Sujet : Re: HHH maps its input to the behavior specified by it --- reaches its halt state --- Which DDD does if HHH(DDD) returns and answer, which it does since it is a decider.
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 10. Aug 2024, 04:15:47
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v96m13$dl6k$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
User-Agent : Mozilla Thunderbird
On 8/9/2024 9:56 PM, Richard Damon wrote:
On 8/9/24 10:43 PM, olcott wrote:
On 8/9/2024 9:35 PM, Richard Damon wrote:
On 8/9/24 10:24 PM, olcott wrote:
On 8/9/2024 8:54 PM, Richard Damon wrote:
On 8/9/24 9:52 PM, olcott wrote:
On 8/9/2024 8:46 PM, Richard Damon wrote:
On 8/9/24 9:25 PM, olcott wrote:
On 8/9/2024 8:05 PM, Richard Damon wrote:
On 8/9/24 8:52 PM, olcott wrote:
>
When we look at every HHH that can possibly exist then
we see that DDD correctly emulated by each one of these
cannot possibly reach its "return" instruction halt state.
>
But ONLY ONE of those actuallu "correctly emulates" the input, and that one isn't a decider.
>
>
In other words you are trying to keep getting away
with the bald-faced lie that when HHH correctly
emulates N instructions of DDD (where N > 0) that
it did not correctly emulate any instructions of DDD.
>
*Give it up you lost you are stuck in repeat mode*
*Give it up you lost you are stuck in repeat mode*
*Give it up you lost you are stuck in repeat mode*
>
>
So, I guess you don't understand English.
>
Where did I say that simulating N instructions correctly is not simulating ANY instructions correctly.
>
>
*Shown above*
"But ONLY ONE of those actuallu "correctly emulates" the input..."
>
>
Right, becuase to correctly emulate, you need to correct emulate EVERY instruction, not just some of them.
>
>
So you defining whole notion simulating termination analyzers
as incorrect even though professor Sipser has agreed that the
simulation need not be complete.
>
No, they just need to do the job right.
>
But it needs to prove that the CORRECT SIMULATION, which would be complete, doesn't ever reach a final state. T
>
void DDD()
{
HHH(DDD);
return;
}
>
You already know that a complete simulation of DDD
by a pure x86 emulator cannot possibly reach its
"return" instruction halt state.
Of course it can. as long as it isn't HHH, and the HHH that DDD was paired with gives an answer.
Your problem is thinking the only simulator allowed is HHH.
When HHH reports that its input cannot possibly reach
the "return" instruction halt state of this input it
is correct. HHH only computes the mapping from its input
to the behavior that this input specifies.
All of your every attempt to rebut this were anchored
in the strawman deception. I am beginning to have no
doubts that you are a deceiver. For your soul's sake
I hope this is an ADD issue and not willful deception.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
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