Re: DD correctly emulated by HHH --- Totally ignoring invalid rebuttals ---PSR---

On 3/9/25 6:32 PM, olcott wrote:

On 3/9/2025 4:43 PM, Richard Damon wrote:

On 3/9/25 3:36 PM, olcott wrote:

On 3/9/2025 2:24 PM, Richard Damon wrote:

On 3/9/25 9:25 AM, olcott wrote:

On 3/9/2025 6:17 AM, Richard Damon wrote:

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:
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The code proves otherwise
https://github.com/plolcott/x86utm/blob/master/Halt7.c

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A program does not prove. In particular, it does not prove that no
different program exists.
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The source code 100% perfectly proves exactly what it
actually does.

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The source code contains a finite sequence of truth preserving steps between axioms and a statement?

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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.
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In other words, the source code does not meet the definition of a proof, so your claim is false.

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Dumb Bunny:
*Proof[0] is anything that shows that X is necessarily true*
*and thus impossibly false*
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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.
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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.
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Proof[math] tries unsuccessfully to inherit from proof[0].
I am stipulating that I have always been referring to proof[0].

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And I am pointing out that it IS the same, it is just that you don't understand that "Show" implies FINITE.
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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].

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or proof[0], since you can not SHOW something "by any means" if those means are not showable due to not being finite.
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You are just proving your stupidity by repeating your disproved claim.
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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.
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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.
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*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.
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But The HHH You are talking about doesn't do a correct simulation, so this statment is not applicable.
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_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]

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WHich is *NOT* a program, as it has an external reference.
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*When we assume that HHH emulates N steps of DD then*
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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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Wrong, because emulaiting for "N Steps" is NOT correctly emulation.
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Correctly emulating N steps is emulating N steps correctly.

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Which is only partially emulating it correctly, and only partially correct is incorrect.
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Everyone here that has sufficient technical competence can
see that for any N steps of DD correctly emulated by HHH
that DD cannot possibly reach its own final state and
terminate normally.
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So? As has been pointed out, since HHH can't do enough steps to get to the actual answer, it never CORRECTLY emulated the input enough to get the answer if it aborts.
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If HHH can see the same pattern that every competent
programmer sees then HHH does not need to emulate DD
more than twice to know that HHH cannot possibly reach
its own final state and terminate normally.
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  The pattern that HHH sees is IDENTICAL to the pattern that HHH1 saw, up to the point it aborts.