XPost: comp.theory, sci.logic, sci.math
On 12/28/2021 8:32 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 12/28/2021 6:32 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 12/27/2021 7:31 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 12/27/2021 9:56 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 12/26/2021 7:37 PM, Ben Bacarisse wrote:
The second point is so significant that you can't say /anything/ about
TMs until you accept that you are wrong. No TM can be built around a >>>>>>>>> copy of another in any reliable way, because in PO-land identical copies
don't behave identically.
All statements you make about TMs are simply junk until you accept that
a TM's behaviour is determined slowly by the input and the state >>>>>>>>> transition function.
When we arrange one copy of a computation to change its behavior on >>>>>>>> the basis of another copy of this computation thus be dependent on the >>>>>>>> second copy whereas this second copy is not dependent on any other >>>>>>>> computation then we can correctly expect different results.
If it's a copy it behaves identically. A TM this does not behave like >>>>>>> the one it is based on is not a copy. To arrange for a copy of a TM to >>>>>>> behave differently, with the same tape, is just Orwellian double speak. >>>>>>> You just need to put your hand up to the mistake and move on.
As long as H has a way to distinguish itself from an identical copy of >>>>>> itself
There is no such way. That's what "identical copy means" for Turing >>>>> machines.
H applied to ⟨Ḧ⟩ ⟨Ḧ⟩ transitions to H.qy and embedded_H applied
to ⟨Ḧ⟩ ⟨Ḧ⟩ transitions to Ḧ.qn, otherwise they both transition to qn.
And since there is no way for H to "distinguish itself", both
transition to qn (or qy).
Since humans can see that Ḧ applied to ⟨Ḧ⟩ never stops running unless
embedded_H aborts its simulation of ⟨Ḧ⟩ ⟨Ḧ⟩, humans can see that a
transition to Ḧ.qn is correct because it does correctly compute the
mapping from ⟨Ḧ⟩ ⟨Ḧ⟩ to Ḧ.qn on the basis of the behavior of UTM(⟨Ḧ⟩,
⟨Ḧ⟩).
Ḧ is built using an identical copy of an other TM (H). Until you
acknowledge that you are wrong about two identical TMs transitioning to
different states despite having identical tapes, it's impossible to
discuss TMs like Ḧ at all.
This is a moot point, the key issue is about embedded_H applied to ⟨Ḧ⟩ >> ⟨Ḧ⟩
But while you believe (and assert) nonsense about embedded copies of
TMs, discussion of them is futile. (Moot has two related meaning these
days. It's not a good word to use.)
Because this is Irrefutable you keep dodging it.
Because this is Irrefutable you keep dodging it.
Because this is Irrefutable you keep dodging it.
Proving that embedded_H at Ĥ.qx correctly maps its inputs ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn on the basis of the behavior of the UTM simulation of these inputs.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
*My criterion measure with Ben's notational conventions*
H.q0 wM w ⊢* H.qy iff UTM(wM, w) halts
H.q0 wM w ⊢* H.qn iff UTM(wM, w) does not halt
We know that H would correctly decide the halt status of its input on
the basis of correctly deciding the halt status of the UTM simulation of
its input.
We know this because a UTM simulation of the Turing machine description
is computationally equivalent to the direct execution of this same
Turing machine.
We know that the copy of H is at Ĥ.qx (AKA embedded_H) applies this
same criterion measure to every instance of embedded_H to any recursive
depth.
We know that this means that embedded_H is examining the behavior of Ĥ
applied to ⟨Ĥ⟩ as if embedded_H was (hypothetically) replaced by a UTM.
We know that the behavior of this hypothetical machine is the criterion
measure for embedded_H.
We know that Ĥ applied to ⟨Ĥ⟩ would never stop running if embedded_H was replaced by a UTM because Ĥ applied to ⟨Ĥ⟩ copies its input then UTM ⟨Ĥ⟩
⟨Ĥ⟩ would repeat this cycle ...
We know that this means that when embedded_H computes the mapping from
its input ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qy or Ĥ.qn on the basis of the UTM simulation of
this input that its transition to Ĥ.qn is correct.
https://www.liarparadox.org/Peter_Linz_HP_315-320.pdf
Linz, Peter 1990. An Introduction to Formal Languages and Automata. Lexington/Toronto: D. C. Heath and Company. (315-320)
--
Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer
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