Re: ChatGPT agrees that HHH refutes the standard halting problem proof method

On 7/1/2025 6:28 AM, Richard Damon wrote:

On 6/30/25 9:26 PM, olcott wrote:

On 6/30/2025 8:10 PM, Richard Damon wrote:

On 6/30/25 1:00 PM, olcott wrote:

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One line of C source-code is a C statement.
HHH simulates six statements of DDD.

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No it doesn't, as that line of C refers to HHH, and to process that line, you need to process ALL the lines in HHH.
>

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Yes this is true.
What the F did you think that I meant by:
>
HHH simulates DDD that calls HHH
that simulates DDD that calls HHH
that simulates DDD that calls HHH
that simulates DDD that calls HHH
that simulates DDD that calls HHH
that simulates DDD that calls HHH...

 Except that isn't what you said HHH does!
 YOu said HHH simulated DDD until it recognizes a non-halting pattern.
 You have omitted this in your "loop"
 

Recursive emulation is not a loop.

It should be:
HHH simulated DDD that calls HHH, until it recognizes a non-halting pattern,
Which results in it simulating HHH simulating DDD until it recognizes a non-halting pattern...
Which results in it simulating HHH simulating DDD until it recognizes a non-halting pattern...
Which results in it simulating HHH simulating DDD until it recognizes a non-halting pattern...
Which results in it simulating HHH simulating DDD until it recognizes a non-halting pattern...
 

It is more precisely accurate the way that you
did it yet too confusing to get the gist of the
idea of recursive emulation.

The problem is when you include that we KNOW that, since the outer HHH *WILL* at some point abort (since you assume that will happen) that this simulated HHH will also do that, and thus make the DDD that called it halting.
 

If you are going to call impossibly reaching its final halt state
halting you might as well call it also makes you breakfast in bed.

Your problem is you didn't CORRECTLY simulate the HHH that DDD calls, as you ERRONEOUSLY assumed that it will not halt in order to claim that you have a non-halting pattern.
 

When N x86 instructions of DDD are simulated
according to the semantics of the x86 language
then N N x86 instructions of DDD are simulated
correctly. This includes HHH simulating itself
simulating DDD at least once.
I don't understand why this is so difficult for
you unless you grossly exaggerated your competence
at programming.

THe problem is whatever criteria is used to abort, is part of the code that is being analyized, and thus you need to take that into account when you try to prove that the pattern is non-halting.
 

Repeat this to yourself 500 times so that you will
remember it by the time you make your next reply.
*DDD correctly simulated by HHH cannot possibly reach*
*its own simulated "return" statement final halt state*
*DDD correctly simulated by HHH cannot possibly reach*
*its own simulated "return" statement final halt state*
*DDD correctly simulated by HHH cannot possibly reach*
*its own simulated "return" statement final halt state*

Your "logic" doesn't understand how programs work and are defined, because your "logic" comes out of your own ignorance of the field.
 

>
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You are just showing you don't understand the basics of how computers and programs work.
>

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Note, "C" doesn't define "instructions", but operations as defined by the abstract machine.
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The operations defined in DDD:
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Fetch the value of DDD
Pass that as a parameter to HHH
Call the funciton HHH,
Perform the operations of function HHH
Return
>

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At the machine language level HHH correctly
simulated four x86 instructions of DDD six times.

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Nope, doesn't simulate the CALL instruction.
>

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Yes it does.

 Then why doesn't it show the x86 instuctions executed?
 Of the sequence points inside of the HHH that it called?
 

>
What the F did you think that I meant by:
HHH simulates DDD that calls HHH
that simulates DDD that calls HHH
that simulates DDD that calls HHH
that simulates DDD that calls HHH
that simulates DDD that calls HHH
that simulates DDD that calls HHH...
>
>

 As I said, that isn't a simulation of HHH, as that isn't what HHH doess, because it LIES about the fact that HHH, as you have defined it, *WILL* abort and return 0, and thus every DDD will halt.
 All you are doing is proving that you don't understand what you are talkinga about, and just refuse to look at the facts, because you are just a pathological liar that has been brainwashed by yourself into unconditionaly believing your own lies.

--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer