Re: Simulating termination analyzers by dummies --- What does halting mean?

On 6/21/24 9:19 AM, olcott wrote:

On 6/21/2024 2:11 AM, Mikko wrote:

On 2024-06-20 15:23:09 +0000, olcott said:
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On 6/20/2024 10:08 AM, Mikko wrote:

On 2024-06-20 05:40:28 +0000, olcott said:
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On 6/20/2024 12:29 AM, Mikko wrote:

On 2024-06-19 14:05:29 +0000, olcott said:
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On 6/19/2024 4:29 AM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 6/18/2024 4:36 PM, Alan Mackenzie wrote:

[ Followup-To: set ]

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In comp.theory olcott <polcott333@gmail.com> wrote:

On 6/18/2024 12:57 PM, joes wrote:

Am Tue, 18 Jun 2024 12:25:44 -0500 schrieb olcott:

On 6/18/2024 12:06 PM, joes wrote:
void DDD()
{
H0(DDD);
}
DDD correctly simulated by any H0 cannot possibly halt.

DDD halts iff H0 halts.

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So H0 returns "doesn't halt" to DDD, which then stops running,
so H0 should have returned "halts".

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This was three messages ago.
I had to make sure that you understood that halting
does not mean stopping for any reason and only includes
the equivalent of terminating normally.

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No.  You're wrong, here.  A turing machine is either running or it's
halted.  There's no third alternative.  If your C programs are not in one
of these two states, they're not equivalent to turing machines.

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Although I agree with this there seems to be nuances of
disagreement across the experts.

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I doubt that very much.  The whole point of turing machines is to remove
ambiguity and unneeded features from the theory of computation. A third
alternative state is unneeded.
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Some people say that a TM can halt in a non-final state.

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People may use different words to express the same facts. What some
people call "halting in a non-final state" is called "rejecting" by
some other people. But the facts are what they are independently of
the words used to express them.

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Ambiguity and vagueness make communication less effective.

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As does use of common words and expressions for uncommon meanings.
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I use C because there are zero gaps in exactly what it means.

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THere are lont of gaps in C. Some are mistakes that are corrected in
technical corrigenda. Others are undefined and implementation defined
behaviour. Your program uses non-standard extensions to C so it does
not communicate well. If also is too big to be a part of a publishable
article.
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*There are zero gaps in the behavior of DDD correctly simulated by HH0*
https://liarparadox.org/HH0_(DDD)_Full_Trace.pdf
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_DDD()
[00002093] 55               push ebp
[00002094] 8bec             mov ebp,esp
[00002096] 6893200000       push 00002093 ; push DDD
[0000209b] e853f4ffff       call 000014f3 ; call HH0
[000020a0] 83c404           add esp,+04
[000020a3] 5d               pop ebp
[000020a4] c3               ret
Size in bytes:(0018) [000020a4]
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Whereas the Linz specification of Ĥ says that embedded_H
does something or other that is totally unspecified:
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When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

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Linz Ĥ is fully defined in terms of H, so its behaviour can be inferred
from the behaviour of H. Therefore Linz can prove about the behaviour of
both Ĥ and H what needs be proven.
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 (a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation
 Two complete simulations show a pair of identical TMD's are simulating a pair of identical inputs.  We can see this thus proving recursive simulation.
 

Except your argument is based on the INCORRECT assumption of what H, and thus embedded_H does.
You are forgetting the consequences of behavior. You have proved that *IF H DOESN'T ABORT ITS SIMULATION* that the behavior is infinite, but that is a FALSE ASSUMPTION, as you later have H abort its simulation.
This is a CLASSICAL example of an unsound argument, making it based on a statement that turns out to be false.
YOu are just provig that you have NO understanding of how logic works.