On 29/02/24 16:34, olcott wrote:
> On 2/29/2024 4:30 AM, immibis wrote:
>> On 28/02/24 23:06, olcott wrote:
>>> H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ *Is not self-contradictory*
>>> Ĥ applied to ⟨Ĥ⟩ contradicts Ĥ.Hq0 applied to ⟨Ĥ⟩ ⟨Ĥ⟩ (itself)
>>
>> Yup, Ĥ is wrong and it does not correctly solve the problem because it
>> is wrong. all possible Ĥs are wrong and therefore the problem cannot
>> be solved.
>
> 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
>
> H ⟨Ĥ⟩ ⟨Ĥ⟩ (in a separate memory space) merely needs to report on
> the actual behavior of Ĥ ⟨Ĥ⟩. We already know that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must
> transition to Ĥ.Hqy or Ĥ.Hqn, H merely needs to see which one.
>

H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy ⇔ Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hqy
H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn ⇔ Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hqn

On 2/29/2024 9:38 AM, immibis wrote:
> On 29/02/24 16:34, olcott wrote:
>> On 2/29/2024 4:30 AM, immibis wrote:
>>> On 28/02/24 23:06, olcott wrote:
>>>> H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ *Is not self-contradictory*
>>>> Ĥ applied to ⟨Ĥ⟩ contradicts Ĥ.Hq0 applied to ⟨Ĥ⟩ ⟨Ĥ⟩ (itself)
>>>
>>> Yup, Ĥ is wrong and it does not correctly solve the problem because
>>> it is wrong. all possible Ĥs are wrong and therefore the problem
>>> cannot be solved.
>>
>> 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
>>
>> H ⟨Ĥ⟩ ⟨Ĥ⟩ (in a separate memory space) merely needs to report on
>> the actual behavior of Ĥ ⟨Ĥ⟩. We already know that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must
>> transition to Ĥ.Hqy or Ĥ.Hqn, H merely needs to see which one.
>>
>
> H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy ⇔ Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hqy
> H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn ⇔ Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hqn

Merely a false assumption.
H and Ĥ are in different memory spaces. H merely needs
to report on what Ĥ actually does and Ĥ must transition
to Ĥ.Hqy or Ĥ.Hqn.

When H1 and H have identical machine language and analyze
the same input machine language H1 gets a different answer
than H because D only calls H in a different memory space
than the memory space that H1 is in.

H and D have a pathological self-reference relationship.
H1 and D DO NOT have a pathological self-reference relationship.
This changes the behavior of H1/D relative to H/D.

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

On 29/02/24 17:04, olcott wrote:
> On 2/29/2024 9:38 AM, immibis wrote:
>> H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy ⇔ Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hqy
>> H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn ⇔ Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hqn
>
> Merely a false assumption.

Show me the shortest subjective Turing machine. I'll wait. There isn't one.