On 3/9/2024 12:17 PM, Richard Damon wrote:
> On 3/9/24 8:53 AM, olcott wrote:
>> On 3/9/2024 10:21 AM, immibis wrote:
>>> On 9/03/24 17:00, olcott wrote:
>>>> On 3/9/2024 1:05 AM, Richard Damon wrote:
>>>>> On 3/8/24 10:40 PM, olcott wrote:
>>>>>> On 3/9/2024 12:25 AM, Richard Damon wrote:
>>>>>>> On 3/8/24 9:55 PM, olcott wrote:
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does
>>>>>>>> not halt
>>>>>>>>
>>>>>>>> Expecting Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to correctly report on the behavior of
>>>>>>>> Ĥ ⟨Ĥ⟩ is a little nuts because Ĥ contradicts Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>>
>>>>>>> But that is the job it signed up for when it tried to call itself
>>>>>>> a Halt Decider.
>>>>>>>
>>>>>>
>>>>>> There is no correct answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly
>>>>>> provide that corresponds to the behavior of this Ĥ ⟨Ĥ⟩.
>>>>>> Therefore it must have a basis for its wrong answer.
>>>>>>
>>>>>>
>>>>>
>>>>> WHY?
>>>>>
>>>>> What is the actual grounds for that statement?
>>>>
>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot just sit around scratching its head it
>>>> must do something. As soon as we have the basis for what
>>>> it will do, then H ⟨Ĥ⟩ ⟨Ĥ⟩ has its basis to correctly
>>>> decide halting.
>>>
>>> H ⟨A⟩ ⟨B⟩ cannot just sit around scratching its head, it must do
>>> something. As soon as we have the basis for what H will do, then we
>>> have a basis to construct an Ĥ where H ⟨Ĥ⟩ ⟨Ĥ⟩ is wrong.
>>>
>>
>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can fool itself yet cannot possibly fool an external H ⟨Ĥ⟩
>> ⟨Ĥ⟩.
>>
>
> Why not?
>
> Are you thinking H will just simulate until H^.H makes its decision?
>
> We know the outcome of that, never answering, and thus failing.

Please switch over to this new post that addresses all of your
objections in advance.
[Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior]

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/9/24 10:36 AM, olcott wrote:
> On 3/9/2024 12:31 PM, Richard Damon wrote:
>> On 3/9/24 9:27 AM, olcott wrote:
>>> On 3/9/2024 11:08 AM, Richard Damon wrote:
>>>> On 3/9/24 7:12 AM, olcott wrote:
>>>>> On 3/9/2024 7:47 AM, immibis wrote:
>>>>>> On 8/03/24 22:34, olcott wrote:
>>>>>>>> And since H^ can "lie" to that embedded H^.H about what its
>>>>>>>> description is, that H can't tell that it is part of an H^
>>>>>>>> computation that is simulating an H^ computation.
>>>>>>>
>>>>>>> That subject must be postponed until after the Olcott refutation
>>>>>>> of the exact Linz proof is either fully accepted by three people
>>>>>>> or actual errors or gaps are found that cannot be addressed or
>>>>>>> corrected.
>>>>>>
>>>>>> It's accepted that the Linz proof doen't work on Olcott machines
>>>>>> because the Linz proof is designed for Turing machines. But you
>>>>>> can't refute the Linz-immibis proof designed for Olcott machines,
>>>>>> where H is lied to about its own description.
>>>>>
>>>>> I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except halt or fail to
>>>>> halt
>>>>> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> could see that.
>>>>>
>>>>
>>>> Because if H simulates "long enough" to see the answer, that "long
>>>> enough" will turn out to be FOREVER, since H^.H will do the same
>>>> thing (BY DEFINITION)
>>>
>>> It is a verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the simulation of
>>> its input and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort the simulation of its input
>>> and verified facts supersede all theories to the contrary.
>>>
>>>
>>
>> Nope.
>
> Please switch over to this new post that addresses all of your
> objections in advance.
> [Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior]
>

Then why did you post here this morning?

