On 2/1/2022 8:48 PM, André G. Isaak wrote:
> On 2022-02-01 19:37, olcott wrote:
>> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>
>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.
>>
>> Sure and so is the "I am going to go to the" part of
>> "I am going to go to the store to buy some ice cream."
>>
>> When you don't cut off what I said in the middle of the sentence then
>> it makes much more sense.
>>
>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to
>> ⟨Ĥ⟩.qn ?
>
> That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you can
> provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate ⟨Ĥ⟩
> applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.

So you are simply being nit picky about my use of terminology.

When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H and
embedded_H correctly determines that its simulated input cannot possibly
reach any final state then embedded_H is necessarily correct to
transition to Ĥ.qn indicating that its simulated input never halts.

--
Copyright 2021 Pete Olcott

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

On 2/1/2022 8:48 PM, Richard Damon wrote:
> On 2/1/22 9:41 PM, olcott wrote:
>> On 2/1/2022 8:21 PM, Richard Damon wrote:
>>> On 2/1/22 9:14 PM, olcott wrote:
>>>> On 2/1/2022 7:55 PM, Richard Damon wrote:
>>>>> On 2/1/22 8:47 PM, olcott wrote:
>>>>>> On 2/1/2022 7:40 PM, Richard Damon wrote:
>>>>>>>
>>>>>>> On 2/1/22 8:03 PM, olcott wrote:
>>>>>>>> On 2/1/2022 6:25 PM, Richard Damon wrote:
>>>>>>>>> On 2/1/22 5:18 PM, olcott wrote:
>>>>>>>>>> On 2/1/2022 4:12 PM, wij wrote:
>>>>>>>>>>> On Wednesday, 2 February 2022 at 05:36:39 UTC+8, olcott wrote:
>>>>>>>>>>>> On 2/1/2022 3:23 PM, wij wrote:
>>>>>>>>>>>>> On Wednesday, 2 February 2022 at 02:37:17 UTC+8, olcott wrote:
>>>>>>>>>>>>>> On 2/1/2022 10:33 AM, wij wrote:
>>>>>>>>>>>>>>> On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:
>>>>>>>>>>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On 1/31/22 11:42 PM, olcott wrote:
>>>>>>>>>>>>>>>>>> On 1/31/2022 10:33 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> On 1/31/22 11:24 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>> On 1/31/2022 10:17 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>> On 1/31/22 10:40 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>> On 1/31/2022 6:41 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>>> On 1/31/22 3:24 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>> On 1/31/2022 2:10 PM, Ben wrote:
>>>>>>>>>>>>>>>>>>>>>>>>> On 1/31/2022 8:06 AM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>> On 1/30/2022 8:20 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>>> On 1/30/22 9:05 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> These statements need the conditions, that
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> H^ goes to
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> H^.Qy/H^.Qn iff H goes to that
>>>>>>>>>>>>>>>>>>>>>>>>>>>>> corresponding state.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an
>>>>>>>>>>>>>>>>>>>>>>>>>>>> input to embedded_H
>>>>>>>>>>>>>>>>>>>>>>>>>>>> in the same way that (5,3) is syntactically
>>>>>>>>>>>>>>>>>>>>>>>>>>>> specified as an
>>>>>>>>>>>>>>>>>>>>>>>>>>>> input to Sum(5,3)
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> Right, and the
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an
>>>>>>>>>>>>>>>>>>>>>>>>>>>> input to
>>>>>>>>>>>>>>>>>>>>>>>>>>>> embedded_H in the same way that (1,2) is NOT
>>>>>>>>>>>>>>>>>>>>>>>>>>>> syntactically
>>>>>>>>>>>>>>>>>>>>>>>>>>>> specified as an input to Sum(5,3)
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> Right, but perhaps you don't understand that
>>>>>>>>>>>>>>>>>>>>>>>>>>> from you above
>>>>>>>>>>>>>>>>>>>>>>>>>>> statement the right answer is based on if
>>>>>>>>>>>>>>>>>>>>>>>>>>> UTM(<H^>,<H^>)
>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts which by the definition of a UTM means
>>>>>>>>>>>>>>>>>>>>>>>>>>> if H^ applied to
>>>>>>>>>>>>>>>>>>>>>>>>>>> <H^> Halts.
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> The biggest reason for your huge mistakes is
>>>>>>>>>>>>>>>>>>>>>>>>>> that you cannot
>>>>>>>>>>>>>>>>>>>>>>>>>> stay sharply focused on a single point. It is
>>>>>>>>>>>>>>>>>>>>>>>>>> as if you either
>>>>>>>>>>>>>>>>>>>>>>>>>> have attention deficit disorder ADD or are
>>>>>>>>>>>>>>>>>>>>>>>>>> addicted to
>>>>>>>>>>>>>>>>>>>>>>>>>> methamphetamine.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input
>>>>>>>>>>>>>>>>>>>>>>>>>> to embedded_H and
>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> After we have mutual agreement on this point
>>>>>>>>>>>>>>>>>>>>>>>>>> we will move on
>>>>>>>>>>>>>>>>>>>>>>>>>> to the points that logically follow from this
>>>>>>>>>>>>>>>>>>>>>>>>>> one.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>> Holy shit try to post something that makes sense.
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Richard does not accept that the input to the
>>>>>>>>>>>>>>>>>>>>>>>> copy of Linz H
>>>>>>>>>>>>>>>>>>>>>>>> embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting
>>>>>>>>>>>>>>>>>>>>>>>> that it is Ĥ ⟨Ĥ⟩.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> No, but apparently you can't understand actual
>>>>>>>>>>>>>>>>>>>>>>> English words.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> The INPUT to H is <H^> <H^> but the CORRECT
>>>>>>>>>>>>>>>>>>>>>>> ANSWER that H must
>>>>>>>>>>>>>>>>>>>>>>> give is based on the behavior of H^ applied to
>>>>>>>>>>>>>>>>>>>>>>> <H^> BECAUSE OF
>>>>>>>>>>>>>>>>>>>>>>> THE DEFINITION of H.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> In other words Sum(3,5) must return the value of
>>>>>>>>>>>>>>>>>>>>>> Sum(7,8)?
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Don't know how you get that from what I said.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Any moron knows that a function is only
>>>>>>>>>>>>>>>>>>>>>> accountable for its actual
>>>>>>>>>>>>>>>>>>>>>> inputs.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> And the actual input to H is <H^> <H^> which MEANS
>>>>>>>>>>>>>>>>>>>>> by the
>>>>>>>>>>>>>>>>>>>>> DEFINITION of the Halting Problem that H is being
>>>>>>>>>>>>>>>>>>>>> asked to decide
>>>>>>>>>>>>>>>>>>>>> on the Halting Status of H^ applied to <H^>
>>>>>>>>>>>>>>>>>>>> No that is not it. That is like saying "by
>>>>>>>>>>>>>>>>>>>> definition" Sum(3,5) is
>>>>>>>>>>>>>>>>>>>> being asked about Sum(7,8).
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Again your RED HERRING.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> H is being asked EXACTLY what it being asked
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> H wM w -> H.Qy if M applied to w Halts, and H.Qn if
>>>>>>>>>>>>>>>>>>> it doesn't
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> AGREED?
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> No that is wrong. embedded_H is being asked:
>>>>>>>>>>>>>>>>>> Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly
>>>>>>>>>>>>>>>>>> transition to ⟨Ĥ⟩.qn ?
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> If you say 'No', then you aren't doing the halting
>>>>>>>>>>>>>>>>> problem, as the
>>>>>>>>>>>>>>>>> requirement I stated is EXACTLY the requirement of the
>>>>>>>>>>>>>>>>> Halting Problem.
>>>>>>>>>>>>>>>> The halting problem is vague on the definition of
>>>>>>>>>>>>>>>> halting, it includes
>>>>>>>>>>>>>>>> that a machine has stopped running and that a machine
>>>>>>>>>>>>>>>> cannot reach its
>>>>>>>>>>>>>>>> final state. My definition only includes the latter.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Sounds like a NDTM.
>>>>>>>>>>>>>> https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> It is not a NDTM, a Turing Machine only actually halts
>>>>>>>>>>>>>> when it reaches
>>>>>>>>>>>>>> its own final state. People not very familiar with this
>>>>>>>>>>>>>> material may get
>>>>>>>>>>>>>> confused and believe that a TM halts when its stops
>>>>>>>>>>>>>> running because its
>>>>>>>>>>>>>> simulation has been aborted. This key distinction is not
>>>>>>>>>>>>>> typically
>>>>>>>>>>>>>> specified in most halting problem proofs.
>>>>>>>>>>>>>> computation that halts … the Turing machine will halt
>>>>>>>>>>>>>> whenever it enters
>>>>>>>>>>>>>> a final state. (Linz:1990:234)
>>>>>>>>>>>>>
>>>>>>>>>>>>> Where did Linz mention 'simulation' and 'abort'?
>>>>>>>>>>>> I have shown how my system directly applies to the actual
>>>>>>>>>>>> halting
>>>>>>>>>>>> problem and it can be understood as correct by anyone that
>>>>>>>>>>>> understands
>>>>>>>>>>>> the halting problem at a much deeper level than rote
>>>>>>>>>>>> memorization.
>>>>>>>>>>>>
>>>>>>>>>>>> The following simplifies the syntax for the definition of
>>>>>>>>>>>> the Linz
>>>>>>>>>>>> Turing machine Ĥ, it is now a single machine with a single
>>>>>>>>>>>> start state.
>>>>>>>>>>>> A copy of Linz H is embedded at Ĥ.qx.
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly
>>>>>>>>>>>> transition to
>>>>>>>>>>>> ⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).
>>>>>>>>>>>>> You are defining POOP [Richard Damon]
>>>>>>>>>>>>> André had recommended many online sites for you to learn or
>>>>>>>>>>>>> test, I forget which posts it is.
>>>>>>>>>>>>> But I think C program is more simpler.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Halting problem undecidability and infinitely nested
>>>>>>>>>>>>>> simulation (V3)
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> --
>>>>>>>>>>>>>> Copyright 2021 Pete Olcott
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Talent hits a target no one else can hit;
>>>>>>>>>>>>>> Genius hits a target no one else can see.
>>>>>>>>>>>>>> Arthur Schopenhauer
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> --
>>>>>>>>>>>> Copyright 2021 Pete Olcott
>>>>>>>>>>>>
>>>>>>>>>>>> Talent hits a target no one else can hit;
>>>>>>>>>>>> Genius hits a target no one else can see.
>>>>>>>>>>>> Arthur Schopenhauer
>>>>>>>>>>>
>>>>>>>>>>> André had recommended many online sites for you to learn or
>>>>>>>>>>> test, I forget which posts it is.
>>>>>>>>>>> Type it into a TM simulator and prove your claim, your words
>>>>>>>>>>> are meaningless.
>>>>>>>>>>
>>>>>>>>>> I have already proved that I know one key fact about halt
>>>>>>>>>> deciders that no one else here seems to know.
>>>>>>>>>>
>>>>>>>>>> No one here understands that because a halt decider is a
>>>>>>>>>> decider that it must compute the mapping from its inputs to an
>>>>>>>>>> accept of reject state on the basis of the actual behavior
>>>>>>>>>> specified by these inputs.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> And the ACTUAL BEHAVIOR of the input <H^> <H^> is EXACTLY the
>>>>>>>>> behavior of H^ applied to <H^> which does Halt if H goes to H.Qn.
>>>>>>>>
>>>>>>>>
>>>>>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly
>>>>>>>> transition to ⟨Ĥ⟩.qn ?
>>>>>>>
>>>>>>> Doesn't matter if embedded_H is not a ACTUAL UTM.
>>>>>>>
>>>>>>
>>>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>>>>    Ĥ copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H simulates ⟨Ĥ1⟩
>>>>>> ⟨Ĥ2⟩
>>>>>>
>>>>>> As soon as embedded_H correctly recognizes this as an infinite
>>>>>> behavior pattern:
>>>>>>
>>>>>> Then these steps would keep repeating:
>>>>>>    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩
>>>>>> ⟨Ĥ3⟩
>>>>>>    Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩
>>>>>> ⟨Ĥ4⟩
>>>>>>    Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩
>>>>>> ⟨Ĥ5⟩...
>>>>>>
>>>>>> Then embedded_H can correctly abort the simulation of its input
>>>>>> and correctly transition to Ĥ.qn.
>>>>>>
>>>>>> The above words can be verified as completely true entirely on the
>>>>>> basis of their meaning.
>>>>>>
>>>>>
>>>>>
>>>>> Nope, proven otherwise.
>>>>>
>>>>
>>>> What I said above is true by logical necessity and you simply aren't
>>>> bright enough to understand this.
>>>>
>>>
>>> Then you can provide a step by step proof of it?
>>>
>>>> If X then Y and if Y then Z and X then Z. There is no way around this.
>>>
>>> And what are your X, Y and Z?
>>>
>>>
>>>>
>>>> If embedded_H correctly recognizes that its input specifies non
>>>> halting behavior then it is necessarily correct for embedded_H to
>>>> report this
>>>> non halting behavior.
>>>
>>> *IF* it correct recognizes. Since there is no pattern in H's
>>> simulation of <H^> <H^> THAT IS a proof of non-halting
>> You must be a liar.
>>
>> Then these steps would keep repeating:
>>     Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
>>     Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
>>     Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩
>> ⟨Ĥ5⟩...
>
> But if H <H^> <H^> -> H.Qn aften N steps, then it is also true that the
> computation H1 <H^> <H^> -> H.Qn after N steps and the pattern ends.

On 2/1/2022 9:24 PM, André G. Isaak wrote:
> On 2022-02-01 19:57, olcott wrote:
>> On 2/1/2022 8:48 PM, André G. Isaak wrote:
>>> On 2022-02-01 19:37, olcott wrote:
>>>> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>>>
>>>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.
>>>>
>>>> Sure and so is the "I am going to go to the" part of
>>>> "I am going to go to the store to buy some ice cream."
>>>>
>>>> When you don't cut off what I said in the middle of the sentence
>>>> then it makes much more sense.
>>>>
>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition
>>>> to ⟨Ĥ⟩.qn ?
>>>
>>> That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you
>>> can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate
>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.
>>
>> So you are simply being nit picky about my use of terminology.
>
> Yes, I insist on terminology being used correctly. And any place where
> you attempt to publish your results will be equally, if not more, nit
> picky.
>

It is fine and good that you help correct my terminology.
What is not fine and good is for you to reject the essence of the gist
of what I am saying entirely on the basis that I did not say it exactly
according to conventions. The is what Ben always did. He never paid any
attention to the actual substance of what I was saying.

>> When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H and
>> embedded_H correctly determines that its simulated input cannot
>> possibly reach any final state then embedded_H is necessarily correct
>> to transition to Ĥ.qn indicating that its simulated input never halts.
>
> But now you've just hidden your meaningless terminological abuse. "Its
> simulated input" is only meaningful when it is construed as meaning the
> simulation of the computation REPRESENTED by the input, i.e. the

Not at all. A simulator simulates a finite string and the actual
behavior of this simulated finite string is the ultimate basis of
whether or not it specifies a finite sequence of configurations.

If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot possibly reach
⟨Ĥ⟩.qn then it is necessarily correct for embedded_H to transition to
Ĥ.qn and nothing else in the universe can possibly refute this.

In much simpler terms if the input to simulating halt decider H never
halts then it is always correct for H to report that its input never halts.

In even simpler terms if you see an actual dog then you are correct to
say: "I saw a dog", even if everyone else in the universe disagrees.

--
Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On 2/1/2022 9:33 PM, André G. Isaak wrote:
> On 2022-02-01 20:08, olcott wrote:
>
>> embedded_H is exactly a UTM with extra features added.
>
> Apparently you don't know what 'exactly' means. embedded_H is not a UTM
> *at* *all*.
>
> If embedded_H were a UTM, then
>
> embedded_H would accept ⟨Ĥ⟩ ⟨Ĥ⟩ if Ĥ accepts ⟨Ĥ⟩

embedded_H merely determines whether or not its input specifies a finite
sequence of configurations. It does not give a rats ass about anything
else in the universe.

--
Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On 2/1/2022 9:58 PM, André G. Isaak wrote:
> On 2022-02-01 20:44, olcott wrote:
>> On 2/1/2022 9:24 PM, André G. Isaak wrote:
>>> On 2022-02-01 19:57, olcott wrote:
>>>> On 2/1/2022 8:48 PM, André G. Isaak wrote:
>>>>> On 2022-02-01 19:37, olcott wrote:
>>>>>> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>>>>>
>>>>>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.
>>>>>>
>>>>>> Sure and so is the "I am going to go to the" part of
>>>>>> "I am going to go to the store to buy some ice cream."
>>>>>>
>>>>>> When you don't cut off what I said in the middle of the sentence
>>>>>> then it makes much more sense.
>>>>>>
>>>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition
>>>>>> to ⟨Ĥ⟩.qn ?
>>>>>
>>>>> That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or
>>>>> you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot
>>>>> simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.
>>>>
>>>> So you are simply being nit picky about my use of terminology.
>>>
>>> Yes, I insist on terminology being used correctly. And any place
>>> where you attempt to publish your results will be equally, if not
>>> more, nit picky.
>>>
>>
>> It is fine and good that you help correct my terminology.
>> What is not fine and good is for you to reject the essence of the gist
>> of what I am saying entirely on the basis that I did not say it
>> exactly according to conventions. The is what Ben always did. He never
>> paid any attention to the actual substance of what I was saying.
>>
>>>> When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H and
>>>> embedded_H correctly determines that its simulated input cannot
>>>> possibly reach any final state then embedded_H is necessarily
>>>> correct to transition to Ĥ.qn indicating that its simulated input
>>>> never halts.
>>>
>>> But now you've just hidden your meaningless terminological abuse.
>>> "Its simulated input" is only meaningful when it is construed as
>>> meaning the simulation of the computation REPRESENTED by the input,
>>> i.e. the
>>
>> Not at all. A simulator simulates a finite string and the actual
>> behavior of this simulated finite string is the ultimate basis of
>> whether or not it specifies a finite sequence of configurations.
>
> No. A simulator simulates a Turing Machine applied to an input. It takes
> as its input a finite string which represents that Turing Machine/Input
> pair. It's completely meaningless to talk about simulating a finite string.
>

It is possible for Turing machines to have blank tapes.

The salient aspect for the Halting problem is whether or not the finite
string machine description specifies a finite or infinite sequence of
configurations. The ultimate basis for determining this is the actual
behavior of the simulated finite string.

Since this equally applies to machines having inputs and machines not
having inputs the distinction relative to inputs is moot.

>> If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot possibly
>> reach ⟨Ĥ⟩.qn then it is necessarily correct for embedded_H to
>> transition to Ĥ.qn and nothing else in the universe can possibly
>> refute this.
>
> Again, you're falling back on your belief that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is
> both meaningful (it isn't) and somehow distinct from H applied to ⟨Ĥ⟩.
>

The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
is the ultimate measure of the halt status of this input.

> What you're doing is essentially the same thing as if I produced a SUM
> TM which, given ⟨3⟩ and ⟨2⟩ as inputs, produced the output ⟨6⟩, and me
> claiming that it is correct because it isn't responsible for adding the
> integers, only the finite strings which SUM was given as inputs, and
> that my algorithm correctly determines that SUM ⟨3⟩⟨2⟩ is ⟨6⟩ despite
> the fact that SUM 3 2 is 5.
>
>> In much simpler terms if the input to simulating halt decider H never
>> halts then it is always correct for H to report that its input never
>> halts.
>
> Inputs don't halt or not halt. Only the TM/input pair which the input
> *describes* can halt or not halt.
>
> André

Inputs to simulating halt deciders are already implicitly specified to
be simulated. We could get verbose and say the simulated inputs to
simulating halt decider:

In much simpler terms if the simulated input to simulating halt decider
H never halts then it is always correct for H to report that its input
never halts.

You and Richard are saying that there are exceptions to this logically
necessary truth. This is like saying that when a dog bites you on the
leg it might not have been a dog and it might not have been your leg
even though it is stipulated that a dog bit you on the leg.

--
Copyright 2021 Pete Olcott

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

On 2/1/2022 10:10 PM, André G. Isaak wrote:
> On 2022-02-01 20:59, olcott wrote:
>> On 2/1/2022 9:33 PM, André G. Isaak wrote:
>>> On 2022-02-01 20:08, olcott wrote:
>>>
>>>> embedded_H is exactly a UTM with extra features added.
>>>
>>> Apparently you don't know what 'exactly' means. embedded_H is not a
>>> UTM *at* *all*.
>>>
>>> If embedded_H were a UTM, then
>>>
>>> embedded_H would accept ⟨Ĥ⟩ ⟨Ĥ⟩ if Ĥ accepts ⟨Ĥ⟩
>
> And in a previous post you accused *me* of snipping part way through?
>

You snipped in the middle of the sentence and then said that the
sentence didn't make sense. I snip down to the most salient point.

When Ĥ is applied to ⟨Ĥ⟩
Ĥ copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩

embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ determines that its simulated input would never reach
its final state ⟨Ĥ⟩.qn on the basis of matching this infinite pattern:

Then these steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...

>> embedded_H merely determines whether or not its input specifies a
>> finite sequence of configurations. It does not give a rats ass about
>> anything else in the universe.
>
> Which again means that it is not a UTM since that is *not* what a UTM
> determines.
>

It is a UTM that has extra features added.

> Being a halt decider and being a UTM are mutually exclusive.
>
> André
>
For all inputs that reach their final state the behavior of embedded_H
is equivalent to a UTM.

--
Copyright 2021 Pete Olcott

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

On 2/2/2022 7:31 AM, olcott wrote:
>> On 2022-02-01 20:59, olcott wrote:
>>>> On 2022-02-01 20:08, olcott wrote

Shut idiot up.

On 2/2/2022 6:22 PM, Richard Damon wrote:
> On 2/2/22 10:20 AM, olcott wrote:
>> On 2/1/2022 9:58 PM, André G. Isaak wrote:
>>> On 2022-02-01 20:44, olcott wrote:
>>>> On 2/1/2022 9:24 PM, André G. Isaak wrote:
>>>>> On 2022-02-01 19:57, olcott wrote:
>>>>>> On 2/1/2022 8:48 PM, André G. Isaak wrote:
>>>>>>> On 2022-02-01 19:37, olcott wrote:
>>>>>>>> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>>>>>>>
>>>>>>>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.
>>>>>>>>
>>>>>>>> Sure and so is the "I am going to go to the" part of
>>>>>>>> "I am going to go to the store to buy some ice cream."
>>>>>>>>
>>>>>>>> When you don't cut off what I said in the middle of the sentence
>>>>>>>> then it makes much more sense.
>>>>>>>>
>>>>>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly
>>>>>>>> transition to ⟨Ĥ⟩.qn ?
>>>>>>>
>>>>>>> That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or
>>>>>>> you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot
>>>>>>> simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.
>>>>>>
>>>>>> So you are simply being nit picky about my use of terminology.
>>>>>
>>>>> Yes, I insist on terminology being used correctly. And any place
>>>>> where you attempt to publish your results will be equally, if not
>>>>> more, nit picky.
>>>>>
>>>>
>>>> It is fine and good that you help correct my terminology.
>>>> What is not fine and good is for you to reject the essence of the
>>>> gist of what I am saying entirely on the basis that I did not say it
>>>> exactly according to conventions. The is what Ben always did. He
>>>> never paid any attention to the actual substance of what I was saying.
>>>>
>>>>>> When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H
>>>>>> and embedded_H correctly determines that its simulated input
>>>>>> cannot possibly reach any final state then embedded_H is
>>>>>> necessarily correct to transition to Ĥ.qn indicating that its
>>>>>> simulated input never halts.
>>>>>
>>>>> But now you've just hidden your meaningless terminological abuse.
>>>>> "Its simulated input" is only meaningful when it is construed as
>>>>> meaning the simulation of the computation REPRESENTED by the input,
>>>>> i.e. the
>>>>
>>>> Not at all. A simulator simulates a finite string and the actual
>>>> behavior of this simulated finite string is the ultimate basis of
>>>> whether or not it specifies a finite sequence of configurations.
>>>
>>> No. A simulator simulates a Turing Machine applied to an input. It
>>> takes as its input a finite string which represents that Turing
>>> Machine/Input pair. It's completely meaningless to talk about
>>> simulating a finite string.
>>>
>>
>> It is possible for Turing machines to have blank tapes.
>>
>> The salient aspect for the Halting problem is whether or not the
>> finite string machine description specifies a finite or infinite
>> sequence of configurations. The ultimate basis for determining this is
>> the actual behavior of the simulated finite string.
>>
>> Since this equally applies to machines having inputs and machines not
>> having inputs the distinction relative to inputs is moot.
>>
>>>> If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot possibly
>>>> reach ⟨Ĥ⟩.qn then it is necessarily correct for embedded_H to
>>>> transition to Ĥ.qn and nothing else in the universe can possibly
>>>> refute this.
>>>
>>> Again, you're falling back on your belief that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is
>>> both meaningful (it isn't) and somehow distinct from H applied to ⟨Ĥ⟩.
>>>
>>
>> The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
>> is the ultimate measure of the halt status of this input.
>
> Which just proves you are not working on the Halting Problem,
No it only proves that you and André don't understand that a halt
decider computes the mapping from the inputs to an accept or reject
state (here is the part that you two don't understand):

On the basis of the actual behavior specified by the actual input.
On the basis of the actual behavior specified by the actual input.
On the basis of the actual behavior specified by the actual input.

--
Copyright 2021 Pete Olcott

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

On 2/2/2022 8:08 PM, Richard Damon wrote:
> On 2/2/22 8:41 PM, olcott wrote:
>> On 2/2/2022 6:22 PM, Richard Damon wrote:
>>> On 2/2/22 10:20 AM, olcott wrote:
>>>> On 2/1/2022 9:58 PM, André G. Isaak wrote:
>>>>> On 2022-02-01 20:44, olcott wrote:
>>>>>> On 2/1/2022 9:24 PM, André G. Isaak wrote:
>>>>>>> On 2022-02-01 19:57, olcott wrote:
>>>>>>>> On 2/1/2022 8:48 PM, André G. Isaak wrote:
>>>>>>>>> On 2022-02-01 19:37, olcott wrote:
>>>>>>>>>> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>>>>>>>>>
>>>>>>>>>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.
>>>>>>>>>>
>>>>>>>>>> Sure and so is the "I am going to go to the" part of
>>>>>>>>>> "I am going to go to the store to buy some ice cream."
>>>>>>>>>>
>>>>>>>>>> When you don't cut off what I said in the middle of the
>>>>>>>>>> sentence then it makes much more sense.
>>>>>>>>>>
>>>>>>>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly
>>>>>>>>>> transition to ⟨Ĥ⟩.qn ?
>>>>>>>>>
>>>>>>>>> That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩
>>>>>>>>> or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You
>>>>>>>>> cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply
>>>>>>>>> ⟨Ĥ⟩ to ⟨Ĥ⟩.
>>>>>>>>
>>>>>>>> So you are simply being nit picky about my use of terminology.
>>>>>>>
>>>>>>> Yes, I insist on terminology being used correctly. And any place
>>>>>>> where you attempt to publish your results will be equally, if not
>>>>>>> more, nit picky.
>>>>>>>
>>>>>>
>>>>>> It is fine and good that you help correct my terminology.
>>>>>> What is not fine and good is for you to reject the essence of the
>>>>>> gist of what I am saying entirely on the basis that I did not say
>>>>>> it exactly according to conventions. The is what Ben always did.
>>>>>> He never paid any attention to the actual substance of what I was
>>>>>> saying.
>>>>>>
>>>>>>>> When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H
>>>>>>>> and embedded_H correctly determines that its simulated input
>>>>>>>> cannot possibly reach any final state then embedded_H is
>>>>>>>> necessarily correct to transition to Ĥ.qn indicating that its
>>>>>>>> simulated input never halts.
>>>>>>>
>>>>>>> But now you've just hidden your meaningless terminological abuse.
>>>>>>> "Its simulated input" is only meaningful when it is construed as
>>>>>>> meaning the simulation of the computation REPRESENTED by the
>>>>>>> input, i.e. the
>>>>>>
>>>>>> Not at all. A simulator simulates a finite string and the actual
>>>>>> behavior of this simulated finite string is the ultimate basis of
>>>>>> whether or not it specifies a finite sequence of configurations.
>>>>>
>>>>> No. A simulator simulates a Turing Machine applied to an input. It
>>>>> takes as its input a finite string which represents that Turing
>>>>> Machine/Input pair. It's completely meaningless to talk about
>>>>> simulating a finite string.
>>>>>
>>>>
>>>> It is possible for Turing machines to have blank tapes.
>>>>
>>>> The salient aspect for the Halting problem is whether or not the
>>>> finite string machine description specifies a finite or infinite
>>>> sequence of configurations. The ultimate basis for determining this
>>>> is the actual behavior of the simulated finite string.
>>>>
>>>> Since this equally applies to machines having inputs and machines
>>>> not having inputs the distinction relative to inputs is moot.
>>>>
>>>>>> If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot possibly
>>>>>> reach ⟨Ĥ⟩.qn then it is necessarily correct for embedded_H to
>>>>>> transition to Ĥ.qn and nothing else in the universe can possibly
>>>>>> refute this.
>>>>>
>>>>> Again, you're falling back on your belief that ⟨Ĥ⟩ applied to ⟨Ĥ⟩
>>>>> is both meaningful (it isn't) and somehow distinct from H applied
>>>>> to ⟨Ĥ⟩.
>>>>>
>>>>
>>>> The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩
>>>> ⟨Ĥ⟩ is the ultimate measure of the halt status of this input.
>>>
>>> Which just proves you are not working on the Halting Problem,
>> No it only proves that you and André don't understand that a halt
>> decider computes the mapping from the inputs to an accept or reject
>> state (here is the part that you two don't understand):
>>
>> On the basis of the actual behavior specified by the actual input.
>> On the basis of the actual behavior specified by the actual input.
>> On the basis of the actual behavior specified by the actual input.
>>
>
> Which is DEFINED by what a the machine the input represents would do,

These words prove themselves true on the basis of their meaning:
The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
is the ultimate measure of the behavior specified by ⟨Ĥ⟩ ⟨Ĥ⟩.

--
Copyright 2021 Pete Olcott

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

On 2/2/2022 10:24 AM, olcott wrote:
> On 2/2/2022 12:19 PM, Greg Smith wrote:
>> On 2/2/2022 7:31 AM, olcott wrote:
>>>> On 2022-02-01 20:59, olcott wrote:
>>
>> Shut up idiot.
>>
>
> User at IP address: 46.165.242.75 must be blocked for abuse

Shut up idiot.

On 2/2/2022 10:42 PM, Richard Damon wrote:
> On 2/2/22 10:50 PM, olcott wrote:
>> On 2/2/2022 9:30 PM, Richard Damon wrote:
>>>
>>> On 2/2/22 10:12 PM, olcott wrote:
>>>> On 2/2/2022 8:57 PM, Richard Damon wrote:
>>>>> On 2/2/22 9:31 PM, olcott wrote:
>>>>>> On 2/2/2022 8:08 PM, Richard Damon wrote:
>>>>>>> On 2/2/22 8:41 PM, olcott wrote:
>>>>>>>> On 2/2/2022 6:22 PM, Richard Damon wrote:
>>>>>>>>> On 2/2/22 10:20 AM, olcott wrote:
>>>>>>>>>> On 2/1/2022 9:58 PM, André G. Isaak wrote:
>>>>>>>>>>> On 2022-02-01 20:44, olcott wrote:
>>>>>>>>>>>> On 2/1/2022 9:24 PM, André G. Isaak wrote:
>>>>>>>>>>>>> On 2022-02-01 19:57, olcott wrote:
>>>>>>>>>>>>>> On 2/1/2022 8:48 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>> On 2022-02-01 19:37, olcott wrote:
>>>>>>>>>>>>>>>> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Sure and so is the "I am going to go to the" part of
>>>>>>>>>>>>>>>> "I am going to go to the store to buy some ice cream."
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> When you don't cut off what I said in the middle of the
>>>>>>>>>>>>>>>> sentence then it makes much more sense.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly
>>>>>>>>>>>>>>>> transition to ⟨Ĥ⟩.qn ?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> That's just as meaningless. You can simulate Ĥ applied to
>>>>>>>>>>>>>>> ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a
>>>>>>>>>>>>>>> simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore
>>>>>>>>>>>>>>> than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> So you are simply being nit picky about my use of
>>>>>>>>>>>>>> terminology.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Yes, I insist on terminology being used correctly. And any
>>>>>>>>>>>>> place where you attempt to publish your results will be
>>>>>>>>>>>>> equally, if not more, nit picky.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> It is fine and good that you help correct my terminology.
>>>>>>>>>>>> What is not fine and good is for you to reject the essence
>>>>>>>>>>>> of the gist of what I am saying entirely on the basis that I
>>>>>>>>>>>> did not say it exactly according to conventions. The is what
>>>>>>>>>>>> Ben always did. He never paid any attention to the actual
>>>>>>>>>>>> substance of what I was saying.
>>>>>>>>>>>>
>>>>>>>>>>>>>> When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider
>>>>>>>>>>>>>> embedded_H and embedded_H correctly determines that its
>>>>>>>>>>>>>> simulated input cannot possibly reach any final state then
>>>>>>>>>>>>>> embedded_H is necessarily correct to transition to Ĥ.qn
>>>>>>>>>>>>>> indicating that its simulated input never halts.
>>>>>>>>>>>>>
>>>>>>>>>>>>> But now you've just hidden your meaningless terminological
>>>>>>>>>>>>> abuse. "Its simulated input" is only meaningful when it is
>>>>>>>>>>>>> construed as meaning the simulation of the computation
>>>>>>>>>>>>> REPRESENTED by the input, i.e. the
>>>>>>>>>>>>
>>>>>>>>>>>> Not at all. A simulator simulates a finite string and the
>>>>>>>>>>>> actual behavior of this simulated finite string is the
>>>>>>>>>>>> ultimate basis of whether or not it specifies a finite
>>>>>>>>>>>> sequence of configurations.
>>>>>>>>>>>
>>>>>>>>>>> No. A simulator simulates a Turing Machine applied to an
>>>>>>>>>>> input. It takes as its input a finite string which represents
>>>>>>>>>>> that Turing Machine/Input pair. It's completely meaningless
>>>>>>>>>>> to talk about simulating a finite string.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> It is possible for Turing machines to have blank tapes.
>>>>>>>>>>
>>>>>>>>>> The salient aspect for the Halting problem is whether or not
>>>>>>>>>> the finite string machine description specifies a finite or
>>>>>>>>>> infinite sequence of configurations. The ultimate basis for
>>>>>>>>>> determining this is the actual behavior of the simulated
>>>>>>>>>> finite string.
>>>>>>>>>>
>>>>>>>>>> Since this equally applies to machines having inputs and
>>>>>>>>>> machines not having inputs the distinction relative to inputs
>>>>>>>>>> is moot.
>>>>>>>>>>
>>>>>>>>>>>> If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot
>>>>>>>>>>>> possibly reach ⟨Ĥ⟩.qn then it is necessarily correct for
>>>>>>>>>>>> embedded_H to transition to Ĥ.qn and nothing else in the
>>>>>>>>>>>> universe can possibly refute this.
>>>>>>>>>>>
>>>>>>>>>>> Again, you're falling back on your belief that ⟨Ĥ⟩ applied to
>>>>>>>>>>> ⟨Ĥ⟩ is both meaningful (it isn't) and somehow distinct from H
>>>>>>>>>>> applied to ⟨Ĥ⟩.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> The behavior of the simulated input when embedded_H applied to
>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt status of this input.
>>>>>>>>>
>>>>>>>>> Which just proves you are not working on the Halting Problem,
>>>>>>>> No it only proves that you and André don't understand that a
>>>>>>>> halt decider computes the mapping from the inputs to an accept
>>>>>>>> or reject state (here is the part that you two don't understand):
>>>>>>>>
>>>>>>>> On the basis of the actual behavior specified by the actual input.
>>>>>>>> On the basis of the actual behavior specified by the actual input.
>>>>>>>> On the basis of the actual behavior specified by the actual input.
>>>>>>>>
>>>>>>>
>>>>>>> Which is DEFINED by what a the machine the input represents would
>>>>>>> do,
>>>>>>
>>>>>> These words prove themselves true on the basis of their meaning:
>>>>>> The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by
>>>>>> embedded_H is the ultimate measure of the behavior specified by
>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>
>>>>>
>>>>> WRONG, which shows you do not actually know the meaning of the words.
>>>> When you disagree that the correct simulation of a machine
>>>> description of a machine is the ultimate measure of the behavior
>>>> specified by this machine description it is just like saying that a
>>>> black cat is not a cat.
>>>>
>>>
>>> The problem is that 'Correct Simulation of a machine description' has
>>> an actual meaning, in that the simulation must match the actual
>>> behavior of the machine whose description it is simulating, RIGHT?
>> It must only do exactly what it actually does, if this does not meet
>> expectations then expectations must be incorrect.
>>
>> Here is what it actually does:
>> These steps would keep repeating:
>>    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
>>    Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
>>    Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩
>> ⟨Ĥ5⟩...
>
> And if that is what it actually does, then H NEVER aborts its simulation
> and thus never give an answer.
>

