Sujet : Re: Peter Olcott here seems to consistently lie about this ---
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theoryDate : 02. Aug 2024, 03:33:28
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <8c7fcb21223d5d1b19e03d484024b02eab5aaadf@i2pn2.org>
References : 1 2
User-Agent : Mozilla Thunderbird
On 8/1/24 10:12 PM, olcott wrote:
*This algorithm is used by all the simulating termination analyzers*
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
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.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
But only for th right definition of "Correctly Simulated" which means of the exact input without aborting.
DDD is correctly emulated by HHH according to the x86
language semantics of DDD and HHH including when DDD
emulates itself emulating DDD
Nope.
Call HHH needs to be followed in the trace by the instructions of HHH
And you "full Trace" printouts are NOT the trace that HHH Makes, but are traces OF HHH doing its decision.
Also, if HHH aborts its simulaiton, it is NOT "Correct" per the x86 lnguage semantics, as those semantics continue to process the code until it reaches a final state at the return in DDD (at least since HHH does decide to abort and return).
*UNTIL*
Which is what makes in not correct, and the decision about it not based on a correct simulation,
HHH correctly determines that never aborting this
emulation would cause DDD and HHH to endlessly repeat.
Nope, HHH determines that the DDD built on a DIFFERENT HHH will not halt, and uses LIES to claim it applies to this DDD
When I say everyone I mean:
Joes, Fred, Richard, Mike, Mikko, Andy, André...
*Excluding only Ben Bacarisse*
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
> I don't think that is the shell game. PO really /has/
> an H (it's trivial to do for this one case) that correctly
> determines that P(P) *would* never stop running *unless*
> aborted.
...
> But H determines (correctly) that D would not halt if
> it were not halted. That much is a truism.
Nope, because you change the input DDD by changing the HHH that it calls, which is invalid.
You are just proving that you don't understand how logic or programing or truth work at all.