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

On 8/20/2024 1:55 PM, joes wrote:

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.
 

A change of one line of code would do this.

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.
 

*This is HHHn and it does emulate itself emulating DDD*
x86utm takes the compiled Halt7.obj file of this c program
https://github.com/plolcott/x86utm/blob/master/Halt7.c
Thus making all of the code of HHH directly available to
DDD and itself. HHH emulates itself emulating DDD.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer