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On 3/9/25 9:25 AM, olcott wrote:Correctly emulating N steps is emulating N steps correctly.On 3/9/2025 6:17 AM, Richard Damon wrote:WHich is *NOT* a program, as it has an external reference.On 3/8/25 10:24 PM, olcott wrote:>On 3/8/2025 9:03 PM, Richard Damon wrote:>On 3/8/25 6:30 PM, olcott wrote:>On 3/8/2025 5:01 PM, dbush wrote:>On 3/8/2025 5:47 PM, olcott wrote:>On 3/8/2025 4:26 PM, dbush wrote:>On 3/8/2025 11:41 AM, olcott wrote:>On 3/8/2025 9:01 AM, dbush wrote:>On 3/8/2025 9:09 AM, olcott wrote:>On 3/8/2025 3:06 AM, Mikko wrote:>On 2025-03-07 15:11:53 +0000, olcott said:>
>The code proves otherwise>
https://github.com/plolcott/x86utm/blob/master/Halt7.c
A program does not prove. In particular, it does not prove that no
different program exists.
>
The source code 100% perfectly proves exactly what it
actually does.
The source code contains a finite sequence of truth preserving steps between axioms and a statement?
The source code 100% completely specifies every single detail
of exactly what it does on each specific input.
Saying that it does not do this is counter-factual.
>
In other words, the source code does not meet the definition of a proof, so your claim is false.
Dumb Bunny:
*Proof[0] is anything that shows that X is necessarily true*
*and thus impossibly false*
>
The source-code in Halt7.c combined with the input to HHH
conclusively proves every detail of the behavior of HHH on
this input. Disagreeing this is either foolish or dishonest.
>
A proof is a finite sequence of truth preserving steps between the axioms of a system and a true statement that show the statement is true.
>
Proof[math] tries unsuccessfully to inherit from proof[0].
I am stipulating that I have always been referring to proof[0].
And I am pointing out that it IS the same, it is just that you don't understand that "Show" implies FINITE.
>
In that single aspect you are correct.
Show that X is definitely true and thus impossibly false
by any means what-so-ever is not proof[math].
or proof[0], since you can not SHOW something "by any means" if those means are not showable due to not being finite.
>>>You are just proving your stupidity by repeating your disproved claim.>
>>>
If you cannot understand the Halt7.c conclusively proves[0]
the actual behavior of HHH(DD) this is merely your lack of
understanding and nothing more.
>
>
Sure I can understand what it does, as Halt7.c shows that the behavior of the input is to HALT since that is what DD will do when main calls it.
>
*WHEN YOU UNDERSTAND THIS THEN YOU KNOW YOU WERE WRONG*
DD correctly emulated by HHH cannot possibly reach
its own "ret" instruction and terminate normally
because DD calls HHH(DD) in recursive emulation.
>
>
But The HHH You are talking about doesn't do a correct simulation, so this statment is not applicable.
>
>
_DD()
[00002133] 55 push ebp ; housekeeping
[00002134] 8bec mov ebp,esp ; housekeeping
[00002136] 51 push ecx ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
[00002141] 83c404 add esp,+04
[00002144] 8945fc mov [ebp-04],eax
[00002147] 837dfc00 cmp dword [ebp-04],+00
[0000214b] 7402 jz 0000214f
[0000214d] ebfe jmp 0000214d
[0000214f] 8b45fc mov eax,[ebp-04]
[00002152] 8be5 mov esp,ebp
[00002154] 5d pop ebp
[00002155] c3 ret
Size in bytes:(0035) [00002155]
>Wrong, because emulaiting for "N Steps" is NOT correctly emulation.
*When we assume that HHH emulates N steps of DD then*
>
DD correctly emulated by HHH cannot possibly reach
its own "ret" instruction and terminate normally
because DD calls HHH(DD) in recursive emulation.
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