On 3/9/2024 12:50 PM, Richard Damon wrote:
> On 3/9/24 10:36 AM, olcott wrote:
>> On 3/9/2024 12:31 PM, Richard Damon wrote:
>>> On 3/9/24 9:27 AM, olcott wrote:
>>>> On 3/9/2024 11:08 AM, Richard Damon wrote:
>>>>> On 3/9/24 7:12 AM, olcott wrote:
>>>>>> On 3/9/2024 7:47 AM, immibis wrote:
>>>>>>> On 8/03/24 22:34, olcott wrote:
>>>>>>>>> And since H^ can "lie" to that embedded H^.H about what its
>>>>>>>>> description is, that H can't tell that it is part of an H^
>>>>>>>>> computation that is simulating an H^ computation.
>>>>>>>>
>>>>>>>> That subject must be postponed until after the Olcott refutation
>>>>>>>> of the exact Linz proof is either fully accepted by three people
>>>>>>>> or actual errors or gaps are found that cannot be addressed or
>>>>>>>> corrected.
>>>>>>>
>>>>>>> It's accepted that the Linz proof doen't work on Olcott machines
>>>>>>> because the Linz proof is designed for Turing machines. But you
>>>>>>> can't refute the Linz-immibis proof designed for Olcott machines,
>>>>>>> where H is lied to about its own description.
>>>>>>
>>>>>> I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except halt or fail to
>>>>>> halt
>>>>>> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> could see that.
>>>>>>
>>>>>
>>>>> Because if H simulates "long enough" to see the answer, that "long
>>>>> enough" will turn out to be FOREVER, since H^.H will do the same
>>>>> thing (BY DEFINITION)
>>>>
>>>> It is a verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the simulation of
>>>> its input and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort the simulation of its input
>>>> and verified facts supersede all theories to the contrary.
>>>>
>>>>
>>>
>>> Nope.
>>
>> Please switch over to this new post that addresses all of your
>> objections in advance.
>> [Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior]
>>
>
> Then why did you post here this morning?

After reviewing all of the posts I came up with
better words that make my point more clearly.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 9/03/24 17:41, olcott wrote:
> On 3/9/2024 9:36 AM, immibis wrote:
>> On 9/03/24 16:12, olcott wrote:
>>> On 3/9/2024 7:47 AM, immibis wrote:
>>>> On 8/03/24 22:34, olcott wrote:
>>>>>> And since H^ can "lie" to that embedded H^.H about what its
>>>>>> description is, that H can't tell that it is part of an H^
>>>>>> computation that is simulating an H^ computation.
>>>>>
>>>>> That subject must be postponed until after the Olcott refutation
>>>>> of the exact Linz proof is either fully accepted by three people
>>>>> or actual errors or gaps are found that cannot be addressed or
>>>>> corrected.
>>>>
>>>> It's accepted that the Linz proof doen't work on Olcott machines
>>>> because the Linz proof is designed for Turing machines. But you
>>>> can't refute the Linz-immibis proof designed for Olcott machines,
>>>> where H is lied to about its own description.
>>>
>>> I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except halt or fail to halt
>>> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> could see that.
>>>
>>
>> I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except for exactly the
>> same thing that H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do.
>
> It is easily proven that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the simulation of
> its input and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort the simulation of its input.

It is easily proven that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> does exactly the same thing
that H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> does, but you continue to dishonestly ignore this
verified fact because you are a liar.

On 9/03/24 17:53, olcott wrote:
> On 3/9/2024 10:21 AM, immibis wrote:
>> On 9/03/24 17:00, olcott wrote:
>>> On 3/9/2024 1:05 AM, Richard Damon wrote:
>>>> On 3/8/24 10:40 PM, olcott wrote:
>>>>> On 3/9/2024 12:25 AM, Richard Damon wrote:
>>>>>> On 3/8/24 9:55 PM, olcott wrote:
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not
>>>>>>> halt
>>>>>>>
>>>>>>> Expecting Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to correctly report on the behavior of
>>>>>>> Ĥ ⟨Ĥ⟩ is a little nuts because Ĥ contradicts Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>
>>>>>> But that is the job it signed up for when it tried to call itself
>>>>>> a Halt Decider.
>>>>>>
>>>>>
>>>>> There is no correct answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly
>>>>> provide that corresponds to the behavior of this Ĥ ⟨Ĥ⟩.
>>>>> Therefore it must have a basis for its wrong answer.
>>>>>
>>>>>
>>>>
>>>> WHY?
>>>>
>>>> What is the actual grounds for that statement?
>>>
>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot just sit around scratching its head it
>>> must do something. As soon as we have the basis for what
>>> it will do, then H ⟨Ĥ⟩ ⟨Ĥ⟩ has its basis to correctly
>>> decide halting.
>>
>> H ⟨A⟩ ⟨B⟩ cannot just sit around scratching its head, it must do
>> something. As soon as we have the basis for what H will do, then we
>> have a basis to construct an Ĥ where H ⟨Ĥ⟩ ⟨Ĥ⟩ is wrong.
>>
>
> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can fool itself yet cannot possibly fool an external H ⟨Ĥ⟩ ⟨Ĥ⟩.
>

