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Oct 25, 2023, 6:25:57 PM10/25/23

to

https://www.liarparadox.org/Linz_Proof.pdf

This Turing Machine description at the top of page 3

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2

Is simplified and clarified to this:

when Ĥ is applied to ⟨Ĥ⟩

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The advantage of the Linz proof is that it simultaneously represents

the entire infinite set of what would otherwise be an infinite set of

H/D pairs by using a single TM template for Ĥ.

*We can say that both yes and no are incorrect answers for every*

*embedded_H that attempts to answer the question*

Does the computation that I am contained within halt?

Any yes/no question that lacks a correct yes/no answer is an incorrect

question. The inability to correctly answer an incorrect question places

no limit on anyone or anything.

On the other hand when embedded_H is trying to answer the question:

Could my input terminate normally?

this question can be answered with a “no”.

--

Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius

hits a target no one else can see." Arthur Schopenhauer

This Turing Machine description at the top of page 3

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2

Is simplified and clarified to this:

when Ĥ is applied to ⟨Ĥ⟩

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The advantage of the Linz proof is that it simultaneously represents

the entire infinite set of what would otherwise be an infinite set of

H/D pairs by using a single TM template for Ĥ.

*We can say that both yes and no are incorrect answers for every*

*embedded_H that attempts to answer the question*

Does the computation that I am contained within halt?

Any yes/no question that lacks a correct yes/no answer is an incorrect

question. The inability to correctly answer an incorrect question places

no limit on anyone or anything.

On the other hand when embedded_H is trying to answer the question:

Could my input terminate normally?

this question can be answered with a “no”.

--

Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius

hits a target no one else can see." Arthur Schopenhauer

Oct 25, 2023, 6:51:35 PM10/25/23

to

On 10/25/23 3:25 PM, olcott wrote:

> https://www.liarparadox.org/Linz_Proof.pdf

>

> This Turing Machine description at the top of page 3

> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞

> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2

No, you are missing key details
> https://www.liarparadox.org/Linz_Proof.pdf

>

> This Turing Machine description at the top of page 3

> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞

> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞ IFF H WM WM went to qy

q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2 IFF H WM WM went to qn

And H WM d is supposed to go to qy if M d will halt, and to qn if it

will not halt in a bounded number of steps to be correct.

>

> Is simplified and clarified to this:

> when Ĥ is applied to ⟨Ĥ⟩

>

> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ IFF H ⟨Ĥ⟩ went to qy

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn IFF H ⟨Ĥ⟩ went to qn

>

> The advantage of the Linz proof is that it simultaneously represents

> the entire infinite set of what would otherwise be an infinite set of

> H/D pairs by using a single TM template for Ĥ.

that H.

>

> *We can say that both yes and no are incorrect answers for every*

> *embedded_H that attempts to answer the question*

> Does the computation that I am contained within halt?

problem with that.

The question was, can you make an H that was correct.

The question that H needs to answer is "Does the computation described

by the input Halt?", and if H goes to qn, the right answer is yes, and

if H goes to qy, the right answer is No, so there is always a correct

answer, its just that H never gives it for the input built by this formula.

>

> Any yes/no question that lacks a correct yes/no answer is an incorrect

> question. The inability to correctly answer an incorrect question places

> no limit on anyone or anything.

The question ISN'T what does H / embedded_H need to return to be right,

but what is the correct answer for the given input.

>

> On the other hand when embedded_H is trying to answer the question:

> Could my input terminate normally?

> this question can be answered with a “no”.

>

In actuality, since your H goes to qn, then the RIGHT answer to the

halting question is YES the input represents a halting computation, so

it should have gone to qn.

You are just continuing to prove your stupidity.

You are just showing that you totally don't understand the nature of the

problem, by the parts you keep leaving out.

Oct 25, 2023, 7:01:37 PM10/25/23

to

On 10/25/2023 5:25 PM, olcott wrote:

> https://www.liarparadox.org/Linz_Proof.pdf

>

> This Turing Machine description at the top of page 3

> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞

> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2

>

> Is simplified and clarified to this:

> when Ĥ is applied to ⟨Ĥ⟩

>

> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

>

> The advantage of the Linz proof is that it simultaneously represents

> the entire infinite set of what would otherwise be an infinite set of

> H/D pairs by using a single TM template for Ĥ.

>

> *We can say that both yes and no are incorrect answers for every*

> *embedded_H that attempts to answer the question*

> Does the computation that I am contained within halt?

