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

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

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...

Which was STIPULATED to end up at H's qn in finite time.

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.

So, does your H have a condition in its simulation to stop its
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)
>

FAIL.

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?

--
Copyright 2022 Pete Olcott

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

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:

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.

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.

--
Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On Thursday, September 22, 2022 at 8:18:20 AM UTC-7, olcott wrote:
>
> 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.
>
You mean:
The input to a halt decider derives the actual behavior specified by this input.

Why do you use so many words?

Incidentally this is not an axiom. It's part of the definition of a halt decider.

And, of course, you have never defined "derives|

On 9/22/2022 4:13 PM, dklei...@gmail.com wrote:
> On Thursday, September 22, 2022 at 8:18:20 AM UTC-7, olcott wrote:
>>
>> 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.
>>
> You mean:
> The input to a halt decider derives the actual behavior specified by this input.
>

The actual behavior of the input is
{shown, measured, derived, demonstrated, seen}
by the correct simulation of this input by the simulating halt decider.

> Why do you use so many words?
>

To short-circuit the dozens of rebuttals that would otherwise occur.

> Incidentally this is not an axiom. It's part of the definition of a halt decider.
>

Hence axiom.

> And, of course, you have never defined "derives|

--
Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

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 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.

olcott <polcott2@gmail.com> writes:
> On 9/22/2022 4:13 PM, dklei...@gmail.com wrote:
[...]
>> Why do you use so many words?
>
> To short-circuit the dozens of rebuttals that would otherwise occur.

How's that workin' out for ya?

--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
Working, but not speaking, for Philips
void Void(void) { Void(); } /* The recursive call of the void */

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
embedded_H never reaches its final state we know that every other
simulation also never reaches its final state, thus this input never halts.

--
Copyright 2022 Pete Olcott

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

On 9/22/2022 6:01 PM, Keith Thompson wrote:
> olcott <polcott2@gmail.com> writes:
>> On 9/22/2022 4:13 PM, dklei...@gmail.com wrote:
> [...]
>>> Why do you use so many words?
>>
>> To short-circuit the dozens of rebuttals that would otherwise occur.
>
> How's that workin' out for ya?
>

It works very poorly with liars.

--
Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On Thursday, September 22, 2022 at 3:24:25 PM UTC-7, olcott wrote:
> On 9/22/2022 4:13 PM, dklei...@gmail.com wrote:
> > Why do you use so many words?
> >
> To short-circuit the dozens of rebuttals that would otherwise occur.
>
I think they are the cause of rebuttals. I suspect that were you to
write laconically you would get lots fewer rebuttals.
>
> > Incidentally this is not an axiom. It's part of the definition of a halt decider.
> >
> Hence axiom.
>
Not in ordinary speech, what do you imagine "axiom" to mean?

On 9/22/2022 7:22 PM, dklei...@gmail.com wrote:
> On Thursday, September 22, 2022 at 3:24:25 PM UTC-7, olcott wrote:
>> On 9/22/2022 4:13 PM, dklei...@gmail.com wrote:
>>> Why do you use so many words?
>>>
>> To short-circuit the dozens of rebuttals that would otherwise occur.
>>
> I think they are the cause of rebuttals. I suspect that were you to
> write laconically you would get lots fewer rebuttals.

I think that most rebuttals are primarily based on the fact that most of
my reviewers only want to argue and don't give a rat's ass about the truth.

If I can back them into a corner and show that their rebuttal also
rejects the notion of a UTM, then then may acknowledge the truth.

>>
>>> Incidentally this is not an axiom. It's part of the definition of a halt decider.
>>>
>> Hence axiom.
>>
> Not in ordinary speech, what do you imagine "axiom" to mean?

Ultimately an axiom is any expression of formal or natural language that
is stipulated to have a Boolean value of true.

Unifying the notion of truth across natural and formal language requires
using the same terminology with the same meanings across formal and
natural language.

