Re: Defining a correct halt decider

Op 07.sep.2024 om 16:35 schreef olcott:

On 9/7/2024 5:31 AM, Fred. Zwarts wrote:

Op 06.sep.2024 om 13:41 schreef olcott:

On 9/6/2024 6:12 AM, Mikko wrote:

On 2024-09-05 13:39:14 +0000, olcott said:
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On 9/5/2024 2:39 AM, Mikko wrote:

On 2024-09-03 13:17:56 +0000, olcott said:
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On 9/3/2024 3:44 AM, Mikko wrote:

On 2024-09-02 16:06:11 +0000, olcott said:
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A correct halt decider is a Turing machine T with one accept state and one reject state such that:
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If T is executed with initial tape contents equal to an encoding of Turing machine X and its initial tape contents Y, and execution of a real machine X with initial tape contents Y eventually halts, the execution of T eventually ends up in the accept state and then stops.
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If T is executed with initial tape contents equal to an encoding of Turing machine X and its initial tape contents Y, and execution of a real machine X with initial tape contents Y does not eventually halt, the execution of T eventually ends up in the reject state and then stops.

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Your "definition" fails to specify "encoding". There is no standard
encoding of Turing machines and tape contents.

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That is why I made the isomorphic x86utm system.
By failing to have such a concrete system all kinds
of false assumptions cannot be refuted.

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If it were isnomorphic the same false assumtipns would apply to both.

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They do yet I cannot provide every single details of
the source-code of the Turing machine because these
details would be too overwhelming.
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So instead every author makes a false assumption that
is simply believed to be true with no sufficient basis
to show that it isn't true.
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Once I prove my point as the x86 level I show how the
same thing applies to the Peter Linz proof.

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Your recent presentations are so far from Linz' proof that they
look totally unrelated.
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I must begin where people are so far no one even understands
the concept of recursive emulation.
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_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
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Show the details of how DDD emulated by HHH
reaches its own machine address 0000217f.
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00002172, 00002173, 00002175, 0000217a calls HHH(DDD)
then
00002172, 00002173, 00002175, 0000217a calls HHH(DDD)...
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WHAT SHOULD THE NEXT STEPS BE?
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That has been told now several times. A correct simulation (as by HHH1, or the unmodified world class simulator)

 What comes next in the above sequence?

I repeat the answer again:
In a correct simulation the next steps (in DDD, ignoring the steps in HHH) are: 0000217f, 00002182, 00002183 and DDD halts. Proven by the correct simulation by the world class simulator and even by HHH1. But HHH fails to reach this part of the simulation, because it aborts too soon. HHH cannot possibly simulate itself correctly up to the end.
Why does olcott each time cut away the answer to this question? Apparently he does not *want* to know it.
Even when the doors hits his face, he does not want to see it.

 Changing the question to answer a different question
is the despicable lie of the strawman deception.
 

Apparently his English is so bad, that he  does not even know what the "strawman deception" is, even after he has read the definition.