In epistemology (theory of knowledge), a self-evident proposition is a
proposition that is known to be true by understanding its meaning
without proof ... (Wikipedia: Self-evidence)

Self-evident(A) Every simulation of input P that never halts unless
simulating halt decider H aborts this simulation <is> a non-halting
computation. This remains true even after H stops simulating P.

Because we know that the only difference in the behavior of a simulating
halt decider and a simulator is that the simulating halt decider stops
simulating some of its inputs we can examine the behavior of these
inputs in a simulator to determine whether or not a simulating halt
decider would stop simulating these inputs.

If the simulation of input P to simulator S would never terminate then
we can know that simulating halt decider H must stop simulating input P.

∃H ∈ Simulating_Halt_Deciders
∀P ∈ Turing_Machine_Descriptions
∀I ∈ Finite_Strings
(UTM(P,I) = ∞) ⊢ (H(P,I) = 0)

In English it says that whenever the input (P,I) to UTM would never halt
then a simulating halt decider correctly decides not halting on this input.

The behavior of the proxy computation forms the halt deciding basis for
the inputs that would otherwise specify the pathological self-reference
error.

http://www.liarparadox.org/Halting_problem_undecidability_and_infinitely_nested_simulation.pdf

Wikipedia contributors. Self-evidence. Wikipedia, The Free Encyclopedia.
May 17, 2021, 19:44 UTC. Available at:
https://en.wikipedia.org/w/index.php?title=Self-evidence&oldid=1023688680.
Accessed May 22, 2021.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 2021-05-22 11:46, olcott wrote:
> In epistemology (theory of knowledge), a self-evident proposition is a
> proposition that is known to be true by understanding its meaning
> without proof ... (Wikipedia: Self-evidence)
>
> Self-evident(A) Every simulation of input P that never halts unless
> simulating halt decider H aborts this simulation <is> a non-halting
> computation. This remains true even after H stops simulating P.
>
> Because we know that the only difference in the behavior of a simulating
> halt decider and a simulator is that the simulating halt decider stops
> simulating some of its inputs we can examine the behavior of these
> inputs in a simulator to determine whether or not a simulating halt
> decider would stop simulating these inputs.

Or, of course, we could just run the computation directly. Your
fascination with simulation is odd to say the least...

Either way, though,

Your H(H_Hat, H_Hat) claims that H_Hat(H_Hat) doesn't halt.

BUT

If we run H_Hat(H_Hat) it *does* halt.

Similarly, if we run

S(H_Hat, H_Hat), it also halts.

You are determined to instead run

S(S_Hat, S_Hat), which admittedly does not halt, but S_Hat(S_Hat) is an
*entirely* *different* computation from H_Hat(H_Hat).

According to your own description "we can examine the behavior of THESE
INPUTS [caps mine] in a simulator to determine whether or not a
simulating halt decider would stop simulating these inputs."

The input to H(H_Hat, H_Hat) is H_Hat, H_Hat. It isn't S_Hat, S_Hat. So
by the very logic you give above you must give your simulator H_Hat,
H_Hat as its input, not S_Hat, S_Hat.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On 5/22/2021 1:04 PM, André G. Isaak wrote:
> On 2021-05-22 11:46, olcott wrote:
>> In epistemology (theory of knowledge), a self-evident proposition is a
>> proposition that is known to be true by understanding its meaning
>> without proof ... (Wikipedia: Self-evidence)
>>
>> Self-evident(A) Every simulation of input P that never halts unless
>> simulating halt decider H aborts this simulation <is> a non-halting
>> computation. This remains true even after H stops simulating P.
>>
>> Because we know that the only difference in the behavior of a
>> simulating halt decider and a simulator is that the simulating halt
>> decider stops simulating some of its inputs we can examine the
>> behavior of these inputs in a simulator to determine whether or not a
>> simulating halt decider would stop simulating these inputs.
>
> Or, of course, we could just run the computation directly. Your
> fascination with simulation is odd to say the least...
>
> Either way, though,
>
> Your H(H_Hat, H_Hat) claims that H_Hat(H_Hat) doesn't halt.
>
> BUT
>
> If we run H_Hat(H_Hat) it *does* halt.
>
> Similarly, if we run
>
> S(H_Hat, H_Hat), it also halts.
>
> You are determined to instead run
>
> S(S_Hat, S_Hat), which admittedly does not halt, but S_Hat(S_Hat) is an
> *entirely* *different* computation from H_Hat(H_Hat).
>
> According to your own description "we can examine the behavior of THESE
> INPUTS [caps mine] in a simulator to determine whether or not a
> simulating halt decider would stop simulating these inputs."
>
> The input to H(H_Hat, H_Hat) is H_Hat, H_Hat. It isn't S_Hat, S_Hat. So
> by the very logic you give above you must give your simulator H_Hat,
> H_Hat as its input, not S_Hat, S_Hat.
>
> André
>

Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn

You have to examine the rest of what I said before the point that I am
making is complete.

