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On 12/03/24 18:46, olcott wrote:*Richard updated his incorrect words*On 3/12/2024 12:36 PM, Richard Damon wrote:Yes it does. It means that if H is a single instance, no matter which one it is, the rest of the sentence is true.On 3/12/24 9:02 AM, olcott wrote:>On 3/11/2024 10:57 PM, Richard Damon wrote:>On 3/11/24 8:37 PM, olcott wrote:>On 3/11/2024 10:31 PM, Richard Damon wrote:>On 3/11/24 7:52 PM, olcott wrote:>On 3/11/2024 9:32 PM, immibis wrote:\On 12/03/24 03:24, olcott wrote:>>Troll detected.>
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Once we understand that either YES or NO is the right answer
Not for this decider/input question: Ĥ.H / ⟨Ĥ⟩ ⟨Ĥ⟩
For that decider/input question both YES and NO are the wrong answer.
The problem that you keeep on missing is that by the point we can ask this question, H and H^ are FULLY CODED, and thus we know their behavirs.
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy // Ĥ applied to ⟨Ĥ⟩ halts
H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn // Ĥ applied to ⟨Ĥ⟩ does not halt
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Since you know that is false why lie?
⊢* specifies an infinite set of encodings.
Nope, read him again, not just skim and assume.
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*Here is the proof that I am correct*
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
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So, H, being an ELEMENT of Turing Machine Deciders, is a SINGLE INSTANCE of it, it is NOT the set itself.
∀ H ∈ Turing_Machine_Deciders *DOES NOT REFER TO A SINGLE INSTANCE*
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