On 8/1/21 9:56 PM, olcott wrote:
> On 8/1/2021 5:54 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 7/31/2021 5:08 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 7/30/2021 2:58 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 7/30/2021 7:39 AM, Ben Bacarisse wrote:
>>>>>>
>>>>>>>> Any chance you will now say if
>>>>>>>>
>>>>>>>>>      Ĥ.qx(⟨Ĥ⟩, ⟨Ĥ⟩)
>>>>>>>>
>>>>>>>> transitions to Ĥ.qn or Ĥ.qy?  If you find this question difficult,
>>>>>>>> please ask for some help in understanding it.
>>>>>>>
>>>>>>> Ĥ.qx(⟨Ĥ⟩, ⟨Ĥ⟩) transitions to Ĥ.qn
>>>>>>
>>>>>> An answer.  Thank you.
>>>>>>
>>>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>
>>>>>> For Ĥ to be "exactly and precisely as in Linz" this, then, is the
>>>>>> clause
>>>>>> that applies to your H and Ĥ:
>>>>>
>>>>> There is no H in the relevant last paragraph of the Linz proof that
>>>>> forms the basis for the Linz conclusion.
>>>> Distraction.  Everything you ignore below is about the proof and refers
>>>> only to Ĥ.
>>>>
>>>>>>>>>>>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>>>>>>>>>>>> if M applied to wM does not halt
>>>>>>
>>>>>> so Ĥ (M) applied to ⟨Ĥ⟩ (wM) does not halt, but you have just told me
>>>>>> that it does.  That is what this full (but abbreviated) state
>>>>>> transition
>>>>>> sequence means:
>>>>>>
>>>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>
>>>>>> Which is it?
>>>>>
>>>>> Ĥ0.q0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then Ĥ0.qx simulates Ĥ1 with the
>>>>> ⟨Ĥ2⟩ copy then
>>>>> Ĥ1.q0 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then Ĥ1.qx simulates Ĥ2 with the
>>>>> ⟨Ĥ3⟩ copy then
>>>>> Ĥ2.q0 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then Ĥ2.qx simulates Ĥ3 with the
>>>>> ⟨Ĥ4⟩ copy then ...
>>>> This is an abuse of the notation (but I know what you mean).  There is
>>>> no Ĥ1 or Ĥ2.  If you think it helps to show which copy of ⟨Ĥ⟩ your
>>>> simulating "decider" is either running and/or currently looking at, you
>>>> need to come up with a notation that does that.
>>>
>>> A better notation is what I have in my PDF actual subscripts but
>>> people here tel me that their newsreader makes sure to totally ignore
>>> posts with HTML so that do even see the post at all.
>>
>> I am happy you have a notation you like.  Are you prepared to address
>> that fact that your H^ is not "as in Linz"?
>>
>>>> At least I know what
>>>> this "math poem" means, because you've been saying this "it's a
>>>> simulator until" stuff for years.
>>>>
>>>>> The outermost Ĥ0.qx correctly decides that its input: (⟨Ĥ1⟩, ⟨Ĥ2⟩)
>>>>> can't possibly ever reach its final state. Then it transitions to
>>>>> Ĥ0.qn causing the outermost Ĥ0 to halt.
>>>> Apart from the bad notation, yes.  All those copies and tests and
>>>> eventual deciding are neatly summed up in the last ⊢* Ĥ.qn of
>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>>> Because the outermost Ĥ0.qx did not decide that Ĥ0 would never halt
>>>>> and it is self evident that its input: (⟨Ĥ1⟩, ⟨Ĥ2⟩) can't possibly
>>>>> ever reach its final state there is no contradiction or paradox and it
>>>>> decided correctly.
>>>> You are free to define "decide correctly" in any way you like provided
>>>> you are honest about it.  But you hooked people in by saying that
>>>> your Ĥ
>>>> is "exactly and precisely as in Linz", and you quoted, even now, what
>>>> Linz has to say about such TMs:
>>>>     Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>     if M applied to wM does not halt
>>>> This is your quote.  You brought it up.  You claimed your Ĥ was as Linz
>>>> states -- that Ĥ.q0 wM ⊢* Ĥ.qn if and only if M applied to wM does not
>>>> halt.  Linz makes no exceptions based on why the transitions from Ĥ.q0
>>>> wM to Ĥ.qn occur.  Linz does not say
>>>>     Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>     if M applied to wM does not halt or if M applied wM only halts
>>>>     because...
>>>> Are you now saying that your TM was not "as in Linz"?  (You should,
>>>> because you've admitted that elsewhere.)
>>>
>>> Ĥ[0] is to be interpreted to mean Ĥ₀
>>>   [0] Means the actual Turing machine and not a TM description.
>>>   [1] Means the first TM description parameter
>>>   [2] Means a copy of the the first TM description parameter
>>>
>>> Now I am saying that when the actual unmodified Linz Ĥ is understood
>>> to have a UTM/Halt-Decider at Ĥ[0].qx that this Ĥ[0].qx does correctly
>>> decide that its input: (⟨Ĥ[1]⟩, ⟨Ĥ[2]⟩) can't possibly ever reach its
>>> final state of Ĥ[1].qn or Ĥ[2].qn, therefore we know that its input
>>> never halts therefore we know that a state transition from Ĥ[0].qx to
>>> Ĥ0.qn is necessarily correct.
>>
>> We all know you are declaring that to be correct.  Here's why your Ĥ is
>> not "as in Linz".  Linz requires that
>>
>>    Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>    if M applied to wM does not halt
>>
>> Any Ĥ that eventually transitions to Ĥ.qn on input wM must do so
>> if, and only if, the encoded M applied to wM does not halt.  But you've
>> given us a case where your Ĥ is not like this:
>>
>>    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> *This did not show up in my other news server so I am pointing it again*
>
> And when we eliminate the fallacy of equivocation error we have
>
> Ĥ[0].q0 ⟨Ĥ⟩ ⊢* Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ ⊢* Ĥ[0].qn
>
> The way that you do it when your twin bother commits a crime that makes
> you guilty.

