On 2/11/24 11:31 PM, olcott wrote:
> On 2/11/2024 9:24 PM, Richard Damon wrote:
>> On 2/11/24 9:56 PM, olcott wrote:
>>> On 2/11/2024 8:24 PM, Richard Damon wrote:
>>>> On 2/11/24 8:48 PM, olcott wrote:
>>>>> On 2/11/2024 7:37 PM, Richard Damon wrote:
>>>>>> On 2/11/24 8:15 PM, olcott wrote:
>>>>>>> On 2/11/2024 6:33 PM, Richard Damon wrote:
>>>>>>>> On 2/11/24 7:11 PM, olcott wrote:
>>>>>>>>> On 2/11/2024 5:25 PM, Richard Damon wrote:
>>>>>>>>>> On 2/11/24 6:02 PM, olcott wrote:
>>>>>>>>>>> On 2/11/2024 3:58 PM, Richard Damon wrote:
>>>>>>>>>>>> On 2/11/24 4:26 PM, olcott wrote:
>>>>>>>>>>>>> On 2/11/2024 12:59 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 2/11/24 1:49 PM, olcott wrote:
>>>>>>>>>>>>>>> On 2/11/2024 12:28 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>> On 2/11/24 10:36 AM, olcott wrote:
>>>>>>>>>>>>>>>>> On 2/11/2024 9:17 AM, Mike Terry wrote:
>>>>>>>>>>>>>>>>>> On 11/02/2024 03:14, immibis wrote:
>>>>>>>>>>>>>>>>>>> On 11/02/24 02:42, Mike Terry wrote:
>>>>>>>>>>>>>>>>>>>> Sometimes he changes the wording, thinking if he
>>>>>>>>>>>>>>>>>>>> does that enough, people will suddenly agree with
>>>>>>>>>>>>>>>>>>>> him. That doesn't mean he agrees his previously
>>>>>>>>>>>>>>>>>>>> worded claims were incorrect!
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> If PO sees someone respond with a load of guff (in
>>>>>>>>>>>>>>>>>>>> PO's opinion) which he doesn't really get, he can't
>>>>>>>>>>>>>>>>>>>> leave that unanswered, since that might suggest to
>>>>>>>>>>>>>>>>>>>> other readers that he's lost some argument.  So he
>>>>>>>>>>>>>>>>>>>> ignores the (in PO's opinion) extraneous complexity
>>>>>>>>>>>>>>>>>>>> and goes back to just posting his core intuition.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> I thought tangent of "D(D) doesn't halt even though
>>>>>>>>>>>>>>>>>>> it seems to halt" was quite amusing. When people were
>>>>>>>>>>>>>>>>>>> pointing out that D(D) does halt (evidence: just run
>>>>>>>>>>>>>>>>>>> it and see) Olcott responded that it does not halt.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Message-ID: <uorkac$1tiu7$1@dont-email.me>
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> "You see, D(D) halts, but it only halts because it
>>>>>>>>>>>>>>>>>>> thinks it does not halt, and if a computation only
>>>>>>>>>>>>>>>>>>> halts because it thinks it does not halt, then it
>>>>>>>>>>>>>>>>>>> DOES NOT COUNT AS HALTING and therefore it was correct!"
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Or something like that.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Yes it's bizarre.  A reader simply wishing to see that
>>>>>>>>>>>>>>>>>> PO's claim is worthless need look no further.  It's
>>>>>>>>>>>>>>>>>> only someone who wants to understand /where/ PO is
>>>>>>>>>>>>>>>>>> going wrong, or to /correct his thinking/ that will be
>>>>>>>>>>>>>>>>>> motivated to dive into all the mucky details...
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> *My 2004 intuitions have proved to be correct and*
>>>>>>>>>>>>>>>>> *two PhD computer science professors elaborate them*
>>>>>>>>>>>>>>>>> *better than I have until now*
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> *The Halting Paradox* Bill Stoddart (2017)
>>>>>>>>>>>>>>>>> https://arxiv.org/abs/1906.05340
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> *Objective and Subjective Specifications* Eric C.R.
>>>>>>>>>>>>>>>>> Hehner (2017)
>>>>>>>>>>>>>>>>> https://www.cs.toronto.edu/~hehner/OSS.pdf
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> *Problems with the Halting Problem* Eric C.R. Hehner
>>>>>>>>>>>>>>>>> (2011)
>>>>>>>>>>>>>>>>> https://www.cs.toronto.edu/~hehner/PHP.pdf
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Alan Turing's Halting Problem is incorrectly formed
>>>>>>>>>>>>>>>>> (PART-TWO) sci.logic
>>>>>>>>>>>>>>>>> *On 6/20/2004 11:31 AM, Peter Olcott wrote*
>>>>>>>>>>>>>>>>>  > PREMISES:
>>>>>>>>>>>>>>>>>  > (1) The Halting Problem was specified in such a way
>>>>>>>>>>>>>>>>> that a solution
>>>>>>>>>>>>>>>>>  > was defined to be impossible.
>>>>>>>>>>>>>>>>>  >
>>>>>>>>>>>>>>>>>  > (2) The set of questions that are defined to not
>>>>>>>>>>>>>>>>> have any possible
>>>>>>>>>>>>>>>>>  > correct answer(s) forms a proper subset of all
>>>>>>>>>>>>>>>>> possible questions.
>>>>>>>>>>>>>>>>>  > …
>>>>>>>>>>>>>>>>>  > CONCLUSION:
>>>>>>>>>>>>>>>>>  > Therefore the Halting Problem is an ill-formed
>>>>>>>>>>>>>>>>> question.
>>>>>>>>>>>>>>>>>  >
>>>>>>>>>>>>>>>>> USENET Message-ID:
>>>>>>>>>>>>>>>>> <kZiBc.103407$Gx4.18142@bgtnsc04-news.ops.worldnet.att.net>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Ĥ applied to ⟨Ĥ⟩ asks: (Olcott 2024)
>>>>>>>>>>>>>>>>> Do you halt on your own Turing Machine Description?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> This is isomorphic to asking whether the Liar Paradox
>>>>>>>>>>>>>>>>> is true or false.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Nope.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Proven otherwise
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Your statement are still just lies.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> That is libelous.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> No, it is a TRUE statement.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> *to say something that is not true in order to deceive*
>>>>>>>>>>>>>>> https://dictionary.cambridge.org/us/dictionary/english/lie
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Which perfectly describes yourself.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> That you call my statement an intentional falsehood
>>>>>>>>>>>>>>> shows a reckless disregard for the truth
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Nope, it is a statement of TRUTH.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> *When you cannot even show that it is a falsehood*
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> That you cannot even show exactly how template Ĥ does not
>>>>>>>>>>>>>>> contradict both Boolean values that embedded_H returns
>>>>>>>>>>>>>>> proves
>>>>>>>>>>>>>>> that you cannot even show that my statement is false.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The falsehood is that you claim to be working on the
>>>>>>>>>>>>>> Halting Problem, but you are not, since you don't use any
>>>>>>>>>>>>>> of the actual requirements of the halting problem.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Your H is not A COMPUTATION.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> your input D is not A REPRESENTATION OF A COMPUTATION.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Thus, your claims are all LIES.
>>>>>>>>>>>>>
>>>>>>>>>>>>> You can't even stay on topic. I never mention H in this
>>>>>>>>>>>>> whole thread. That you are responding to things that I
>>>>>>>>>>>>> didn't even say is a reckless disregard for the truth.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Really?
>>>>>>>>>>>>
>>>>>>>>>>>> Lying again.
>>>>>>>>>>>>
>>>>>>>>>>>> Look up at the quotes:
>>>>>>>>>>>>
>>>>>>>>>>>>  >>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>>>>>>>>>>  >>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>  >>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> The fact that you don't even check what you are talking
>>>>>>>>>>>> about shows your stupidity.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> That IS NOT H IT IS CALLED H-HAT PAY ATTENTION
>>>>>>>>>>> That IS NOT H IT IS CALLED H-HAT PAY ATTENTION
>>>>>>>>>>> That IS NOT H IT IS CALLED H-HAT PAY ATTENTION
>>>>>>>>>>> That IS NOT H IT IS CALLED H-HAT PAY ATTENTION
>>>>>>>>>>
>>>>>>>>>> But Ĥ uses its copy of H that you call embedded_H.
