Sujet : Re: key error in all the proofs --- Mike's correction
De : news.dead.person.stones (at) *nospam* darjeeling.plus.com (Mike Terry)
Groupes : comp.theoryDate : 14. Aug 2024, 21:56:36
Autres entêtes
Message-ID : <LcucnRYb5ZiYhyD7nZ2dnZfqnPednZ2d@brightview.co.uk>
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On 14/08/2024 18:45, olcott wrote:
On 8/14/2024 11:31 AM, joes wrote:
Am Wed, 14 Aug 2024 08:42:33 -0500 schrieb olcott:
On 8/14/2024 2:30 AM, Mikko wrote:
On 2024-08-13 13:30:08 +0000, olcott said:
On 8/13/2024 6:23 AM, Richard Damon wrote:
On 8/12/24 11:45 PM, olcott wrote:
>
*DDD correctly emulated by HHH cannot possibly reach its* *own
"return" instruction final halt state, thus never halts*
>
Which is only correct if HHH actuallly does a complete and correct
emulation, or the behavior DDD (but not the emulation of DDD by HHH)
will reach that return.
>
A complete emulation of a non-terminating input has always been a
contradiction in terms.
HHH correctly predicts that a correct and unlimited emulation of DDD
by HHH cannot possibly reach its own "return" instruction final halt
state.
>
That is not a meaningful prediction because a complete and unlimited
emulation of DDD by HHH never happens.
>
A complete emulation is not required to correctly predict that a
complete emulation would never halt.
What do we care about a complete simulation? HHH isn't doing one.
>
Please go read how Mike corrected you.
Lol, dude... I mentioned nothing about complete/incomplete simulations.
But while we're here - a complete simulation of input D() would clearly halt. You have seen that yourself, e.g. with main() calling DDD(), or UTM(DDD), or HHH1(DDD). [All of those simulate DDD to completion and see DDD return. What I said earlier was that HHH(DDD) does not simulate DDD to completion, which I think everyone recognises - it aborts before DDD() halts.
Mike.
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