Except for Turing Machines, you do every thing your 'twin brother' does,
so if he commits a crime, so did you, so you ARE guilty.

>
> Ĥ[0].q0 ⟨Ĥ[1]⟩ only halts because the simulation of the
> the input to Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩ was aborted.
>
> (a) The simulation of the input to Ĥ[0].qx ⟨Ĥ[1]⟩ ⟨Ĥ[2]⟩
> was aborted because neither ⟨Ĥ[1]⟩ nor ⟨Ĥ[2]⟩ could
> ever possibly reach their final states.
>
> (b) Because neither ⟨Ĥ[1]⟩ nor ⟨Ĥ[2]⟩ could ever possibly reach
> their final states we know that they never halt.
>
> (c) Because they never halt we know that Ĥ[0].qx correctly decided
> that its input never halts.
>
> This is the same as: X > Y & Y > Z therefore X > Z
> I do seem to have a correct chain of inference.
> I do not think that you can point to any error in
> this chain of inference.
>
> In an honest dialogue when you would disagree that a
> chain of inference derives a correct conclusion you
> would point to a specific error in this chain of inference.
>
> What is the error in (a)(b)(c) ?

Because that isn't the definition of Halting. The definition is what the
ACTUAL machine does, and by the definition of a UTM what the simulaion
by a REAL UTM would do. A simulator that aborts its simulation is NOT a
UTM -- PERIOD.

The fact that H^[1] is aborted before it reaches the state that H^[0] is
KNOWN AND ADMITTED to reach, we know that H^[1] WILL do exactly the same
thing if allowed to continue.

The fact that H make the error and aborts its simulation too soon
doesn't change this fact.

THe fact that UTM((H^[1]), (H^[2])) will halts (since this IS the
identical computation to H^[0]((H^[1]) since all (H^[i]) are identical.

>
>
>> Here we can see that Ĥ applied to ⟨Ĥ⟩ halts.  You can call your Ĥ's
>> behaviour "correct".  You can call it anything you like.  But it's not
>> "as in Linz".  It does not say anything about Linz's proof.  It does not
>> do anything people would call impossible or even interesting.
>>
>> Presumably, you will simply explain, yet again, why you choose to call
>> it correct.  You might even, yet again, quote the symbols from Linz that
>> don't apply to your Ĥ in order to make you posts seem relevant.
>>
>
>