Re: DDD correctly emulated by HHH cannot possibly halt Because HHH that correctly simulates does not Halt

On 7/11/24 11:30 PM, olcott wrote:

On 7/11/2024 10:18 PM, Richard Damon wrote:

On 7/11/24 10:28 PM, olcott wrote:

On 7/11/2024 9:08 PM, Richard Damon wrote:

On 7/11/24 10:51 AM, olcott wrote:

On 7/10/2024 8:50 PM, Richard Damon wrote:

On 7/10/24 9:21 PM, olcott wrote:

On 7/10/2024 8:11 PM, Richard Damon wrote:

On 7/10/24 9:01 PM, olcott wrote:

On 7/10/2024 7:37 PM, Richard Damon wrote:

On 7/10/24 8:24 PM, olcott wrote:

On 7/10/2024 7:01 PM, Richard Damon wrote:

On 7/10/24 9:41 AM, olcott wrote:

On 7/10/2024 8:27 AM, joes wrote:

Am Tue, 09 Jul 2024 23:19:25 -0500 schrieb olcott:

On 7/9/2024 11:01 PM, joes wrote:
  > That means that HHH doesn't return, in particular that it doesn't
  > abort.
DDD correctly emulated by any pure function HHH that correctly emulates
1 to ∞ steps of DDD can't make it past the above line of code no matter
what.

That line being the call to itself -> it can't simulate itself.
>

*DDD NEVER HALTS*

DDD ONLY calls HHH...
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ lines of DDD can't make it
to the second line of DDD no matter what.
>
>

>
Nope, DDD does if HHH(DDD) returns.
>

>
You have a dead cat in your driveway does not mean that
you have a peanut butter sandwich on your front porch.
It has taken you at least 1000 messages to see that.
>
DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ lines of DDD can't make it
to the second line of DDD no matter what.

>
WRONG, you don't seem to understand the difference between DDD and HHH's emualtion of it.
>
>

>
Would you bet your immortal soul that DDD simulated
by HHH (as provided above) would terminate normally?
>

>
That is a ambiguous statement, showing your attempt at deciet.
>

>
We stipulate that the only measure of a correct emulation
is the semantics of the x86 programming language. By this
measure when 1 to ∞ steps of DDD are correctly emulated
by each pure function x86 emulator HHH (of the infinite
set of every HHH that can possibly exist) then DDD cannot
possibly reach its own machine address of 00002174 and halt.

>
And thus you stipulate that you are a LIAR.
>
By the semantic of the x86 programming language, the only correct simulation is a FULL simulation

In other words you are trying to get away with the lie that
when 1 step of DDD is correctly emulated that 0 steps of DDD
are correctly emulated.
>
Repent of this lie or risk damnation.
>

>
WHAT LIE?
>

>
When 1,2,3... ∞ steps of DDD are correctly emulated by
HHH it is a lie to say that this many instructions were
not correctly emulated and you know it.
>

>
But only N instructions "correctly emulated" is NOT a CORRECT emulaition of the instructions of DDD/HHH
>

 I didn't limit it to N. Is this your ADD? I say 1 to infinity steps !!!
I didn't limit it to N. Is this your ADD? I say 1 to infinity steps !!!
I didn't limit it to N. Is this your ADD? I say 1 to infinity steps !!!
I didn't limit it to N. Is this your ADD? I say 1 to infinity steps !!!
I didn't limit it to N. Is this your ADD? I say 1 to infinity steps !!!
 

So, I guess NONE of them ever stop before reaching the end, if none of them stop before that.
So ALL of them will run forever an never abort
So NONE of them ever answer
and ALL of them fail to be a decider
so NONE of them correctly answer
and EVREYTHING you said has been about HHH being a decider is just a LIE.
showing you know NOTHING of which you talk about.
So, 1,2,3,... ∞ means the go until they reach a final state, and thus you talk about it being ok to abort was just lies that you have seen the light now.
So, I guess you just flush your theory down the toilet as