Re: Anyone that disagrees with this is not telling the truth --- V5

On 8/20/24 9:09 AM, olcott wrote:

On 8/19/2024 11:02 PM, Richard Damon wrote:

On 8/19/24 11:50 PM, olcott wrote:

On 8/19/2024 10:32 PM, Richard Damon wrote:

On 8/19/24 10:47 PM, olcott wrote:

*Everything that is not expressly stated below is*
*specified as unspecified*

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Looks like you still have this same condition.
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I thought you said you removed it.
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void DDD()
{
   HHH(DDD);
   return;
}
>
_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]
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*It is a basic fact that DDD emulated by HHH according to*
*the semantics of the x86 language cannot possibly stop*
*running unless aborted* (out of memory error excluded)

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But it can't emulate DDD correctly past 4 instructions, since the 5th instruciton to emulate doesn't exist.
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And, you can't include the memory that holds HHH, as you mention HHHn below, so that changes, but DDD, so the input doesn't and thus is CAN'T be part of the input.
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X = DDD emulated by HHH∞ according to the semantics of the x86 language
Y = HHH∞ never aborts its emulation of DDD
Z = DDD never stops running
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The above claim boils down to this: (X ∧ Y) ↔ Z

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And neither X or Y are possible.
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x86utm takes the compiled Halt7.obj file of this c program
https://github.com/plolcott/x86utm/blob/master/Halt7.c
Thus making all of the code of HHH directly available to
DDD and itself. HHH emulates itself emulating DDD.

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Which is irrelevent and a LIE as if HHHn is part of the input, that input needs to be DDDn
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And, in fact,
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Since, you have just explicitly introduced that all of HHHn is available to HHHn when it emulates its input, that DDD must actually be DDDn as it changes.
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Thus, your ACTUAL claim needs to be more like:
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X = DDD∞ emulated by HHH∞ according to the semantics of the x86 language
Y = HHH∞ never aborts its emulation of DDD∞
Z = DDD∞ never stops running
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The above claim boils down to this: (X ∧ Y) ↔ Z
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Yes that is correct.

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So, you only prove that the DDD∞ that calls the HHH∞ is non-halting.
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Not any of the other DDDn
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Your problem is that for any other DDDn / HHHn, you don't have Y so you don't have Z.
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void EEE()
{
   HERE: goto HERE;
}
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HHHn correctly predicts the behavior of DDD the same
way that HHHn correctly predicts the behavior of EEE.
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Nope, HHHn can form a valid inductive proof of the input.
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It can't for DDDn, since when we move to HHHn+1 we no longer have DDDn but DDDn+1, which is a different input.
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You already agreed that (X ∧ Y) ↔ Z is correct.
Did you do an infinite trace in your mind?

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But only for DDD∞, not any of the other ones.
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If you can do it and I can do it then HHH can
do this same sort of thing. Computations are
not inherently dumber than human minds.
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But HHHn isn't given DDD∞ as its input, so that doesn't matter.
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HHHn is given DDDn as its input,
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Remeber, since you said that the input to HHH includes all the memory, if that differs, it is a DIFFERENT input, and needs to be so marked.
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You are just admittig that you are just stupid and think two things that are different are the same.
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 *attempts to use misdirection to weasel word around this are dismissed*
*attempts to use misdirection to weasel word around this are dismissed*
*attempts to use misdirection to weasel word around this are dismissed*
 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then
 

Right, so the decider needs top be able to show that its exact input will not halt. Not just that the partial emulation by the decider if it aborts does not reach the final state, but the FULL behavior of the EXACT input does not halt. And if the input was based on the decider as HHH/DDD was, it is based on the FINAL decider that you chose, and thus HHH needs to show that the full behavior of the DDD that calls this HHH that decides to abort and return would not halt.
Since DDD halts if HHH aborts and returns, HHH can not corretly abort its emulation based on DDD never halting, since THIS DDD will halt because this HHH does abort and return (at least if it thinks that condition was meet), so HHH is just wrong about it being meet.
YOu surely don't think that Professor Sipser thought that the decider should answer about an input it wasn't given, the OTHER DDD built on that OTHER HHH that doesn't about,
If you do, you are just showing your stupidity, and if you understood this, you are showing your duplicity.