If H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to state qn then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ also transitions
to state H.qn. If it does not do this then you did not follow the
instructions properly when you wrote Ĥ.

On 9/03/24 19:36, olcott wrote:
>
> Please switch over to this new post that addresses all of your
> objections in advance.
> [Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior]
>

All you proved is that you want them to have different behaviour. You
didn't prove that they actually do.

On 3/9/2024 2:49 PM, immibis wrote:
> On 9/03/24 17:53, olcott wrote:
>> On 3/9/2024 10:21 AM, immibis wrote:
>>> On 9/03/24 17:00, olcott wrote:
>>>> On 3/9/2024 1:05 AM, Richard Damon wrote:
>>>>> On 3/8/24 10:40 PM, olcott wrote:
>>>>>> On 3/9/2024 12:25 AM, Richard Damon wrote:
>>>>>>> On 3/8/24 9:55 PM, olcott wrote:
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does
>>>>>>>> not halt
>>>>>>>>
>>>>>>>> Expecting Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to correctly report on the behavior of
>>>>>>>> Ĥ ⟨Ĥ⟩ is a little nuts because Ĥ contradicts Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>>
>>>>>>> But that is the job it signed up for when it tried to call itself
>>>>>>> a Halt Decider.
>>>>>>>
>>>>>>
>>>>>> There is no correct answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly
>>>>>> provide that corresponds to the behavior of this Ĥ ⟨Ĥ⟩.
>>>>>> Therefore it must have a basis for its wrong answer.
>>>>>>
>>>>>>
>>>>>
>>>>> WHY?
>>>>>
>>>>> What is the actual grounds for that statement?
>>>>
>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot just sit around scratching its head it
>>>> must do something. As soon as we have the basis for what
>>>> it will do, then H ⟨Ĥ⟩ ⟨Ĥ⟩ has its basis to correctly
>>>> decide halting.
>>>
>>> H ⟨A⟩ ⟨B⟩ cannot just sit around scratching its head, it must do
>>> something. As soon as we have the basis for what H will do, then we
>>> have a basis to construct an Ĥ where H ⟨Ĥ⟩ ⟨Ĥ⟩ is wrong.
>>>
>>
>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can fool itself yet cannot possibly fool an external H ⟨Ĥ⟩
>> ⟨Ĥ⟩.
>>
>
> If H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to state qn then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ also transitions
> to state H.qn. If it does not do this then you did not follow the
> instructions properly when you wrote Ĥ.

*Please see my new post that explains these things more clearly*
[Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior]

