On 2024-03-16 14:46:42 +0000, olcott said:

> On 3/16/2024 6:51 AM, Mikko wrote:
>> On 2024-03-15 16:31:15 +0000, olcott said:
>>
>>> On 3/15/2024 5:31 AM, Mikko wrote:
>>>> On 2024-03-14 20:32:37 +0000, olcott said:
>>>>
>>>>> On 3/14/2024 7:19 AM, Mikko wrote:
>>>>>> On 2024-03-13 17:19:06 +0000, olcott said:
>>>>>>
>>>>>>> On 3/13/2024 12:10 PM, Mikko wrote:
>>>>>>>> On 2024-03-12 15:06:49 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 3/12/2024 4:32 AM, Mikko wrote:
>>>>>>>>>> On 2024-03-11 15:06:10 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>> On 3/11/2024 5:56 AM, Mikko wrote:
>>>>>>>>>>>> On 2024-03-11 01:24:55 +0000, olcott said:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 3/10/2024 7:40 PM, immibis wrote:
>>>>>>>>>>>>>> On 10/03/24 20:16, olcott wrote:
>>>>>>>>>>>>>>> On 3/10/2024 2:09 PM, immibis wrote:
>>>>>>>>>>>>>>>> On 10/03/24 19:32, olcott wrote:
>>>>>>>>>>>>>>>>> On 3/10/2024 1:08 PM, immibis wrote:
>>>>>>>>>>>>>>>>>> On 10/03/24 18:17, olcott wrote:
>>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as unsound.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not sound?
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Both H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:
>>>>>>>>>>>>>>>>>>>>> (a) Their input halts H.qy
>>>>>>>>>>>>>>>>>>>>> (b) Their input fails to halt or has a pathological
>>>>>>>>>>>>>>>>>>>>> relationship to itself H.qn.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> But the "Pathological Relationship" is ALLOWED.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as unsound
>>>>>>>>>>>>>>>>>>> expressly disallowing the "Pathological Relationship".
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not real Turing machines?
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> I am only claiming that both H and Ĥ.H correctly say YES
>>>>>>>>>>>>>>>>>>> when their input halts and correctly say NOT YES otherwise.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> well the halting problem requires them to correctly say NO, so you
>>>>>>>>>>>>>>>>>> haven't solved it
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> All decision problem instances of program/input such that both
>>>>>>>>>>>>>>>>> yes and no are the wrong answer toss out the input as invalid.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> all decision problems are defined so that all instances are valid or
>>>>>>>>>>>>>>>> else they are not defined properly
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Not in the case of Russell's Paradox.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> And now we are back to: Every Turing machine and input pair defines an
>>>>>>>>>>>>>> execution sequence. Every sequence is either finite or infinite.
>>>>>>>>>>>>>> Therefore it is well-defined and there is no paradox.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Can you show me a Turing machine that specifies a sequence of
>>>>>>>>>>>>>> configurations that is not finite or infinite?
>>>>>>>>>>>>>
>>>>>>>>>>>>> When we construe every yes/no question that cannot possibly
>>>>>>>>>>>>> have a correct yes/no answer as an incorrect question
>>>>>>>>>>>>>
>>>>>>>>>>>>> then we must correspondingly construe every decider/input
>>>>>>>>>>>>> pair that has no correct yes/no answer as invalid input.
>>>>>>>>>>>>
>>>>>>>>>>>> Apparently the answer is "no" as no such Turing machine is shown
>>>>>>>>>>>> or mentioned above.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> *Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>>>>>>>>> This proves that there is something wrong with the question.
>>>>>>>>>>
>>>>>>>>>> No, it doesn't. It proves that there is something wrong with
>>>>>>>>>> every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩. Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets
>>>>>>>>>> the wrong answer the other answer is right.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> *Your reasoning is just like this reasoning*
>>>>>>>>> "This sentence is not true." is not true and that makes it true.
>>>>>>>>> You are stopping right there and not seeing the infinite cycle.
>>>>>>>>>
>>>>>>>>> *Please see my new post for a complete elaboration of this*
>>>>>>>>> [Proving my 2004 claim that some decider/input pairs are incorrect questions]
>>>>>>>>
>>>>>>>> Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer the other answer is right.
>>>>>>>
>>>>>>> *You have already affirmed otherwise*
>>>>>>>
>>>>>>> On 3/13/2024 11:45 AM, Mikko wrote:
>>>>>>>  > On 2024-03-12 15:31:58 +0000, olcott said:
>>>>>>>  >> *Formalized*
>>>>>>>  >> ∀ H ∈ Turing_Machine_Deciders
>>>>>>>  >> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>>  >> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>  >
>>>>>>>  > Yes, this is one way to state Linz' conclusion.
>>>>>>
>>>>>> That says nothing about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ nor about correctness of
>>>>>> the other answer. Therefore "otherwise" is not affirmed there.
>>>>>>
>>>>>
>>>>> *The conventional spec is proved to be wrong for these instances*
>>>>> 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
>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
>>>>
>>>> Nice to see that you don't disagree with my observation about your mistake.
>>>>
>>>
>>> Best selling author of Theory of Computation textbooks:
>>> *Introduction To The Theory Of Computation 3RD, by sipser*
>>> https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>>>
>>> Date 10/13/2022 11:29:23 AM
>>> *MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
>>> (He has neither reviewed nor agreed to anything else in this paper)
>>> (a) If simulating halt decider H correctly simulates its input D until
>>> H correctly determines that its simulated D would never stop running
>>> unless aborted then
>>> (b) H can abort its simulation of D and correctly report that D
>>> specifies a non-halting sequence of configurations.
>>>
>>> *When we apply the abort criteria* (elaborated above)
>>> Will you halt if you never abort your simulation?
>>> *Then H(D,D) is proven to meet this criteria*
>>>
>>> This same thing applies to: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>> ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct according to the above criteria.
>>>
>>> *Lets stick with H(D,D)==0 is correct according to the above criteria*
>>> until we have mutual agreement on that because:
>>>
>>> I cannot afford to tolerate the [change the subject]
>>> form of rebuttal that wasted 15 years with Ben Bacarisse.
>>
>> Anyway it is nice to see that you agree even if that may be
>> a change of subject.
>>
>
> I have no idea what you are saying.
>
> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
> that H correctly determined that it had to abort the simulation
> of its input to prevent the infinite execution of this input.

What would H return if it incorrecly detected that it had to
abort the simulation of its input to prevent the infinite execution
of this input?

--
Mikko