From: wyniijj5@gmail.com
On Sat, 2025-07-26 at 21:43 -0500, olcott wrote:
> On 7/26/2025 8:30 PM, Richard Damon wrote:
> > On 7/26/25 7:43 PM, olcott wrote:
> > > On 7/26/2025 6:35 PM, Richard Damon wrote:
> > > > On 7/26/25 7:08 PM, olcott wrote:
> > > > > On 7/26/2025 5:49 PM, olcott wrote:
> > > > > > On 7/26/2025 2:58 PM, olcott wrote:
> > > > > > > On 7/26/2025 2:52 PM, Mr Flibble wrote:
> > > > > > > > On Sat, 26 Jul 2025 14:26:27 -0500, olcott wrote:
> > > > > > > >
> > > > > > > > > On 7/26/2025 1:30 PM, Alan Mackenzie wrote:
> > > > > > > > > > In comp.theory olcott wrote:
> > > > > > > > > >
> > > > > > > > > > > The error of all of the halting problem proofs is that
they
> > > > > > > > > > > require a
> > > > > > > > > > > Turing machine halt decider to report on the behavior of
a
> > > > > > > > > > > directly
> > > > > > > > > > > executed Turing machine.
> > > > > > > > > >
> > > > > > > > > > > It is common knowledge that no Turing machine decider
can take
> > > > > > > > > > > another
> > > > > > > > > > > directly executing Turing machine as an input, thus the
above
> > > > > > > > > > > requirement is not precisely correct.
> > > > > > > > > >
> > > > > > > > > > > When we correct the error of this incorrect requirement
it
> > > > > > > > > > > becomes a
> > > > > > > > > > > Turing machine decider indirectly reports on the
behavior of a
> > > > > > > > > > > directly executing Turing machine through the proxy of a
> > > > > > > > > > > finite string
> > > > > > > > > > > description of this machine.
> > > > > > > > > >
> > > > > > > > > > > Now I have proven and corrected the error of all of the
halting
> > > > > > > > > > > problem proofs.
> > > > > > > > > >
> > > > > > > > > > No you haven't, the subject matter is too far beyond your
> > > > > > > > > > intellectual
> > > > > > > > > > capacity.
> > > > > > > > > >
> > > > > > > > > >
> > > > > > > > > It only seems to you that I lack understanding because you
are
> > > > > > > > > so sure
> > > > > > > > > that I must be wrong that you make sure to totally ignore
the
> > > > > > > > > subtle
> > > > > > > > > nuances of meaning that proves I am correct.
> > > > > > > > >
> > > > > > > > > No Turing machine based (at least partial) halt decider can
> > > > > > > > > possibly
> > > > > > > > > *directly* report on the behavior of any directly executing
Turing
> > > > > > > > > machine. The best that any of them can possibly do is
> > > > > > > > > indirectly report
> > > > > > > > > on this behavior through the proxy of a finite string machine
> > > > > > > > > description.
> > > > > > > >
> > > > > > > > Partial decidability is not a hard problem.
> > > > > > > >
> > > > > > > > /Flibble
> > > > > > >
> > > > > > > My point is that all of the halting problem proofs
> > > > > > > are wrong when they require a Turing machine decider
> > > > > > > H to report on the behavior of machine M on input i
> > > > > > > because machine M is not in the domain of any Turing
> > > > > > > machine decider. Only finite strings such as ⟨M⟩ the
> > > > > > > Turing machine description of machine M are its
> > > > > > > domain.
> > > > > > >
> > > > > >
> > > > > > Definition of Turing Machine Ĥ
> > > > > > Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
> > > > > > if Ĥ applied to ⟨Ĥ⟩ halts, and //
incorrect requirement
> > > > > > Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> > > > > > if Ĥ applied to ⟨Ĥ⟩ does not halt. // incorrect
requirement
> > > > > >
> > > > > > (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⟩ ...
> > > > > >
> > > > > > The fact that the correctly simulated input
> > > > > > specifies recursive simulation prevents the
> > > > > > simulated ⟨Ĥ⟩ from ever reaching its simulated
> > > > > > final halt state of ⟨Ĥ.qn⟩, thus specifies non-termination.
> > > > > >
> > > > > > This is not contradicted by the fact that
> > > > > > Ĥ applied to ⟨Ĥ⟩ halts because Ĥ is outside of
> > > > > > the domain of every Turing machine computed function.
> > > > > >
> > > > >
> > > > > In the atypical case where the behavior of the simulation
> > > > > of an input to a potential halt decider disagrees with the
> > > > > behavior of the direct execution of the underlying machine
> > > > > (because this input calls this same simulating decider) it
> > > > > is the behavior of the input that rules because deciders
> > > > > compute the mapping for their inputs.
> > > > >
> > > >
> > > > Nope, just more of your lies.
> > > >
> > > > The behavior of an input to a halt decider is DEFINED in all cases to
> > > > be the behavior of the machine the input represents,
> > >
> > > Yet I have conclusively proven otherwise and
> > > you are too stupid to understand the proof.
> >
> > No, because you proof needs to call different inputs the same or partial
> > simulaiton to be correct.
> >
>
> When HHH(DDD) simulates DDD it also simulates itself
> simulating DDD because DDD calls HHH(DDD).
>
> When HHH1(DDD) simulates DDD DOES NOT simulate itself
> simulating DDD because DDD DOES NOT CALL HHH1(DDD).
>
> For three fucking years everyone here pretended that
> they could NOT fucking see that.
It is you who proved yourself an idiot, worse, a liar, EVERYDAY.
olcott's claim form H(D)=0 is correct, H(D)=1 is correct, both are correct...
'I' was talking about HH,HH2,HHH, DD,DDD,...not H(D)!! ... numerous.
And recently, 'I' was not refuting HP. HP is correct. 'I' was refuting Linz's
proof, and HHH(DD)=1 is correct!! (Undecidable and HHH(DD)=1 are both correct!)
A couple days before, you have shown again you don't understand basic logic
(AND,IF,...)
You cannot construct a TM that compute the length of its input.
Your understanding of C/Assembly is shown very low, I never saw anyone is
lower.
No one in internet I ever saw is lower than yours. Keep blind yourself,
'genius'.
--- SoupGate-Win32 v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
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