Re: HHH(DDD) computes the mapping from its input to HHH emulating itself emulating DDD --- anyone that says otherwise is a liar

On 11/25/24 11:08 PM, olcott wrote:

On 11/24/2024 11:18 AM, Richard Damon wrote:

On 11/24/24 9:30 AM, olcott wrote:

On 11/23/2024 11:54 AM, Richard Damon wrote:

On 11/23/24 11:54 AM, olcott wrote:

On 11/23/2024 9:35 AM, Richard Damon wrote:

On 11/23/24 10:15 AM, olcott wrote:

On 11/23/2024 9:02 AM, Richard Damon wrote:

On 11/23/24 9:04 AM, olcott wrote:

On 11/23/2024 1:59 AM, Mikko wrote:

On 2024-11-22 16:45:52 +0000, olcott said:
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On 11/22/2024 2:30 AM, Mikko wrote:

On 2024-11-21 15:32:38 +0000, olcott said:
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On 11/21/2024 3:12 AM, Mikko wrote:

On 2024-11-20 22:03:43 +0000, olcott said:
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On 11/20/2024 3:53 AM, Mikko wrote:

On 2024-11-20 03:23:12 +0000, olcott said:
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On 11/19/2024 4:12 AM, Mikko wrote:

On 2024-11-18 20:42:02 +0000, olcott said:
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On 11/18/2024 3:41 AM, Mikko wrote:

The "the mapping" on the subject line is not correct. The subject line
does not specify which mapping and there is no larger context that could
specify that. Therefore it should be "a mapping".
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On 2024-11-17 18:36:17 +0000, olcott said:
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void DDD()
{
   HHH(DDD);
   return;
}
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_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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DDD emulated by any encoding of HHH that emulates N
to infinity number of steps of DDD cannot possibly
reach its "return" instruction final halt state.

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Because it cannot reach the instructions before tha return.
Because it cannot reach the instruction after the HHH call.
Because it cannot reach return instruction of HHH.
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This applies to every DDD emulated by any HHH no
matter the recursive depth of emulation. Thus it is
a verified fact that the input to HHH never halts.

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That is too vague to be regareded true or false. It is perfectly possibe
to define two programs and call them DDD and HHH

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What a jackass. DDD and HHH have been fully specified
for many months.

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They are specified in a way that makes your "every DDD" and "any DDD"
bad (perhaps even incorrect) use of Common language.
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I specify the infinite sets with each element numbered
on the top of page 2 of my paper. Back in April of 2023
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https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D

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You have also specifed that HHH is the program in your GitHub repository.
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Should I assume that you must be lying about
this because you did not quote where I did this?

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No, you may assume that I was confused by your lack of clarity and
in particular by your bad choice of names.
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If you clearly state that HHH is not the function HHH that you have
in your GitHub repository then I needn't to consider the possiblity
that you just triying to deceive by equivcation.
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HHH is one concrete example of an infinite set of instances
such that DDD is emulated by HHH N times.

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That sentence says that there is only one HHH, contradicting your
earlier statement that HHH is a generic term for every member of some
set.
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You seem to be a damned liar: "infinite set of instances"

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You mean you lied when you said "one concrete example"?
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One element of an infinite set does not say there
is no infinite set. Is says there is an infinite set.
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But one element of an infinite set is not the infinite set.
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You are just showing that your logic is based on proven incorrect set theory.
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IF HHH is an ELEMENT of the set, then it is that one element for the entire evaluation,

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Liar:
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A proof by induction consists of two cases. The first, the base case,
proves the statement for n=0 without assuming any knowledge of
other cases. The second case, the induction step, proves that if the
statement holds for any given case n=k, then it must also hold for
the next case n=k+1. These two steps establish that the statement
holds for every natural number n. The base case does not necessarily
begin with n=0, but often with n=1, and possibly with any fixed natural
number n=N, establishing the truth of the statement for all natural
numbers n ≥ N.
https://en.wikipedia.org/wiki/Mathematical_induction
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And when have you ever provided such a proof for your statement?
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NOWHERE
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Your problem is you don't even have a logical basis to express your statements in, so you can't do an induction on them.
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So, you are just demonstrating that your "logic" is based on the meaningless use of buzzwords that you don't understand, but can parrot their unlearned meaning, but have no idea how to actually use.
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*As you already admitted below*
when N steps of DDD are emulated by HHH
DDD cannot reach past its call to HHH (statement)

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But that was for the DDD that INCLUDED HHH as part of it, which you have now made clear is NOT what you consider DDD to be. And for that case DDD[n] calls HHH[n] (where HHH[n] is the version of HHH that does only n steps of emulation) and while we can say that HHH[n[ does not emulate DDD[n] to its final state, that property is NOT a property of of DDD[n], but of HHH[n] and DDD[n] as its input.

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That every DDD[n] calls its HHH[n] in recursive emulation
conclusively proves that no DDD[n] emulated by HHH[n] halts,
thus each HHH[n] is correct to reject its input as non halting.

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But every HHH[n] aborts its emulaton and returns, and thus DDD[n] halts, and thus HHH is INCORRECT to call its input non-halting.
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*You are a stupid liar*
You know that halting means reaching a final state and you
know that no input to HHH can possibly reach its final state.
So you aren't just a liar, you are a stupid one.
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 And you should know that "Halting" is a property of Turing Machines / Computations / Progrzms / completely defined function and the like ONLY.