--
Copyright 2022 Pete Olcott

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

On 9/22/22 8:56 PM, olcott wrote:
> On 9/22/2022 7:22 PM, dklei...@gmail.com wrote:
>> On Thursday, September 22, 2022 at 3:24:25 PM UTC-7, olcott wrote:
>>> On 9/22/2022 4:13 PM, dklei...@gmail.com wrote:
>>>> Why do you use so many words?
>>>>
>>> To short-circuit the dozens of rebuttals that would otherwise occur.
>>>
>> I think they are the cause of rebuttals. I suspect that were you to
>> write laconically you would get lots fewer rebuttals.
>
> I think that most rebuttals are primarily based on the fact that most of
> my reviewers only want to argue and don't give a rat's ass about the truth.
>
> If I can back them into a corner and show that their rebuttal also
> rejects the notion of a UTM, then then may acknowledge the truth.

But YOU are the one that is backed into the corner.

YOUR LIES have been exposed.

>
>
>>>
>>>> Incidentally this is not an axiom. It's part of the definition of a
>>>> halt decider.
>>>>
>>> Hence axiom.
>>>
>> Not in ordinary speech, what do you imagine "axiom" to mean?
>
> Ultimately an axiom is any expression of formal or natural language that
> is stipulated to have a Boolean value of true.
>
> Unifying the notion of truth across natural and formal language requires
> using the same terminology with the same meanings across formal and
> natural language.
>

Which means you need to adapt your "natural" language terms to be
restricted to the FORMAL language definitions.

Note, FORMAL logic has a very specific set of rules, you can't argue
that something is correct in that formal system if you inject concepts
that don't match there meaning in the formal system.

Note, "Logic" in Natural Language is basically DOOMED to failure, as
Natural Language is almost always filled with inconsistencies that break
the logic if you try to use it too formally.

On 9/22/22 7:28 PM, olcott wrote:
> 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
> embedded_H never reaches its final state we know that every other
> simulation also never reaches its final state, thus this input never halts.
>
>

Who says it doesn't?

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.

On 9/22/2022 8:17 PM, Richard Damon wrote:
>
> On 9/22/22 7:28 PM, olcott wrote:
>> 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
>> embedded_H never reaches its final state we know that every other
>> simulation also never reaches its final state, thus this input never
>> halts.
>>
>>
>
> Who says it doesn't?
>
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.

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.

--

Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On 9/22/22 10:35 PM, olcott wrote:
> On 9/22/2022 8:17 PM, Richard Damon wrote:
>>
>> On 9/22/22 7:28 PM, olcott wrote:
>>> 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
>>> embedded_H never reaches its final state we know that every other
>>> simulation also never reaches its final state, thus this input never
>>> halts.
>>>
>>>
>>
>> Who says it doesn't?
>>
> 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⟩.

Almost, the COMPLETELY (and correctly) simulated input to H would reach
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⟩.
>

Almost, the COMPLETELY (and correctly) simulate input to H would NEVER,
after an unbounded number of steps

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

But then this "input" isn't H^ any more, so the logic is incorrect.

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⟩.

Since the input you are looking at isn't H^ anymore, your logic is unsound.

>
> From this we can determine that
> The input to H never halts, thus is correctly determined to be non-halting.

No, you have proved that something that wasn't the input to H to be
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 are just proving your stupditiy.

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.

On Thursday, September 22, 2022 at 5:56:30 PM UTC-7, olcott wrote:
> On 9/22/2022 7:22 PM, dklei...@gmail.com wrote:
> > On Thursday, September 22, 2022 at 3:24:25 PM UTC-7, olcott wrote:
> >> On 9/22/2022 4:13 PM, dklei...@gmail.com wrote:
> >>> Why do you use so many words?
> >>>
> >> To short-circuit the dozens of rebuttals that would otherwise occur.
> >>
> > I think they are the cause of rebuttals. I suspect that were you to
> > write laconically you would get lots fewer rebuttals.
> I think that most rebuttals are primarily based on the fact that most of
> my reviewers only want to argue and don't give a rat's ass about the truth.
>
> If I can back them into a corner and show that their rebuttal also
> rejects the notion of a UTM, then then may acknowledge the truth.
> >>
> >>> Incidentally this is not an axiom. It's part of the definition of a halt decider.
> >>>
> >> Hence axiom.
> >>
> > Not in ordinary speech, what do you imagine "axiom" to mean?
> Ultimately an axiom is any expression of formal or natural language that
> is stipulated to have a Boolean value of true.
>
> Unifying the notion of truth across natural and formal language requires
> using the same terminology with the same meanings across formal and
> natural language.