When we accept the idea of a pure function and that computations
> https://www.liarparadox.org/Linz_Proof.pdf

>

> This Turing Machine description at the top of page 3

> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞

> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2

>

> Is simplified and clarified to this:

> when Ĥ is applied to ⟨Ĥ⟩

>

> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

>

> The advantage of the Linz proof is that it simultaneously represents

> the entire infinite set of what would otherwise be an infinite set of

> H/D pairs by using a single TM template for Ĥ.

>

> *We can say that both yes and no are incorrect answers for every*

> *embedded_H that attempts to answer the question*

> Does the computation that I am contained within halt?

must be pure functions then we know that it would be incorrect

for H to report on the computation that itself is contained within.

embedded_H cannot even see the computation that itself is contained

within it can only see the behavior of its actual input.

Oct 25, 2023, 8:27:07 PM10/25/23

to

On 10/25/23 4:01 PM, olcott wrote:

> On 10/25/2023 5:25 PM, olcott wrote:

>> https://www.liarparadox.org/Linz_Proof.pdf

>>

>> This Turing Machine description at the top of page 3

>> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞

>> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2

>>

>> Is simplified and clarified to this:

>> when Ĥ is applied to ⟨Ĥ⟩

>>

>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

>>

>> The advantage of the Linz proof is that it simultaneously represents

>> the entire infinite set of what would otherwise be an infinite set of

>> H/D pairs by using a single TM template for Ĥ.

>>

>> *We can say that both yes and no are incorrect answers for every*

>> *embedded_H that attempts to answer the question*

>> Does the computation that I am contained within halt?

>

>

> When we accept the idea of a pure function and that computations

> must be pure functions then we know that it would be incorrect

> for H to report on the computation that itself is contained within.

Right, it is impossible to write an input that tries to actually use the
> On 10/25/2023 5:25 PM, olcott wrote:

>> https://www.liarparadox.org/Linz_Proof.pdf

>>

>> This Turing Machine description at the top of page 3

>> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞

>> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2

>>

>> Is simplified and clarified to this:

>> when Ĥ is applied to ⟨Ĥ⟩

>>

>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

>>

>> The advantage of the Linz proof is that it simultaneously represents

>> the entire infinite set of what would otherwise be an infinite set of

>> H/D pairs by using a single TM template for Ĥ.

>>

>> *We can say that both yes and no are incorrect answers for every*

>> *embedded_H that attempts to answer the question*

>> Does the computation that I am contained within halt?

>

>

> When we accept the idea of a pure function and that computations

> must be pure functions then we know that it would be incorrect

> for H to report on the computation that itself is contained within.

decider that is deciding on it, as such an input would not represent an

actual computation.

Thus, "the input" that the decider needs to decide on is always the

specific input of that problem, built on a specific decider, and that

input thus always has a defined behavior, and thus a correct answer.

Thus we START choosing a decider, and then we can built an input, and

that input WILL have a correct answer, it just is a fact that the

decider doesn't give it.

The fact the actual question, "Does the computation represented by the

input halt?" actually has a correct answer for every possible version of

this problem, means it is a valid question.

The fact that no decider can give the right answer for some particular

input, means that no decider is actually correct, and thus the problme

is uncomputable.

>

> embedded_H cannot even see the computation that itself is contained

> within it can only see the behavior of its actual input.

>

And the fact that the computation specified by the input just happens to

match itself is thus irrelevent to the validity of the question, since

the problem statement is to try to handle ALL inputs, and this input is

a member of the input set.

You keep on trying to make the input dependent on the decider it is run

on, THAT is incorrect. The input is based on the decider that it was

built to prove wrong, and is thus a FIXED input, with an answer.

Your problem seems to be that you want to limit the behavior that H is

required to answer on to be something that it can actually compute.

You try to make Ĥ to not have a "correct" answer, but Ĥ isn't a decider,

so it doesn't need to have any particular answer. It just has defined

behavior, that H needs to correctly predict for *H* to be correct.

H is the machine that needs to have the "correct" behavior to be right,

and that behavior is fixed by the code the defines H.

The fact that we can't possibly write an H that gives the correct answer

to this particular input just means that the question is uncomputable,

not "invalid".

