On 3/23/22 7:20 PM, olcott wrote:
On 3/23/2022 6:04 PM, Richard Damon wrote:
On 3/23/22 9:09 AM, olcott wrote:
On 3/23/2022 6:19 AM, Richard Damon wrote:
On 3/23/22 12:00 AM, olcott wrote:You know that a decider only computes the mapping from its input
On 3/22/2022 10:37 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/22/2022 9:32 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 3/21/2022 10:22 PM, Ben Bacarisse wro0te:
olcott <NoOne@NoWhere.com> writes:
A copy of Linz H is embedded at Ĥ.qx as a simulating halt >>>>>>>>>>>> decider (SHD).But for your "PO-machines":
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ >>>>>>>>>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach
its final state.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>>>>>>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never
reach its
final state.
"Ĥ.qx maps ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn
corresponds to
H maps ⟨Ĥ⟩ ⟨Ĥ⟩ to H.qy"
and
"The copy of H at Ĥ.qx correctly decides that its input >>>>>>>>>>> never halts.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input
halts"
so this has nothing to do with Linz. He is talking about Turing >>>>>>>>>>> machines.
The Linz conclusion only pertains to the behavior the copy of H >>>>>>>>>> embedded within Ĥ applied to ⟨Ĥ⟩ ⟨Ĥ⟩.
Everything Linz says, everything, is predicated on what a
Turing machine
is. Unlike Turing machines, your machines are magic --
identical state
transition functions can entail different configuration
sequences for
the same input. Nothing you say has any relevance to Linz's >>>>>>>>> Turing
machines until you categorically repudiate this nonsense.
That your only rebuttal to what I say now is dredging up what I >>>>>>>> said
many months ago proves that you are being dishonest.
You said this:
"The copy of H at Ĥ.qx correctly decides that its input never >>>>>>> halts.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decides that its input halts"
If ⟨H⟩ and ⟨Ĥ⟩ were identical finite strings then they must derive
the same result. They are not identical final strings.
four days ago and you haven't retracted it. Until you do, when you >>>>>>> write Ĥ your readers must assume that you are referring to something >>>>>>> about which this quote applies.
What's more, for your remarks to have any bearing on Linz's Ĥ you >>>>>>> must
not only repudiate what you said, you must accept the converse,
i.e. that if
Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* Ĥ.qn
then
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.qn
So, do you retract what you said and accept this fact about
Linz's H and
Ĥ?
You you continue to say that you believe that a decider must
report on its own behavior when you already know damn well that a
decider only computes the mapping from its inputs to its own final >>>>>> state.
A Decider must report on its own behavior (or the behavior of a
copy of it) if that is what the input asks for.
finite strings to its own final state thus you know that you lie
what you say that a decider must compute the mapping from a
non-finite sting non-input.
WHY LIE ? WHY LIE ? WHY LIE ? WHY LIE ?
I don't, but you seem to like to.
I never said that H needs to compute a mapping of anything but what
has been given as an input.
The only thing H needs to compute the mapping of is <H^> <H^>, which
is EXACTLY the string on its input.
The problem which you don't seem to understand is that the MAPPING it
needs to try and compute (and which is not guaranteed to BE
Computable), is the Behavior of the machine it represents, H^ applied
to <H^>, as that is the mapping of the Halting Function.
That is a non finite string non input, so you lied.
<H^> <H^> is a finite string, which is what needs to be mapped, so you
are just lying.
the MAPPING it needs to try and compute (and which is
not guaranteed to BE Computable), is the Behavior of
the machine it represents, H^ applied to <H^>,
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