Re: The execution trace of HHH1(DDD) shows the divergence

Am Sat, 07 Jun 2025 11:02:55 -0500 schrieb olcott:

On 6/7/2025 10:54 AM, wij wrote:

On Sat, 2025-06-07 at 10:35 -0500, olcott wrote:

On 6/7/2025 10:31 AM, wij wrote:

On Sat, 2025-06-07 at 09:57 -0500, olcott wrote:

On 6/7/2025 9:54 AM, wij wrote:

On Sat, 2025-06-07 at 09:32 -0500, olcott wrote:

The execution trace of HHH1(DDD) shows the divergence of DDD
emulated by HHH from DDD emulated by HHH1.
Shows that DDD emulated by HHH and DDD emulated by HHH1 diverges
as soon as HHH begins emulating itself emulating DDD.
>
*From the execution trace of HHH1(DDD) shown below*
DDD emulated by HHH1              DDD emulated by HHH [00002183]
push ebp               [00002183] push ebp [00002184] mov
ebp,esp            [00002184] mov ebp,esp [00002186] push 00002183
; DDD    [00002186] push 00002183 ; DDD [0000218b] call 000015c3 ;
HHH    [0000218b] call 000015c3 ; HHH *HHH1 emulates DDD once then
HHH emulates DDD once, these match*

What do you mean by "then" and "once"? That implies completion and
succession, however we are only ever simulating a single call, and
only HHH1 simulates any returns of DDD/HHH.
HHH1 simulates DDD completely, HHH recurses and aborts *inside*. HHH
does not simulate DDD once, it only enters the call, but never exits.

The next instruction of DDD that HHH emulates is at the machine
address of 00002183.
The next instruction of DDD that HHH1 emulates is at the machine
address of 00002190.

At those addresses we have the first instruction of DDD and the one after
the call, respectively.


[main -> HHH1(DDD) -> HHH(DDD) -> HHH(DDD)]

New slave_stack at:198d21  DDD emulated by HHH *This is the
beginning of the divergence of the behavior*
*HHH is emulating itself emulating DDD, HHH1 never does that*

HHH1 does simulate HHH simulating DDD.

The HP is asking for such a H that H(D)==1 iff D() halts.
You are always solving POO Problem.
>

int main()
{
     DDD(); // The HHH(DDD) that DDD calls cannot report
}          // on the behavior of its caller.

>
That is what the HP theorem says, the halting decider is not
possible.
>

The HP theorem never bothered to notice that it has contradictory
axioms. HHH(DDD) IS NOT ALLOWED TO REPORT ON THE BEHAVIOR OF ITS
CALLER.

Which axiom contradicts which?

Nope. It you who don't understand English.
 

The theory of computation does not allow a halt decider to report on the
behavior of its caller. Cite the chapter and verse where it does allow
this.

And lo, no program shall be forbidden to call the mighty halt decider.
Recursions 22:22

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.