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/9/24 10:45 AM, olcott wrote:
> On 3/9/2024 12:17 PM, Richard Damon wrote:
>> On 3/9/24 8:53 AM, olcott wrote:
>>> On 3/9/2024 10:21 AM, immibis wrote:
>>>> On 9/03/24 17:00, olcott wrote:
>>>>> On 3/9/2024 1:05 AM, Richard Damon wrote:
>>>>>> On 3/8/24 10:40 PM, olcott wrote:
>>>>>>> On 3/9/2024 12:25 AM, Richard Damon wrote:
>>>>>>>> On 3/8/24 9:55 PM, olcott wrote:
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does
>>>>>>>>> not halt
>>>>>>>>>
>>>>>>>>> Expecting Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to correctly report on the behavior of
>>>>>>>>> Ĥ ⟨Ĥ⟩ is a little nuts because Ĥ contradicts Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>>>
>>>>>>>> But that is the job it signed up for when it tried to call
>>>>>>>> itself a Halt Decider.
>>>>>>>>
>>>>>>>
>>>>>>> There is no correct answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly
>>>>>>> provide that corresponds to the behavior of this Ĥ ⟨Ĥ⟩.
>>>>>>> Therefore it must have a basis for its wrong answer.
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> WHY?
>>>>>>
>>>>>> What is the actual grounds for that statement?
>>>>>
>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot just sit around scratching its head it
>>>>> must do something. As soon as we have the basis for what
>>>>> it will do, then H ⟨Ĥ⟩ ⟨Ĥ⟩ has its basis to correctly
>>>>> decide halting.
>>>>
>>>> H ⟨A⟩ ⟨B⟩ cannot just sit around scratching its head, it must do
>>>> something. As soon as we have the basis for what H will do, then we
>>>> have a basis to construct an Ĥ where H ⟨Ĥ⟩ ⟨Ĥ⟩ is wrong.
>>>>
>>>
>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can fool itself yet cannot possibly fool an external H
>>> ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>
>>
>> Why not?
>>
>> Are you thinking H will just simulate until H^.H makes its decision?
>>
>> We know the outcome of that, never answering, and thus failing.
>
> Please switch over to this new post that addresses all of your
> objections in advance.
> [Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior]
>

Then why did you post your reply here?

On 3/9/2024 2:53 PM, immibis wrote:
> On 9/03/24 19:36, olcott wrote:
>>
>> Please switch over to this new post that addresses all of your
>> objections in advance.
>> [Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior]
>>
>
> All you proved is that you want them to have different behaviour. You
> didn't prove that they actually do.

Then you could point out the specific mistake in this new post.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/9/2024 3:06 PM, Richard Damon wrote:
> On 3/9/24 10:45 AM, olcott wrote:
>> On 3/9/2024 12:17 PM, Richard Damon wrote:
>>> On 3/9/24 8:53 AM, olcott wrote:
>>>> On 3/9/2024 10:21 AM, immibis wrote:
>>>>> On 9/03/24 17:00, olcott wrote:
>>>>>> On 3/9/2024 1:05 AM, Richard Damon wrote:
>>>>>>> On 3/8/24 10:40 PM, olcott wrote:
>>>>>>>> On 3/9/2024 12:25 AM, Richard Damon wrote:
>>>>>>>>> On 3/8/24 9:55 PM, olcott wrote:
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does
>>>>>>>>>> not halt
>>>>>>>>>>
>>>>>>>>>> Expecting Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to correctly report on the behavior of
>>>>>>>>>> Ĥ ⟨Ĥ⟩ is a little nuts because Ĥ contradicts Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>>>>
>>>>>>>>> But that is the job it signed up for when it tried to call
>>>>>>>>> itself a Halt Decider.
>>>>>>>>>
>>>>>>>>
>>>>>>>> There is no correct answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly
>>>>>>>> provide that corresponds to the behavior of this Ĥ ⟨Ĥ⟩.
>>>>>>>> Therefore it must have a basis for its wrong answer.
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>> WHY?
>>>>>>>
>>>>>>> What is the actual grounds for that statement?
>>>>>>
>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot just sit around scratching its head it
>>>>>> must do something. As soon as we have the basis for what
>>>>>> it will do, then H ⟨Ĥ⟩ ⟨Ĥ⟩ has its basis to correctly
>>>>>> decide halting.
>>>>>
>>>>> H ⟨A⟩ ⟨B⟩ cannot just sit around scratching its head, it must do
>>>>> something. As soon as we have the basis for what H will do, then we
>>>>> have a basis to construct an Ĥ where H ⟨Ĥ⟩ ⟨Ĥ⟩ is wrong.
>>>>>
>>>>
>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can fool itself yet cannot possibly fool an external H
>>>> ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>
>>>
>>> Why not?
>>>
>>> Are you thinking H will just simulate until H^.H makes its decision?
>>>
>>> We know the outcome of that, never answering, and thus failing.
>>
>> Please switch over to this new post that addresses all of your
>> objections in advance.
>> [Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior]
>>
>
> Then why did you post your reply here?

