Re: D correctly simulated by H proved for THREE YEARS --- finite string transformation rules

On 6/12/24 7:53 PM, olcott wrote:

On 6/12/2024 6:41 PM, Richard Damon wrote:

On 6/12/24 7:12 PM, olcott wrote:

On 6/12/2024 5:59 PM, Richard Damon wrote:

On 6/12/24 12:50 PM, olcott wrote:

On 6/12/2024 6:33 AM, Richard Damon wrote:

On 6/11/24 11:34 PM, olcott wrote:

On 6/11/2024 9:32 PM, Richard Damon wrote:

On 6/11/24 8:57 PM, olcott wrote:

On 6/11/2024 7:20 PM, Python wrote:

Le 12/06/2024 à 01:23, olcott a écrit :
...

It turns out that by the generic definition of a decider
what the directly executed D(D) does is not any of the
business of H.

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LOL
>

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There are no finite string transformations from the input
to H to the behavior of D(D), thus the behavior of D(D)
is irrelevant.
>

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Of course there is.
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That is exactly what the definition of a UTM is.
>

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Unless we are as concrete as the x86 language truth slips
though the cracks of vagueness.
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Show each step of DDD correctly simulated by HH such that
DDD terminates normally.

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WHy? I never claimed that to be true.
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The lack of finding a couter example doesn't prove that no counter example exists, it might just not be discovered.
>

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On 5/29/2021 2:26 PM, olcott wrote:
https://groups.google.com/g/comp.theory/c/dTvIY5NX6b4/m/cHR2ZPgPBAAJ
>
*This is still Truthmaker Maximalism*
The actual behavior of the input to H(D,D) is the truthmaker
for halt decider H.

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Which is DEFINED to be the behavior of the program described by the input when directly run.
>

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*If that was true then you could show ALL OF THE DETAILED STEPS*
*of the mapping that H(D,D) computes to derive that behavior*
>

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NO, because I never said that H COULD do that computation. That is whythe question is DOES THERE EXIST a machine that can do it.
>

 There are no finite string transformation rules that H(D,D)
can use to transform its finite string input into the behavior
that you expect.

So?
There is a mapping, and thus the question is VALID.
That there is no finite steps to compute that mapping meaning that the Halting Function is just uncomputable.
That is a perfectly fine result.

 Halt Deciders are only allowed to report on the behavior that
they can map their finite string input to.
 

Nope, you have the issues backwards.
They only CAN report on what they can map with a finite algortithm.
To be a Halt Decider, they MUST try to compute the defined mapping, which, as you state, can not be done

You are finally admitting that you are expecting H to report
on something that it cannot see and not report on that it does see.

Yep.
To be a Halt decider, it must try to compute something that isn't defined by a finite algorithm. Since, it turns out we can actually PROVE that there is no finite algorithm that CAN do it, we can conclude, as the Theorem states, that the Halting Functino is uncomputable.
Nothing wrong with that.

 _D()
[00000cfc](01) 55          push ebp
[00000cfd](02) 8bec        mov ebp,esp
[00000cff](03) 8b4508      mov eax,[ebp+08]
[00000d02](01) 50          push eax       ; push D
[00000d03](03) 8b4d08      mov ecx,[ebp+08]
[00000d06](01) 51          push ecx       ; push D
[00000d07](05) e800feffff  call 00000b0c  ; call H
[00000d0c](03) 83c408      add esp,+08
[00000d0f](02) 85c0        test eax,eax
[00000d11](02) 7404        jz 00000d17
[00000d13](02) 33c0        xor eax,eax
[00000d15](02) eb05        jmp 00000d1c
[00000d17](05) b801000000  mov eax,00000001
[00000d1c](01) 5d          pop ebp
[00000d1d](01) c3          ret
Size in bytes:(0034) [00000d1d]