Oct 25, 2023, 8:55:09 PM10/25/23

to

On 10/25/2023 6:01 PM, olcott wrote:

> On 10/25/2023 5:25 PM, olcott wrote:

>> https://www.liarparadox.org/Linz_Proof.pdf

>>

>> This Turing Machine description at the top of page 3

>> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞

>> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2

>>

>> Is simplified and clarified to this:

>> when Ĥ is applied to ⟨Ĥ⟩

>>

>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

>>

>> The advantage of the Linz proof is that it simultaneously represents

>> the entire infinite set of what would otherwise be an infinite set of

>> H/D pairs by using a single TM template for Ĥ.

>>

>> *We can say that both yes and no are incorrect answers for every*

>> *embedded_H that attempts to answer the question*

>> Does the computation that I am contained within halt?

>

>

> When we accept the idea of a pure function and that computations

> must be pure functions then we know that it would be incorrect

> for H to report on the computation that itself is contained within.

>

> embedded_H cannot even see the computation that itself is contained

> within it can only see the behavior of its actual input.

> On 10/25/2023 5:25 PM, olcott wrote:

>> https://www.liarparadox.org/Linz_Proof.pdf

>>

>> This Turing Machine description at the top of page 3

>> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞

>> q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2

>>

>> Is simplified and clarified to this:

>> when Ĥ is applied to ⟨Ĥ⟩

>>

>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

>>

>> The advantage of the Linz proof is that it simultaneously represents

>> the entire infinite set of what would otherwise be an infinite set of

>> H/D pairs by using a single TM template for Ĥ.

>>

>> *We can say that both yes and no are incorrect answers for every*

>> *embedded_H that attempts to answer the question*

>> Does the computation that I am contained within halt?

>

>

> When we accept the idea of a pure function and that computations

> must be pure functions then we know that it would be incorrect

> for H to report on the computation that itself is contained within.

>

> embedded_H cannot even see the computation that itself is contained

> within it can only see the behavior of its actual input.

Right, it needs to answer about the computataion specified by the input.

Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it

is contained within.

Oct 25, 2023, 9:45:15 PM10/25/23

to

So, it IS the criteria the answer, to be correct, needs to be based on.

What else would the answer to:

"Does the computation reprented by the input Halt?" refer to?

What other computation does the input represent?

I think you understand that you have been lying about this for years.

That, or you are admitting that you are a total idiot, as it HAS been

pointed out to you many times.

Oct 25, 2023, 10:07:06 PM10/25/23

to

input because the computation specified by the input

includes that Ĥ is invoking embedded_H recursively.

Oct 25, 2023, 10:13:12 PM10/25/23

to

execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report

that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.

Oct 25, 2023, 10:49:00 PM10/25/23

to

call.

Ĥ is using a copy of H that needs to determine (to be correct) what the

input that it is give will do when run.

You are just talking about your INCORRECT reframing, that isn't actually

equivalent, since it has Ĥ using the decider deciding on it instead of

the decider that it is supposed to be built on.

Oct 25, 2023, 10:49:07 PM10/25/23

to

Don't you understand what the words mean.

If your name is Peter, and so is your boss, if someone asks you the name

of your boss, would it be incorrect to answer Peter because that

happened to also be your name?

You clearly don't understand what you are talking about.

Oct 25, 2023, 11:01:33 PM10/25/23

to

Since it is WRONG for embedded_H to report on the behavior of the direct

execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report

that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.

execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report

that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.

Oct 25, 2023, 11:14:02 PM10/25/23

to

H MUST report on the behavior of its input, NO MATTER WHAT THAT INPUT IS.

There is no "rule" that it can't report on the direct execution of Ĥ

⟨Ĥ⟩, where are you getting that?

It can't see "itself", but the input isn't "itself", it is just a

machine that has a copy of itself.

There is no way to "code" a Turing Machine to some how know if it is

embedded in another machine, or what that machine is if it is.

It is wrong to make it change its mind based on who is using it, and

embedded_H is Identical to H, then if it is incorrect to ask it about

the input (Ĥ) ⟨Ĥ⟩ which represents Ĥ ⟨Ĥ⟩ then it must be invalid to ask

H about that either, and thus H has specifically said it isn't a proper

Halt Decider, since a Halt Decider needs to be able to answer about ANY

computation.

So, you are just admitting you have been lying that H actually is a Halt

Decider, or even a partial Halt Decier that handles the pathological

case, since that is PRECISELY the case you say it can't handle.

Thus, you are admitting you life has been a waste because your argument

is based on your lies.

Oct 25, 2023, 11:50:11 PM10/25/23

to

People that are not stupid idiots realize that ad

hominem attacks from them directed at others make

them look very foolish.

