Re: ZFC solution to incorrect questions: reject them

On 3/12/2024 9:28 PM, Richard Damon wrote:

On 3/12/24 4:31 PM, olcott wrote:

On 3/12/2024 6:11 PM, Richard Damon wrote:

On 3/12/24 3:53 PM, olcott wrote:

On 3/12/2024 5:30 PM, Richard Damon wrote:

On 3/12/24 2:34 PM, olcott wrote:

On 3/12/2024 4:23 PM, Richard Damon wrote:

On 3/12/24 1:11 PM, olcott wrote:

On 3/12/2024 2:40 PM, Richard Damon wrote:

On 3/12/24 12:02 PM, olcott wrote:

On 3/12/2024 1:31 PM, immibis wrote:

On 12/03/24 19:12, olcott wrote:

∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions  |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)

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And it can be a different TMD to each H.
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When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer

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Once we understand that either YES or NO is the right answer, the whole rebuttal is tossed out as invalid and incorrect.
>

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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
BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩

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No, because a given H will only go to one of the answers. THAT will be wrong, and the other one right.
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∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions  |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.

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You mean a pair of DIFFERENT machines. Any difference is different.

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Every decider/input pair (referenced in the above set) has a
corresponding decider/input pair that only differs by the return
value of its decider.

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Nope.
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∀ H ∈ Turing_Machines_Returning_Boolean
∃ TMD ∈ Turing_Machine_Descriptions  |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
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Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.

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That isn't in the set above.
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That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
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Nope, since both aren't in the set selected.
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When they are deciders that must get the correct answer both
of them are not in the set.

 *IF* they are correct decider.
 WHen we select from all Turing Machine Deciders, there is no requirement that any of them get any particular answer right.
 So, ALL deciders are in the set that we cycle through and apply the following logic to ALL of them.
 Each is them paired with an input that it will get wrong, and the existance of the input was what as just proven, the ^ template
 

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When they are Turing_Machines_Returning_Boolean the this
set inherently includes identical pairs that only differ
by return value.

 But in the step of select and input that they will get wrong, they will be givne DIFFERENT inputs.
 

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You just don't understand what that statement is saying.
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I've expalined it, but it seems over you head.
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No the problem is that you are not paying attention.

 No, you keep on making STUPID mistakes, like thinking that select a input that the machine will get wrong needs to be the same for two differnt machines.
  

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For Every H, we show we can find at least one input (chosen just for that machine) that it will get wrong.
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When we use machine templates then we can see instances of
the same machine that only differs by return value where both
get the wrong answer on the same input. By same input I mean
the same finite string of numerical values.
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 But if they returned differnt values, they will have different descriptions.
 Otherwise, how could a UTM get the right answer, since it only gets the description.

We can get around all of this stuff by simply using this criteria:
Date 10/13/2022 11:29:23 AM
*MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
(He has neither reviewed nor agreed to anything else in this paper)
(a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running unless aborted then
(b) H can abort its simulation of D and correctly report that D specifies a non-halting sequence of configurations.
*When we apply this criteria* (elaborated above)
Will you halt if you never abort your simulation?
*Then the halting problem is conquered*
When two different machines implementing this criteria
get different results from identical inputs then we
know that Pathological Self-Reference has been detected.
We don't even need to know that for:
*denial-of-service-attack detection*
*NO always means reject as unsafe*
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer