On 12/18/2021 3:04 PM, André G. Isaak wrote:
> On 2021-12-18 13:44, olcott wrote:
>> On 12/18/2021 2:07 PM, Richard Damon wrote:
>>>
>>> On 12/18/21 2:28 PM, olcott wrote:
>>>> On 12/18/2021 5:06 AM, Richard Damon wrote:
>>>>> On 12/17/21 11:29 PM, olcott wrote:
>>>>>> On 12/17/2021 10:18 PM, Richard Damon wrote:
>>>>>>> On 12/17/21 10:47 PM, olcott wrote:
>>>>>>>> On 12/17/2021 9:34 PM, Richard Damon wrote:
>>>>>>>>> On 12/17/21 9:59 PM, olcott wrote:
>>>>>>>>>> On 12/17/2021 8:44 PM, Richard Damon wrote:
>>>>>>>>>>> On 12/17/21 9:26 PM, olcott wrote:
>>>>>>>>>>>> On 12/17/2021 7:50 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 12/17/21 7:54 PM, olcott wrote:
>>>>>>>>>>>>>> On 12/17/2021 6:43 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On 12/17/2021 3:25 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> A function f with domain D is said to be
>>>>>>>>>>>>>>>>>> Turing-computable
>>>>>>>>>>>>>>>>>> or just computable if there exists some Turing machine
>>>>>>>>>>>>>>>>>> M = (Q, Σ, Γ, δ, q0, □, F) such that q0 w ⊢* Mqff(w),
>>>>>>>>>>>>>>>>>> qf ∈ F
>>>>>>>>>>>>>>>>>> for all w ∈ D (Linz:1990:243)
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> We all know your cut and paste skills are top notch.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Olcott paraphrase of above machine definition: Machine
>>>>>>>>>>>>>>>>>> M begins at
>>>>>>>>>>>>>>>>>> start state q0 on input w and transitions to qf as a
>>>>>>>>>>>>>>>>>> function of input
>>>>>>>>>>>>>>>>>> w.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> But that's not a correct paraphrase.  As for the jumble
>>>>>>>>>>>>>>>>> of words that is
>>>>>>>>>>>>>>>>> "the computable function copy of H at Ĥ.qx", well, I
>>>>>>>>>>>>>>>>> despair.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I take this to mean that you have no idea what the
>>>>>>>>>>>>>>>> behavior of Ĥ
>>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ would be:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You should, in general, take me to mean what I say, with
>>>>>>>>>>>>>>> as little
>>>>>>>>>>>>>>> interpretation as you can muster.  Your track record in
>>>>>>>>>>>>>>> paraphrasing me
>>>>>>>>>>>>>>> is abysmal.  Is my claim that I despair at your jumble of
>>>>>>>>>>>>>>> words in some
>>>>>>>>>>>>>>> way unclear?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Since this is equivalent to Linz notation you can't say
>>>>>>>>>>>>>> that my question is not clear?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy
>>>>>>>>>>>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> What would the behavior of Ĥ applied to ⟨Ĥ⟩ be if we
>>>>>>>>>>>>>> assume that Ĥ.qx acts exactly as if it was UTM(⟨Ĥ⟩, ⟨Ĥ⟩) ???
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> So, if you want to DEFINE that you H IS the same as UTM,
>>>>>>>>>>>>> then you have shown that H(<H^>,<H^>) nevers returns an
>>>>>>>>>>>>> answer, so while H^(<H^>) is FOR THIS H, non-halting, it
>>>>>>>>>>>>> doesn't matter as this H never answers, and so it wrong
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Yes that it correct. I don't think that Ben understands
>>>>>>>>>>>> these things well enough to understand that.
>>>>>>>>>>>>
>>>>>>>>>>>>> Hypotheticals which change the contents of H^
>>>>>>>>>>>>
>>>>>>>>>>>> All hypotheticals are imaginary and have no effect on the
>>>>>>>>>>>> actual world.
>>>>>>>>>>>>
>>>>>>>>>>>>> (which include its copy of H) don't say anything about the
>>>>>>>>>>>>> original H^ that was built from a different H.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ // infinite loop added
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>
>>>>>>>>>>>> When we put the infinite loop back in Ĥ applied to ⟨Ĥ⟩ still
>>>>>>>>>>>> never stops running because it still specified infinitely
>>>>>>>>>>>> nested simulation. It never reaches the infinite loop.
>>>>>>>>>>>
>>>>>>>>>>> So, which H do you want to try to claim to be your H, that is
>>>>>>>>>>> supposedly correct?
>>>>>>>>>>>
>>>>>>>>>>> The one that gets stuck in the infinite loop, and thus
>>>>>>>>>>> doesn't answer, or the one that DOES abort the simulation,
>>>>>>>>>>> and thus return the answer to H^ which lets H^ be halting,
>>>>>>>>>>> and thus H was wrong?