So that you would not simply ignore this new post
and understand that my subsequent dialogue will be there.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/9/2024 3:09 PM, olcott wrote:
> On 3/9/2024 3:06 PM, Richard Damon wrote:
>> On 3/9/24 10:45 AM, olcott wrote:
>>> On 3/9/2024 12:17 PM, Richard Damon wrote:
>>>> On 3/9/24 8:53 AM, olcott wrote:
>>>>> On 3/9/2024 10:21 AM, immibis wrote:
>>>>>> On 9/03/24 17:00, olcott wrote:
>>>>>>> On 3/9/2024 1:05 AM, Richard Damon wrote:
>>>>>>>> On 3/8/24 10:40 PM, olcott wrote:
>>>>>>>>> On 3/9/2024 12:25 AM, Richard Damon wrote:
>>>>>>>>>> On 3/8/24 9:55 PM, olcott wrote:
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does
>>>>>>>>>>> not halt
>>>>>>>>>>>
>>>>>>>>>>> Expecting Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to correctly report on the behavior of
>>>>>>>>>>> Ĥ ⟨Ĥ⟩ is a little nuts because Ĥ contradicts Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>>>>>
>>>>>>>>>> But that is the job it signed up for when it tried to call
>>>>>>>>>> itself a Halt Decider.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> There is no correct answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly
>>>>>>>>> provide that corresponds to the behavior of this Ĥ ⟨Ĥ⟩.
>>>>>>>>> Therefore it must have a basis for its wrong answer.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>
>>>>>>>> WHY?
>>>>>>>>
>>>>>>>> What is the actual grounds for that statement?
>>>>>>>
>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot just sit around scratching its head it
>>>>>>> must do something. As soon as we have the basis for what
>>>>>>> it will do, then H ⟨Ĥ⟩ ⟨Ĥ⟩ has its basis to correctly
>>>>>>> decide halting.
>>>>>>
>>>>>> H ⟨A⟩ ⟨B⟩ cannot just sit around scratching its head, it must do
>>>>>> something. As soon as we have the basis for what H will do, then
>>>>>> we have a basis to construct an Ĥ where H ⟨Ĥ⟩ ⟨Ĥ⟩ is wrong.
>>>>>>
>>>>>
>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can fool itself yet cannot possibly fool an external H
>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>
>>>>
>>>> Why not?
>>>>
>>>> Are you thinking H will just simulate until H^.H makes its decision?
>>>>
>>>> We know the outcome of that, never answering, and thus failing.
>>>
>>> Please switch over to this new post that addresses all of your
>>> objections in advance.
>>> [Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior]
>>>
>>
>> Then why did you post your reply here?
>
> So that you would not simply ignore this new post
> and understand that my subsequent dialogue will be there.
>

I am going to reopen this post in its brand new thread
when I finish working out the details.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/10/2024 7:49 AM, Mikko wrote:
> On 2024-03-09 18:22:53 +0000, Richard Damon said:
>
>> On 3/9/24 9:39 AM, olcott wrote:
>>> On 3/9/2024 11:12 AM, Richard Damon wrote:
>>>> On 3/9/24 8:41 AM, olcott wrote:
>>>>> On 3/9/2024 9:36 AM, immibis wrote:
>>>>>> On 9/03/24 16:12, olcott wrote:
>>>>>>> On 3/9/2024 7:47 AM, immibis wrote:
>>>>>>>> On 8/03/24 22:34, olcott wrote:
>>>>>>>>>> And since H^ can "lie" to that embedded H^.H about what its
>>>>>>>>>> description is, that H can't tell that it is part of an H^
>>>>>>>>>> computation that is simulating an H^ computation.
>>>>>>>>>
>>>>>>>>> That subject must be postponed until after the Olcott refutation
>>>>>>>>> of the exact Linz proof is either fully accepted by three people
>>>>>>>>> or actual errors or gaps are found that cannot be addressed or
>>>>>>>>> corrected.
>>>>>>>>
>>>>>>>> It's accepted that the Linz proof doen't work on Olcott machines
>>>>>>>> because the Linz proof is designed for Turing machines. But you
>>>>>>>> can't refute the Linz-immibis proof designed for Olcott
>>>>>>>> machines, where H is lied to about its own description.
>>>>>>>
>>>>>>> I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except halt or fail
>>>>>>> to halt
>>>>>>> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> could see that.
>>>>>>>
>>>>>>
>>>>>> I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except for exactly the
>>>>>> same thing that H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do.
>>>>>
>>>>> It is easily proven that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the simulation of
>>>>> its input and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort the simulation of its input.
>>>>
>>>> How?
>>>
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>
>> Comment are SPECIFICATION, not actual behavior until existance of a
>> conforming H is proven
>
> To me they do not look like a specification but a false statement of
> the actual behaviour (for some non-conforming H).
>

H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy // Ĥ applied to ⟨Ĥ⟩ halts
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn // Ĥ applied to ⟨Ĥ⟩ does not halt

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt

Unlike anyone else has ever done my simulating termination
analyzers can always detect when their input will cause
themselves to never terminate. I have demonstrated this
for the Halting Problem's pathological input:

Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
(1) H applied ⟨Ĥ⟩ ⟨Ĥ⟩ simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(2) which begins at simulated ⟨Ĥ.q0⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process

The earliest point when Turing machine H can detect the repeating
state of Ĥ ⟨Ĥ⟩ is when Ĥ first reaches (c) where it would begin
simulating a copy of itself with a copy of its input.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 3/10/24 7:59 AM, olcott wrote:
> On 3/10/2024 7:49 AM, Mikko wrote:
>> On 2024-03-09 18:22:53 +0000, Richard Damon said:
>>
>>> On 3/9/24 9:39 AM, olcott wrote:
>>>> On 3/9/2024 11:12 AM, Richard Damon wrote:
>>>>> On 3/9/24 8:41 AM, olcott wrote:
>>>>>> On 3/9/2024 9:36 AM, immibis wrote:
>>>>>>> On 9/03/24 16:12, olcott wrote:
>>>>>>>> On 3/9/2024 7:47 AM, immibis wrote:
>>>>>>>>> On 8/03/24 22:34, olcott wrote:
>>>>>>>>>>> And since H^ can "lie" to that embedded H^.H about what its
>>>>>>>>>>> description is, that H can't tell that it is part of an H^
>>>>>>>>>>> computation that is simulating an H^ computation.
>>>>>>>>>>
>>>>>>>>>> That subject must be postponed until after the Olcott refutation
>>>>>>>>>> of the exact Linz proof is either fully accepted by three people
>>>>>>>>>> or actual errors or gaps are found that cannot be addressed or
>>>>>>>>>> corrected.
>>>>>>>>>
>>>>>>>>> It's accepted that the Linz proof doen't work on Olcott
>>>>>>>>> machines because the Linz proof is designed for Turing
>>>>>>>>> machines. But you can't refute the Linz-immibis proof designed
>>>>>>>>> for Olcott machines, where H is lied to about its own description.
>>>>>>>>
>>>>>>>> I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except halt or fail
>>>>>>>> to halt
>>>>>>>> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> could see that.
>>>>>>>>
>>>>>>>
>>>>>>> I am not sure what Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do except for exactly
>>>>>>> the same thing that H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> would do.
>>>>>>
>>>>>> It is easily proven that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the simulation of
>>>>>> its input and H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort the simulation of its input.
>>>>>
>>>>> How?
>>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>
>>> Comment are SPECIFICATION, not actual behavior until existance of a
>>> conforming H is proven
>>
>> To me they do not look like a specification but a false statement of
>> the actual behaviour (for some non-conforming H).
>>
>
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy  // Ĥ applied to ⟨Ĥ⟩ halts
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn  // Ĥ applied to ⟨Ĥ⟩ does not halt
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>
> Unlike anyone else has ever done my simulating termination
> analyzers can always detect when their input will cause
> themselves to never terminate. I have demonstrated this
> for the Halting Problem's pathological input:

HOW?

You assume H is a halt decider, so it can be a Halt Decider.

>
> Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
> (1) H applied ⟨Ĥ⟩ ⟨Ĥ⟩ simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
> (2) which begins at simulated ⟨Ĥ.q0⟩
> (a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
> (b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
> (c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
>
> The earliest point when Turing machine H can detect the repeating
> state of Ĥ ⟨Ĥ⟩ is when Ĥ first reaches (c) where it would begin
> simulating a copy of itself with a copy of its input.
>

Right, and then so does H^.H and thus H and H^ both say it will not halt
when it halts.