On 2/1/22 10:22 AM, 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.
>

No, it is NOT 'Vague', a machine will EITHER stop running because it
will reach a final state, or it can NEVER reach such a state.

Please show a machine that doesn't reach its final state but also
doesn't run forever?

You seem to think that it is possible for a machine to be in some middle
state.

Please provide an example of such a machine.

Note, the definition is stated the way it is because a simulator that
aborts its simulation does NOT indicate either of the cases and does not
provide evidence of the Halting state of a computation.

> The halting problem does not bother to mention the requirement that
> because all halt deciders are deciders they are only accountable for
> computing the mapping from their finite string inputs to an accept or
> reject state on the basis of the actual behavior specified by this input.

But if they do not compute the mapping per the definition, they are NOT
'Halt Deciders', that is your problem, what you are doing is trying to
define a POOP decider can call it a Halt Decider.

You are not ALLOWED to change the definiton of Halting, when you try, it
just means you logic is unsound and doesn't prove anything, because it
si based on a false premise.

PERIOD.

>
> The halting problem does not specifically examine simulating halt
> deciders, none-the-less the behavior of a correctly simulated machine
> description is known to be equivalent to the behavior of the direct
> execution of this same machine.

Right, NON-ABORTED simuluation, and UNSTOPPED execution. So an H that
aborts or 'debug steps' does NOT prove hon-halting.

PERIOD.

FAIL.

>
> Since a simulating halt decider is merely a UTM for simulated inputs
> that reach their final state when a simulating halt decider correctly
> determines that its simulated its input cannot possibly reach its final
> state this is complete proof that this simulated input never halts.
>

No, it is NOT a UTM if it aborts its simulation for ANY reason other
than the machine reached a final statee.

Even if it correctly predicts that its input is non-halting, a UTM does
not abort its simulation, as BY DEFINITION, it behaves the same as the
machine it is simulating, so if that is non-halting, then the UTM must
be too.

PERIOD.

DEFINITIOPN.

FAIL.