On 2/2/2022 11:09 PM, Richard Damon wrote:
> On 2/2/22 11:50 PM, olcott wrote:
>> On 2/2/2022 10:42 PM, Richard Damon wrote:
>>> On 2/2/22 10:50 PM, olcott wrote:
>>>> On 2/2/2022 9:30 PM, Richard Damon wrote:
>>>>>
>>>>> On 2/2/22 10:12 PM, olcott wrote:
>>>>>> On 2/2/2022 8:57 PM, Richard Damon wrote:
>>>>>>> On 2/2/22 9:31 PM, olcott wrote:
>>>>>>>> On 2/2/2022 8:08 PM, Richard Damon wrote:
>>>>>>>>> On 2/2/22 8:41 PM, olcott wrote:
>>>>>>>>>> On 2/2/2022 6:22 PM, Richard Damon wrote:
>>>>>>>>>>> On 2/2/22 10:20 AM, olcott wrote:
>>>>>>>>>>>> On 2/1/2022 9:58 PM, André G. Isaak wrote:
>>>>>>>>>>>>> On 2022-02-01 20:44, olcott wrote:
>>>>>>>>>>>>>> On 2/1/2022 9:24 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>> On 2022-02-01 19:57, olcott wrote:
>>>>>>>>>>>>>>>> On 2/1/2022 8:48 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>> On 2022-02-01 19:37, olcott wrote:
>>>>>>>>>>>>>>>>>> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Sure and so is the "I am going to go to the" part of
>>>>>>>>>>>>>>>>>> "I am going to go to the store to buy some ice cream."
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> When you don't cut off what I said in the middle of
>>>>>>>>>>>>>>>>>> the sentence then it makes much more sense.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H
>>>>>>>>>>>>>>>>>> possibly transition to ⟨Ĥ⟩.qn ?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> That's just as meaningless. You can simulate Ĥ applied
>>>>>>>>>>>>>>>>> to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a
>>>>>>>>>>>>>>>>> simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩
>>>>>>>>>>>>>>>>> anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> So you are simply being nit picky about my use of
>>>>>>>>>>>>>>>> terminology.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Yes, I insist on terminology being used correctly. And
>>>>>>>>>>>>>>> any place where you attempt to publish your results will
>>>>>>>>>>>>>>> be equally, if not more, nit picky.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> It is fine and good that you help correct my terminology.
>>>>>>>>>>>>>> What is not fine and good is for you to reject the essence
>>>>>>>>>>>>>> of the gist of what I am saying entirely on the basis that
>>>>>>>>>>>>>> I did not say it exactly according to conventions. The is
>>>>>>>>>>>>>> what Ben always did. He never paid any attention to the
>>>>>>>>>>>>>> actual substance of what I was saying.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider
>>>>>>>>>>>>>>>> embedded_H and embedded_H correctly determines that its
>>>>>>>>>>>>>>>> simulated input cannot possibly reach any final state
>>>>>>>>>>>>>>>> then embedded_H is necessarily correct to transition to
>>>>>>>>>>>>>>>> Ĥ.qn indicating that its simulated input never halts.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> But now you've just hidden your meaningless
>>>>>>>>>>>>>>> terminological abuse. "Its simulated input" is only
>>>>>>>>>>>>>>> meaningful when it is construed as meaning the simulation
>>>>>>>>>>>>>>> of the computation REPRESENTED by the input, i.e. the
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Not at all. A simulator simulates a finite string and the
>>>>>>>>>>>>>> actual behavior of this simulated finite string is the
>>>>>>>>>>>>>> ultimate basis of whether or not it specifies a finite
>>>>>>>>>>>>>> sequence of configurations.
>>>>>>>>>>>>>
>>>>>>>>>>>>> No. A simulator simulates a Turing Machine applied to an
>>>>>>>>>>>>> input. It takes as its input a finite string which
>>>>>>>>>>>>> represents that Turing Machine/Input pair. It's completely
>>>>>>>>>>>>> meaningless to talk about simulating a finite string.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> It is possible for Turing machines to have blank tapes.
>>>>>>>>>>>>
>>>>>>>>>>>> The salient aspect for the Halting problem is whether or not
>>>>>>>>>>>> the finite string machine description specifies a finite or
>>>>>>>>>>>> infinite sequence of configurations. The ultimate basis for
>>>>>>>>>>>> determining this is the actual behavior of the simulated
>>>>>>>>>>>> finite string.
>>>>>>>>>>>>
>>>>>>>>>>>> Since this equally applies to machines having inputs and
>>>>>>>>>>>> machines not having inputs the distinction relative to
>>>>>>>>>>>> inputs is moot.
>>>>>>>>>>>>
>>>>>>>>>>>>>> If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot
>>>>>>>>>>>>>> possibly reach ⟨Ĥ⟩.qn then it is necessarily correct for
>>>>>>>>>>>>>> embedded_H to transition to Ĥ.qn and nothing else in the
>>>>>>>>>>>>>> universe can possibly refute this.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Again, you're falling back on your belief that ⟨Ĥ⟩ applied
>>>>>>>>>>>>> to ⟨Ĥ⟩ is both meaningful (it isn't) and somehow distinct
>>>>>>>>>>>>> from H applied to ⟨Ĥ⟩.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> The behavior of the simulated input when embedded_H applied
>>>>>>>>>>>> to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt status of
>>>>>>>>>>>> this input.
>>>>>>>>>>>
>>>>>>>>>>> Which just proves you are not working on the Halting Problem,
>>>>>>>>>> No it only proves that you and André don't understand that a
>>>>>>>>>> halt decider computes the mapping from the inputs to an accept
>>>>>>>>>> or reject state (here is the part that you two don't understand):
>>>>>>>>>>
>>>>>>>>>> On the basis of the actual behavior specified by the actual
>>>>>>>>>> input.
>>>>>>>>>> On the basis of the actual behavior specified by the actual
>>>>>>>>>> input.
>>>>>>>>>> On the basis of the actual behavior specified by the actual
>>>>>>>>>> input.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Which is DEFINED by what a the machine the input represents
>>>>>>>>> would do,
>>>>>>>>
>>>>>>>> These words prove themselves true on the basis of their meaning:
>>>>>>>> The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by
>>>>>>>> embedded_H is the ultimate measure of the behavior specified by
>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>>>
>>>>>>>
>>>>>>> WRONG, which shows you do not actually know the meaning of the
>>>>>>> words.
>>>>>> When you disagree that the correct simulation of a machine
>>>>>> description of a machine is the ultimate measure of the behavior
>>>>>> specified by this machine description it is just like saying that
>>>>>> a black cat is not a cat.
>>>>>>
>>>>>
>>>>> The problem is that 'Correct Simulation of a machine description'
>>>>> has an actual meaning, in that the simulation must match the actual
>>>>> behavior of the machine whose description it is simulating, RIGHT?
>>>> It must only do exactly what it actually does, if this does not meet
>>>> expectations then expectations must be incorrect.
>>>>
>>>> Here is what it actually does:
>>>> These steps would keep repeating:
>>>>    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
>>>>    Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
>>>>    Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩
>>>> ⟨Ĥ5⟩...
>>>
>>> And if that is what it actually does, then H NEVER aborts its
>>> simulation and thus never give an answer.
>>>
>>
>> When embedded_H correctly matches the above infinite sequence this
>> conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot
>> possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).
>>
>>
>
> Excepts as previously said, that pattern only exists if H never aborts.
>
That is not true. The pattern exists for at least any finite number of
steps where it can be recognized. The three iterations shown above are
plenty enough for it to be recogized.

On 2/3/2022 6:15 AM, olcott wrote:
> On 2/3/2022 5:50 AM, Richard Damon wrote:
>> On 2/3/22 12:24 AM, olcott wrote:
>>> On 2/2/2022 11:09 PM, Richard Damon wrote:
>>>> On 2/2/22 11:50 PM, olcott wrote:
>>>>> On 2/2/2022 10:42 PM, Richard Damon wrote:
>>>>>> On 2/2/22 10:50 PM, olcott wrote:
>>>>>>> On 2/2/2022 9:30 PM, Richard Damon wrote:
>>>>>>>>
>>>>>>>> On 2/2/22 10:12 PM, olcott wrote:
>>>>>>>>> On 2/2/2022 8:57 PM, Richard Damon wrote:
>>>>>>>>>> On 2/2/22 9:31 PM, olcott wrote:
>>>>>>>>>>> On 2/2/2022 8:08 PM, Richard Damon wrote:
>>>>>>>>>>>> On 2/2/22 8:41 PM, olcott wrote:
>>>>>>>>>>>>> On 2/2/2022 6:22 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 2/2/22 10:20 AM, olcott wrote:
>>>>>>>>>>>>>>> On 2/1/2022 9:58 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>> On 2022-02-01 20:44, olcott wrote:
>>>>>>>>>>>>>>>>> On 2/1/2022 9:24 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>> On 2022-02-01 19:57, olcott wrote:
>>>>>>>>>>>>>>>>>>> On 2/1/2022 8:48 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>>> On 2022-02-01 19:37, olcott wrote:
>>>>>>>>>>>>>>>>>>>>> On 2/1/2022 8:08 PM, André G. Isaak wrote:

Shut up idiot.