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>> The fact that for the case that H is identical to UTM gives
>>>>>>>>>>>>> a non-halting H^ doesn't provide ANY evidence that H^ built
>>>>>>>>>>>>> from an H that answers is non-halting.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> When we understand that the hypothetical pure simulation of
>>>>>>>>>>>> the input to any simulating halt decider does provide the
>>>>>>>>>>>> correct halt status criterion measure for every input then
>>>>>>>>>>>> we understand that this is correct: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>
>>>>>>>>>>> No, because H isn't supposed to answer about the
>>>>>>>>>>> 'Hypothetcial' input based on a different H than the one that
>>>>>>>>>>> H^ is actually built on, but the actual input that it was
>>>>>>>>>>> given which is an H^ built on the H that you claim to be the
>>>>>>>>>>> correct answering one.
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Any input that must have its simulation aborted to prevent
>>>>>>>>>>>> the infinite execution of this input is an input that never
>>>>>>>>>>>> halts.
>>>>>>>>>>>
>>>>>>>>>>> No. Pure LIE.
>>>>>>>>>>>
>>>>>>>>>>> Halting is based on what the computation ACTUALLY DOES.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> No one ever thought this aspect through before. Linz rejects
>>>>>>>>>> simulation before ever fully examining it.
>>>>>>>>>
>>>>>>>>> No, he doesn't, he doesn't distinguish at all about how the
>>>>>>>>> halt decider works, as it just doesn't matter. All that matters
>>>>>>>>> is that H needs to be a DEFINED machine, and he looks at the
>>>>>>>>> answer it gives, by how ever it does it.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Every input that never halts when its pure simulation by H
>>>>>>>>>> never halts is an input that never halts.
>>>>>>>>>>
>>>>>>>>>> The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach its final
>>>>>>>>>> state when 0 to ∞ steps of this input are simulated by H.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> But it does when given to a real UTM. UTM applied to <H^> <H^>
>>>>>>>>> will halt if H applied to <H^> <H^> goes to H.qn.
>>>>>>>>>
>>>>>>>>> You need to FIRST define your H, and once you do you are NOT
>>>>>>>>> allowed to change it in determining what H^ does.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>> Your definition is just how you seem to befine your POOP.
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> The meaning of those words proves that those words are correct.
>>>>>>>>>>>
>>>>>>>>>>> LIE.
>>>>>>>>>>>
>>>>>>>>>>> The meaning of Halting is that the thing ACTUALLY being
>>>>>>>>>>> decided on reaches a final Halting state.
>>>>>>>>>>>
>>>>>>>>>>> Since the ACUTAL H^ applied to <H^> does reach its final
>>>>>>>>>>> haltibng state,
>>>>>>>>>>
>>>>>>>>>> Because you know that TM's cannot have TM's for inputs it
>>>>>>>>>> seems quite stupid that you say that Ĥ.qx is applied to Ĥ.
>>>>>>>>>
>>>>>>>>> Who said H^.qx is applied to H^. WHERE DID I SAY THAT?
>>>>>>>>>
>>>>>>>> When everyone says
>>>>>>>>
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>>>>
>>>>>>>> Instead of saying
>>>>>>>> if simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>>>>>>>>
>>>>>>>> They are saying that Ĥ.qx is being applied to a TM and not a TM
>>>>>>>> description.
>>>>>>>
>>>>>>> No, they are appling the definition of a CORRECT Halt Decider.
>>>>>>>
>>>>>>>>
>>>>>>>> Ĥ.qx is NOT applied to Ĥ
>>>>>>>> Ĥ.qx is applied to ⟨Ĥ⟩
>>>>>>>
>>>>>>> Right, but if H is a correcg halt decider (as is your claim) than
>>>>>>> H (and just H^.qx) must go to qy if H^ applied to <H^> halts and
>>>>>>> to qn if it doesn't.
>>>>>>>
>>>>>> That is under the false assumption that the behavior of the input
>>>>>> is the same when executed inside the halt decider as when it is
>>>>>> executed outside the halt decider.
>>>>>>
>>>>>
>>>>> But if they are different, that says that either the decider is NOT
>>>>> being an accurate simulator or the input is not a representation of
>>>>> an actual Algorithm.
>>>>>
>>>>
>>>> When the input to H is intentionally defined to have different
>>>> behavior inside of H than outside of H then this behavior will
>>>> necessarily vary.
>>>
>>> But H^ DOESN'T have different behavior inside of H than outside, and
>>> thus neither does its representation.
>>>
>>> H^ ALWAYS does the opposite of what the copy of H says it will do,
>>> whether it is being decided by H or not.
>>>
>>
>> The actual behavior of UTM(⟨Ĥ⟩ ⟨Ĥ⟩) of the actual input to to Ĥ.qx is
>> the only thing that this computable function is accountable for.
>>
>> Actual computer scientists would know this.
>
> Actual computer scientists would know that the claim that there is a
> computable function at Ĥ.qx is incoherent nonsense.
>
> André
>