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Sep 21, 2022, 2:22:18 PMSep 21

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computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

When the copy of Linz H that is embedded within Linz Ĥ is a simulating

halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated ⟨Ĥ⟩ would never

reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞ steps of correct

simulation.

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

I added this material from the Peter Linz text to address the Turing

machine proofs. I paraphrased the Linz encoding because the Linz version

has two start states.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy

If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would reach its own final

state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own

final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Ĥ.q0 copies its input then invokes its embedded copy of the original

Linz H...

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Because we can see that replacing H with a UTM would cause Ĥ to become

stuck in infinitely recursive simulation we can see that the input

correctly simulated by H would never reach the final state of this

simulated input, thus never halts.

Linz, Peter 1990. An Introduction to Formal Languages and Automata.

Lexington/Toronto: D. C. Heath and Company. (317-320)

--

Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;

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

a final state. (Linz:1990:234)

When the copy of Linz H that is embedded within Linz Ĥ is a simulating

halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated ⟨Ĥ⟩ would never

reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞ steps of correct

simulation.

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

I added this material from the Peter Linz text to address the Turing

machine proofs. I paraphrased the Linz encoding because the Linz version

has two start states.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy

If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would reach its own final

state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own

final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Ĥ.q0 copies its input then invokes its embedded copy of the original

Linz H...

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Because we can see that replacing H with a UTM would cause Ĥ to become

stuck in infinitely recursive simulation we can see that the input

correctly simulated by H would never reach the final state of this

simulated input, thus never halts.

Linz, Peter 1990. An Introduction to Formal Languages and Automata.

Lexington/Toronto: D. C. Heath and Company. (317-320)

--

Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;

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

Sep 21, 2022, 7:04:29 PMSep 21

to

On 9/21/22 2:22 PM, olcott wrote:

> computation that halts … the Turing machine will halt whenever it enters

> a final state. (Linz:1990:234)

>

> When the copy of Linz H that is embedded within Linz Ĥ is a simulating

> halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated ⟨Ĥ⟩ would never

> reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞ steps of correct

> simulation.

reach the final state if H <H^> <H^> goes to H.Qn

If H <H^> <H^> doesn't go to H.Qn or H.Qy, then it fails to be a decider.

We see this at the path of the CORRECT (and thus complete) simulation if

the input to H will be: (here <H^> is wM from

The input to H <H^> <H^> represents H^ <H^>

that starts at q0 <H^> then goes to

H^.q0 <H^> <H^> after duplicating the input.

this is where the copy of H's q0 <H^> <H^> is

since we know that H <H^> <H^> is ending up at is y1 qn y2 we know the

copy will end up at H^ y1 qn y2

And this is defined as a final state.

Please point out the error here.

>

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

>

> I added this material from the Peter Linz text to address the Turing

> machine proofs. I paraphrased the Linz encoding because the Linz version

> has two start states.

>

> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy

> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would reach its own final

> state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

>

> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own

> final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

>

> Ĥ.q0 copies its input then invokes its embedded copy of the original

> Linz H...

Thus we KNOW that this H can't just unconditionally simulate its input,

or it does get stuck in this loop.

>

> computation that halts … the Turing machine will halt whenever it enters

> a final state. (Linz:1990:234)

>

> Because we can see that replacing H with a UTM would cause Ĥ to become

> stuck in infinitely recursive simulation we can see that the input

> correctly simulated by H would never reach the final state of this

> simulated input, thus never halts.

simulaiton and return its answer to the machine it is embedded in, and

thus H^ doesn't get stuck in an infinite loop, or does H get stuck in

this loop and NEVER give an answer.

>

> Linz, Peter 1990. An Introduction to Formal Languages and Automata.

> Lexington/Toronto: D. C. Heath and Company. (317-320)

>

Sep 21, 2022, 7:53:22 PMSep 21

to

On 9/21/2022 6:04 PM, Richard Damon wrote:

>

> On 9/21/22 2:22 PM, olcott wrote:

>> computation that halts … the Turing machine will halt whenever it

>> enters a final state. (Linz:1990:234)

>>

>> When the copy of Linz H that is embedded within Linz Ĥ is a simulating

>> halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated ⟨Ĥ⟩ would

>> never reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞ steps of

>> correct simulation.

>

> WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL

> reach the final state if H <H^> <H^> goes to H.Qn

>

Yet another attempt to get away with the strawman deception.
>

> On 9/21/22 2:22 PM, olcott wrote:

>> computation that halts … the Turing machine will halt whenever it

>> enters a final state. (Linz:1990:234)

>>

>> When the copy of Linz H that is embedded within Linz Ĥ is a simulating

>> halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated ⟨Ĥ⟩ would

>> never reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞ steps of

>> correct simulation.