OK. But using a word eccentrically will gain you no friends and
make your writing harder to understand.

On Thursday, September 22, 2022 at 6:13:27 PM UTC-7, richar...@gmail.com wrote:
>
> Note, "Logic" in Natural Language is basically DOOMED to failure, as
> Natural Language is almost always filled with inconsistencies that break
> the logic if you try to use it too formally.
>
Natural Language works in a way that remains mysterious. But we are
certain it works very unlike logic, formal or informal . However it is
flexible enough that logic can be expressed in natural language if
done carefully enough.

On 9/23/22 1:03 AM, dklei...@gmail.com wrote:
> On Thursday, September 22, 2022 at 6:13:27 PM UTC-7, richar...@gmail.com wrote:
>>
>> Note, "Logic" in Natural Language is basically DOOMED to failure, as
>> Natural Language is almost always filled with inconsistencies that break
>> the logic if you try to use it too formally.
>>
> Natural Language works in a way that remains mysterious. But we are
> certain it works very unlike logic, formal or informal . However it is
> flexible enough that logic can be expressed in natural language if
> done carefully enough.

Logic requires a formalism that Natural Language doesn't naturally obey.
Yes, you can use Natural Language in logic, but only by restricting it
with formalisms, at which point it isn't "Natural Language" anymore.

The problem comes to the fact that in Natural Language, words no longer
have precise meanings, which are needed to do the logic.

You can't use statements like "All Birds are Animals" as a categorical
statement, because the words "Birds" and "Animals", in Natural Langugee
include meanings that make this statement false (or perhaps true in a
very different way).

On 2022-09-23 11:12:37 +0000, Richard Damon said:

> On 9/23/22 1:03 AM, dklei...@gmail.com wrote:
>> On Thursday, September 22, 2022 at 6:13:27 PM UTC-7, richar...@gmail.com wrote:
>>>
>>> Note, "Logic" in Natural Language is basically DOOMED to failure, as
>>> Natural Language is almost always filled with inconsistencies that break
>>> the logic if you try to use it too formally.
>>>
>> Natural Language works in a way that remains mysterious. But we are
>> certain it works very unlike logic, formal or informal . However it is
>> flexible enough that logic can be expressed in natural language if
>> done carefully enough.
>
> Logic requires a formalism that Natural Language doesn't naturally
> obey. Yes, you can use Natural Language in logic, but only by
> restricting it with formalisms, at which point it isn't "Natural
> Language" anymore.
>
> The problem comes to the fact that in Natural Language, words no longer
> have precise meanings, which are needed to do the logic.
>
> You can't use statements like "All Birds are Animals" as a categorical
> statement, because the words "Birds" and "Animals", in Natural Langugee
> include meanings that make this statement false (or perhaps true in a
> very different way).

Logic in Natural Language:

Peter is an ass.
An ass has long ears.
Therefore, Peter has long ears.

Mikko

On 9/22/2022 11:58 PM, dklei...@gmail.com wrote:
> On Thursday, September 22, 2022 at 5:56:30 PM UTC-7, olcott wrote:
>> On 9/22/2022 7:22 PM, dklei...@gmail.com wrote:
>>> On Thursday, September 22, 2022 at 3:24:25 PM UTC-7, olcott wrote:
>>>> On 9/22/2022 4:13 PM, dklei...@gmail.com wrote:
>>>>> Why do you use so many words?
>>>>>
>>>> To short-circuit the dozens of rebuttals that would otherwise occur.
>>>>
>>> I think they are the cause of rebuttals. I suspect that were you to
>>> write laconically you would get lots fewer rebuttals.
>> I think that most rebuttals are primarily based on the fact that most of
>> my reviewers only want to argue and don't give a rat's ass about the truth.
>>
>> If I can back them into a corner and show that their rebuttal also
>> rejects the notion of a UTM, then then may acknowledge the truth.
>>>>
>>>>> Incidentally this is not an axiom. It's part of the definition of a halt decider.
>>>>>
>>>> Hence axiom.
>>>>
>>> Not in ordinary speech, what do you imagine "axiom" to mean?
>> Ultimately an axiom is any expression of formal or natural language that
>> is stipulated to have a Boolean value of true.
>>
>> Unifying the notion of truth across natural and formal language requires
>> using the same terminology with the same meanings across formal and
>> natural language.
>
> OK. But using a word eccentrically will gain you no friends and
> make your writing harder to understand.

