Re: behavior and description --- All of the code is here it is just hard to find (No its not, no HHH given)

On 8/7/24 8:59 AM, olcott wrote:

On 8/7/2024 6:25 AM, Richard Damon wrote:

On 8/6/24 11:47 PM, olcott wrote:

On 8/6/2024 10:21 PM, Richard Damon wrote:

On 8/6/24 11:10 PM, olcott wrote:

On 8/6/2024 10:00 PM, Richard Damon wrote:

On 8/6/24 10:49 PM, olcott wrote:

On 8/6/2024 9:27 PM, Richard Damon wrote:

On 8/6/24 9:48 PM, olcott wrote:

On 8/6/2024 8:38 PM, Richard Damon wrote:

On 8/6/24 1:16 PM, olcott wrote:

On 8/6/2024 12:02 PM, joes wrote:

Am Tue, 06 Aug 2024 09:43:30 -0500 schrieb olcott:

Understanding that DDD correctly simulated by HHH cannot possibly reach
its own "return" instruction is a mandatory prerequisite to further
discussion.

>

There is nothing to discuss after agreeing with your conclusion.
>

Everyone remains convinced that HHH must report on the behavior of the
computation that itself is contained within and not the behavior that
its finite string input specifies.

>

The construction is not recursive if the description does not describe
the surrounding computation. And that behaviour cannot depend on the
decider, as they should all give the same answer.
>

>
That is far too vague.
>
DDD correctly emulated by HHH according to the semantics
of the x86 programming language specifies a single exact
sequence of state changes. None of these state changes
ends up at the x86 machine language address of the "ret"
instruction of DDD.
>

>
Which would be meaningful if HHH actual did a correct emulation of the

>
HHH does emulate the exact sequence that the machine code
of DDD specifies. This has been conclusively proven by
the execution traces that the two instances of HHH provide.

>
Nope, because it didn't emulate the call instruction properly.
>

>
It is proved that it does emulate the call instruction
properly by the correct execution trace of the second
DDD derived by the second HHH.

>
Nope, just proves you don't know what you are talking about.
>

>
Because your rebuttals have always been pure bluster

>
No, YOUR rebuttal have been pure bluster that have avoided answering my challenges, thus effectively ADMITTING that you arte just a liar.
>

you have only used double-talk and misdirection to dodge
pointing out any mistake in the following:

>
How about the trace that HHH generates shouldn't start with the code of main, since that isn't what HHH starts simulating, thus your claims about the trace generate by HHH can not be supported by something elses trace.
>

>
That is a very ignorant thing to say.

>
NO, it is the truth.
>

>

Second, the CORRECT emulation of a "call" instruciton should show the instruction of the routine called.
>

>
The extra 200 pages prove to be too confusing. The
fact that after DDD calls HHH we do get the correct
execution trace that would be derived from this call
proves that the second HHH was called.

>
Nope, they show you wrong.
>
HOW can the emulation that HHH is doing show code that HHH doesn't look at?
>
Your failure to answer (may times from the past) just prove that you KNOW that you have been lying, but don't want to face it.
>
Sorry, you are just proving your stupidity
>

>
I will highlight all of the code so that you
can see the each instruction both HHH are shown
and both DDD are shown. I will use four different
colors.

>
Since the trace isn't the right trace, that doesn't mean anything.
>
And, where are the colors, you said there would be colors.
>
Did you not know that this group is text only?
>

>
It really have already been 100% totally proven in
that after DDD calls HHH we to get the correct
execution trace of DDD emulated by this second HHH
>

Thus, a "Call 000015d2" *MUST* be followed by the instruction AT 000015d2, and not "Begin Local Halt Decider Simulation"
>
Since you have just been repeating this smae ERROR over and over, you are just proving that you just don't care about the truth, but are just repeating your same LIES over and over and over, securing your place in Gehenna.
>

 >>>> _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]
>
_main()
[00002192] 55         push ebp
[00002193] 8bec       mov ebp,esp
[00002195] 6872210000 push 00002172 ; push DDD
[0000219a] e833f4ffff call 000015d2 ; call HHH(DDD)
[0000219f] 83c404     add esp,+04
[000021a2] 50         push eax
[000021a3] 6843070000 push 00000743
[000021a8] e8b5e5ffff call 00000762
[000021ad] 83c408     add esp,+08
[000021b0] 33c0       xor eax,eax
[000021b2] 5d         pop ebp
[000021b3] c3         ret
Size in bytes:(0034) [000021b3]
>
  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[00002192][00103820][00000000] 55         push ebp
[00002193][00103820][00000000] 8bec       mov ebp,esp
[00002195][0010381c][00002172] 6872210000 push 00002172 ; push DDD
[0000219a][00103818][0000219f] e833f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:1038c4
>
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55         push ebp      ; housekeeping
[00002173][001138bc][001138c0] 8bec       mov ebp,esp   ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec

>
This execution trace proves that the second HHH did
emulate the second DDD correctly:
>

>
Nope, where is the fact that that either HHH has the power to stop any of this from happening by aborting there simulation of it?
>

 I am going to stop right here because you
seem to fail to understand this point:

So, what you do think I don't understand about those?
Those rules are talking about the EXECUTION of PROGRAMs.
To be a PROGRAM, DDD needs to include the HHH that it calls, or none of that applies, as it isn't a program.
A

 On 8/2/2024 11:32 PM, Jeff Barnett wrote:
 > ...In some formulations, there are specific states
 >    defined as "halting states" and the machine only
 >    halts if either the start state is a halt state...
  > ...these and many other definitions all have
 >    equivalent computing prowess...
 https://liarparadox.org/HHH(DDD)_Full_Trace.pdf
 

Omitting that conditionality makes the trace just an incorrect emulation by any defintion.
>

 void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}
 It does not.

Sure it does.
Or are you saying that HHH doesn't CONDITIONALLY emulate its input, and thus it can't ever abort it??

*If we did not divide the tester from the test subject*
then Infinite_Recursion() would be determined to halt
on the basis that HHH halts.
 

No, because the tester has no effect on the behavior of that program.
When it decides on a program that calls itself, it needs to look at the actual code it was given, even if it includes ITSELF, or the programmer is just a LIAR,.

Instead of that we define reaching the "return"
halt state of Infinite_Recursion() as halting.
 

Right, and you can prove from a partial emualtion of that Infinite_Recursion, that if you gave the exact same input, to a different emulator, like a real UTM, that said emulator would never reach a final state.
You can't do that with DDD and HHH, as the input DDD, INCLUDEES the HHH that it was pair with, and if you give that DDD/HHH pairing to a UTM, it will trace DDD calling HHH(DDD) then HHH emulating that copy of DDD for a bit, then it deciding to abort, and returning to DDD and then DDD reaching that FINAL STATE of the return instuction.
Since we have this as the actual correct emulation of the input if HHH EVER aborts as returns, if its code does that, it is just wrong.
It is just incorrect to argue about a DDD that calls some other HHH, that it doesn't, as that is just a lIE.