On 2/3/2022 5:36 PM, Richard Damon wrote:
> On 2/3/22 9:15 AM, olcott wrote:
>> On 2/3/2022 5:50 AM, Richard Damon wrote:
>>> On 2/3/22 12:24 AM, olcott wrote:
>>>> On 2/2/2022 11:09 PM, Richard Damon wrote:
>>>>> On 2/2/22 11:50 PM, olcott wrote:
>>>>>> On 2/2/2022 10:42 PM, Richard Damon wrote:
>>>>>>> On 2/2/22 10:50 PM, olcott wrote:
>>>>>>>> On 2/2/2022 9:30 PM, Richard Damon wrote:
>>>>>>>>>
>>>>>>>>> On 2/2/22 10:12 PM, olcott wrote:
>>>>>>>>>> On 2/2/2022 8:57 PM, Richard Damon wrote:
>>>>>>>>>>> On 2/2/22 9:31 PM, olcott wrote:
>>>>>>>>>>>> On 2/2/2022 8:08 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 2/2/22 8:41 PM, olcott wrote:
>>>>>>>>>>>>>> On 2/2/2022 6:22 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 2/2/22 10:20 AM, olcott wrote:
>>>>>>>>>>>>>>>> On 2/1/2022 9:58 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>> On 2022-02-01 20:44, olcott wrote:
>>>>>>>>>>>>>>>>>> On 2/1/2022 9:24 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>> On 2022-02-01 19:57, olcott wrote:
>>>>>>>>>>>>>>>>>>>> On 2/1/2022 8:48 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>>>> On 2022-02-01 19:37, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Sure and so is the "I am going to go to the" part of
>>>>>>>>>>>>>>>>>>>>>> "I am going to go to the store to buy some ice
>>>>>>>>>>>>>>>>>>>>>> cream."
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> When you don't cut off what I said in the middle
>>>>>>>>>>>>>>>>>>>>>> of the sentence then it makes much more sense.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H
>>>>>>>>>>>>>>>>>>>>>> possibly transition to ⟨Ĥ⟩.qn ?
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> That's just as meaningless. You can simulate Ĥ
>>>>>>>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the
>>>>>>>>>>>>>>>>>>>>> input to a simulator. You cannot simulate ⟨Ĥ⟩
>>>>>>>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> So you are simply being nit picky about my use of
>>>>>>>>>>>>>>>>>>>> terminology.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Yes, I insist on terminology being used correctly.
>>>>>>>>>>>>>>>>>>> And any place where you attempt to publish your
>>>>>>>>>>>>>>>>>>> results will be equally, if not more, nit picky.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> It is fine and good that you help correct my terminology.
>>>>>>>>>>>>>>>>>> What is not fine and good is for you to reject the
>>>>>>>>>>>>>>>>>> essence of the gist of what I am saying entirely on
>>>>>>>>>>>>>>>>>> the basis that I did not say it exactly according to
>>>>>>>>>>>>>>>>>> conventions. The is what Ben always did. He never paid
>>>>>>>>>>>>>>>>>> any attention to the actual substance of what I was
>>>>>>>>>>>>>>>>>> saying.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider
>>>>>>>>>>>>>>>>>>>> embedded_H and embedded_H correctly determines that
>>>>>>>>>>>>>>>>>>>> its simulated input cannot possibly reach any final
>>>>>>>>>>>>>>>>>>>> state then embedded_H is necessarily correct to
>>>>>>>>>>>>>>>>>>>> transition to Ĥ.qn indicating that its simulated
>>>>>>>>>>>>>>>>>>>> input never halts.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> But now you've just hidden your meaningless
>>>>>>>>>>>>>>>>>>> terminological abuse. "Its simulated input" is only
>>>>>>>>>>>>>>>>>>> meaningful when it is construed as meaning the
>>>>>>>>>>>>>>>>>>> simulation of the computation REPRESENTED by the
>>>>>>>>>>>>>>>>>>> input, i.e. the
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Not at all. A simulator simulates a finite string and
>>>>>>>>>>>>>>>>>> the actual behavior of this simulated finite string is
>>>>>>>>>>>>>>>>>> the ultimate basis of whether or not it specifies a
>>>>>>>>>>>>>>>>>> finite sequence of configurations.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> No. A simulator simulates a Turing Machine applied to
>>>>>>>>>>>>>>>>> an input. It takes as its input a finite string which
>>>>>>>>>>>>>>>>> represents that Turing Machine/Input pair. It's
>>>>>>>>>>>>>>>>> completely meaningless to talk about simulating a
>>>>>>>>>>>>>>>>> finite string.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> It is possible for Turing machines to have blank tapes.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The salient aspect for the Halting problem is whether or
>>>>>>>>>>>>>>>> not the finite string machine description specifies a
>>>>>>>>>>>>>>>> finite or infinite sequence of configurations. The
>>>>>>>>>>>>>>>> ultimate basis for determining this is the actual
>>>>>>>>>>>>>>>> behavior of the simulated finite string.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Since this equally applies to machines having inputs and
>>>>>>>>>>>>>>>> machines not having inputs the distinction relative to
>>>>>>>>>>>>>>>> inputs is moot.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
>>>>>>>>>>>>>>>>>> cannot possibly reach ⟨Ĥ⟩.qn then it is necessarily
>>>>>>>>>>>>>>>>>> correct for embedded_H to transition to Ĥ.qn and
>>>>>>>>>>>>>>>>>> nothing else in the universe can possibly refute this.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Again, you're falling back on your belief that ⟨Ĥ⟩
>>>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ is both meaningful (it isn't) and
>>>>>>>>>>>>>>>>> somehow distinct from H applied to ⟨Ĥ⟩.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The behavior of the simulated input when embedded_H
>>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt
>>>>>>>>>>>>>>>> status of this input.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Which just proves you are not working on the Halting
>>>>>>>>>>>>>>> Problem,
>>>>>>>>>>>>>> No it only proves that you and André don't understand that
>>>>>>>>>>>>>> a halt decider computes the mapping from the inputs to an
>>>>>>>>>>>>>> accept or reject state (here is the part that you two
>>>>>>>>>>>>>> don't understand):
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On the basis of the actual behavior specified by the
>>>>>>>>>>>>>> actual input.
>>>>>>>>>>>>>> On the basis of the actual behavior specified by the
>>>>>>>>>>>>>> actual input.
>>>>>>>>>>>>>> On the basis of the actual behavior specified by the
>>>>>>>>>>>>>> actual input.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Which is DEFINED by what a the machine the input represents
>>>>>>>>>>>>> would do,
>>>>>>>>>>>>
>>>>>>>>>>>> These words prove themselves true on the basis of their
>>>>>>>>>>>> meaning:
>>>>>>>>>>>> The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by
>>>>>>>>>>>> embedded_H is the ultimate measure of the behavior specified
>>>>>>>>>>>> by ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> WRONG, which shows you do not actually know the meaning of
>>>>>>>>>>> the words.
>>>>>>>>>> When you disagree that the correct simulation of a machine
>>>>>>>>>> description of a machine is the ultimate measure of the
>>>>>>>>>> behavior specified by this machine description it is just like
>>>>>>>>>> saying that a black cat is not a cat.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> The problem is that 'Correct Simulation of a machine
>>>>>>>>> description' has an actual meaning, in that the simulation must
>>>>>>>>> match the actual behavior of the machine whose description it
>>>>>>>>> is simulating, RIGHT?
>>>>>>>> It must only do exactly what it actually does, if this does not
>>>>>>>> meet expectations then expectations must be incorrect.
>>>>>>>>
>>>>>>>> Here is what it actually does:
>>>>>>>> These steps would keep repeating:
>>>>>>>>    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates
>>>>>>>> ⟨Ĥ2⟩ ⟨Ĥ3⟩
>>>>>>>>    Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates
>>>>>>>> ⟨Ĥ3⟩ ⟨Ĥ4⟩
>>>>>>>>    Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates
>>>>>>>> ⟨Ĥ4⟩ ⟨Ĥ5⟩...
>>>>>>>
>>>>>>> And if that is what it actually does, then H NEVER aborts its
>>>>>>> simulation and thus never give an answer.
>>>>>>>
>>>>>>
>>>>>> When embedded_H correctly matches the above infinite sequence this
>>>>>> conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot
>>>>>> possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).
>>>>>>
>>>>>>
>>>>>
>>>>> Excepts as previously said, that pattern only exists if H never
>>>>> aborts.
>>>>>
>>>> That is not true. The pattern exists for at least any finite number
>>>> of steps where it can be recognized. The three iterations shown
>>>> above are plenty enough for it to be recogized.
>>>>
>>>
>>> But if it only exists for a finite number of steps (till it is
>>> recognized)
>>
>> Then embedded_H has conclusively proved that its simulated ⟨Ĥ⟩ ⟨Ĥ⟩
>> cannot possibly ever reach ⟨Ĥ⟩.qn even in an infinite number of
>> simulated steps thus meeting the Linz definition of a sequence of
>> configurations that never halt.
>>
>> computation that halts … the Turing machine will halt whenever it
>> enters a final state. (Linz:1990:234)
>>
>
> No, it hasn't, because I just showed you that if H -> H.Qn then the
> computation H^ <H^> Halts.
>

On 2/3/2022 5:50 AM, Richard Damon wrote:
> On 2/3/22 12:24 AM, olcott wrote:
>> On 2/2/2022 11:09 PM, Richard Damon wrote:
>>> On 2/2/22 11:50 PM, olcott wrote:
>>>> On 2/2/2022 10:42 PM, Richard Damon wrote:
>>>>> On 2/2/22 10:50 PM, olcott wrote:
>>>>>> On 2/2/2022 9:30 PM, Richard Damon wrote:
>>>>>>>
>>>>>>> On 2/2/22 10:12 PM, olcott wrote:
>>>>>>>> On 2/2/2022 8:57 PM, Richard Damon wrote:
>>>>>>>>> On 2/2/22 9:31 PM, olcott wrote:
>>>>>>>>>> On 2/2/2022 8:08 PM, Richard Damon wrote:
>>>>>>>>>>> On 2/2/22 8:41 PM, olcott wrote:
>>>>>>>>>>>> On 2/2/2022 6:22 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 2/2/22 10:20 AM, olcott wrote:
>>>>>>>>>>>>>> On 2/1/2022 9:58 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>> On 2022-02-01 20:44, olcott wrote:
>>>>>>>>>>>>>>>> On 2/1/2022 9:24 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>> On 2022-02-01 19:57, olcott wrote:
>>>>>>>>>>>>>>>>>> On 2/1/2022 8:48 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>> On 2022-02-01 19:37, olcott wrote:
>>>>>>>>>>>>>>>>>>>> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Sure and so is the "I am going to go to the" part of
>>>>>>>>>>>>>>>>>>>> "I am going to go to the store to buy some ice cream."
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> When you don't cut off what I said in the middle of
>>>>>>>>>>>>>>>>>>>> the sentence then it makes much more sense.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H
>>>>>>>>>>>>>>>>>>>> possibly transition to ⟨Ĥ⟩.qn ?
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> That's just as meaningless. You can simulate Ĥ
>>>>>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the
>>>>>>>>>>>>>>>>>>> input to a simulator. You cannot simulate ⟨Ĥ⟩ applied
>>>>>>>>>>>>>>>>>>> to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> So you are simply being nit picky about my use of
>>>>>>>>>>>>>>>>>> terminology.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Yes, I insist on terminology being used correctly. And
>>>>>>>>>>>>>>>>> any place where you attempt to publish your results
>>>>>>>>>>>>>>>>> will be equally, if not more, nit picky.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> It is fine and good that you help correct my terminology.
>>>>>>>>>>>>>>>> What is not fine and good is for you to reject the
>>>>>>>>>>>>>>>> essence of the gist of what I am saying entirely on the
>>>>>>>>>>>>>>>> basis that I did not say it exactly according to
>>>>>>>>>>>>>>>> conventions. The is what Ben always did. He never paid
>>>>>>>>>>>>>>>> any attention to the actual substance of what I was saying.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider
>>>>>>>>>>>>>>>>>> embedded_H and embedded_H correctly determines that
>>>>>>>>>>>>>>>>>> its simulated input cannot possibly reach any final
>>>>>>>>>>>>>>>>>> state then embedded_H is necessarily correct to
>>>>>>>>>>>>>>>>>> transition to Ĥ.qn indicating that its simulated input
>>>>>>>>>>>>>>>>>> never halts.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> But now you've just hidden your meaningless
>>>>>>>>>>>>>>>>> terminological abuse. "Its simulated input" is only
>>>>>>>>>>>>>>>>> meaningful when it is construed as meaning the
>>>>>>>>>>>>>>>>> simulation of the computation REPRESENTED by the input,
>>>>>>>>>>>>>>>>> i.e. the
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Not at all. A simulator simulates a finite string and
>>>>>>>>>>>>>>>> the actual behavior of this simulated finite string is
>>>>>>>>>>>>>>>> the ultimate basis of whether or not it specifies a
>>>>>>>>>>>>>>>> finite sequence of configurations.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> No. A simulator simulates a Turing Machine applied to an
>>>>>>>>>>>>>>> input. It takes as its input a finite string which
>>>>>>>>>>>>>>> represents that Turing Machine/Input pair. It's
>>>>>>>>>>>>>>> completely meaningless to talk about simulating a finite
>>>>>>>>>>>>>>> string.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> It is possible for Turing machines to have blank tapes.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The salient aspect for the Halting problem is whether or
>>>>>>>>>>>>>> not the finite string machine description specifies a
>>>>>>>>>>>>>> finite or infinite sequence of configurations. The
>>>>>>>>>>>>>> ultimate basis for determining this is the actual behavior
>>>>>>>>>>>>>> of the simulated finite string.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Since this equally applies to machines having inputs and
>>>>>>>>>>>>>> machines not having inputs the distinction relative to
>>>>>>>>>>>>>> inputs is moot.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
>>>>>>>>>>>>>>>> cannot possibly reach ⟨Ĥ⟩.qn then it is necessarily
>>>>>>>>>>>>>>>> correct for embedded_H to transition to Ĥ.qn and nothing
>>>>>>>>>>>>>>>> else in the universe can possibly refute this.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Again, you're falling back on your belief that ⟨Ĥ⟩
>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ is both meaningful (it isn't) and somehow
>>>>>>>>>>>>>>> distinct from H applied to ⟨Ĥ⟩.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The behavior of the simulated input when embedded_H
>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt
>>>>>>>>>>>>>> status of this input.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Which just proves you are not working on the Halting Problem,
>>>>>>>>>>>> No it only proves that you and André don't understand that a
>>>>>>>>>>>> halt decider computes the mapping from the inputs to an
>>>>>>>>>>>> accept or reject state (here is the part that you two don't
>>>>>>>>>>>> understand):
>>>>>>>>>>>>
>>>>>>>>>>>> On the basis of the actual behavior specified by the actual
>>>>>>>>>>>> input.
>>>>>>>>>>>> On the basis of the actual behavior specified by the actual
>>>>>>>>>>>> input.
>>>>>>>>>>>> On the basis of the actual behavior specified by the actual
>>>>>>>>>>>> input.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Which is DEFINED by what a the machine the input represents
>>>>>>>>>>> would do,
>>>>>>>>>>
>>>>>>>>>> These words prove themselves true on the basis of their meaning:
>>>>>>>>>> The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by
>>>>>>>>>> embedded_H is the ultimate measure of the behavior specified
>>>>>>>>>> by ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> WRONG, which shows you do not actually know the meaning of the
>>>>>>>>> words.
>>>>>>>> When you disagree that the correct simulation of a machine
>>>>>>>> description of a machine is the ultimate measure of the behavior
>>>>>>>> specified by this machine description it is just like saying
>>>>>>>> that a black cat is not a cat.
>>>>>>>>
>>>>>>>
>>>>>>> The problem is that 'Correct Simulation of a machine description'
>>>>>>> has an actual meaning, in that the simulation must match the
>>>>>>> actual behavior of the machine whose description it is
>>>>>>> simulating, RIGHT?
>>>>>> It must only do exactly what it actually does, if this does not
>>>>>> meet expectations then expectations must be incorrect.
>>>>>>
>>>>>> Here is what it actually does:
>>>>>> These steps would keep repeating:
>>>>>>    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩
>>>>>> ⟨Ĥ3⟩
>>>>>>    Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩
>>>>>> ⟨Ĥ4⟩
>>>>>>    Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩
>>>>>> ⟨Ĥ5⟩...
>>>>>
>>>>> And if that is what it actually does, then H NEVER aborts its
>>>>> simulation and thus never give an answer.
>>>>>
>>>>
>>>> When embedded_H correctly matches the above infinite sequence this
>>>> conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot
>>>> possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).
>>>>
>>>>
>>>
>>> Excepts as previously said, that pattern only exists if H never aborts.
>>>
>> That is not true. The pattern exists for at least any finite number of
>> steps where it can be recognized. The three iterations shown above are
>> plenty enough for it to be recogized.
>>
>
> But if it only exists for a finite number of steps (till it is
> recognized)

On 2/3/2022 5:50 AM, Richard Damon wrote:
> On 2/3/22 12:24 AM, olcott wrote:
>> On 2/2/2022 11:09 PM, Richard Damon wrote:
>>> On 2/2/22 11:50 PM, olcott wrote:
>>>> On 2/2/2022 10:42 PM, Richard Damon wrote:
>>>>> On 2/2/22 10:50 PM, olcott wrote:

>>>>>> Here is what it actually does:
>>>>>> These steps would keep repeating:
>>>>>>    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩
>>>>>> ⟨Ĥ3⟩
>>>>>>    Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩
>>>>>> ⟨Ĥ4⟩
>>>>>>    Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩
>>>>>> ⟨Ĥ5⟩...
>>>>>
>>>>> And if that is what it actually does, then H NEVER aborts its
>>>>> simulation and thus never give an answer.
>>>>>
>>>>
>>>> When embedded_H correctly matches the above infinite sequence this
>>>> conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot
>>>> possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).
>>>>
>>>
>>> Excepts as previously said, that pattern only exists if H never aborts.
>>>
>> That is not true. The pattern exists for at least any finite number of
>> steps where it can be recognized. The three iterations shown above are
>> plenty enough for it to be recogized.
>>
>
> But if it only exists for a finite number of steps (till it is
> recognized)

We are discussing the point in the execution of embedded_H where it has
just correctly matched an infinite behavior pattern while it was doing
its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩

At this point right here embedded_H has complete proof that the correct
simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn.

--
Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

On 2/3/2022 9:40 PM, Richard Damon wrote:
> On 2/3/22 10:10 PM, olcott wrote:
>> On 2/3/2022 5:50 AM, Richard Damon wrote:
>>> On 2/3/22 12:24 AM, olcott wrote:
>>>> On 2/2/2022 11:09 PM, Richard Damon wrote:
>>>>> On 2/2/22 11:50 PM, olcott wrote:
>>>>>> On 2/2/2022 10:42 PM, Richard Damon wrote:
>>>>>>> On 2/2/22 10:50 PM, olcott wrote:
>>
>>
>>>>>>>> Here is what it actually does:
>>>>>>>> These steps would keep repeating:
>>>>>>>>    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates
>>>>>>>> ⟨Ĥ2⟩ ⟨Ĥ3⟩
>>>>>>>>    Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates
>>>>>>>> ⟨Ĥ3⟩ ⟨Ĥ4⟩
>>>>>>>>    Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates
>>>>>>>> ⟨Ĥ4⟩ ⟨Ĥ5⟩...
>>>>>>>
>>>>>>> And if that is what it actually does, then H NEVER aborts its
>>>>>>> simulation and thus never give an answer.
>>>>>>>
>>>>>>
>>>>>> When embedded_H correctly matches the above infinite sequence this
>>>>>> conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot
>>>>>> possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).
>>>>>>
>>>>>
>>>>> Excepts as previously said, that pattern only exists if H never
>>>>> aborts.
>>>>>
>>>> That is not true. The pattern exists for at least any finite number
>>>> of steps where it can be recognized. The three iterations shown
>>>> above are plenty enough for it to be recogized.
>>>>
>>>
>>> But if it only exists for a finite number of steps (till it is
>>> recognized)
>>
>> We are discussing the point in the execution of embedded_H where it
>> has just correctly matched an infinite behavior pattern while it was
>> doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩
>>
>
> No, you are CLAIMING (incorrectly) that it has made that determination.

THIS IS WHAT YOU HAVE BEEN DISAGREEING WITH:
As soon as an infinite behavior pattern is correctly recognized in a
finite number of steps then it is definitely correct for embedded_H to
transition to Ĥ.qn.

--
Copyright 2021 Pete Olcott

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

On 2/3/2022 10:20 PM, Richard Damon wrote:
> On 2/3/22 10:56 PM, olcott wrote:
>> On 2/3/2022 9:40 PM, Richard Damon wrote:
>>> On 2/3/22 10:10 PM, olcott wrote:
>>>> On 2/3/2022 5:50 AM, Richard Damon wrote:
>>>>> On 2/3/22 12:24 AM, olcott wrote:
>>>>>> On 2/2/2022 11:09 PM, Richard Damon wrote:
>>>>>>> On 2/2/22 11:50 PM, olcott wrote:
>>>>>>>> On 2/2/2022 10:42 PM, Richard Damon wrote:
>>>>>>>>> On 2/2/22 10:50 PM, olcott wrote:
>>>>
>>>>
>>>>>>>>>> Here is what it actually does:
>>>>>>>>>> These steps would keep repeating:
>>>>>>>>>>    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates
>>>>>>>>>> ⟨Ĥ2⟩ ⟨Ĥ3⟩
>>>>>>>>>>    Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates
>>>>>>>>>> ⟨Ĥ3⟩ ⟨Ĥ4⟩
>>>>>>>>>>    Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates
>>>>>>>>>> ⟨Ĥ4⟩ ⟨Ĥ5⟩...
>>>>>>>>>
>>>>>>>>> And if that is what it actually does, then H NEVER aborts its
>>>>>>>>> simulation and thus never give an answer.
>>>>>>>>>
>>>>>>>>
>>>>>>>> When embedded_H correctly matches the above infinite sequence
>>>>>>>> this conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>>>>> cannot possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any
>>>>>>>> UTM).
>>>>>>>>
>>>>>>>
>>>>>>> Excepts as previously said, that pattern only exists if H never
>>>>>>> aborts.
>>>>>>>
>>>>>> That is not true. The pattern exists for at least any finite
>>>>>> number of steps where it can be recognized. The three iterations
>>>>>> shown above are plenty enough for it to be recogized.
>>>>>>
>>>>>
>>>>> But if it only exists for a finite number of steps (till it is
>>>>> recognized)
>>>>
>>>> We are discussing the point in the execution of embedded_H where it
>>>> has just correctly matched an infinite behavior pattern while it was
>>>> doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>
>>>
>>> No, you are CLAIMING (incorrectly) that it has made that determination.
>>
>> THIS IS WHAT YOU HAVE BEEN DISAGREEING WITH:
>> As soon as an infinite behavior pattern is correctly recognized in a
>> finite number of steps then it is definitely correct for embedded_H to
>> transition to Ĥ.qn.
>>
>>
>
> Except that such a pattern in H^ is a Fairy Dust Powered Unicorn,
So in other words the concept of logical necessity is so far over your
head that you cannot begin to fathom it.