My purpose is to mathematically formalize the notion of Truth, Tarski
"proved" this is impossible on the basis that he could not prove that
the liar paradox is true.

When we understand that for the entire body of analytical truth all of
correct reasoning only has two sources of truth:

(1) Expressions of natural or formal language that are defined to be
true. I called these axioms.

(2) Expressions that are derived by applying truth preserving operations
to (1) or (2).

Then all of undecidable propositions are understood to simply not be
truth bearers.

--
Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On 9/23/2022 12:03 AM, dklei...@gmail.com wrote:
> On Thursday, September 22, 2022 at 6:13:27 PM UTC-7, richar...@gmail.com wrote:
>>
>> Note, "Logic" in Natural Language is basically DOOMED to failure, as
>> Natural Language is almost always filled with inconsistencies that break
>> the logic if you try to use it too formally.
>>
> Natural Language works in a way that remains mysterious. But we are
> certain it works very unlike logic, formal or informal . However it is
> flexible enough that logic can be expressed in natural language if
> done carefully enough.

Simply make a meta-language rich enough to fully express natural
language semantics. Richard Montague did a very good job with this.
I created Minimal Type Theory as the basis for doing this.

https://www.researchgate.net/publication/331859461_Minimal_Type_Theory_YACC_BNF

--
Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On 9/22/2022 10:09 PM, Richard Damon wrote:
> On 9/22/22 10:35 PM, olcott wrote:
>> On 9/22/2022 8:17 PM, Richard Damon wrote:
>>>
>>> On 9/22/22 7:28 PM, olcott wrote:
>>>> 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 embedded_H never reaches its final state we know that every other
>>>> simulation also never reaches its final state, thus this input never
>>>> halts.
>>>>
>>>>
>>>
>>> Who says it doesn't?
>>>
>> 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⟩.
>
> Almost, the COMPLETELY (and correctly) simulated input to H would reach
> its own final state.

If it is not simulated by H then it is not input to H.

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⟩.

--
Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On 9/23/2022 7:30 AM, Mikko wrote:
> On 2022-09-23 11:12:37 +0000, Richard Damon said:
>
>> On 9/23/22 1:03 AM, dklei...@gmail.com wrote:
>>> On Thursday, September 22, 2022 at 6:13:27 PM UTC-7,
>>> richar...@gmail.com wrote:
>>>>
>>>> Note, "Logic" in Natural Language is basically DOOMED to failure, as
>>>> Natural Language is almost always filled with inconsistencies that
>>>> break
>>>> the logic if you try to use it too formally.
>>>>
>>> Natural Language works in a way that remains mysterious. But we are
>>> certain it works very unlike logic, formal or informal . However it is
>>> flexible enough that logic can be expressed in natural language if
>>> done carefully enough.
>>
>> Logic requires a formalism that Natural Language doesn't naturally
>> obey. Yes, you can use Natural Language in logic, but only by
>> restricting it with formalisms, at which point it isn't "Natural
>> Language" anymore.
>>
>> The problem comes to the fact that in Natural Language, words no
>> longer have precise meanings, which are needed to do the logic.
>>
>> You can't use statements like "All Birds are Animals" as a categorical
>> statement, because the words "Birds" and "Animals", in Natural
>> Langugee include meanings that make this statement false (or perhaps
>> true in a very different way).
>
> Logic in Natural Language:
>
> Peter is an ass.
> An ass has long ears.
> Therefore, Peter has long ears.
>
> Mikko
>

The Cyc formalization of natural language uses globally unique
identifiers GUIDs for each sense meaning of each word, thus the fallacy
of equivocation error is not possible in the Cyc system.

https://en.wikipedia.org/wiki/Cyc

--
Copyright 2022 Pete Olcott

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