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On 11/3/24 9:39 AM, olcott wrote:Yes this is exactly correct. I don't understand>You have that somewhat backwards. It *CAN* only do what it can compute.
That is why I used to fully defined semantics of the x86
language to make this 100% perfectly unequivocal.
>
A few lines of x86 code express complex algorithms
succinctly enough that human minds are not totally
overwhelmed by far too much tedious detail.
>It is not pspecified>
in the usual formulation of the problem. Also note that
the behaviour exists before those strings so "describe"
should be and usually is used instead of "specify". The
use of latter may give the false impression that the behaviour
is determined by those strings.
>
In order for any machine to compute the mapping from
a finite string it must to so entirely on the basis
of the actual finite string and its specified semantics.
The mapping is not required to *BE* computable.
>Right, and it must CORRECTLY determine what an unbounded emulation of that input would do, even if its own programming only lets it emulate a part of that.
The finite string input to HHH specifies that HHH
MUST EMULATE ITSELF emulating DDD.
DDD emulated by HHH specifies that HHH will>But the semantics of the string haven't changed, as the string needs to contain all the details of how the machine it is looking at will work.
The finite string input to HHH1 specifies that HHH1
MUST NOT EMULATE ITSELF emulating DDD.
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>>You have cause and effect backwards.
Unless HHH rejects its input DDD as non halting the
executed DDD never stops running. This itself proves
that HHH is correct and that DDD is not the same
instance as the one that HHH rejected.
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