Re: Who here understands that the last paragraph is Necessarily true?

On 7/19/24 10:23 AM, olcott wrote:

On 7/19/2024 2:57 AM, Mikko wrote:

On 2024-07-17 13:27:08 +0000, olcott said:
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On 7/17/2024 2:43 AM, Mikko wrote:

On 2024-07-16 18:24:49 +0000, olcott said:
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On 7/16/2024 3:12 AM, Mikko wrote:

On 2024-07-15 02:33:28 +0000, olcott said:
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On 7/14/2024 9:04 PM, Richard Damon wrote:

On 7/14/24 9:27 PM, olcott wrote:

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Any input that must be aborted to prevent the non termination
of simulating termination analyzer HHH necessarily specifies
non-halting behavior or it would never need to be aborted.

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Excpet, as I have shown, it doesn't.
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Your problem is you keep on ILEGALLY changing the input in your argument because you have misdefined what the input is.
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_DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]
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The input *is* the machine address of this finite
string of bytes: 558bec6863210000e853f4ffff83c4045dc3

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You have already said that a decider is not allowed to answer anything
other than its input. Now you say that the the program at 15c3 is not
a part of the input. Therefore a decider is not allowed consider it
even to the extent to decide whether it ever returns. But without that
knowledge it is not possible to determine whether DDD halts.
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It maps the finite string 558bec6863210000e853f4ffff83c4045dc3
to non-halting behavior because this finite string calls HHH(DDD)
in recursive simulation.

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That mapping is not a part of the finite string and not a part of the
problem specification.

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decider/input pairs <are> a key element of the specification.

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 Computable functions are the formalized analogue of the intuitive notion of algorithms, in the sense that a function is computable if there exists an algorithm that can do the job of the function, i.e. given an input of the function domain it can return the corresponding output. https://en.wikipedia.org/wiki/Computable_function
 

Right, but not all functions are computable, so you can't assume the existance of a program to get the answer.

Not of any specification of any interesting problem.
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 Everyone here seems to think that they can stupidly ignore
the fact that an input calls its own decider and make pretend
that this pathological relationship does not exist.

No, you need to take it FULLY into account, since the program the input represents WILL HALT since the copy of the decider it calls WILL RETUN 0, since that IS the behavior of you claimed to be correct decider.

 

The finite string does not reveal what is the
effect of calling whatever that address happens to contain.

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A simulating termination analyzer proves this.

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Irrelevant, as you just said it is not a part of the input.
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 It is not part of the input in that we already know that
HHH halts and we only need to find out whether or not DDD halts.
 

But since HHH halts, it is a NECCESARY fact that the DDD that just clll that HHH and retuns will halt halt.
It is IMPOSSIBLE for this DDD to not halt when it calls a HHH that halts.
HHH doesn't know this, and makes a INVALID assumption in its logic because its programmer is just an idiot, but that doesn't affect ther VERIFIABLE FACT that the actual behavior, as defined by the x86 interpreation of ALL the code of DDD will halt.
Now, that fact that you try to exclude some of the necessary code from being part of DDD, just means you are just a damned lying idiot that doesn't understand the requirements for an input to be a representation of a PROGRAM, because you made yourself INTENTIONALLY into an IGNORANT IDIOT in the field.
Either the problem is invalid, of DDD will halt if HHH halts.

The
behaviour of HHH is specified outside of the input. Therefore your
"decider" decides about a non-input, which you said is not allowed.

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HHH is not allowed to report on the behavior of it actual self
in its own directly executed process. HHH is allowed to report on
the effect of the behavior of the simulation of itself simulating DDD.

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Now you said that it is allowed to report on a non-input.
Earlier you have said that it is not allowed to report on a non-input.
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 Not the same. It cannot report on its actual self as a directly
executed process. I can report on a copy of itself that it being
emulating in a different process.
 

But it can report on the representation of itself it has been given.
If you claim it isn't in the input representation, then yolu LIED that the input is the representatio of the PROGRAM DDD, which just shows you are just an lying ignorant idiot about computer science.