Until we have a mutual understanding on all the points in this thread
you will not be able to understand how it provides the correct basis for
the embedded halt decider at state Ĥ.qx to correctly decide not halting
on its input: ([Ĥ],[Ĥ]).

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/22/21 1:46 PM, olcott wrote:
> In epistemology (theory of knowledge), a self-evident proposition is a
> proposition that is known to be true by understanding its meaning
> without proof ... (Wikipedia: Self-evidence)

Again, your trying to get things by that you want by claiming you don't
need to prove them. We will not clarifications needed to definitions as
we go.

>
> Self-evident(A) Every simulation of input P that never halts unless
> simulating halt decider H aborts this simulation <is> a non-halting
> computation. This remains true even after H stops simulating P.

Yes, If H is simulating <[P],[I]>, and it can PROVE that P(I) will not
halt, it can abort the simlulation of <[P],{I]> and doing so doesn't
make P(I) halting.

This action DOES change the value of Halting for the computation that H
is in, as if it did not stop the simulation, H would be non-halting, but
now it is Halting.

Important note, as this means any proof that H uses, that passes through
a copy of something like H that simulates and might abort, H in proving
the non-Halting of P, needs to take that action into count.

(Your 'proof' of H^ fails here).

>
> Because we know that the only difference in the behavior of a simulating
> halt decider and a simulator is that the simulating halt decider stops
> simulating some of its inputs we can examine the behavior of these
> inputs in a simulator to determine whether or not a simulating halt
> decider would stop simulating these inputs.

Incorrect. BECAUSE the simulating Halt decider can convert a non-halting
computation that the simulator would be part of into a Halting
Computaton (note, not of the machine being simulated, but of the machine
DOING the simulation) we need to take this into account when deciding a
machine that does this.

>
> If the simulation of input P to simulator S would never terminate then
> we can know that simulating halt decider H must stop simulating input P.
>
> ∃H ∈ Simulating_Halt_Deciders
> ∀P ∈ Turing_Machine_Descriptions
> ∀I ∈ Finite_Strings
>   (UTM(P,I) = ∞) ⊢ (H(P,I) = 0)

Yes, If a UTM of a machine wouldn't halt then the Halt decider should
return 0, and if it does halt, then the Halt Decider needs to return a 1.

Note you CAN'T say ∃H ∈ Simulating_Halt_Deciders, because we don't know
if such a machine can actual exist (In fact, we know it can't). We can
say that any H that will be accurate needs to meet this rule.

Note, this does NOT say that UTM(P,I) is in any way equivalent to H(P,I)

>
> In English it says that whenever the input (P,I) to UTM would never halt
> then a simulating halt decider correctly decides not halting on this input.
>
> The behavior of the proxy computation forms the halt deciding basis for
> the inputs that would otherwise specify the pathological self-reference
> error.
>
> http://www.liarparadox.org/Halting_problem_undecidability_and_infinitely_nested_simulation.pdf

And this is where you fall of the rails, no where have you been able to
actually prove the H^ is non-halting because you keep neglecting that a
Halt Decider ISN'T that same as a simulator as far as the machine it is
part of. As I pointed out above, the aborting of the simulation doesn't
change the nature of the machine it was simulating, but it DOES change
tthe nature of the machine DOING the simulation, even if they are copies
of the same machine.

The replacement of the Halt Decider for the Simulator only applies if
the Halt Decider doesn't abort the simulation. If it does, at the least
you need to take into account that it might abort, so it IS a
conditional. You fail to do this so your proof is flawed.

By this logic, If I halt decide on a machine that halt decides an
infinite loop, the outer halt decider can claim that it is a non-halting
behavior as it sees the infinite pattern in its trace, even though the
inner halt decider will abort the simulation and return the non-halting
decision.

>
>
> Wikipedia contributors. Self-evidence. Wikipedia, The Free Encyclopedia.
> May 17, 2021, 19:44 UTC. Available at:
> https://en.wikipedia.org/w/index.php?title=Self-evidence&oldid=1023688680.
> Accessed May 22, 2021.
>

So, you are making a claim to fame by REMOVING an example from a page?

On 5/22/2021 2:05 PM, Richard Damon wrote:
> On 5/22/21 1:46 PM, olcott wrote:
>> In epistemology (theory of knowledge), a self-evident proposition is a
>> proposition that is known to be true by understanding its meaning
>> without proof ... (Wikipedia: Self-evidence)
>
> Again, your trying to get things by that you want by claiming you don't
> need to prove them. We will not clarifications needed to definitions as
> we go.
>
>>
>> Self-evident(A) Every simulation of input P that never halts unless
>> simulating halt decider H aborts this simulation <is> a non-halting
>> computation. This remains true even after H stops simulating P.
>
> Yes, If H is simulating <[P],[I]>, and it can PROVE that P(I) will not
> halt, it can abort the simlulation of <[P],{I]> and doing so doesn't
> make P(I) halting.
>
> This action DOES change the value of Halting for the computation that H
> is in, as if it did not stop the simulation, H would be non-halting, but
> now it is Halting.
>
> Important note, as this means any proof that H uses, that passes through
> a copy of something like H that simulates and might abort, H in proving
> the non-Halting of P, needs to take that action into count.
>
> (Your 'proof' of H^ fails here).
>
>>
>> Because we know that the only difference in the behavior of a simulating
>> halt decider and a simulator is that the simulating halt decider stops
>> simulating some of its inputs we can examine the behavior of these
>> inputs in a simulator to determine whether or not a simulating halt
>> decider would stop simulating these inputs.
>>
>> If the simulation of input P to simulator S would never terminate
>> then we can know that simulating halt decider H must stop simulating
>> input P.