When we know that we have a black cat then we know that we have a cat
and you dishonestly disagree.

--
Copyright 2021 Pete Olcott

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

On 2/3/2022 11:02 PM, Richard Damon wrote:
>
> On 2/3/22 11:45 PM, olcott wrote:
>> On 2/3/2022 10:20 PM, Richard Damon wrote:
>>> On 2/3/22 10:56 PM, olcott wrote:
>>>> On 2/3/2022 9:40 PM, Richard Damon wrote:
>>>>> On 2/3/22 10:10 PM, olcott wrote:
>>>>>> On 2/3/2022 5:50 AM, Richard Damon wrote:
>>>>>>> On 2/3/22 12:24 AM, olcott wrote:
>>>>>>>> On 2/2/2022 11:09 PM, Richard Damon wrote:
>>>>>>>>> On 2/2/22 11:50 PM, olcott wrote:
>>>>>>>>>> On 2/2/2022 10:42 PM, Richard Damon wrote:
>>>>>>>>>>> On 2/2/22 10:50 PM, olcott wrote:
>>>>>>>>>
>>>>>>>> That is not true. The pattern exists for at least any finite
>>>>>>>> number of steps where it can be recognized. The three iterations
>>>>>>>> shown above are plenty enough for it to be recogized.
>>>>>>>>
>>>>>>>
>>>>>>> But if it only exists for a finite number of steps (till it is
>>>>>>> recognized)
>>>>>>
>>>>>> We are discussing the point in the execution of embedded_H where
>>>>>> it has just correctly matched an infinite behavior pattern while
>>>>>> it was doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>>>
>>>>>
>>>>> No, you are CLAIMING (incorrectly) that it has made that
>>>>> determination.
>>>>
>>>> THIS IS WHAT YOU HAVE BEEN DISAGREEING WITH:
>>>> As soon as an infinite behavior pattern is correctly recognized in a
>>>> finite number of steps then it is definitely correct for embedded_H
>>>> to transition to Ĥ.qn.
>>>>
>>>>
>>>
>>> Except that such a pattern in H^ is a Fairy Dust Powered Unicorn,
>> So in other words the concept of logical necessity is so far over your
>> head that you cannot begin to fathom it.
>
> WHY is it logicallyt necessary that the pattern you have presupposed to
> exist to actually exist?
>
> As I agreed, **IF** H could find such a pattern, it would be correct to
> abort and go to H.Qn,

Great, yet it took far too long to get agreement on a statement that is
true by logical necessity.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

You already agreed that these steps would repeat if there was a UTM at
Ĥ.qx instead of embedded_H:

These steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...

--
Copyright 2021 Pete Olcott

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

On 2/3/2022 11:39 PM, Richard Damon wrote:
> On 2/4/22 12:12 AM, olcott wrote:
>> On 2/3/2022 11:02 PM, Richard Damon wrote:
>>>
>>> On 2/3/22 11:45 PM, olcott wrote:
>>>> On 2/3/2022 10:20 PM, Richard Damon wrote:
>>>>> On 2/3/22 10:56 PM, olcott wrote:
>>>>>> On 2/3/2022 9:40 PM, Richard Damon wrote:
>>>>>>> On 2/3/22 10:10 PM, olcott wrote:
>>>>>>>> On 2/3/2022 5:50 AM, Richard Damon wrote:
>>>>>>>>> On 2/3/22 12:24 AM, olcott wrote:
>>>>>>>>>> On 2/2/2022 11:09 PM, Richard Damon wrote:
>>>>>>>>>>> On 2/2/22 11:50 PM, olcott wrote:
>>>>>>>>>>>> On 2/2/2022 10:42 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 2/2/22 10:50 PM, olcott wrote:
>>>>>>>>>>>
>>>>>>>>>> That is not true. The pattern exists for at least any finite
>>>>>>>>>> number of steps where it can be recognized. The three
>>>>>>>>>> iterations shown above are plenty enough for it to be recogized.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> But if it only exists for a finite number of steps (till it is
>>>>>>>>> recognized)
>>>>>>>>
>>>>>>>> We are discussing the point in the execution of embedded_H where
>>>>>>>> it has just correctly matched an infinite behavior pattern while
>>>>>>>> it was doing its correct simulation of the first N steps of ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>>>>>
>>>>>>>
>>>>>>> No, you are CLAIMING (incorrectly) that it has made that
>>>>>>> determination.
>>>>>>
>>>>>> THIS IS WHAT YOU HAVE BEEN DISAGREEING WITH:
>>>>>> As soon as an infinite behavior pattern is correctly recognized in
>>>>>> a finite number of steps then it is definitely correct for
>>>>>> embedded_H to transition to Ĥ.qn.
>>>>>>
>>>>>>
>>>>>
>>>>> Except that such a pattern in H^ is a Fairy Dust Powered Unicorn,
>>>> So in other words the concept of logical necessity is so far over
>>>> your head that you cannot begin to fathom it.
>>>
>>> WHY is it logicallyt necessary that the pattern you have presupposed
>>> to exist to actually exist?
>>>
>>> As I agreed, **IF** H could find such a pattern, it would be correct
>>> to abort and go to H.Qn,
>>
>> Great, yet it took far too long to get agreement on a statement that
>> is true by logical necessity.
>>
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>> You already agreed that these steps would repeat if there was a UTM at
>> Ĥ.qx instead of embedded_H:
>>
>> These steps would keep repeating:
>> Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
>> Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
>> Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...
>>
>
> Right, and if there WAS a UTM at H^.Qx then H is 'just' a UTM, and could
> never abort its simulation (as if it did, it wouldn't be a UTM) and thus
> H never answers, and FAILS.

Since you just now agreed that the above is an infinitely repeating
pattern if Ĥ.qx was a UTM, (and you only saw three repetitions) then
embedded_H could simulate the exact same three repetitions of nested
simulations and see the same pattern that you saw.

--
Copyright 2021 Pete Olcott

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

On 2/1/2022 9:33 PM, André G. Isaak wrote:
> On 2022-02-01 20:08, olcott wrote:
>
>> embedded_H is exactly a UTM with extra features added.
>
> Apparently you don't know what 'exactly' means. embedded_H is not a UTM
> *at* *all*.
>
> If embedded_H were a UTM, then
>
> embedded_H would accept ⟨Ĥ⟩ ⟨Ĥ⟩ if Ĥ accepts ⟨Ĥ⟩

If ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by embedded_H cannot possibly ever transition to
⟨Ĥ⟩.qn then it fails to meet the Linz definition:

computation that halts … the Turing machine will halt whenever it enters
a final state. (Linz:1990:234) thus the simulated ⟨Ĥ⟩ ⟨Ĥ⟩ does not halt.

This makes it necessarily correct for embedded_H to transition to Ĥ.qn.

--
Copyright 2021 Pete Olcott

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

Ukraine cruise missiles that reach Moscow & Stalingrad// Give Taiwan cruise missiles that reach Beijing everytime NK fires a missile// tit for tat is the only way to stop bullies// Science Council Rules Earth.

Diplomacy never works on madmen. We learned that lesson with Hitler. The only thing that works on madmen is force for force.

Give Ukraine cruise missiles that reach Moscow & Stalingrad// Give Taiwan cruise missiles that reach Beijing everytime NK fires a missile// tit for tat is the only way to stop bullies.

I read that India makes the world's finest cruise missiles, send a shipment up north to Ukraine.

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

That is right Pete, we cannot see Moscow from Cambridge University campus

Sainsbury's Cambridge deluded crackpot as Olcott, and Harry Cliff's LHCb insane electron of atoms as 0.5MeV.
>
2> Olcott can David Sainsbury, Peter Johnstone, Imre Leader, Gabriel Paternain ever do a geometry proof of Fundamental Theorem of Calculus, or is that totally foreign to them? Mind you, not a limit analysis hornswaggle for that is not geometry, limit analysis is not even a math proof for anyone can analysis things, analysis this post and only math hypocrites would think it is a proof.
> Olcott can Heine, Hine, Hobson, Hope-Coles, Howie, ever ask the question, which is the atom's real electron, the muon stuck inside a 840MeV proton torus doing the Faraday law or the 0.5MeV particle that AP calls the Dirac magnetic monopole. Or does that thought fly way too above their heads?
>
> Pete Olcott spam says his spam is copyrighted, fresh from UK dust bins

