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On 2024-07-27 14:08:10 +0000, olcott said:When we compute the mapping from the input to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
On 7/27/2024 2:21 AM, Mikko wrote:And even more simplified semantics.On 2024-07-26 14:08:11 +0000, olcott said:>
>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:>
>On 7/23/2024 1:32 AM, 0 wrote:>On 2024-07-22 13:46:21 +0000, olcott said:In this case we have two x86utm machines that are identical
>On 7/22/2024 2:57 AM, Mikko wrote:>On 2024-07-21 13:34:40 +0000, olcott said:>
>On 7/21/2024 4:34 AM, Mikko wrote:>On 2024-07-20 13:11:03 +0000, olcott said:>
>On 7/20/2024 3:21 AM, Mikko wrote:>On 2024-07-19 14:08:24 +0000, olcott said:>
>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.
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.
>
*Because this is true I don't understand how you are not simply lying*
int main
{
DDD();
}
>
Calls HHH(DDD) that must abort the emulation of its input
or {HHH, emulated DDD and executed DDD} never stop running.
You are the lying one.
>
If HHH(DDD) abrots its simulation and returns true it is correct as a
halt decider for DDD really halts.
>
(b) We know that a decider is not allowed to report on the behavior
computation that itself is contained within.
No, we don't. There is no such prohibition.
>
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.
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.
>Therefore It is not allowed to report on its own behavior.>
Anyway, that does not follow. The theory of Turing machines does not
prohibit anything.
>Another different TM can take the TM description of this>
machine and thus accurately report on its actual behavior.
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.
>
except that DDD calls HHH and DDD does not call HHH1.
That DDD calls HHH and DDD does not call HHH1 is not a difference
between two unnamed turing machines.
>
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.
That you can say all sorts stupid things does not mean that it be a
good idea to do so.
>
Some of the properties you assert exsit actually do exist, some don't.
>When Ĥ is applied to ⟨Ĥ⟩>
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The above is merely simplified syntax for the top of page 3
https://www.liarparadox.org/Linz_Proof.pdf
The above is the whole original Linz proof.
(a) Ĥ copies its input ⟨Ĥ⟩The above is an obvious tight loop of (d), (e), (f), and (g).
(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
>
You are supposed to evaluate the above as a contiguous
sequence of moves such that non-halting behavior is
identified.
Its relevance (it any) to the topic of the discussion is not
obvious.
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