On 6/25/21 11:07 PM, olcott wrote:
> On 6/25/2021 9:07 PM, Richard Damon wrote:
>> On 6/25/21 9:46 PM, olcott wrote:
>>> On 6/25/2021 8:37 PM, Richard Damon wrote:
>>>> On 6/25/21 9:01 PM, olcott wrote:
>>>>> On 6/25/2021 7:40 PM, Richard Damon wrote:
>>>>>> On 6/25/21 6:56 PM, olcott wrote:
>>>>>>> On 6/25/2021 4:59 PM, Richard Damon wrote:
>>>>>>>> On 6/25/21 4:40 PM, olcott wrote:
>>>>>>>>> On 6/25/2021 3:11 PM, Richard Damon wrote:
>>>>>>>>>> On 6/25/21 2:50 PM, olcott wrote:
>>>>>>>>>>> On 6/25/2021 12:14 PM, Richard Damon wrote:
>>>>>>>>>>>
>>>>>>>>>>>> WRONG. P is DEFINED based on H. If you Hypothetically create
>>>>>>>>>>>> a P
>>>>>>>>>>>> that
>>>>>>>>>>>> doesn't follow that form, then you are hypothetically creating
>>>>>>>>>>>> nonsense
>>>>>>>>>>>> and can't use it to for anything logical.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Of every possible encoding of simulating partial halt decider H
>>>>>>>>>>> that can
>>>>>>>>>>> possibly exist  H*, if H* never aborts the simulation of its
>>>>>>>>>>> input
>>>>>>>>>>> results in the infinite execution of the invocation of of P(P)
>>>>>>>>>>> then a
>>>>>>>>>>> simulating halt decider H that does abort its simulation of this
>>>>>>>>>>> input
>>>>>>>>>>> does correctly decide that this input does specify the never
>>>>>>>>>>> halting
>>>>>>>>>>> behavior of an infinite chain of invocations.
>>>>>>>>>>
>>>>>>>>>> Yes, if H* is an element of the set of non-aborting deciders
>>>>>>>>>> (Hn), P
>>>>>>>>>> will result in infinite recursion,
>>>>>>>>>
>>>>>>>>> Which logically entails beyond all possible doubt that the set of
>>>>>>>>> encodings of simulating partial halt deciders H2* that do abort
>>>>>>>>> the
>>>>>>>>> simulation of the (P,P) input would correctly report that this
>>>>>>>>> input
>>>>>>>>> never halts.
>>>>>>>>
>>>>>>>> WHY?
>>>>>>>>
>>>>>>>
>>>>>>> Axiom(1) Every computation that never halts unless its simulation is
>>>>>>> aborted is a computation that never halts. This verified as true on
>>>>>>> the
>>>>>>> basis of the meaning of its words.
>>>>>>
>>>>>> WRONG.
>>>>>>
>>>>>> Your test does not match the plain meaning of the words, as has been
>>>>>> explained many times.
>>>>>>
>>>>>
>>>>> Those words may be over your head, yet several others understand that
>>>>> they are necessarily correct.
>>>>
>>>> I have seen NO ONE agree to your interpretation of it. The plain
>>>> meaning
>>>> is that if it can be shown that if the given instance of the simulator
>>>> simulating a given input doesn't stop its simulation that this
>>>> simulation will run forevr, then the machine that is being simulated
>>>> can
>>>> be corrected decided as non-Halting.
>>>>
>>>> An more formal way to say that is if UTM(P,I) is non-halting then it is
>>>> correct for H(P,I) to return the non-halting result.
>>>>
>>>> This actually follows since UTM(P,I) will be non-halting if and only if
>>>> P(I) is non-halting by the definition of a UTM, so that statement is
>>>> trivially proven.
>>>>
>>>> Your interpretation, where even if a copy of the algorithm of H is
>>>> included in P and that included copy needs to abort the simulation of
>>>> the copy of the machine that it was given, can be PROVEN wrong, as even
>>>> you have shown that P(P) in this case does Halt, thus your claimed
>>>> correct answer is wrong by the definition of the problem.
>>>>
>>>> Only if you define that your answer isn't actually supposed to be the
>>>> answer to the halting problem can you justify your answer to be
>>>> correct,
>>>> but then you proof doesn't achieve the goal you claim.
>>>>
>>>>>
>>>>>> Note, it is easy to show that your interpretation is wrong since even
>>>>>> you admit that Linz H^, now called P by you will come to its end and
>>>>>> halt when given it own representation as its input, and thus is BY
>>>>>> DEFINITION a Halting Computation, Thus the H deciding it didn't
>>>>>> need to
>>>>>> abort its execution to get the wrong answer of Non-Halting.
>>>>>>
>>>>>
>>>>> Because at least one invocation of the infinite invocation chain
>>>>> specified by P(P) had to be terminated to prevent the infinite
>>>>> execution
>>>>> of this infinite invocation chain it is confirmed beyond all possible
>>>>> doubt that P(P) specifies an invocation chain.
>>>>
>>>> WRONG. Given that we have an H that can answer H(P,P) because it knows
>>>> at least enough to terminate the pattern you describe, then when we run
>>>> P(P) then because the H within it also knows to abort this sequence
>>>> (since it is built on the same algorithm) this P is NOT part of an
>>>> infinite chain of execution, and thus its H can return its (wrong)
>>>> answer to it and that P can then Halt.
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> P(P) specifies in infinite invocation sequence that is terminated on its
>>> third invocation of H(P,P).
>>>
>>> Now I have told this this several hundred times.
>>>
>>>
>>
>> WRONG.
>>
>> P(P) starts.
>>
>> Calls H(P,P)
>>
>> H starts the simulation.
>>
>> H simulates P starting
>>
>> H simulates P calling H
>>
>> H simulates H starting its simulation
>>
>> H simulates H simulating P starting
>>
>> H simulates H simulating P calling H
>>
>> The first H about here detects what it THINKS is an infinite execution
>>
>> THe first H aborts its simulation
>>
>> The first H returns its answer (Non-Halting) to its caller
>>
>> P then Halts
>>
>> Showing P is a Halting Computation.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>
> As you already admitted P ONLY halts because some H aborts some P
> otherwise P never ever halts.
>

Yes, P halts because the H it contains terminated the simulation of
another copy of its description.

YOUR problem is that you think that actually means something, it doesn't

Halting is DEFINED as the ultimate operation of the machine. Did it
finish or not. The fact that in its path to finishing it did a
simulation that it terminated doesn't affect that.

Your problem is that you have latched onto a misinterpretation of a run
that can be used to decide if you need to abort a simulation to answer
about it. (Key, it defines what is correctly terminating, not Halting).

Your mind has lost all contact with reality since the fact that you
admit that P(P) does come to its final halt somehow doesn't tell you
that P(P) is a computation that halts. Why is Halting not Halting?

Ultimately, I think the issue is that you don't really care about this
problem, only that it inconveniently proves something that you don't
want to accept as true, so you just assume it can't be. You are
incapable of actually dealing with the truth that truth itself is more
complicated than you want it to be.