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On 5/10/2025 12:27 AM, wij wrote:Only when the simulation ignores the most relevant part of the input and halts before it can see it. Even then the only difference is that all steps that are simulated are exactly the same as the direct execution and there is no reason to think that the behaviour specified in the input would be different in the following steps.On Sat, 2025-05-10 at 00:19 -0500, olcott wrote:The notion that the behavior specified by the finiteOn 5/10/2025 12:13 AM, wij wrote:>On Sat, 2025-05-10 at 00:06 -0500, olcott wrote:>>>When mathematical mapping is properly understood>
it will be known that functions computed by models
of computation must transform their input into
outputs according to the specific steps of an
algorithm.
>
_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]
>
For example HHH(DDD) only correctly map to the
behavior that its input actually specifies by correctly
emulating DDD according to the rules of the x86 language.
>
This causes the first four instructions of DDD
to be emulated followed by HHH emulating itself
emulating the first three instructions of DDD.
>
It is right at this recursive simulation just
before HHH(DDD) is called again that HHH recognizes
the repeating pattern and rejects DDD.
Yes, but you still did not answer the question: Is POOH exactly about HP?
>
>>>>> H(D)=1 if D() halt.
>>>>> H(D)=0 if D() not halt.
>
Right now it is mostly about proving the
above requirements are is mistaken.
>
Why is the requirement invalid?
>
H(D)=1 if D() halt.
H(D)=0 if D() not halt.
>
string input to a simulating termination analyzer
does sometimes differ from the behavior of its direct
execution. It is a provably different sequence of steps.
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