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On 3/12/24 4:31 PM, olcott wrote:We can get around all of this stuff by simply using this criteria:On 3/12/2024 6:11 PM, Richard Damon wrote:*IF* they are correct decider.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:∀ H ∈ Turing_Machines_Returning_BooleanOn 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)
And it can be a different TMD to each H.
>When we disallow decider/input pairs that are incorrect>
questions where both YES and NO are the wrong answer
Once we understand that either YES or NO is the right answer, the whole rebuttal is tossed out as invalid and incorrect.
>
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⟩
No, because a given H will only go to one of the answers. THAT will be wrong, and the other one right.
>
∀ 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.
You mean a pair of DIFFERENT machines. Any difference is different.
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.
Nope.
>
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>
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.
That isn't in the set above.
>>Nope, since both aren't in the set selected.
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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>
>
When they are deciders that must get the correct answer both
of them are not in the set.
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
>But in the step of select and input that they will get wrong, they will be givne DIFFERENT inputs.
When they are Turing_Machines_Returning_Boolean the this
set inherently includes identical pairs that only differ
by return value.
>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.You just don't understand what that statement is saying.No the problem is that you are not paying attention.
>
I've expalined it, but it seems over you head.
>
>But if they returned differnt values, they will have different descriptions.For Every H, we show we can find at least one input (chosen just for that machine) that it will get wrong.When we use machine templates then we can see instances of
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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.
>
Otherwise, how could a UTM get the right answer, since it only gets the description.
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