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On 5/25/2025 3:05 PM, dbush wrote:False, as the fixed algorithm DDD (i.e. the simulated input) simulated by UTM halts. It's just that the fixed algorithm HHH does not simulate that far.On 5/25/2025 3:56 PM, olcott wrote:*VERFIED FACT**Mike understood this perfectly*>
On 5/14/2025 7:36 PM, Mike Terry wrote:
--------- Sipser quote -----
If simulating halt decider H correctly simulates its input D until H
correctly determines that its simulated D would never stop running
unless aborted then H can abort its simulation of D and correctly
report that D specifies a non-halting sequence of configurations.
----------------------------
>
we can easily interpret that as saying exactly what I said a SHD
does above. It tells PO that in the tight loop example, H correctly
simulates as far as [A], at which point it correctly determines that
"its simulated input would never stop running unless aborted", so
it can decide "non-halting".
>
All correct and natural, and no deliberately
false premises to mislead PO.
>
On 5/14/2025 7:36 PM, Mike Terry wrote:
https://al.howardknight.net/? STYPE=msgid&MSGI=%3C1003cu5%242p3g1%241%40dont-email.me%3E
And you dishonestly left out the part that immediately follows where he states that you are wrong:
>
Mike Terry Proves ---
How the requirements that Professor Sipser agreed to are exactly met
PO's problem is his misinterpretation of "its simulated input would never stop running unless aborted".Ridiculously COUNTER-FACTUALIn the case of his HHH/DD, the simulated input (DD) /does/ stop running if simulated far enough, but
void DDD()Category error, as the fixed algorithm HHH simulates for a fixed number of steps and aborts in violation of the x86 language.
{
HHH(DDD);
return;
}
At exactly what point does DDD simulated by
HHH reach its simulated "return" statement?
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