>

> WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL

> reach the final state if H <H^> <H^> goes to H.Qn

>

The question is: Can the correctly simulated ⟨Ĥ⟩ reaches its own

simulated final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn?

Sep 21, 2022, 8:17:02 PMSep 21

to

On 9/21/22 7:53 PM, olcott wrote:

> On 9/21/2022 6:04 PM, Richard Damon wrote:

>>

>> On 9/21/22 2:22 PM, olcott wrote:

>>> computation that halts … the Turing machine will halt whenever it

>>> enters a final state. (Linz:1990:234)

>>>

>>> When the copy of Linz H that is embedded within Linz Ĥ is a

>>> simulating halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated

>>> ⟨Ĥ⟩ would never reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞

>>> steps of correct simulation.

>>

>> WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL

>> reach the final state if H <H^> <H^> goes to H.Qn

>>

>

> Yet another attempt to get away with the strawman deception.

>

> The question is: Can the correctly simulated ⟨Ĥ⟩ reaches its own

> simulated final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn?

>

No, that just shows that you don't understand the definiton of a Halt
> On 9/21/2022 6:04 PM, Richard Damon wrote:

>>

>> On 9/21/22 2:22 PM, olcott wrote:

>>> computation that halts … the Turing machine will halt whenever it

>>> enters a final state. (Linz:1990:234)

>>>

>>> When the copy of Linz H that is embedded within Linz Ĥ is a

>>> simulating halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated

>>> ⟨Ĥ⟩ would never reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞

>>> steps of correct simulation.

>>

>> WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL

>> reach the final state if H <H^> <H^> goes to H.Qn

>>

>

> Yet another attempt to get away with the strawman deception.

>

> The question is: Can the correctly simulated ⟨Ĥ⟩ reaches its own

> simulated final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn?

>

Decider:

H wM w -> Qy Iff M w Halts, and Qn iff M w will never halt.

In English, H needs to decide if the machine/input that its input

describes will halt when run as an independent machine.

It is the behavior of the machine the input represents, executed as an

independent machine that is the key.

NOT "can H simulate its input to a final state?"

Just proves you are a ignorant pathological lying idiot that doesn't

know what you are talking about.

FAIL.

Sep 22, 2022, 11:18:21 AMSep 22

to

On 9/21/2022 7:16 PM, Richard Damon wrote:

> On 9/21/22 7:53 PM, olcott wrote:

>> On 9/21/2022 6:04 PM, Richard Damon wrote:

>>>

>>> On 9/21/22 2:22 PM, olcott wrote:

>>>> computation that halts … the Turing machine will halt whenever it

>>>> enters a final state. (Linz:1990:234)

>>>>

>>>> When the copy of Linz H that is embedded within Linz Ĥ is a

>>>> simulating halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated

>>>> ⟨Ĥ⟩ would never reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞

>>>> steps of correct simulation.

>>>

>>> WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL

>>> reach the final state if H <H^> <H^> goes to H.Qn

>>>

>>

>> Yet another attempt to get away with the strawman deception.

>>

>> The question is: Can the correctly simulated ⟨Ĥ⟩ reaches its own

>> simulated final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn?

>>

>

> No, that just shows that you don't understand the definiton of a Halt

> Decider:

It is an axiom that the correctly simulated input to a halt decider
> On 9/21/22 7:53 PM, olcott wrote:

>> On 9/21/2022 6:04 PM, Richard Damon wrote:

>>>

>>> On 9/21/22 2:22 PM, olcott wrote:

>>>> computation that halts … the Turing machine will halt whenever it

>>>> enters a final state. (Linz:1990:234)

>>>>

>>>> When the copy of Linz H that is embedded within Linz Ĥ is a

>>>> simulating halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated

>>>> ⟨Ĥ⟩ would never reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞

>>>> steps of correct simulation.

>>>

>>> WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL

>>> reach the final state if H <H^> <H^> goes to H.Qn

>>>

>>

>> Yet another attempt to get away with the strawman deception.

>>

>> The question is: Can the correctly simulated ⟨Ĥ⟩ reaches its own

>> simulated final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn?

>>

>

> No, that just shows that you don't understand the definiton of a Halt

> Decider:

derives the actual behavior specified by this input otherwise the notion

of a UTM is baselessly rejected out-of-hand.

Sep 22, 2022, 6:35:38 PMSep 22

to

On 9/22/22 11:18 AM, olcott wrote:

> On 9/21/2022 7:16 PM, Richard Damon wrote:

>> On 9/21/22 7:53 PM, olcott wrote:

>>> On 9/21/2022 6:04 PM, Richard Damon wrote:

>>>>

>>>> On 9/21/22 2:22 PM, olcott wrote:

>>>>> computation that halts … the Turing machine will halt whenever it

>>>>> enters a final state. (Linz:1990:234)

>>>>>

>>>>> When the copy of Linz H that is embedded within Linz Ĥ is a

>>>>> simulating halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated

>>>>> ⟨Ĥ⟩ would never reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞

>>>>> steps of correct simulation.

>>>>

>>>> WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL

>>>> reach the final state if H <H^> <H^> goes to H.Qn

>>>>

>>>

>>> Yet another attempt to get away with the strawman deception.

>>>

>>> The question is: Can the correctly simulated ⟨Ĥ⟩ reaches its own

>>> simulated final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn?

>>>

>>

>> No, that just shows that you don't understand the definiton of a Halt

>> Decider:

>

> It is an axiom that the correctly simulated input to a halt decider

> derives the actual behavior specified by this input otherwise the notion

> of a UTM is baselessly rejected out-of-hand.

>

>

The correctly AND COMPLETELY simulate input to a halt decider.
> On 9/21/2022 7:16 PM, Richard Damon wrote:

>> On 9/21/22 7:53 PM, olcott wrote:

>>> On 9/21/2022 6:04 PM, Richard Damon wrote:

>>>>

>>>> On 9/21/22 2:22 PM, olcott wrote:

>>>>> computation that halts … the Turing machine will halt whenever it

>>>>> enters a final state. (Linz:1990:234)

>>>>>

>>>>> When the copy of Linz H that is embedded within Linz Ĥ is a

>>>>> simulating halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated

>>>>> ⟨Ĥ⟩ would never reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to ∞

>>>>> steps of correct simulation.

>>>>

>>>> WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL

>>>> reach the final state if H <H^> <H^> goes to H.Qn

>>>>

>>>

>>> Yet another attempt to get away with the strawman deception.

>>>

>>> The question is: Can the correctly simulated ⟨Ĥ⟩ reaches its own

>>> simulated final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn?

>>>

>>

>> No, that just shows that you don't understand the definiton of a Halt

>> Decider:

>

> It is an axiom that the correctly simulated input to a halt decider

> derives the actual behavior specified by this input otherwise the notion

> of a UTM is baselessly rejected out-of-hand.

>

>

Since H doesn't do one (at least not the one that gives an answer), we

need to look at the UTM simulation of the input.

If H(P,P) returns 0, then UTM(P,P) will halt.

THIS HAS BEEN PROVEN.

Thus your statememt above just PROVES that ANY H that returns 0 for

H(P,P) where P has been built on that H is incorrect.

YOU HAVE FAILED.

Sep 22, 2022, 7:28:35 PMSep 22

to

On 9/22/2022 5:35 PM, Richard Damon wrote:

> On 9/22/22 11:18 AM, olcott wrote:

>> On 9/21/2022 7:16 PM, Richard Damon wrote:

>>> On 9/21/22 7:53 PM, olcott wrote:

>>>> On 9/21/2022 6:04 PM, Richard Damon wrote:

>>>>>

>>>>> On 9/21/22 2:22 PM, olcott wrote:

>>>>>> computation that halts … the Turing machine will halt whenever it

>>>>>> enters a final state. (Linz:1990:234)

>>>>>>

>>>>>> When the copy of Linz H that is embedded within Linz Ĥ is a

>>>>>> simulating halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated

>>>>>> ⟨Ĥ⟩ would never reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to

>>>>>> ∞ steps of correct simulation.

>>>>>

>>>>> WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL

>>>>> reach the final state if H <H^> <H^> goes to H.Qn

>>>>>

>>>>

>>>> Yet another attempt to get away with the strawman deception.

>>>>

>>>> The question is: Can the correctly simulated ⟨Ĥ⟩ reaches its own

>>>> simulated final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn?

>>>>

>>>

>>> No, that just shows that you don't understand the definiton of a Halt

>>> Decider:

>>

>> It is an axiom that the correctly simulated input to a halt decider

>> derives the actual behavior specified by this input otherwise the

>> notion of a UTM is baselessly rejected out-of-hand.

>>

>>

>

> The correctly AND COMPLETELY simulate input to a halt decider.

Since we can see that the correctly and completely simulated input to
> On 9/22/22 11:18 AM, olcott wrote:

>> On 9/21/2022 7:16 PM, Richard Damon wrote:

>>> On 9/21/22 7:53 PM, olcott wrote:

>>>> On 9/21/2022 6:04 PM, Richard Damon wrote:

>>>>>

>>>>> On 9/21/22 2:22 PM, olcott wrote:

>>>>>> computation that halts … the Turing machine will halt whenever it

>>>>>> enters a final state. (Linz:1990:234)

>>>>>>

>>>>>> When the copy of Linz H that is embedded within Linz Ĥ is a

>>>>>> simulating halt decider applied to ⟨Ĥ⟩ ⟨Ĥ⟩ the correctly simulated

>>>>>> ⟨Ĥ⟩ would never reach its final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn in 1 to

>>>>>> ∞ steps of correct simulation.

