 4:31 PM, olcott wrote:
>> On 3/12/2024 6:11 PM, Richard Damon wrote:
>>> On 3/12/24 3:53 PM, olcott wrote:
>>>> On 3/12/2024 5:30 PM, Richard Damon wrote:
>>>>> 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.
>>>>>
>>>> ∀ H ∈ Turing_Machines_Returning_Boolean
>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>
>>>> Every H/TMD pair (referenced in the above set) has a
>>>> corresponding H/TMD pair that only differs by the return
>>>> value of its Boolean_TM.
>>>
>>> That isn't in the set above.
>>>
>>>>
>>>> That both of these H/TMD 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.
>>>>
>>>>
>>> Nope, since both aren't in the set selected.
>>>
>>
>> When they are deciders that must get the correct answer both
>> of them are not in the set.
>
> *IF* they are correct decider.
>
> WHen we select from all Turing Machine Deciders, there is no requirement
> that any of them get any particular answer right.
>
> So, ALL deciders are in the set that we cycle through and apply the
> following logic to ALL of them.
>
> Each is them paired with an input that it will get wrong, and the
> existance of the input was what as just proven, the ^ template
>
>>
>> When they are Turing_Machines_Returning_Boolean the this
>> set inherently includes identical pairs that only differ
>> by return value.
>
> But in the step of select and input that they will get wrong, they will
> be givne DIFFERENT inputs.
>
>>
>>> You just don't understand what that statement is saying.
>>>
>>> I've expalined it, but it seems over you head.
>>>
>> No the problem is that you are not paying attention.
>
> No, you keep on making STUPID mistakes, like thinking that select a
> input that the machine will get wrong needs to be the same for two
> differnt machines.
>
>
>
>>
>>> For Every H, we show we can find at least one input (chosen just for
>>> that machine) that it will get wrong.
>>>
>> When we use machine templates then we can see instances of
>> the same machine that only differs by return value where both
>> get the wrong answer on the same input. By same input I mean
>> the same finite string of numerical values.
>>
>
> But if they returned differnt values, they will have different
> descriptions.
>
> Otherwise, how could a UTM get the right answer, since it only gets the
> description.

We can get around all of this stuff by simply using this criteria:
Date 10/13/2022 11:29:23 AM
*MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
(He has neither reviewed nor agreed to anything else in this paper)
(a) If simulating halt decider H correctly simulates its input D until H
correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.

*When we apply this criteria* (elaborated above)
Will you halt if you never abort your simulation?
*Then the halting problem is conquered*

When two different machines implementing this criteria
get different results from identical inputs then we
know that Pathological Self-Reference has been detected.

We don't even need to know that for:
*denial-of-service-attack detection*
*NO always means reject as unsafe*

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