Re: Even Google AI Overview understands me now --- HHH(DDD)==0

On 10/7/2024 7:48 PM, Richard Damon wrote:

On 10/7/24 7:36 PM, olcott wrote:

On 10/7/2024 6:04 PM, Richard Damon wrote:

On 10/7/24 6:57 PM, olcott wrote:

On 10/7/2024 5:39 PM, Richard Damon wrote:

On 10/7/24 8:08 AM, olcott wrote:

On 10/7/2024 6:02 AM, Richard Damon wrote:

On 10/6/24 10:05 PM, olcott wrote:

On 10/6/2024 8:29 PM, Richard Damon wrote:

On 10/6/24 8:05 PM, olcott wrote:

On 10/6/2024 5:36 PM, Richard Damon wrote:

On 10/6/24 3:05 PM, olcott wrote:

On 10/6/2024 1:52 PM, Richard Damon wrote:

On 10/6/24 2:32 PM, olcott wrote:

On 10/6/2024 1:13 PM, Richard Damon wrote:

On 10/6/24 2:03 PM, olcott wrote:

On 10/6/2024 12:29 PM, Richard Damon wrote:

On 10/6/24 1:07 PM, olcott wrote:

On 10/6/2024 11:59 AM, Richard Damon wrote:

On 10/6/24 8:39 AM, olcott wrote:

>
DDD emulated by each corresponding HHH that can possibly
exist never returns. Each of these HHH emulators that does
return 0 correctly reports the above non-halting behavior.
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No, the DDD return (if the HHH(DDD) gives an answer), just after the HHH that emulated them gave up.
>

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DDD emulated by each corresponding HHH that can possibly
exist never returns.
>
DDD emulated by each corresponding HHH
DDD emulated by each corresponding HHH
DDD emulated by each corresponding HHH
DDD emulated by each corresponding HHH

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Which, as you have been told but seems to be above your head means that the execution of DDD,

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gets to ignore the fact that DDD was defined to
have a pathological relationship with HHH that
HHH cannot ignore.
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No, that isn't ignoring it, but taking into account that since HHH is defined to be a specific program, it has specific behavior.
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The behavior of the executed DDD after the emulated
DDD has already been aborted is different than the
behavior of the emulated DDD that must be aborted.

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Nope, it is the exact same code on the exact same data, and thus does the exact same behavior.
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The execution trace proves that the executed DDD has
different behavior that need not be aborted because
emulated DDD must be an is aborted.

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Nope, whst instruction ACTUALLY EMULATE showed a different behavior than the executed DDD?
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All you do is look at a DIFFERENT INPUT which is just a lie, since that isn't the DDD that HHH was given (since the PROGRAM DDD includes the all the exact code of the HHH that it calls, thus you can't change it to hypothosze a diffferent non- aborting HHH)
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No one can be stupid enough to think that:
MUST BE ABORTED
   is exactly the same as
NEED NOT BE ABORTED
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Who said otherwise.
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The directly executed DDD need not be aborted.
DDD emulated by HHH must be aborted, thus
proving that their behavior IS NOT THE SAME.
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No, the design of HHH does abort its emulation, because if you had a DIFFERENT HHH, which would be given a DIFFERENT DDD (since DDD includes the HHH that it is calling) it would fail worse at the task at the meta- level by not answering.
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That you are not addressing my points seems to be over your head.
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No, the fact that I *AM* adddressing your points and pointing out your error just proves that you are nothing but a stupid idiot.
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That you don't even try to point out an error in what I say, proves that you don't actually care about what is right, but that you just want to blindly hold on to your position. The fact that you consistantly snip out much of the arguement shows that you know you are defeated, but still insist on your WRONG position.
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Halting is a property of PROGRAMS.

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void DDD()
{
   HHH(DDD);
   return;
}
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Terminating is a property of finite string machine descriptions.
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And, for the PROGRAM DDD, must include the FULL decription  of the HHH that it calls.
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It does and the source-code specifies that it does
yet this is simply over-your-head.
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https://github.com/plolcott/x86utm/blob/master/Halt7.c
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But that isn't the finite string you are claiming above.
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When you include the code of HHH in DDD, then when you hypotosize HHH not aborting, that hypothetical HHH is still given the DDD that calls the HHH that DOES, and your hypothetical HHH proves that this HHH is wrong.
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No it continues to be you fail to pay complete attention
to every detail of every words that I said.
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*THE FOLLOWING REMAINS TRUE NO MATTER WHAT HHH DOES*
DDD emulated by each corresponding HHH that can possibly
exist never returns.
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 No, because *DDD*, the one that was emulated by each of the HHH, will HALT if that HHH(DDD) returns 0, just after its HHH aborted its emulaiton.
 

PLEASE PLAY 100% COMPLETE ATTENTION TO THESE EXACT WORDS
DDD emulated by each corresponding HHH
that can possibly exist never returns.
Your incorrect paraphrase of my words proves that
you are either clueless or deliberately deceptive.
I am going with clueless.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer