On 6/1/2024 2:52 AM, Mikko wrote:"Except". So it is not an UTM.
On 2024-05-31 15:35:18 +0000, olcott said:
*A quick summary of the reasoning provided below*There is no restrictions on what embedded_H is allowed to report on.
The LHS is behavior that embedded_H is allowed to report on.
embedded_H is only allowed to report on the behavior that its finite
string Turing Machine Description specifies to a UTM.
embedded_H <is> a UTM except that it stops simulating and reports
non-halting as soon as it correctly recognizes a non-halting behavior
pattern that is specified by its input.
When Ĥ is applied to ⟨Ĥ⟩I can't read that notation. What is H^ and what does it look like?
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d)
embedded_H is not allowed to be applied to Ĥ ⟨Ĥ⟩ because inputs can only be finite strings and Ĥ is not a finite string. This means
that embedded_H is not allowed to report on its own actual behavior.
embedded_H only allowed to report on the behavior specified by its^
finite string input. That behavior never stops running for 1 to ∞ steps
of ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H.
The only reauirement is that embedded_H has the same transitionLinz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ derives a different result than embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩.
rules as H. Therefore embedded_H reports the same as H, whether
allowed or not.
This is because the in the latter case embedded_H must determine that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly stop running
after 1 to ∞ steps of correct simulation. Thus embedded_H meets its
abort simulation criteria.
The former case of Linz H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ can see that embedded_H
has already aborted its simulation, thus it never reaches its own
abort criteria.
It is only because everyone since 1936 has rejected simulation
OUT-OF-HAND without review that no one ever noticed this before.
The RHS is behavior that embedded_H NOT is allowed to report on.
The LHS and the RHS specify different behaviors.
You have not shown anything with behaviours as LHS and RHS.
Please to not reply here instead reply at the end of my proof
after all of the steps have been presented.
Not a reasonable request. Correctness of a step of proof does not
depend on what follows. If one step is erroneous the rest is
irrelevant.
On 6/3/2024 10:16 AM, Mikko wrote:mistakes.
On 2024-06-03 12:36:00 +0000, olcott said:
On 6/3/2024 3:01 AM, Mikko wrote:
On 2024-06-02 13:21:56 +0000, olcott said:
On 6/2/2024 2:42 AM, Mikko wrote:
On 2024-06-01 19:26:55 +0000, olcott said:
On 6/1/2024 1:52 PM, joes wrote:
Am Sat, 01 Jun 2024 09:37:01 -0500 schrieb olcott:
On 6/1/2024 2:52 AM, Mikko wrote:
On 2024-05-31 15:35:18 +0000, olcott said:
I am so sure that I must be right that I do not notice any actual
The only "mistake" that anyone ever pointed out is essentially "that'sNo, your mistake is the contradictory requirements for H:
not the way that I memorized it".
When the behavior of DD correctly simulated by HH is not the same as the behavior of the directly executed DD(DD)That's not an assumption. That's the definition of simulation: doing
they assume the simulation must be wrong even though it it an easily
verified fact that it is correct.
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