Re: HHH(DDD) is correct to reject its input as non-halting --- EVIDENCE THAT I AM CORRECT

Op 30.jun.2025 om 19:22 schreef olcott:

On 6/30/2025 2:40 AM, Fred. Zwarts wrote:

Op 29.jun.2025 om 15:46 schreef olcott:

On 6/29/2025 5:38 AM, Fred. Zwarts wrote:

Op 28.jun.2025 om 15:02 schreef olcott:

On 6/28/2025 3:50 AM, Fred. Zwarts wrote:

Op 28.jun.2025 om 01:30 schreef olcott:

On 6/26/2025 4:16 AM, Fred. Zwarts wrote:

Op 25.jun.2025 om 16:09 schreef olcott:

On 6/25/2025 2:59 AM, Fred. Zwarts wrote:

Op 24.jun.2025 om 16:06 schreef olcott:

>
None of the code in HHH can possibly cause DDD correctly
simulated by HHH to reach its own simulated "return" statement.

>
Yes, exactly, that is the bug.
>

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Recursive emulation is only a tiny bit more complicated
than recursion yet no one here seems to have a clue.
Do you know what recursion is?
(If you don't that would explain a lot)

As usual irrelevant claims without evidence. No rebuttal.

>
Ah so you don't know what recursion is.

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As usual a false claim without evidence.
>

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HHH has a bug that makes that it does not recognise the halting behaviour of the program specified in the input.

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If you don't even know what recursion is then
you are totally unqualified to review these things.

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And since the condition in the 'if' fails, the conclusion is not true.
>

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Even a beginner can see that the input is a pointer to code, including the code to abort and halt. But HHH is programmed to ignore the conditional branch instructions, when simulating itself, so it thinks that there is an infinite loop when there are only a finite number of recursions.
But Olcott does not understand that not all recursions are infinite.

>
When the measure is whether or not DDD correctly
simulated by HHH can possibly reach its own "return"
instruction final halt state nothing inside HHH can
possibly have any effect on this.
>
That you don't know this proves that you are unqualified
to review my work.

The failure of HHH is an incorrect measure for the halting behaviour specified in the input.
That you do not understand this explains your invalid claims.
The halting behaviour of the input can be analysed by several other methods and they show that HHH is incorrect in its analysis.
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>

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No Turing machine can ever report on the behavior of
any directly executing Turing Machine because no TM
can ever take another directly executing Turing Machine
as its input.

There is no need to report about another Turing Machine. It only needs to report about its input.

 Then the fact that DDD() halts does not contradict the
fact that the input to HHH(DDD) specifies non-halting
behavior because the directly executed DDD() is outside
of the domain of HHH.
 

A usual claims  without evidence.
There is no contradiction, because the input for the aborting HHH specifies a DDD that calls the aborting HHH, which makes that the input specifies a halting program. Of course. The semantics of the x86 language allow for only one behaviour. Therefore, the specification agrees with the direct execution. Direct execution is only a proof for what is specified in the input.
That HHH is programmed with bugs, so that it aborts before it reaches that final halt state, does not change the specification of a halting program.
The non-halting property is a property of the simulator, not of the program specified in the input.