>>>>>>>>>>
>>>>>>>>>> So, H is there.
>>>>>>>>>
>>>>>>>>> No H is not there you should see the screwing thing
>>>>>>>>> that Linz calls embedded_H: Ĥq0 (a second start state)
>>>>>>>>>
>>>>>>>>> Top of page 3
>>>>>>>>> https://www.liarparadox.org/Linz_Proof.pdf
>>>>>>>>>
>>>>>>>>
>>>>>>>> So, you don't understand that
>>>>>>>
>>>>>>> I understand that when Linz is all done with all of the
>>>>>>> transformations and actually specifies the completed Ĥ
>>>>>>> that he calls embedded_H *Ĥq0* on the top of page 3.
>>>>>>>
>>>>>>
>>>>>> Nope, you THINK you understand that, but you don't actually
>>>>>> understand what he is doing.
>>>>>>
>>>>>> He has no "embedded_H", and that Ĥq0 is the name of of the state
>>>>>> in Ĥ that is the beginning state for H, not ALL the states of H.
>>>>>
>>>>> Figure 12.2 shows that he merely appends the infinite loop to qy
>>>>> thus calling his Ĥq0 (second start state) embedded_H is perfectly
>>>>> accurate.
>>>>>
>>>>
>>>> No, because H was the states q0, qy, and qn and all the "omited"
>>>> details represented by the squiggly lines. The whole of Figure 12.1
>>>>
>>>> Figure 12.2 shows the changes going from H to H' (the adding of the
>>>> infinite loop to qy)
>>>>
>>>> Calling ONE state of a full machine as "The Machine" is just an error.
>>>>
>>>> You are just showing your total ignorance.
>>>
>>> embedded_H refers to all of the states specified in his H template.
>>> two of these states have merely been renamed indicating that they
>>> have been integrated into Ĥ because H has been embedded in Ĥ.
>>
>> Then why did you say it was just Ĥq0 that represented embedded_H?
>>
>>>
>>> His two start states are wrong as Ben agreed to years ago.
>>> No TM ever has two q0 states.
>>
>> No, there is q0 and Ĥq0
>>
>> Admittedly, his nomenclature is a bit awkward.
>>
>> The execution of the Machine Ĥ starts at the state q0 (as he defines
>> q0 to be the starting state of all machines) with the expresion
>>
>> q0 Wm
>>
>> Which means at state q0, with a tape contents of Wm
>>
>> It then goes through the steps to duplicate the tape and ends up at
>> state Ĥq0 which is the Ĥ machine equivalent of H's q0, at an expression
>>
>> Ĥq0 Wm Wm
>>
>> Meaning at state Ĥq0 with tape contents Wm Wm
>>
>> and then (depending on the decision that H makes) to either
>>
>> Ĥ∞
>>
>
> All of the above is correct.
>
> *This improves on his notation making is more clear*
> q0 ⟨M⟩ ⊢* Ĥq0  ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qy ∞  // YES for M applied to ⟨M⟩ halts
Which is for H saying that M applied to (M) Halts, which if M is Ĥ it
doesn't

> q0 ⟨M⟩ ⊢* Ĥq0  ⟨M⟩ ⟨M⟩ ⊢* Ĥ.qn    // NO for M applied to ⟨M⟩ does not halt
Which is for H sayng that M applied to (M) Halts, which if M is Ĥ it
does Halt.

So, which every H you happen to try to use, it is wrong (and the two H's
got DIFFERENT input, so that doesn't say that either question didn't
have a correct answer, just not one that H gave.

So, this proves that NO H will correctly answer the halting question for
the input representing the specific computation built from the template
applied to that particular H.

>
>> indicating that Ĥ doesn't halt or
>>
>> Ĥ y0 qn y1
>>
>> indicating it goes to state Ĥqn with the tape possible having symbols
>> y0 before the head, and y1 after the head.
>>
>>>
>>> My name embedded_H is much easier to understand than his
>>> second q0 state.
>>>
>>
>>
>