The above <is> a self-evident truth, thus impossibly incorrect.

> Incorrect. BECAUSE the simulating Halt decider can convert a non-halting
> computation that the simulator would be part of into a Halting
> Computaton (note, not of the machine being simulated, but of the machine
> DOING the simulation) we need to take this into account when deciding a
> machine that does this.
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/22/21 3:24 PM, olcott wrote:
> The above <is> a self-evident truth, thus impossibly incorrect.

You might try actually saying something about what I said rather than
just reiterate your position. Just repeating yourself tend to imply that
you can't actually rebut what was said.

I will note that there is a name for people who see things that no one
else sees. And a name for system that can prove a statement and its
opposite.

Even YOU have agreed that H^([H^]) does Halt when run, and there is the
definition/Truism that a computation that halts is a Halting Computation.

On 5/22/2021 3:42 PM, Richard Damon wrote:
> On 5/22/21 3:24 PM, olcott wrote:
>> The above <is> a self-evident truth, thus impossibly incorrect.
>
> You might try actually saying something about what I said rather than
> just reiterate your position. Just repeating yourself tend to imply that
> you can't actually rebut what was said.
>
> I will note that there is a name for people who see things that no one
> else sees.

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

> And a name for system that can prove a statement and its
> opposite.
>

Inconsistent.

> Even YOU have agreed that H^([H^]) does Halt when run, and there is the
> definition/Truism that a computation that halts is a Halting Computation.
>
We have to go through all the steps one-at-a-time. That you insist on
skipping to the end blocks your path to understanding the truth.

That you insist on skipping to the end prevents you from paying enough
attention.

We must have mutual agreement on every single line of V8 before you will
understand. This may require a few more versions of improved wording.

That I am not a very good communicator makes improved wordings a pure
matter of trial-and-error.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 2021-05-22 12:15, olcott wrote:
> On 5/22/2021 1:04 PM, André G. Isaak wrote:
>> On 2021-05-22 11:46, olcott wrote:
>>> In epistemology (theory of knowledge), a self-evident proposition is
>>> a proposition that is known to be true by understanding its meaning
>>> without proof ... (Wikipedia: Self-evidence)
>>>
>>> Self-evident(A) Every simulation of input P that never halts unless
>>> simulating halt decider H aborts this simulation <is> a non-halting
>>> computation. This remains true even after H stops simulating P.
>>>
>>> Because we know that the only difference in the behavior of a
>>> simulating halt decider and a simulator is that the simulating halt
>>> decider stops simulating some of its inputs we can examine the
>>> behavior of these inputs in a simulator to determine whether or not a
>>> simulating halt decider would stop simulating these inputs.
>>
>> Or, of course, we could just run the computation directly. Your
>> fascination with simulation is odd to say the least...
>>
>> Either way, though,
>>
>> Your H(H_Hat, H_Hat) claims that H_Hat(H_Hat) doesn't halt.
>>
>> BUT
>>
>> If we run H_Hat(H_Hat) it *does* halt.
>>
>> Similarly, if we run
>>
>> S(H_Hat, H_Hat), it also halts.
>>
>> You are determined to instead run
>>
>> S(S_Hat, S_Hat), which admittedly does not halt, but S_Hat(S_Hat) is
>> an *entirely* *different* computation from H_Hat(H_Hat).
>>
>> According to your own description "we can examine the behavior of
>> THESE INPUTS [caps mine] in a simulator to determine whether or not a
>> simulating halt decider would stop simulating these inputs."
>>
>> The input to H(H_Hat, H_Hat) is H_Hat, H_Hat. It isn't S_Hat, S_Hat.
>> So by the very logic you give above you must give your simulator
>> H_Hat, H_Hat as its input, not S_Hat, S_Hat.
>>
>> André
>>
>
> Ĥ.q0  wM  ⊢*  Ĥ.qx  wM  wM  ⊢*  Ĥ.qy  ∞
> Ĥ.q0  wM  ⊢*  Ĥ.qx  wM  wM  ⊢*  Ĥ.qn

If the above is intended to clarify some point you should perhaps
indicate exactly what that point is.

> You have to examine the rest of what I said before the point that I am
> making is complete.

