Re: Defining a correct halt decider

On 9/6/24 7:41 AM, olcott wrote:

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.
 _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]
 Show the details of how DDD emulated by HHH
reaches its own machine address 0000217f.
 00002172, 00002173, 00002175, 0000217a calls HHH(DDD)
then
00002172, 00002173, 00002175, 0000217a calls HHH(DDD)...
 WHAT SHOULD THE NEXT STEPS BE?
 

As you have been told MANY times, the step after the 0000217a call HHH, should have been the emulation of the instruction at 000015d2, the first instruction of the subroutine HHH that is part of the program DDD.
That you don't understand what a "correct emulation" is, what the semantics of the x86 instructions set is, what a "program" is, or anything related just proves your ignorance and inability to learn basic facts.
Sorry, you are just PROVING that fact, and that you are NOTHING but a pathetic ignorant pathological lying idiot.