Re: Flat out dishonest or totally ignorant? Liar ??? Maybe...

On 7/3/24 10:41 PM, olcott wrote:

On 7/3/2024 9:34 PM, Richard Damon wrote:

On 7/3/24 10:05 PM, olcott wrote:

On 7/3/2024 8:47 PM, Richard Damon wrote:

On 7/3/24 9:26 PM, olcott wrote:

On 7/3/2024 8:17 PM, Richard Damon wrote:

On 7/3/24 8:40 PM, olcott wrote:

On 7/3/2024 6:18 PM, Richard Damon wrote:

On 7/3/24 10:19 AM, olcott wrote:

On 7/3/2024 9:11 AM, joes wrote:

Am Tue, 02 Jul 2024 22:55:12 -0500 schrieb olcott:

On 7/2/2024 10:50 PM, joes wrote:

Am Tue, 02 Jul 2024 14:46:38 -0500 schrieb olcott:

On 7/2/2024 2:17 PM, Fred. Zwarts wrote:

Op 02.jul.2024 om 21:00 schreef olcott:

On 7/2/2024 1:42 PM, Fred. Zwarts wrote:

Op 02.jul.2024 om 14:22 schreef olcott:

On 7/2/2024 3:22 AM, Fred. Zwarts wrote:

Op 02.jul.2024 om 03:25 schreef olcott:

>

HHH repeats the process twice and aborts too soon.

>
DDD is correctly emulated by any HHH that can exist which calls this
emulated HHH(DDD) to repeat the process until aborted (which may be
never).

Whatever HHH does, it does not run forever but aborts.
>

HHH halts on input DDD.
DDD correctly simulated by HHH cannot possibly halt.

WTF? It only calls HHH, which you just said halts.
>

>
An aborted simulation does not count as halting.

>
And doesn't show non-halting either.
>

Reaching it own machine address 00002183 counts as halting.
DDD correctly simulated by HHH cannot possibly do that.

>
But HHH doesn't DO a "Correct Simulation" that can show that, it only does a PARTIAL simulation.
>

>
<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>
>
until H correctly determines
until H correctly determines
until H correctly determines
until H correctly determines
until H correctly determines
until H correctly determines
until H correctly determines

>
Which it doesn't.
>

>
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
>

>
Nope, just double talk.
>
H never CORRECTLY determined that a CORRECT SIMULATION (which means one that matchs the behavior of the machine represented by the input) would never halt, sinc ehta tmachine halts.
>

>
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
>
OK so it is not your ADD you continue to insist that
you can disagree with the x86 language that conclusively
proves that DDD correctly simulated by HHH cannot
possibly get past machine instruction 0000217a.

>
No, the claim is that it isn't the simulation by HHH that determines the actual behavior of the input, but the detailed semantics of the x86 instruciton set of the WHOLE input (which includes this particular HHH which you say DOES abort its simulation and return).

>
You are not paying close enough attention to the exact words
that I am saying. DDD cannot possibly reach past its own
machine address 0000217a no matter what the Hell that HHH does.
>
>

>
OF course it can. HHH might not be able to simulate it getting there,

 Because the machine instructions provided to HHH specify that
it is not possible to go there. HHH uses an x86 emulator with
two decades of development effort.

No, the exact same instructions that HHH used when called by main are in the presentation to HHH by DDD, and since the first one returned, so will the COMPLETE (which is what the semantic require) emulation of those exact same x86 instructions of HHH to also return, and thus DDD will return.
The fact that HHH gives up part way on the inner emulation it is doing doesn't affect the semantics of the instructions that were there.

 Correctly simulated means OBEYS THE X86 SEMANTICS.
Correctly simulated DOES NOT MEAN do what Richard expects.
 

Right, so if Main calling HHH(DDD) causes HHH to return to main, the x86 instruction in HHH, which are ALSO presented to HHH, will cause the COMPLETE emulation of that code (not the partial one by HHH) to return.
You are just stuck in your lies.
I guess you don't understand that programs are exactly what they are and do exactly what they do, and just because one emulator stooped its emulation doesn't change that behavior.
You keep on trying to pull a "shell game" and attempt to change HHH and DDD between tests, that is just lying.