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

On 2024-07-27 14:08:10 +0000, olcott said:

On 7/27/2024 2:21 AM, Mikko wrote:

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:
 

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.
 

In this case we have two x86utm machines that are identical
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.

And even more simplified semantics.

(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
 You are supposed to evaluate the above as a contiguous
sequence of moves such that non-halting behavior is
identified.

The above is an obvious tight loop of (d), (e), (f), and (g).
Its relevance (it any) to the topic of the discussion is not
obvious.
--
Mikko