>>>>>

>>>>> WRONG. The CORRECT (and thus COMPLETE) simulation of the input WILL

>>>>> reach the final state if H <H^> <H^> goes to H.Qn

>>>>>

>>>>

>>>> Yet another attempt to get away with the strawman deception.

>>>>

>>>> The question is: Can the correctly simulated ⟨Ĥ⟩ reaches its own

>>>> simulated final state of ⟨Ĥ⟩.qy or ⟨Ĥ⟩.qn?

>>>>

>>>

>>> No, that just shows that you don't understand the definiton of a Halt

>>> Decider:

>>

>> It is an axiom that the correctly simulated input to a halt decider

>> derives the actual behavior specified by this input otherwise the

>> notion of a UTM is baselessly rejected out-of-hand.

>>

>>

>

> The correctly AND COMPLETELY simulate input to a halt decider.

embedded_H never reaches its final state we know that every other

simulation also never reaches its final state, thus this input never halts.

Sep 22, 2022, 9:18:01 PMSep 22

to

You have posted the CORRECT and COMPLETE simulation of the input to H /

embedded_H and it reaches a final state.

Remember, a simulation that is aborted does not show that the simultion

would be non-halting.

You need to run an INDEPENDENT simulation of that input by a UTM, and

that has been shown to be halting.

You keep on looking at the WRONG input, because you change the

definition of H abd thus make your "PROOF" just a LIE.

You are proving your ignorance of the subject, that you only know how to

lie, and that you are just an idiot for thinking people can't see

through your deciet.

FAIL.

Sep 22, 2022, 10:35:11 PMSep 22

to

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy

If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would reach its own final

state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own

final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

When we replace H with a UTM we can see that Ĥ never stops running.
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would reach its own final

state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own

final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

From this we can determine

For every definition of simulating halt decider of H none of the

correctly simulated inputs ever reach their own final state ⟨Ĥ.qy⟩ or

⟨Ĥ.qn⟩.

From this we can determine that

The input to H never halts, thus is correctly determined to be non-halting.

That I have to keep repeating these things to you would seem to indicate

that you either have brain damage or are dishonest, another alternative

is that these things are much more difficult for you than they are for me.

Sep 22, 2022, 11:09:46 PMSep 22

to

its own final state.

>

> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own

> final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

>

after an unbounded number of steps

> When we replace H with a UTM we can see that Ĥ never stops running.

You are just proving your ignorance, stupidity, and deceit.

FFFFF A IIIII L

F A A I L

F A A I L

FFFFF AAAAA I L

F A A I L

F A A I L

F A A IIIII LLLLL

Unless you are going to claim that your H actually IS a UTM, but then

all you have done is show that your H will never answer, and thus also fail,

>

> From this we can determine

> For every definition of simulating halt decider of H none of the

> correctly simulated inputs ever reach their own final state ⟨Ĥ.qy⟩ or

> ⟨Ĥ.qn⟩.

>

> From this we can determine that

> The input to H never halts, thus is correctly determined to be non-halting.

non-halting, and that you are an idiot.

>

> That I have to keep repeating these things to you would seem to indicate

> that you either have brain damage or are dishonest, another alternative

> is that these things are much more difficult for you than they are for me.

>

You logic is based on fallacious arguements, and incorrect operation.

We have shown you the CORRECT complete simulation of the ACTUAL input to

H, and that halts.

Sep 23, 2022, 11:30:17 AMSep 23

to

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy

⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Sep 23, 2022, 6:43:55 PMSep 23

to

The "Input" to H is EXACTLY the finite string.

The behavior of the string being defined as the UTN Simulation is a

precisely defined property.

Note, by YOUR definition, it is IMPOSSIBLE to even define a Halt Decider

by the required definition, so they are simple proven to not exist.

FAIL.

>

> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy

> When ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by H reaches its own final state of

> ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

>

> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

> When ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by H would never reach its own final

> state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

>

>

that you aren't working on the Halting Problem.

This means you are aditting that you have wasted the last 18 years of

your life.

YOU HAVE FAILED.

There is NO requirement that the simulation be by the decider.

Adding that mean you are just talkig about your POOP, not halting.

Oct 18, 2022, 4:30:12 PMOct 18

to

On 9/21/2022 1:22 PM, olcott wrote:

> computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)

>

>

> computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)

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