Re: Anyone that disagrees with this is not telling the truth --- V5

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Sujet : Re: Anyone that disagrees with this is not telling the truth --- V5
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theory
Date : 20. Aug 2024, 19:55:08
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <fe16a31bc47457c310e545d61f7969f57a60b56f@i2pn2.org>
References : 1 2 3 4 5 6 7
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Tue, 20 Aug 2024 08:18:57 -0500 schrieb olcott:
On 8/20/2024 5:29 AM, Fred. Zwarts wrote:
Op 20.aug.2024 om 06:33 schreef olcott:
On 8/19/2024 11:02 PM, Richard Damon wrote:
On 8/19/24 11:50 PM, olcott wrote:
On 8/19/2024 10:32 PM, Richard Damon wrote:
On 8/19/24 10:47 PM, olcott wrote:

But HHHn isn't given DDD∞ as its input, so that doesn't matter.
All of the DDD have identical bytes it is only the HHH that varies.
HHHn(DDD) predicts the behavior of HHH∞(DDD).
Not all HHH can be at the same memory at the same time.
Counter factual. HHH∞ is hypothetical thus takes no memory.
HHH and DDD remains at the same physical machine address locations.
HHH_oo can be implemented.

When HHHn is in the memory, then DDD calls HHHn, not HHH∞.
When HHHn is doing the simulation, HHHn is in that memory, therefore,
it should simulate HHHn, not HHH∞.
They cannot be at the same memory location at the same time, unless you
are cheating with the Root variable to switch between HHHn and HHH∞,
which causes HHHn to process the non-input HHH∞ instead of the input
HHHn.
HHH∞ is hypothetical thus takes no memory. HHHn(DDD) predicts the
behavior of a hypothetical HHH∞(DDD) as described below
HHHn should simulate itself, and HHH_oo should also be simulated by
itself.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.

Date Sujet#  Auteur
6 Jul 25 o 

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