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

On 8/20/24 4:07 PM, olcott wrote:

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*

No, it emulates DDDn, not just DDD, as DDD varies on the decider it calls, which is the decider that needs to get it right.
It is just a LIE to try to say that all the HHHns get the "same" input, becuase the input contains HHHn, and they are different.

 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.