Re: Who knows that DDD correctly simulated by HHH cannot possibly reach its own return instruction final state?

On 8/5/24 11:21 PM, olcott wrote:

On 8/5/2024 10:12 PM, Mike Terry wrote:

On 06/08/2024 03:25, olcott wrote:

On 8/5/2024 8:32 PM, Richard Damon wrote:

On 8/5/24 8:07 PM, olcott wrote:

On 8/5/2024 5:59 PM, Richard Damon wrote:

On 8/5/24 9:49 AM, olcott wrote:

On 8/5/2024 2:39 AM, Mikko wrote:

On 2024-08-04 18:59:03 +0000, olcott said:
>

On 8/4/2024 1:51 PM, Richard Damon wrote:

On 8/4/24 9:53 AM, olcott wrote:

On 8/4/2024 1:22 AM, Fred. Zwarts wrote:

Op 03.aug.2024 om 18:35 schreef olcott:

 >>>> ∞ instructions of DDD correctly emulated by HHH[∞] never

reach their own "return" instruction final state.
>
So you are saying that the infinite one does?
>

>
Dreaming again of HHH that does not abort? Dreams are no substitute for facts.
The HHH that aborts and halts, halts. A tautology.

>
void DDD()
{
   HHH(DDD);
   return;
}
>
That is the right answer to the wrong question.
I am asking whether or not DDD emulated by HHH
reaches its "return" instruction.

>
But the "DDD emulated by HHH" is the program DDD above,

>
When I say DDD emulated by HHH I mean at any level of
emulation and not and direct execution.

>
If you mean anything other than what the words mean you wihout
a definition in the beginning of the same message then it is
not reasonable to expect anyone to understand what you mean.
Instead people may think that you mean what you say or that
you don't know what you are saying.
>

>
If you don't understand what the word "emulate" means look it up.
>
DDD (above) cannot possibly reach its own "return" instruction halt
state when its machine code is correctly emulated by HHH.
>

>
Only because an HHH that does so never returns to anybody.
>

>
Do you really not understand that recursive emulation <is>
isomorphic to infinite recursion?
>

>
Not when the emulation is conditional.
>

>
Infinite_Recursion() meets the exact same condition that DDD
emulated by HHH makes and you know this. Since you are so
persistently trying to get away contradicting the semantics
of the x86 language the time is coming where there is zero
doubt that this is an honest mistake.
>
Ben does correctly understand that the first half of the Sipser
approved criteria is met. Even Mike finally admitted this.

>
I don't recall doing that.  Please provide a reference for this.
>

 On 8/2/2024 8:19 PM, Mike Terry wrote:
 > It's easy enough to say "PO has his own criterion for
 > halting, which is materially different from the HP condition,
 > and so we all agree PO is correct by his own criterion...
 

(Of course, everything depends on what you take Sipser's quote to be saying.  I choose to interpret it as I'm pretty confident that Sipser intended, under which the first half is mpst certainly NOT met!)
>
>
Mike.
>
>

 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then
      H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
 void DDD()
{
   HHH(DDD);
   return;
}
 It is certainly the case that DDD correctly simulated by any
HHH cannot possibly stop running unless aborted.
 I don't see how any expert in the C language can deny that
with a straight face. Four have affirmed it. Two of these
four have masters degrees in computer science.
 

The problem is that this only works with the correct definition of "Correctly Simulated" but not YOUR definition of Correctly Simulated.
By the correct defintition, that statement only applies to the one DDD built by the one HHH that actually does the actually correct simulation and never aborts. And that means that all the other HHHs that do abort, can not use that property on the DDD that calls them,
By YOUR definition, which tries to accept partial simulations as "correct", it just isn't true. The Simulation by HHH of DDD, won't reach the return statement, but the PROGRAM DDD that the at HHH only partially simulated, will continue past that point and reach the return.
Your intentional confusion of the partial simulation of DDD by HHH with the actual behavior of the full program DDD that the HHH only partially simulated just shows how much of a pathological liar you are.
So, You are hoisted on your own petard, and proved to be just a pathological liar.
Your claim shows that right definition, that you then refuse to use, or shows that the definiton you are using just isn't the right definition.