Following your lead, I was stopping at the first mistake. But, nothing
which followed the portion I responded to in any way altered the point I
was making. The remainder was as follows:

> If the simulation of input P to simulator S would never terminate then
> we can know that simulating halt decider H must stop simulating input P.
>
> ∃H ∈ Simulating_Halt_Deciders
> ∀P ∈ Turing_Machine_Descriptions
> ∀I ∈ Finite_Strings
> (UTM(P,I) = ∞) ⊢ (H(P,I) = 0)
>
> In English it says that whenever the input (P,I) to UTM would never halt
> then a simulating halt decider correctly decides not halting on this
input.

You've written some largely meaningless gibberish which, even under the
most charitable interpretation, doesn't remotely state what you claim
that it say 'In English'.

But, since you are determined to include these meaningless 'Formalisms'
without actually attempting to learn how any of these symbols are
actually used, let me offer a few suggestions.

Since you obviously include such gibberish to make your claims look more
'mathy', you should keep the following points in mind:

(1) While it's perfectly acceptable to use Latin letters in 'equations',
you should always include at least a few Greek letters. That makes it
look more mathy. But avoid letters like omicron or capital Eta since
they don't really look all that Greek.

(2) For things that are supposed to represent your central contribution,
maybe use a completely novel symbol to make it look REALLY important.
The Elder Futhark and Sumero-Akkadian Cuneiform are both highly
underutilized in math, so these would be excellent choices. Though in
your specific case you might also want to consider the Unicode Emoticons
block.

(3) You've got some parentheses in the above, but parentheses are fairly
boring. You should use at lease a few other types of delimiters. It
doesn't really matter which ones since their purpose is entirely
aesthetic. Just pick the ones that look best. Non-matching delimiters
like bra-kets are especially good.

(4) Putting in something that requires vertical stacking -- a summation,
for example -- always makes things look highly technical. Avoid using
capital Sigma though, since that's boring. Again, pick an entirely new
symbol. Maybe a rupee symbol or something like that.

I hope these pointers help.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On 5/22/2021 1:04 PM, André G. Isaak wrote:
> On 2021-05-22 11:46, olcott wrote:
>> In epistemology (theory of knowledge), a self-evident proposition is a
>> proposition that is known to be true by understanding its meaning
>> without proof ... (Wikipedia: Self-evidence)
>>
>> Self-evident(A) Every simulation of input P that never halts unless
>> simulating halt decider H aborts this simulation <is> a non-halting
>> computation. This remains true even after H stops simulating P.
>>
>> Because we know that the only difference in the behavior of a
>> simulating halt decider and a simulator is that the simulating halt
>> decider stops simulating some of its inputs we can examine the
>> behavior of these inputs in a simulator to determine whether or not a
>> simulating halt decider would stop simulating these inputs.
>

If the simulation of input P to simulator S would never terminate then
we can know that simulating halt decider H must stop simulating input P.

> Or, of course, we could just run the computation directly. Your
> fascination with simulation is odd to say the least...

You stopped just before the most important sentence.
Do you agree with the above or not?

If you do not agree then I really need to know exactly which aspects
that you disagree with so that I can improve the wording.

Because I am a relatively terrible communicator I cannot possibly form
clear words without a long and grueling trial-and-error approach.

The rest of what you say lacks the correct foundational basis of an
understanding of the above words.

>
> Either way, though,
>
> Your H(H_Hat, H_Hat) claims that H_Hat(H_Hat) doesn't halt.
>
> BUT
>
> If we run H_Hat(H_Hat) it *does* halt.
>
> Similarly, if we run
>
> S(H_Hat, H_Hat), it also halts.
>
> You are determined to instead run
>
> S(S_Hat, S_Hat), which admittedly does not halt, but S_Hat(S_Hat) is an
> *entirely* *different* computation from H_Hat(H_Hat).
>
> According to your own description "we can examine the behavior of THESE
> INPUTS [caps mine] in a simulator to determine whether or not a
> simulating halt decider would stop simulating these inputs."
>
> The input to H(H_Hat, H_Hat) is H_Hat, H_Hat. It isn't S_Hat, S_Hat. So
> by the very logic you give above you must give your simulator H_Hat,
> H_Hat as its input, not S_Hat, S_Hat.
>
> André
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

André G. Isaak <agisaak@gm.invalid> writes:

> But, since you are determined to include these meaningless
> 'Formalisms' without actually attempting to learn how any of these
> symbols are actually used, let me offer a few suggestions.
>
> Since you obviously include such gibberish to make your claims look
> more 'mathy', you should keep the following points in mind:
>
> (1) While it's perfectly acceptable to use Latin letters in
> 'equations', you should always include at least a few Greek
> letters. That makes it look more mathy. But avoid letters like omicron
> or capital Eta since they don't really look all that Greek.
>
> (2) For things that are supposed to represent your central
> contribution, maybe use a completely novel symbol to make it look
> REALLY important. The Elder Futhark and Sumero-Akkadian Cuneiform are
> both highly underutilized in math, so these would be excellent
> choices. Though in your specific case you might also want to consider
> the Unicode Emoticons block.
>
> (3) You've got some parentheses in the above, but parentheses are
> fairly boring. You should use at lease a few other types of
> delimiters. It doesn't really matter which ones since their purpose is
> entirely aesthetic. Just pick the ones that look best. Non-matching
> delimiters like bra-kets are especially good.
>
> (4) Putting in something that requires vertical stacking -- a
> summation, for example -- always makes things look highly
> technical. Avoid using capital Sigma though, since that's
> boring. Again, pick an entirely new symbol. Maybe a rupee symbol or
> something like that.
>
> I hope these pointers help.

