Re: DDD correctly emulated by HHH is Correctly rejected as non-halting V2

On 7/26/2024 3:45 AM, Mikko wrote:

On 2024-07-24 13:33:55 +0000, olcott said:
 

On 7/24/2024 3:57 AM, Mikko wrote:

On 2024-07-23 13:31:35 +0000, olcott said:
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On 7/23/2024 1:32 AM, 0 wrote:

On 2024-07-22 13:46:21 +0000, olcott said:
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On 7/22/2024 2:57 AM, Mikko wrote:

On 2024-07-21 13:34:40 +0000, olcott said:
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On 7/21/2024 4:34 AM, Mikko wrote:

On 2024-07-20 13:11:03 +0000, olcott said:
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On 7/20/2024 3:21 AM, Mikko wrote:

On 2024-07-19 14:08:24 +0000, olcott said:
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When we use your incorrect reasoning we would conclude
that Infinite_Loop() is not an infinite loop because it
only repeats until aborted and is aborted.

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You and your HHH can reason or at least conclude correctly about
Infinite_Loop but not about DDD. Possibly because it prefers to
say "no", which is correct about Infinte_loop but not about DDD.
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*Because this is true I don't understand how you are not simply lying*
int main
{
   DDD();
}
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Calls HHH(DDD) that must abort the emulation of its input
or {HHH, emulated DDD and executed DDD} never stop running.

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You are the lying one.
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If HHH(DDD) abrots its simulation and returns true it is correct as a
halt decider for DDD really halts.
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(b) We know that a decider is not allowed to report on the behavior
computation that itself is contained within.

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No, we don't. There is no such prohibition.
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Turing machines never take actual Turing machines as inputs.
They only take finite strings as inputs and an actual executing
Turing machine is not itself a finite string.

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The definition of a Turing machine does not say that a Turing machine
is not a finite string. It is an abstract mathematical object without
a specification of its exact nature. It could be a set or a finite
string. Its exact nature is not relevant to the theory of computation,
which only cares about certain properties of Turing machines.
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Therefore It is not allowed to report on its own behavior.

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Anyway, that does not follow. The theory of Turing machines does not
prohibit anything.
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Another different TM can take the TM description of this
machine and thus accurately report on its actual behavior.

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If a Turing machine can take a description of a TM as its input
or as a part of its input it can also take its own description.
Every Turing machine can be given its own description as input
but a Turing machine may interprete it as something else.
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In this case we have two x86utm machines that are identical
except that DDD calls HHH and DDD does not call HHH1.

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That DDD calls HHH and DDD does not call HHH1 is not a difference
between two unnamed turing machines.
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The same thing happens at the Peter Linz Turing Machine level
I will provide that more difficult example if and only if you
prove that you understand this one.

 However, Peter Linz does not call taht same thing a difference.
 

We can call everything "late for dinner" with a unique integer
index and the properties that I assert exist still exist.
When Ĥ is applied to ⟨Ĥ⟩
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) with one more level of simulation
Two complete simulations show a pair of identical TMD's are
simulating a pair of identical inputs. We can see this thus
proving recursive simulation.
When embedded_H is accountable for the behavior of its
input and not accountable for the behavior of the
computation that it is contained within then embedded_H
is necessarily correct to transition to its own Ĥ.qn state.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer