Re: This function proves that only the outermost HHH examines the execution trace

On 7/28/24 7:54 PM, olcott wrote:

On 7/28/2024 4:23 AM, Mikko wrote:

On 2024-07-27 18:20:19 +0000, olcott said:
>

On 7/27/2024 1:14 PM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:
>

Stopping running is not the same as halting.
DDD emulated by HHH stops running when its emulation has been aborted.
This is not the same as reaching its ret instruction and terminating
normally (AKA halting).

>
I think you're wrong, here.  All your C programs are a stand in for
turing machines.  A turing machine is either running or halted. There is
no third state "aborted".

>
Until you take the conventional ideas of
(a) UTM
(b) TM Description
(c) Decider
and combine them together to become a simulating partial halt decider.

>
You also need the conventional ideas of halting and halt decider.
The latter is largely a combination of the conventional ideas of
decider and halting but also involves the conventional of
prediction, so you need that, too.
>
Although the conventional idea of testing is not relevant to the construction of a simulating partial halt decider it is helpful to presentation of the
result, especially if your target audience contains software engineers. If your target audience is mainly mathematicians the convnetional idea of proofs is more useful because in that case most of your presentation must be proofs.
>

 My ideas must be anchored in fully specified running software
otherwise the false assumptions made by computer science people
remain hidden.

And that software PROVES your ideas wrong.

 Even when I slap them in the face with proven facts they deny
these proven facts on the basis of their indoctrination.
 Even Mike is trying to get away with saying that DDD correctly
emulated by HHH according to the semantics specified by the
machine code of DDD and the machine code of HHH when DDD calls
HHH(DDD) in recursive emulation is incorrectly emulated.
 HHH(DDD) has the exact same pattern as Infinite_Recursion()
where there are no conditional branch instructions that would
prevent the first three instructions of Infinite_Recursion()
from endlessly repeating.

How do you say that?
Infinite_Recursion calls itself and HHH can trace a full loop of the cycle to see no conditionals, allowing the inductive proof to be formed.
DDD calls HHH(DDD), which, isn't the same as calling DDD.
If HHH does an ACTUAL x86 emulation, it will  emulate into HHH and NEVER see the actual path of execution get back to DDD, as HHH never actually "jumps" or "calls" DDD, but just emulates it, which is only a logical action at a different level of interpretation.
If you relax your definitions, and allow HHH to be just a functional emulator, then it needs to know that HHH is a CONDITIONAL emulator (if it isn't then it can't ever abort to answer) and as such, that conditional gets inserted before EVERY instruction that HHH emulates, and that conditional breaks the inductive proof, unless you can show that it is never taken (which again breaks HHH as a decider0.
So, all you have shown is that you are so stupid that you think that a condtional emulation is "EXACTLY" the same thing as a direct call, even after it has been clearly pointed out to you, showing that you are nothing but a pathetic ignorant patholgocial lying idiot with a total and reckless disregard for the truth.

 void Infinite_Recursion()
{
   Infinite_Recursion();
}
 _Infinite_Recursion()
[0000215a] 55         push ebp      ; 1st line
[0000215b] 8bec       mov ebp,esp   ; 2nd line
[0000215d] e8f8ffffff call 0000215a ; 3rd line
[00002162] 5d         pop ebp
[00002163] c3         ret
Size in bytes:(0010) [00002163]
 Begin Local Halt Decider Simulation   Execution Trace Stored at:113934
[0000215a][00113924][00113928] 55         push ebp      ; 1st line
[0000215b][00113924][00113928] 8bec       mov ebp,esp   ; 2nd line
[0000215d][00113920][00002162] e8f8ffffff call 0000215a ; 3rd line
[0000215a][0011391c][00113924] 55         push ebp      ; 1st line
[0000215b][0011391c][00113924] 8bec       mov ebp,esp   ; 2nd line
[0000215d][00113918][00002162] e8f8ffffff call 0000215a ; 3rd line
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
 =====
 void DDD()
{
   HHH(DDD);
}
 _DDD()
[00002177] 55               push ebp      ; 1st line
[00002178] 8bec             mov ebp,esp   ; 2nd line
[0000217a] 6877210000       push 00002177 ; push DDD
[0000217f] e853f4ffff       call 000015d7 ; call HHH
[00002184] 83c404           add esp,+04
[00002187] 5d               pop ebp
[00002188] c3               ret
Size in bytes:(0018) [00002188]
 // executed HHH emulates 1st instance of DDD
New slave_stack at:10388d
Begin Local Halt Decider Simulation   Execution Trace Stored at:113895
[00002177][00113885][00113889] 55         push ebp      ; 1st line
[00002178][00113885][00113889] 8bec       mov ebp,esp   ; 2nd line
[0000217a][00113881][00002177] 6877210000 push 00002177 ; push DDD
[0000217f][0011387d][00002184] e853f4ffff call 000015d7 ; call HHH
 // emulated HHH emulates 2nd instance of DDD

Which ISN'T a correct x86 simulation, and for a conditional emulaiton you need to add:

New slave_stack at:14e2b5

simulated HHH decides not to abort simulation yet.

[00002177][0015e2ad][0015e2b1] 55         push ebp      ; 1st line

aimulated HHH decides not to abort simulation yet.

[00002178][0015e2ad][0015e2b1] 8bec       mov ebp,esp   ; 2nd line

simulated HHH decides not to abort simulation yet.

[0000217a][0015e2a9][00002177] 6877210000 push 00002177 ; push DDD

simulatd HHH decides not to abort simulation yet.

[0000217f][0015e2a5][00002184] e853f4ffff call 000015d7 ; call HHH

Local Halt Decider: Infinite Recursion Detected Simulation Stopped
 

But, since we have all those conditionals, that on later cycles of the loop might turn into decisions to abort, just like the outer one did, the conclusion is INCORRECT.
Sorry, you are just proving you don't understand what you are talking about.