Re: Simulation vs. Execution in the Halting Problem

On 6/8/25 2:00 AM, olcott wrote:

On 6/8/2025 12:49 AM, Mikko wrote:

On 2025-06-04 16:27:48 +0000, olcott said:
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On 6/4/2025 2:32 AM, Mikko wrote:

On 2025-06-03 20:28:36 +0000, olcott said:
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On 6/3/2025 2:55 AM, Mikko wrote:

On 2025-06-02 15:23:15 +0000, olcott said:
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On 6/2/2025 1:56 AM, Mikko wrote:

On 2025-06-01 21:41:36 +0000, olcott said:
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On 6/1/2025 6:30 AM, Mikko wrote:

On 2025-05-30 15:41:59 +0000, olcott said:
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On 5/30/2025 3:45 AM, Mikko wrote:

On 2025-05-29 18:10:39 +0000, olcott said:
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On 5/29/2025 12:34 PM, Mr Flibble wrote:

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🧠 Simulation vs. Execution in the Halting Problem
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In the classical framework of computation theory (Turing machines),
simulation is not equivalent to execution, though they can approximate one
another.

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To the best of my knowledge a simulated input
always has the exact same behavior as the directly
executed input unless this simulated input calls
its own simulator.

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The simulation of the behaviour should be equivalent to the real
behaviour.

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That is the same as saying a function with infinite
recursion must have the same behavior as a function
without infinite recursion.

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A function does not have a behaviour. A function has a value for
every argument in its domain.
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A function is not recursive. A definition of a function can be
recursive. There may be another way to define the same function
without recursion.
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A definition of a function may use infinite recursion if it is also
defined how that infinite recursion defines a value.
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Anyway, from the meaning of "simulation" follows that a simulation
of a behaviour is (at least in some sense) similar to the real
behaviour. Otherwise no simulation has happened.
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void DDD()
{
   HHH(DDD);
   return;
}
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The *input* to simulating termination analyzer HHH(DDD)
specifies recursive simulation that can never reach its
*simulated "return" instruction final halt state*

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It does not matter whether a particular simulation does or does not
reach its "return" instruction.

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It completely matters. DDD correctly simulated by HHH
proves the exact behavior that the input to HHH(DDD)
actually specifies.

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It proves nothing without a proof that DDD is correctly simulated by HHH.

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I have shown that proof too many times and people
denied the very obvious verified facts of it.

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You have never shown any proof of anything. But a verifiable and verified
fact is that DDD halts. An obvious conseqence of that fact is that every
report that means 'DDD does not halt' is wrong.

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When I provide proof that you cannot understand
this does not mean that I did not provide proof.

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Yes, it does.

 What I just said is a truism, tautology, self-evident truth.

Only with the correct meaning of the words, which you are not actually using.

 In epistemology (theory of knowledge), a self-evident
proposition is a proposition that is known to be true
by understanding its meaning without proof...
https://en.wikipedia.org/wiki/Self-evidence
 

Which is meaningless in Formal Systems, as any such statement would either BE an axiom of the system (as defintions are) or will be provable from them.
All you are doing is showing you don't even understand what sort of logic system you are working in.
Formal logic systems do not allow for the word games of General Philosophy, that is what makes them "Formal".
All you are doing is proving you have no idea what you are actually talking about, but that you are trying to earn your PhD, the Pile it Higher and Deeper degree.
Sorry, you are showing your stupidity.

 

To understand a proof does not require any skills other
than proof checking, which is a Turing computable function.
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