Re: Two dozen people were simply wrong --- Try to prove otherwise --- pinned down

On 5/31/2024 9:25 PM, Richard Damon wrote:

On 5/31/24 10:08 PM, olcott wrote:

On 5/31/2024 8:35 PM, Richard Damon wrote:

On 5/31/24 9:10 PM, olcott wrote:

On 5/31/2024 7:39 PM, Richard Damon wrote:

On 5/31/24 7:57 PM, olcott wrote:

On 5/31/2024 6:33 PM, Richard Damon wrote:

On 5/31/24 6:54 PM, olcott wrote:

On 5/31/2024 5:46 PM, Richard Damon wrote:

On 5/31/24 6:08 PM, olcott wrote:

On 5/31/2024 4:36 PM, Richard Damon wrote:

On 5/31/24 10:10 AM, olcott wrote:

On 5/31/2024 6:16 AM, Richard Damon wrote:

On 5/30/24 11:27 PM, olcott wrote:

Try and show how HH using an x86 emulator can correctly emulate
the following x86 machine code such that DD reaches its own
machine address 00001c47.

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Why should I, since that isn't what I was saying.
>

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*To me that looks like you know that*
*you have been busted in a lie and are backing down*

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no, YOU are LYING RIGHT HERE AND NOW.
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Prove that I said that the simulation by HH made it there, or admit to being a DAMNED LIAR.
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What I have been saying is the the DIRECT EXDCUTION of DD, and the CORRECT (and complete) simulation of the input to HH by an actual UTM will get there.
>

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That has always been the dishonest dodge strawman deception
CHANGE-THE-SUBJECT fake rebuttal regarding
the behavior of DD correctly simulated by pure function HH.

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But it is your talking about the "correctly simulated by HH" that is the dishonest dodge,

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Try and show how HH using an x86 emulator can correctly emulate
the following x86 machine code such that DD reaches its own
machine address 00001c47.

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Never said it could. But haven't looked hard enough to be willing to say it can't, but then, who cares, it doesn't say a thing about the real halting problem, since H's simulation isn't "correct" by a definition that relates simulation to non-halting behavior,
>

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"...the Turing machine will halt whenever it enters a final state."
Linz(1990:234)

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Right, and that is talking about runnig the Turing Machine, not simulating a representation of it.
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DD correctly simulated by HH cannot possibly reach its own simulated
final state. This is conclusively proven beyond all possible doubt
by the x86 machine code of DD.

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Depends on the specification of HH, as has beeen shown.
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You can lie about this and try to get away with changing the subject.
What you cannot do is show that it is not true.

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But I don't try to claim one side of the other, as it isn't relevent.
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*Showing that it is relevant is a whole other different subject that*
*we can get to as soon as you quit your deception on this subject*

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But if it isn't relevent, why are we looking at it.
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It seems you just WANT to waste you time trying to convince people about something they don't care about.
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*Showing that it is relevant is a whole other different subject that*
*we can get to as soon as you quit your deception on this subject*
>
*Showing that it is relevant is a whole other different subject that*
*we can get to as soon as you quit your deception on this subject*
>

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*If DD correctly simulated by HH can't possibly reach its own*
*final state then DD correctly simulated by HH is non-halting*

Nope, Where did that definition say ANYTHING about one machine simulationg another.
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Trying to get away with saying that you don't "believe in" UTMs
can't possibly fool anyone that knows what UTMs are.
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But I DO beleive in UTMs, and know exact what they do.
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Your embedded_H can NOT actually be a UTM, or neither H nor embedded_H are deciders.
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PERIOD,.
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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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When embedded_H <is> a UTM then ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H
cannot possibly reach its own simulated final state of ⟨Ĥ.qn⟩. In this
case embedded_H fails to be a decider, however it also proves:

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And then, while THAT H^ is non-halting, H isn't a decider.
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that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach its
own simulated final state of ⟨Ĥ.qn⟩ for any embedded_H based on a UTM
that only simulates some finite sequence of steps.

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But that only apply *if* embedded_H (and thus H) *IS* a UTM, if it aborts its simuliation, then its
>

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"...the Turing machine will halt whenever it enters a final state."
Linz(1990:234)
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⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach its
own simulated final state of ⟨Ĥ.qn⟩

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And and embedded_H that actually correct simulates the input, by Computaiton Theory definitions is the UTM that, by definition, doesn't abort its simulation.
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*The input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H SPECIFIES non-halting behavior*
*The input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H SPECIFIES non-halting behavior*
*The input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H SPECIFIES non-halting behavior*

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Only if embedded_H and H are ACTUALLY UTMs,

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*AS LONG AS 1 to ∞ steps of*
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach their
own simulated final state of ⟨Ĥ.qn⟩
THEN *The input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H SPECIFIES non-halting behavior*

 And each of those is of a DIFFERENT input,

Every element of the infinite set of embedded_H / ⟨Ĥ⟩ has this same
property
1 to ∞ steps of ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach their own simulated final state of ⟨Ĥ.qn⟩
How the Hell did you think you could keep getting away with
that ridiculous shell game? No one here is that much of a rube!
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer