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On 3/9/24 9:49 PM, olcott wrote:Yes.On 3/9/2024 11:36 PM, Richard Damon wrote:And if it does, as I said below, so will H^.H when it is run.On 3/9/24 9:14 PM, olcott wrote:Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ haltsOn 3/9/2024 10:55 PM, Richard Damon wrote:>On 3/9/24 8:30 PM, olcott wrote:>On 3/9/2024 7:40 PM, immibis wrote:>On 10/03/24 02:37, olcott wrote:>On 3/9/2024 7:32 PM, immibis wrote:>On 10/03/24 02:29, olcott wrote:>On 3/9/2024 7:24 PM, immibis wrote:>On 10/03/24 01:30, olcott wrote:>On 3/9/2024 6:24 PM, immibis wrote:>On 10/03/24 01:22, olcott wrote:>On 3/9/2024 5:57 PM, immibis wrote:>On 10/03/24 00:26, olcott wrote:>On 3/9/2024 5:10 PM, immibis wrote:>On 9/03/24 23:22, olcott wrote:>On 3/9/2024 3:50 PM, immibis wrote:>On 9/03/24 22:34, olcott wrote:>>>
What criteria would you use so that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ knows what
wrong answer to provide?
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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that H ⟨Ĥ⟩ uses.
Simulating halt deciders must make sure that they themselves
do not get stuck in infinite execution. This means that they
must abort every simulation that cannot possibly otherwise halt.
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This requires Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation and does not
require H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts
its simulation.
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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does simulate itself in recursive simulation H ⟨Ĥ⟩ ⟨Ĥ⟩
does not simulate itself in recursive simulation.
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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that H ⟨Ĥ⟩ uses.
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*Only because Ĥ.H is embedded within Ĥ and H is not*
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
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You dishonestly ignored that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same OBJECTIVE criteria that H ⟨Ĥ⟩ uses.
The above is true no matter what criteria that is used
as long as H is a simulating halt decider.
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Objective criteria cannot vary based on who the subject is. They are objective. The answer to different people is the same answer if the criteria are objective.
It is objectively true that Ĥ.H can get stuck in recursive
simulation because Ĥ copies its input thus never runs
out of params.
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It is objectively true that Ĥ cannot possibly get stuck
in recursive because H does not copy its input thus runs
out of params.
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Wrong. Dead wrong. Stupidly wrong. So wrong that a dead monkey could do better. Write the Olcott machine (not x86utm) code for Ĥ and I would show you.
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
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Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
That's not a verified fact, that's just something you want to be true.
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∞ means infinite loop. Infinite loop doesn't halt. You see how stupid it is, to say that an infinite loop halts?
>Execution trace of Ĥ applied to ⟨Ĥ⟩>
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ BECAUSE IT IS PRECISELY IDENTICAL TO STEPS B AND C:
> (b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
> (c) which begins at Ĥ's own simulated ⟨Ĥ.q0⟩ to repeat the process
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*Yes and the key step of copying its input is left out so*
*H ⟨Ĥ⟩ ⟨Ĥ⟩ runs out of params and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ never runs out of params*
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that isn't how any of this works. Do you even know what words mean?
(b) and (c) are not the same as (1) and (2)
Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
(1) H applied ⟨Ĥ⟩ ⟨Ĥ⟩ simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(2) which begins at simulated ⟨Ĥ.q0⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
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This means that Turing machine H ⟨Ĥ⟩ ⟨Ĥ⟩ can see one more execution
trace of Ĥ ⟨Ĥ⟩ than its simulated Turing machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see.
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Nope, your just being stuupid, perhaps intentionally.
(c) just moves around to its simulation of a
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(a) H^.q0 (H^)
H^ then makes a copy of its inp
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(b) H^.H (H^) (H^) == (1) H (H^) (H^)
The algorithm of H begins a simulation of its input, watching the behaior of H^ (H^)
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(c) = (2)
Which begins at the simulation of H^.q0 (H^)
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(d = sim a) = (sim a)
Ths Simulated H^.q0 (H^) makes a copy of its input
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(e = sim b) = (sim b)
The Simulated H^.H (H^) (H^) has is H begin the simulation of its input ...
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and so on.
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Both machine see EXACTLY the same level of details.
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Yes, the top level H is farther along at any given time then its simulated machine, and that is H's problem, it has to act before it sees how its simulation will respond to its copy of its actions.
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Thus, if it stops, it needs to make its decision "blind" and not with an idea of how the machine it is simulating will perform.
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If it doesn't stop, the level of recursion just keeps growing and no answer ever comes out.
The earliest point that H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly see to abort
its simulation is immediately before Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would begin
its simulation. Right before its cycle repeats the first time.
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So?
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If it DOES abort there, then so will H^.H when it gets to that point in its simulation, which will be AFTER The point that H has stopped simulating it, so H doesn't know what H^ will do.
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Thus, if H DOES abort there, we presume from your previous answer it will think the input will not halt and answer qn.
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Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts right after Ĥ.Hq0 before it simulates ⟨Ĥ⟩ ⟨Ĥ⟩.
Remember, the first to lines, the comment is the requirement, not neccesarily what actually happens, and is determined by H. If H says Halting, then H^ ends up at H^.qy and looping. If H says non-halting, H^ ends up a H.qn and halts.--
You seem to think that H's partial simulation is what actually happens and then the machine just disappears.
The ACTUAL test to see if H is correct, is to look at the ACTUAL DIRECT EXECUTION of the computation described by the input, which is
H^ (H^) (H^)
That machine's H algorithm will ALSO abort its simulation and return the same answer as H does and thus H will be wrong.
You attempts to rewrite the criteria from the objective standard of the Halting Problem to your category error filled subjective standard is just a LIE, and shows your utter stupidity and ignorance about what you are trying to talk about.
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>But, since H^.H will do the same thing when we execute it, its H^.H will do the same thing and abort its simulation when it sees it get to that same point, and abort and go to qn, and then H^ wil Halt, making H wrong.>
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If you change your mind, and H is now going to double think and say H^.H will also abort, and will say non-Halting, I will says Halting and go to qy
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Then, since H^.H has that same code with the same double think, when it also abort there, it will also think, this H^.H will also abort is simulation and go to qn, so I should go to qy, so H^ ends up in qy and does an infinite loop.
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Thus, H is again wrong, because H^ is defined to exactly mimic the algorithm of H, and the act contrary to it, so NOTHING H can do can out smart it.
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Turing Completeness requires that this ability be present, since it can be done with Turing Machine, otherwise there is demonstrated a machine that can be built with Turing Machines, but not that other system, the H^ to any H.
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We have gone over this before, but you keep on confusing the partial simulation of H^ with the actual execution of it. We evaluate H based on what the actual execution of H^ will be, which goes past the point that H simulated it (if it aborts, and if it doesn't, it never answers)
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