Made my day! (Though you owe me a new keyboard.)

--
Ben.

On 5/22/21 5:02 PM, olcott wrote:
> On 5/22/2021 3:42 PM, Richard Damon wrote:
>> On 5/22/21 3:24 PM, olcott wrote:
>>> The above <is> a self-evident truth, thus impossibly incorrect.
>>
>> You might try actually saying something about what I said rather than
>> just reiterate your position. Just repeating yourself tend to imply that
>> you can't actually rebut what was said.
>>
>> I will note that there is a name for people who see things that no one
>> else sees.
>
> Talent hits a target no one else can hit; Genius hits a target no one
> else can see. Arthur Schopenhauer

Natural Talent never really is, but is always developed with work in the
field. You have shown enough LACK of understanding in this topic (and
even SCORN for knowing about it) that I was pointing at the OTHER end of
that spectrum.

I will also point out that once Genius hits the mark, it can show it to
others, and explain it. It is impossible to be a Genius in a field you
don't understand.

>
>> And a name for system that can prove a statement and its
>> opposite.
>>
>
> Inconsistent.

Yes, like your system that allows H^(H^) to be both a Halting
computation and a Non-Halting Computation. Note, ALL you have done is to
try and create a system that can maybe justify you calling the
non-halting decision correct. You haven't done ANYTHING about the fact
that the Halting answer is easily proved. If you only answer is that you
have proved non-halting is right, then all you have done is proved you
system is inconsistent.

>
>> Even YOU have agreed that H^([H^]) does Halt when run, and there is the
>> definition/Truism that a computation that halts is a Halting Computation.
>>
> We have to go through all the steps one-at-a-time. That you insist on
> skipping to the end blocks your path to understanding the truth.

A proof that doesn't allow a challenger to look ahead isn't sound. If it
was, you could point to the mistake that I was making. Point to the
logical fallacy in what I am saying. Your 'proof' seems to be based on
setting up a frame work with one set of definitions and then stepping
though a tortured set of steps to let you twist those definitions to
mean something else. That is why you don't want the jump ahead, it
doesn't give you the length or rope to twist your meaning.

If you REALLY had a proof, and I mean a REAL proof and not just a
rhetorical argument, you could just lay it out. Start from REAL
fundamental principles, and not the 'Truism' that aren't true, or
'self-evident' things that need long explanations to get someot=ne to
maybe agree to them.

>
> That you insist on skipping to the end prevents you from paying enough
> attention.
>
> We must have mutual agreement on every single line of V8 before you will
> understand. This may require a few more versions of improved wording.
>
> That I am not a very good communicator makes improved wordings a pure
> matter of trial-and-error.
>

LOGICAL PROOFS don't need 'communication skill'. A True Analytical
Logical proof is self explanatory. You start from the basic truths that
people agree on, and established rules of logic. You then move step by
step to your conclusion.

The major flaw in your argument is that fundamentally you want to
redefine fundamental terms that you aren't allowed to redefine and stay
in the same theory.

Going back to the inconsistency, Unless you can point to an ACTUAL
logical error in Linz, you can't 'disprove it' by showing a counter
example, all that does is show that something is inconsistent.

You have tried arguments about him not noticing an infinite recursion
when you build your decider a ceratin way. The problem is that there
ISN'T an infinite recursion in the system if H can answer the question,
so you are stuck trying create a smoke screen.

Actually, Linz doesn't CARE if there is an infinite recursion possible
or actual in the design of the Halt Decider, because he NEVER needs to
actually look at how it does its job. The Halt Decider can only do one
of 4 things when given H^(H^).

It can answer Halting, at which point H^ will be Non-Halting.

It can answer Non-Halting, at which point H^ will be Halting

It can not answer but Halt ('aborting its caller') but then it didn't
answer so failed to be a decider.

It can not answer by going into an infinite loop, but then it didn't
answer so failed to be a decider.

Because it is a Computation, it MUST do the same thing to all uses of it
with that input.

It has NO behavior where it is right, so it is shown that for ANY Halt
Decider (since there were NO limits put on the design of H other than it
be a Halt Decider) we can make a machine that it can't get right, so
therefore no Halt Decider can get all problems right.

Unless you find an actual flaw in the proof, you haven't disproved Linz,
just proved that you have created either a different problem because you
redefined some base term (at which point who cares about Olcott-Halting)
or you have proved you logic system to be inconsistent.

I've mentioned this before, but I don't think you have understood it.

Even if you COULD prove that H decided H^ right, you haven't disproved
Linz et al, all you have done is proved that either you are in some
other problem space, so it doesn't matter, you have made an error in
your logic that no one has found yet, or you have just proved that you
have made you system inconsistent.

In my mind, I think it is all three, so I have had fun working in what
they call a 'Target Rich' environment.

On 2021-05-22 16:09, olcott wrote:
> On 5/22/2021 1:04 PM, André G. Isaak wrote:
>> On 2021-05-22 11:46, olcott wrote:
>>> In epistemology (theory of knowledge), a self-evident proposition is
>>> a proposition that is known to be true by understanding its meaning
>>> without proof ... (Wikipedia: Self-evidence)
>>>
>>> Self-evident(A) Every simulation of input P that never halts unless
>>> simulating halt decider H aborts this simulation <is> a non-halting
>>> computation. This remains true even after H stops simulating P.
>>>
>>> Because we know that the only difference in the behavior of a
>>> simulating halt decider and a simulator is that the simulating halt
>>> decider stops simulating some of its inputs we can examine the
>>> behavior of these inputs in a simulator to determine whether or not a
>>> simulating halt decider would stop simulating these inputs.
>>
>
> If the simulation of input P to simulator S would never terminate then
> we can know that simulating halt decider H must stop simulating input P.
>
>> Or, of course, we could just run the computation directly. Your
>> fascination with simulation is odd to say the least...
>
> You stopped just before the most important sentence.
> Do you agree with the above or not?

I already answered this. Why not actually read my post?

If you're looking for a simple unqualified yes/no, you're not going to
get one for the simple reason that you frequently write one thing and
then it becomes apparent that you mean something entirely different. I'm
not going to give a simple yes/no if I can't be sure exactly what it is
that I am agreeing to.

> If you do not agree then I really need to know exactly which aspects
> that you disagree with so that I can improve the wording.

Which was explained in what I wrote in my post. You just need to
actually read it.

André

> Because I am a relatively terrible communicator I cannot possibly form
> clear words without a long and grueling trial-and-error approach.
>
> The rest of what you say lacks the correct foundational basis of an
> understanding of the above words.
>
>>
>> Either way, though,
>>
>> Your H(H_Hat, H_Hat) claims that H_Hat(H_Hat) doesn't halt.
>>
>> BUT
>>
>> If we run H_Hat(H_Hat) it *does* halt.
>>
>> Similarly, if we run
>>
>> S(H_Hat, H_Hat), it also halts.
>>
>> You are determined to instead run
>>
>> S(S_Hat, S_Hat), which admittedly does not halt, but S_Hat(S_Hat) is
>> an *entirely* *different* computation from H_Hat(H_Hat).
>>
>> According to your own description "we can examine the behavior of
>> THESE INPUTS [caps mine] in a simulator to determine whether or not a
>> simulating halt decider would stop simulating these inputs."
>>
>> The input to H(H_Hat, H_Hat) is H_Hat, H_Hat. It isn't S_Hat, S_Hat.
>> So by the very logic you give above you must give your simulator
>> H_Hat, H_Hat as its input, not S_Hat, S_Hat.
>>
>> André
>>
>
>

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On 5/22/2021 1:04 PM, André G. Isaak wrote:
> On 2021-05-22 11:46, olcott wrote:
>> In epistemology (theory of knowledge), a self-evident proposition is a
>> proposition that is known to be true by understanding its meaning
>> without proof ... (Wikipedia: Self-evidence)
>>
>> Self-evident(A) Every simulation of input P that never halts unless
>> simulating halt decider H aborts this simulation <is> a non-halting
>> computation. This remains true even after H stops simulating P.
>>
>> Because we know that the only difference in the behavior of a
>> simulating halt decider and a simulator is that the simulating halt
>> decider stops simulating some of its inputs we can examine the
>> behavior of these inputs in a simulator to determine whether or not a
>> simulating halt decider would stop simulating these inputs.
>
> Or, of course, we could just run the computation directly. Your
> fascination with simulation is odd to say the least...
>
> Either way, though,
>
> Your H(H_Hat, H_Hat) claims that H_Hat(H_Hat) doesn't halt.
>
This is not any part of the focused analysis that I have listed below.

> BUT
>
> If we run H_Hat(H_Hat) it *does* halt.
>
This is not any part of the focused analysis that I have listed below.

> Similarly, if we run
>
> S(H_Hat, H_Hat), it also halts.
>
This is not any part of the focused analysis that I have listed below.

> You are determined to instead run
>
> S(S_Hat, S_Hat), which admittedly does not halt, but S_Hat(S_Hat) is an
> *entirely* *different* computation from H_Hat(H_Hat).

This <is> part of the focused analysis listed below
Yes it <is> a different computation.
It is not an entirely different computation.

The <only> difference is shown below in (a).
The <only> difference is shown below in (a).
The <only> difference is shown below in (a).

> According to your own description "we can examine the behavior of THESE
> INPUTS [caps mine] in a simulator to determine whether or not a
> simulating halt decider would stop simulating these inputs."
>
> The input to H(H_Hat, H_Hat) is H_Hat, H_Hat. It isn't S_Hat, S_Hat. So
> by the very logic you give above you must give your simulator H_Hat,
> H_Hat as its input, not S_Hat, S_Hat.
>
> André
>

I have changed the wording and now have specifically named machines.

(a) The only difference between a simulating halt decider and a
UTM/simulator is that the former stops its simulation at some point.

(b) Where a computation P has an embedded simulating halt decider H we
create proxy computation P2 replacing every instance of H with simulator S.

The following two points are the key basis of my whole proof, so far
they have proved far too subtle for anyone here to begin to grasp.

(c) Because of (a) when P2 is a halting computation then H does not need
to stop P and H decides halting.

(d) Because of (a) when P2 is an infinite computation then H must stop P
and H decides not halting.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On Saturday, May 22, 2021 at 1:46:13 PM UTC-4, olcott wrote:
> In epistemology (theory of knowledge), a self-evident proposition is a
> proposition that is known to be true by understanding its meaning
> without proof ... (Wikipedia: Self-evidence)
>
> Self-evident(A) Every simulation of input P that never halts unless
> simulating halt decider H aborts this simulation <is> a non-halting
> computation. This remains true even after H stops simulating P.
>
> Because we know that the only difference in the behavior of a simulating
> halt decider and a simulator is that the simulating halt decider stops
> simulating some of its inputs we can examine the behavior of these
> inputs in a simulator to determine whether or not a simulating halt
> decider would stop simulating these inputs.
>
> If the simulation of input P to simulator S would never terminate then
> we can know that simulating halt decider H must stop simulating input P.
>
> ∃H ∈ Simulating_Halt_Deciders
> ∀P ∈ Turing_Machine_Descriptions
> ∀I ∈ Finite_Strings
> (UTM(P,I) = ∞) ⊢ (H(P,I) = 0)
>
> In English it says that whenever the input (P,I) to UTM would never halt
> then a simulating halt decider correctly decides not halting on this input.
>
> The behavior of the proxy computation forms the halt deciding basis for
> the inputs that would otherwise specify the pathological self-reference
> error.
>
> http://www.liarparadox.org/Halting_problem_undecidability_and_infinitely_nested_simulation.pdf
>
> Wikipedia contributors. Self-evidence. Wikipedia, The Free Encyclopedia.
> May 17, 2021, 19:44 UTC. Available at:
> https://en.wikipedia.org/w/index.php?title=Self-evidence&oldid=1023688680.
> Accessed May 22, 2021.
>
> --
> Copyright 2021 Pete Olcott
>
> "Great spirits have always encountered violent opposition from mediocre
> minds." Einstein
There are numerous proofs of the Halting Problem.
Are you simply maintaining that a particular proof is flawed or that all proofs are flawed?
If it is just one, then you should look at the other ones.
For example, the halting set is not recursive. It is Epsilon 1 not Epsilon 0. (Kleene Arithmetic Hierarchy.)
Turing 1931 is particularly simple. Any flaws in it?

On 5/23/2021 10:31 AM, Charlie-Boo wrote:
> On Saturday, May 22, 2021 at 1:46:13 PM UTC-4, olcott wrote:
>> In epistemology (theory of knowledge), a self-evident proposition is a
>> proposition that is known to be true by understanding its meaning
>> without proof ... (Wikipedia: Self-evidence)
>>
>> Self-evident(A) Every simulation of input P that never halts unless
>> simulating halt decider H aborts this simulation <is> a non-halting
>> computation. This remains true even after H stops simulating P.
>>
>> Because we know that the only difference in the behavior of a simulating
>> halt decider and a simulator is that the simulating halt decider stops
>> simulating some of its inputs we can examine the behavior of these
>> inputs in a simulator to determine whether or not a simulating halt
>> decider would stop simulating these inputs.
>>
>> If the simulation of input P to simulator S would never terminate then
>> we can know that simulating halt decider H must stop simulating input P.
>>
>> ∃H ∈ Simulating_Halt_Deciders
>> ∀P ∈ Turing_Machine_Descriptions
>> ∀I ∈ Finite_Strings
>> (UTM(P,I) = ∞) ⊢ (H(P,I) = 0)
>>
>> In English it says that whenever the input (P,I) to UTM would never halt
>> then a simulating halt decider correctly decides not halting on this input.
>>
>> The behavior of the proxy computation forms the halt deciding basis for
>> the inputs that would otherwise specify the pathological self-reference
>> error.
>>
>> http://www.liarparadox.org/Halting_problem_undecidability_and_infinitely_nested_simulation.pdf
>>
>> Wikipedia contributors. Self-evidence. Wikipedia, The Free Encyclopedia.
>> May 17, 2021, 19:44 UTC. Available at:
>> https://en.wikipedia.org/w/index.php?title=Self-evidence&oldid=1023688680.
>> Accessed May 22, 2021.
>>
>> --
>> Copyright 2021 Pete Olcott
>>
>> "Great spirits have always encountered violent opposition from mediocre
>> minds." Einstein
> There are numerous proofs of the Halting Problem.
> Are you simply maintaining that a particular proof is flawed or that all proofs are flawed?
> If it is just one, then you should look at the other ones.
> For example, the halting set is not recursive. It is Epsilon 1 not Epsilon 0. (Kleene Arithmetic Hierarchy.)
> Turing 1931 is particularly simple. Any flaws in it?
>

I am currently referring to all the proofs that follow the Sipser, Linz,
and Kozen model specified on page 7 of my paper.

I refer to the Linz proof only because it is the simplest one that
specifies all of the key details in terms of precise state transitions.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

On 5/23/21 11:45 AM, olcott wrote:
> On 5/23/2021 10:31 AM, Charlie-Boo wrote:
>> There are numerous proofs of the Halting Problem.
>> Are you simply maintaining that a particular proof is flawed or that
>> all proofs are flawed?
>> If it is just one, then you should look at the other ones.
>> For example, the halting set is not recursive.  It is Epsilon 1 not
>> Epsilon 0. (Kleene Arithmetic Hierarchy.)
>> Turing 1931 is particularly simple.  Any flaws in it?
>>
>
> I am currently referring to all the proofs that follow the Sipser, Linz,
> and Kozen model specified on page 7 of my paper.
>
> I refer to the Linz proof only because it is the simplest one that
> specifies all of the key details in terms of precise state transitions.
>

And, has been pointed out, 'refuting' just one proof doesn't actually
get you anything. As long as it is provable that a Universally Correct
Halt Decider can't exist, then you still have the problem of the all the
proofs that use that fact that counter your idea that all Truth is Provable.

In fact, since all you seem to be aiming to do is show that under some
special conditions H can correctly decide on H^, but don't actually show
an error in the logic of Linz, you haven't actually refuted the proof,
just shown that the logic system with your special conditions is
inconsistent.

This doesn't really seem to be much of a goal to aim your life work at,
the creation of another inconsistent logic system.

Richard Damon <Richard@damon-family.org> wrote:
> On 5/23/21 11:45 AM, olcott wrote:
>> On 5/23/2021 10:31 AM, Charlie-Boo wrote:
>>> There are numerous proofs of the Halting Problem.
>>> Are you simply maintaining that a particular proof is flawed or that
>>> all proofs are flawed?
>>> If it is just one, then you should look at the other ones.
>>> For example, the halting set is not recursive. It is Epsilon 1 not
>>> Epsilon 0. (Kleene Arithmetic Hierarchy.)
>>> Turing 1931 is particularly simple. Any flaws in it?
>> I am currently referring to all the proofs that follow the Sipser, Linz,
>> and Kozen model specified on page 7 of my paper.
>> I refer to the Linz proof only because it is the simplest one that
>> specifies all of the key details in terms of precise state transitions.
> And, has been pointed out, 'refuting' just one proof doesn't actually
> get you anything. As long as it is provable that a Universally Correct
> Halt Decider can't exist, then you still have the problem of the all the
> proofs that use that fact that counter your idea that all Truth is Provable.
....

Crank science usually proceeds in phases.
The initial step is the major crank claim that unknowingly claims
to refute most of what is known in some specialist area as well
as some significant amoung of related areas.
It is various pointed out to The Hero that many things are wrong
with their idea and they start to read and discover they need
prove other things that refute their thesis must be wrong as well.
Then we enter the phase of creeping denial which generally
ends in proofs that 1+1 != 2.

In climate science e.g. the first claim is the climate isnt changing.
Then it's the climate isn't changing now.
Then it's the climate is changing but it can't be caused by
anything people do.
Then it's something is wrong about the theory of greenhouse gases.
Then something is wrong with radiation physics.
Then something is wrong with physic.
Then something is wrong with science.
Then something is wrong with math.
&ct.

--
Kruger and Dunning argue that for a given skill, incompetent people will:
1. tend to overestimate their own level of skill;
2. fail to recognize genuine skill in others;
3. fail to recognize the extremity of their inadequacy;
4. recognize and acknowledge their own previous lack of skill, only if
they can be trained to substantially improve [their own performance].

Dunning later drew an analogy with anosognosia in which a person who
suffers a physical disability because of brain injury seems unaware of
or denies the existence of the disability, even for dramatic
impairments such as blindness or paralysis.

Dunning & Kruger & others concluded that the root cause is that, in
contrast to high performers, "poor performers do not learn from
feedback suggesting a need to improve".

-- Ehrlinger, Joyce; Johnson, Kerri; Banner, Matthew; Dunning, David;
Kruger, Justin (2008).
"Why the unskilled are unaware: Further explorations of (absent)
self-insight among the incompetent".
Organizational Behavior and Human Decision Processes 105 (105): 98-121.