AD Hominem attacks on others is a direct admission

that one has run out of reasoning.

Oct 25, 2023, 11:56:25 PM10/25/23

to

Since it is WRONG for embedded_H to report on the behavior of the direct

execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report

that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.

The computation that embedded_H is contained within is
execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report

that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.

not the input and does not have the behavior of the input.

Oct 26, 2023, 12:04:27 AM10/26/23

to

>

> People that are not stupid idiots realize that ad

> hominem attacks from them directed at others make

> them look very foolish.

>

> AD Hominem attacks on others is a direct admission

> that one has run out of reasoning.

>

Just shows how stupid you are.

Note, that ISN'T an ad hominem, because I am pointing out your stupidity

as being shown by the words you used, NOT claim you are wrong because

you are stupid.

You are just PROVING your stupidity.

Note, you can't even quote the statement you want to accuse me with.

I guess it is because you know it will show that you don't actually have

any ground to stand on.

As you continue to use invalid arguements, you are just proving your

ignorance of the system.

Sorry that you are so dumb.

Oct 26, 2023, 12:10:11 AM10/26/23

to

an admission that you don't actually have any valid arguement, but are

just engaged in a "Big Lie" campaign and don't actually care about what

is true.

Igt CAN'T be wrong for embedded_H to report on the behavior of the

direct execution of Ĥ ⟨Ĥ⟩ if that is what the input represents. So, your

arguement is just based on LIES.

If the input is not a representation of this particular comuptation that

embedded_H is contained within, namely Ĥ ⟨Ĥ⟩, then your input is formed

incorrectly, and thus you are again shown to be a LIAR.

if (Ĥ) ⟨Ĥ⟩ is not the representation of Ĥ ⟨Ĥ⟩, then you are just shown

to be a LIAR in that you claim you have built the system by the rules of

the proof.

I guess you are just admitting you are just a stupid lying idiot because

you say what you have claimed can't be true, therefore you are calling

yourself a liar.

Oct 26, 2023, 12:25:06 AM10/26/23

to

of the stupidest things to do within the context

of fields having professional decorum.

Oct 26, 2023, 12:33:50 AM10/26/23

to

support the insults, just proves they are correct.

Your refusal to quote them, implies you agree with that.

Oct 26, 2023, 12:34:12 AM10/26/23

to

function of their inputs and thus are not allowed to report on

the computation that they are contained within also understand

that embedded_H is not allowed to report that the direct execution

of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.

eventually get this.

Oct 26, 2023, 1:06:31 AM10/26/23

to

called by Ĥ to change its behavior on the input (Ĥ) (Ĥ)?

The fact that is CAN'T means that you 'arguement' that the fact that the

input matches the machine it is part of is just a big fat LIE

The fat that you don't quote the message that point out your errors, but

you just repeat your erroneous claims show that you know tha tyou

arguments don't actually answer the problems reveiled.

You are just proving your stupidity.

Oct 26, 2023, 11:45:48 AM10/26/23

to

IF H / embedded_H is a pure function, how can it tell that it is being

called by Ĥ to change its behavior on the input (Ĥ) (Ĥ)?

It cannot, that is why it must rely on ⟨Ĥ⟩ correctly
called by Ĥ to change its behavior on the input (Ĥ) (Ĥ)?

simulated by embedded_H as its only valid measure.

embedded_H is expressly not allowed to consider that

Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.

Oct 26, 2023, 2:54:37 PM10/26/23

to

Where is the "rule" that says it can't consider that case? It *MUST*

consider that case or it violates the "All Inputs" requirement.

How can that possibly affect the behavior of the code that already

exists in H / embedded_H.

It only CAN use what it can compute, but the key is that it turns out

this is insufficient to actually be able to answer the question.

If embedded_H needs to rely on its correct simulation of the input, then

it needs to correctly simulate that input, which means its simulation

needs to reflect what the direct execution of the program the input

specifies will do. Since that program halts, since you claim H(D,D)

returns 0, that simulation must indicate the input is Halting, or the

simulation is incorrect, and embedded_H needs to blaim its own failing

to correctly simulate.

You want to ban the correct answer, because it shows that the

computation is impossible to do, but that it part of the question, *IS*

the computatoin possible.

You are just showing you don't understand that basics of computability,

or Truth, in your unfounded assumption that all quetions need to be

computable, or Truths provable.

Your illogical adherence to this false premise is just making you whole

logic system unsound and inconsistant.

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