Re: Defining a correct simulating halt decider

On 9/9/24 2:21 PM, olcott wrote:

On 9/8/2024 9:56 AM, Mikko wrote:

On 2024-09-07 13:56:02 +0000, olcott said:
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On 9/7/2024 3:27 AM, Mikko wrote:

On 2024-09-06 11:42:48 +0000, olcott said:
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On 9/6/2024 6:19 AM, Mikko wrote:

On 2024-09-05 13:24:20 +0000, olcott said:
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On 9/5/2024 2:34 AM, Mikko wrote:

On 2024-09-03 13:00:50 +0000, olcott said:
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On 9/3/2024 5:25 AM, Mikko wrote:

On 2024-09-02 16:38:03 +0000, olcott said:
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A halt decider is a Turing machine that computes
the mapping from its finite string input to the
behavior that this finite string specifies.

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A halt decider needn't compute the full behaviour, only whether
that behaviour is finite or infinite.
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void DDD()
{
   HHH(DDD);
   return;
}
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New slave_stack at:1038c4
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55         push ebp      ; housekeeping
[00002173][001138bc][001138c0] 8bec       mov ebp,esp   ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec
[00002172][0015e2e4][0015e2e8] 55         push ebp      ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec       mov ebp,esp   ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
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Hence  HHH(DDD)==0 is correct

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Nice to see that you don't disagree with what said.
Unvortunately I can't agree with what you say.
HHH terminates,

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os DDD obviously terminates, too. No valid

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DDD emulated by HHH never reaches it final halt state.

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If that iis true it means that HHH called by DDD does not return
and therefore is not a ceicder.
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The directly executed HHH is a decider.

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If the called HHH behaves differently from the direcly executed HHH
then the DDD is not relevant to classic proofs of the impossibility
of a halting decider.
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If you can't show encoding rules that permit the encoidng of the
behaviour of the directly executed DDD to HHH then HHH is not a
halting decider.
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I SHOW THE ACTUAL EXECUTION TRACE AND EVERYONE DISAGREES WITH IT.

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There are no encoding rules in the actual execution trace.
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 The x86 execution trace is encoded in the x86 language.
Why do you insist on lying about this?
 

Right, which means it show EXACTLY what the processor will execute. That means a call to HHH does the code of the function HHH, which is how it emulated the input DDD, not the results of the emulation of DDD, which shows that it has conditionals in it, and thus can, (and will) abort its emulation to return.
You just lie to yourself because you can't face the truth.