> Olcott, why cannot Haniff, Heavens-Ward, Heine, Hine, Hobson, Hope-Coles, Howie, Hughes, Irvine, Jardine ask the question which is the atom's real electron-- muon or 0.5MeV particle? Is it because they cannot even do logic correctly with their 2 OR 1 = 3 with AND as subtraction?
> >
> > Olcott why does Cambridge Univ Stephen J. Toope, David Sainsbury, Peter Johnstone, Imre Leader, Gabriel Paternain keep teaching Boole error filled logic of 2 OR 1 = 3 with AND as subtraction, and never a geometry proof of Fundamental Theorem of Calculus and these crazies still think the slant cut in cone is a ellipse when in fact it is a Oval. Why brainwash and pollute more students like Pete Olcott who is crazy enough as it is.
> >
> > Olcott why is noone in Cambridge physics able to ask the question which is the atom's true real electron-- muon or 0.5MeV particle? Sargent, Saunders, Saxena, Schneider, Scott, Scrivener, Sebastian, Simmons, Simons, Sirringhaus, Smith?? Do they not have a brain to ask a simple question????
> >
> > 3rd published book
> >
> > AP's Proof-Ellipse was never a Conic Section // Math proof series, book 1 Kindle Edition
> > by Archimedes Plutonium (Author)
> >
> > Ever since Ancient Greek Times it was thought the slant cut into a cone is the ellipse. That was false. For the slant cut in every cone is a Oval, never an Ellipse. This book is a proof that the slant cut is a oval, never the ellipse. A slant cut into the Cylinder is in fact a ellipse, but never in a cone.
> >
> > Length: 21 pages
> >
> > File Size: 1620 KB
> > Print Length: 21 pages
> > Publication Date: March 11, 2019
> > Sold by: Amazon Digital Services LLC
> > Language: English
> > ASIN: B07PLSDQWC
> > Text-to-Speech: Enabled
> > X-Ray: Not Enabled
> > Word Wise: Not Enabled
> > Lending: Enabled
> > Enhanced Typesetting: Enabled
> >
> >
> > #8-2, 11th published book
> >
> > World's First Geometry Proof of Fundamental Theorem of Calculus// Math proof series, book 2 Kindle Edition
> > by Archimedes Plutonium (Author)
> >
> > Last revision was 19May2021. This is AP's 11th published book of science.
> > Preface:
> > Actually my title is too modest, for the proof that lies within this book makes it the World's First Valid Proof of Fundamental Theorem of Calculus, for in my modesty, I just wanted to emphasis that calculus was geometry and needed a geometry proof. Not being modest, there has never been a valid proof of FTC until AP's 2015 proof. This also implies that only a geometry proof of FTC constitutes a valid proof of FTC.
> >
> > Calculus needs a geometry proof of Fundamental Theorem of Calculus. But none could ever be obtained in Old Math so long as they had a huge mass of mistakes, errors, fakes and con-artist trickery such as the "limit analysis". To give a Geometry Proof of Fundamental Theorem of Calculus requires math be cleaned-up and cleaned-out of most of math's mistakes and errors. So in a sense, a Geometry FTC proof is a exercise in Consistency of all of Mathematics. In order to prove a FTC geometry proof, requires throwing out the error filled mess of Old Math. Can the Reals be the true numbers of mathematics if the Reals cannot deliver a Geometry proof of FTC? Can the functions that are not polynomial functions allow us to give a Geometry proof of FTC? Can a Coordinate System in 2D have 4 quadrants and still give a Geometry proof of FTC? Can a equation of mathematics with a number that is _not a positive decimal Grid Number_ all alone on the right side of the equation, at all times, allow us to give a Geometry proof of the FTC?
> >
> > Cover Picture: Is my hand written, one page geometry proof of the Fundamental Theorem of Calculus, the world's first geometry proof of FTC, 2013-2015, by AP.
> >
> > Length: 137 pages
> >
> > Product details
> > ASIN : B07PQTNHMY
> > Publication date : March 14, 2019
> > Language : English
> > File size : 1307 KB
> > Text-to-Speech : Enabled
> > Screen Reader : Supported
> > Enhanced typesetting : Enabled
> > X-Ray : Not Enabled
> > Word Wise : Not Enabled
> > Print length : 137 pages
> > Lending : Enabled
> > Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
> > #2 in 45-Minute Science & Math Short Reads
> > #134 in Calculus (Books)
> > #20 in Calculus (Kindle Store)
> >
> > 5th published book
> >
> > Suspend all College Classes in Logic, until they Fix their Errors // Teaching True Logic series, book 1 Kindle Edition
> > by Archimedes Plutonium (Author)
> >
> > Last revision was 29Mar2021. This is AP's 5th published book of science..
> > Preface:
> > First comes Logic-- think straight and clear which many logic and math professors are deaf dumb and blind to, and simply refuse to recognize and fix their errors.
> >
> > The single biggest error of Old Logic of Boole and Jevons was their "AND" and "OR" connectors. They got them mixed up and turned around. For their logic ends up being that of 3 OR 2 = 5 with 3 AND 2 = either 3 or 2 but never 5, when even the local village idiot knows that 3 AND 2 = 5 (addition) with 3 OR 2 = either 3 or 2 (subtraction). The AND connector in Logic stems from the idea, the mechanism involved, that given a series of statements, if just one of those many statements has a true truth value, then the entire string of statements is overall true, and thus AND truth table is truly TTTF and never TFFF. And secondly, their error of the If->Then conditional. I need to make it clear enough to the reader why the true Truth Table of IF --> Then requires a U for unknown or uncertain with a probability outcome for F --> T = U and F --> F = U. Some smart readers would know that the reason for the U is because without the U, Logic has no means of division by 0 which is undefined in mathematics. You cannot have a Logic that is less than mathematics. A logic that is impoverished and cannot do a "undefined for division by 0 in mathematics". The true logic must be able to have the fact that division by 0 is undefined. True logic is larger than all of mathematics, and must be able to fetch any piece of mathematics from out of Logic itself. So another word for U is undefined. And this is the crux of why Reductio ad Absurdum cannot be a proof method of mathematics, for a starting falsehood in a mathematics proof can only lead to a probability end conclusion.
> >
> > My corrections of Old Logic have a history that dates before 1993, sometime around 1991, I realized the Euclid proof of infinitude of primes was illogical, sadly sadly wrong, in that the newly formed number by "multiply the lot and add 1" was necessarily a new prime in the indirect proof method. So that my history of fixing Old Logic starts in 1991, but comes to a synthesis of correcting all four of the connectors of Equal/not, And, Or, If->Then, by 2015.
> >
> > Cover picture: some may complain my covers are less in quality, but I have a good reason for those covers-- I would like covers of math or logic to show the teacher's own handwriting as if he were back in the classroom writing on the blackboard or an overhead projector.
> >
> >
> > Length: 72 pages
> >
> > File Size: 773 KB
> > Print Length: 72 pages
> > Publication Date: March 12, 2019
> > Sold by: Amazon Digital Services LLC
> > Language: English
> > ASIN: B07PMB69F5
> > Text-to-Speech: Enabled
> > X-Ray: 
Not Enabled 

> > Word Wise: Not Enabled
> > Lending: Enabled
> > Screen Reader: Supported
> > Enhanced Typesetting: Enabled
> > 

> >
> >
> > #6-2, 27th published book
> >
> > Correcting Reductio Ad Absurdum// Teaching True Logic series, book 2 Kindle Edition
> > by Archimedes Plutonium (Author)
> >
> >
> > Last revision was 9NOV2020. This is AP's 27th published book.
> >
> > Preface:
> > These are the TRUE Truth Tables of the 4 connectors of Logic
> >
> > Equal+Not
> > T = T = T
> > T = ~F = T
> > F = ~T = T
> > F = F = T
> >
> > If--> then
> > T --> T = T
> > T --> F = F
> > F --> T = U (unknown or uncertain)
> > F --> F = U (unknown or uncertain)
> >
> > And
> > T & T = T
> > T & F = T
> > F & T = T
> > F & F = F
> >
> >
> > Or
> > T or T = F
> > T or F = T
> > F or T = T
> > F or F = F
> >
> > Those can be analyzed as being Equal+Not is multiplication. If-->then is division. And is addition and Or is subtraction in mathematics. Now I need to emphasis this error of Old Logic, the If->Then conditional. I need to make it clear enough to the reader why the true Truth Table of IF --> Then requires a U for unknown or uncertain with a probability outcome for F --> T = U and F --> F = U. Some smart readers would know that the reason for the U is because without the U, Logic has no means of division by 0 which is undefined in mathematics. You cannot have a Logic that is less than mathematics. A logic that is impoverished and cannot do a "undefined for division by 0 in mathematics". The true logic must be able to have the fact that division by 0 is undefined. True logic is larger than all of mathematics, and must be able to fetch any piece of mathematics from out of Logic itself.. So another word for U is undefined. And this is the crux of why Reductio ad Absurdum cannot be a proof method of mathematics, for a starting falsehood in a mathematics proof can only lead to a probability unknown, undefined end conclusion.
> >
> > Now in Old Logic they had for Reductio Ad Absurdum as displayed by this schematic:
> >
> > | | ~p
> > | |---
> > | | .
> > | | .
> > | | q
> > | | .
> > | | .
> > | | ~q
> > | p
> >
> > Which is fine except for the error of not indicating the end conclusion of "p" is only a probability of being true, not guaranteed as true. And this is the huge huge error that mathematicians have fallen victim of. For the Reductio Ad Absurdum is not a proof method for mathematics, it is probability of being true or false. Math works on guaranteed truth, not probability. This textbook is written to fix that error.
> > Length: 86 pages
> >
> > Product details
> > • ASIN : B07Q18GQ7S
> > • Publication date : March 23, 2019
> > • Language : English
> > • File size : 1178 KB
> > • Text-to-Speech : Enabled
> > • Enhanced typesetting : Enabled
> > • X-Ray : Not Enabled
> > • Word Wise : Not Enabled
> > • Print length : 86 pages
> > • Lending : Enabled
> > • Best Sellers Rank: #346,875 in Kindle Store (See Top 100 in Kindle Store)
> > ◦ #28 in Logic (Kindle Store)
> > ◦ #95 in Two-Hour Science & Math Short Reads
> > ◦ #217 in Mathematical Logic
> > •
> >
> >
> > True Chemistry: Chemistry Series, book 1 Kindle Edition
> > by Archimedes Plutonium (Author)
> >
> > Physics and chemistry made a mistake in 1897 for they thought that J.J. Thomson's small particle of 0.5MeV was the electron of atoms. By 2017, Archimedes Plutonium discovered that the rest mass of 940 for neutron and proton was really 9 x 105MeV with a small sigma-error. Meaning that the real proton is 840MeV, real electron is 105 MeV= muon, and that little particle Thomson discovered was in fact the Dirac magnetic monopole. Dirac circa 1930s was looking for a magnetic monopole, and sadly, Dirac passed away before 2017, because if he had lived to 2017, he would have seen his long sought for magnetic monopole which is everywhere.
> >
> > Cover picture: shows 3 isomers of CO2 and the O2 molecule.
> >
> > Length: 1150 pages
> >
> >
> > Product details
> > • File Size : 2167 KB
> > • ASIN : B07PLVMMSZ
> > • Publication Date : March 11, 2019
> > • Word Wise : Enabled
> > • Print Length : 1150 pages
> > • Language: : English
> > • Text-to-Speech : Not enabled
> > • Enhanced Typesetting : Enabled
> > • X-Ray : Not Enabled
> > • Lending : Enabled
> > Amazon Best Sellers Rank: #590,212 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
> > #181 in General Chemistry & Reference
> > #1324 in General Chemistry
> > #1656 in Physics (Kindle Store)
> >
> >
> > Cambridge professors insane about Logic turns students like Pete Olcott insane also.
> >
> > Cambridge Physics Dept
> >
> > Ahnert, Alai, Alexander, Allison, Ansorge, Atature, Barker, Barnes, Bartlett, Batley, Baumberg, Bohndiek, Bowman, Brown, Buscher, Butler, Campbell Carilli, Carter, Castelnovo, Challis, Chalut, Chaudhri, Chin, Ciccarelli, Cicuta, Cole, Cooper, Cowburn, Credgington, Cross, Croze, Deschler, Donald, Duffett-Smith, Dutton, Eiser, Ellis, Euser, Field, Flynn, Ford, Friend, Gibson, Green, Greenham, Gripaios, Grosche, Guck, Gull, Haniff, Heavens-Ward, Heine, Hine, Hobson, Hope-Coles, Howie, Hughes, Irvine, Jardine, Jenkins, Jones, Josephson, Keyser, Khmeinitskii, King, Kotlyar, Lamacraft, Lasenby, Lester, Longair, Lonzarich, Maiolino, Marshall, Martin, Mitov, Morris, Mortimer, Moller, Needs, Norman, Nunnenkamp, Padman,Parker, Patel, Payne, Pepper, Phillips, Pramauro, Queloz, Rao, Richer, Riley, Ritchie, Sargent, Saunders, Saxena, Schneider, Scott, Scrivener, Sebastian, Simmons, Simons, Sirringhaus, Smith, Sutherland, Taylor, Teichmann, Terentjev, Thomson, Verrechia, Walker, Ward, Warner, Weale, Webber, Whyles, Withington.
> >
> > Cambridge Math Dept
> >
> > Alan Baker
> > Bela Bollobas
> > Darwin Smith
> > John Coates
> > Timothy Gowers
> > Peter Johnstone
> > Imre Leader
> > Gabriel Paternain
> >
> > Can any-one at Cambridge start correcting the error filled Boole, Jevons, Russell, Whitehead, Godel, Wittgenstein, all failures of logic and logical reasoning, include Cantor and his tripe of undefined infinity, an infinity without a borderline between finite and infinite.
> >
> > Cambridge, you no longer are a premiere University but a school that fosters and shelters losers of logical reasoning.
> 