Liste des Groupes | Revenir à s logic |
On 5/29/2024 8:02 PM, Richard Damon wrote:Of course it is.On 5/29/24 8:53 PM, olcott wrote:*No that is never the case*On 5/29/2024 7:47 PM, Richard Damon wrote:>On 5/29/24 8:21 PM, olcott wrote:>On 5/29/2024 7:09 PM, Richard Damon wrote:>On 5/29/24 8:01 PM, olcott wrote:>On 5/29/2024 6:47 PM, Richard Damon wrote:>>*Formalizing the Linz Proof structure*>
∃H ∈ Turing_Machines
∀x ∈ Turing_Machines_Descriptions
∀y ∈ Finite_Strings
such that H(x,y) = Halts(x,y)
>
And since NO H, can get right the H^ built to contradict IT, that claim is proven false.
>
YOU KEEP TRYING TO GET AWAY WITH CHANGING THE SUBJECT
THE ABOVE FORMALIZATION IS CORRECT
>
How?
>
The above is the question that Linz asks and the he gets
an answer of no, no such H exists.
>
>
So, you now agree with Linz. Good.
>
I said that Linz says that. The point is that the Linz
template examines an infinite set of Turing Machine / input
pairs the same way my H/D template references an infinite set
of C function / input pairs.
>
The difference is, In Linz's formulation, each machine is INDIVIDUALLY EVALUTED with its inputs,
When Ĥ is applied to ⟨Ĥ⟩Why do you say that?
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The entire category of every decider/input pair is examined ALL AT ONCE.
No one is dumb enough to look at each element of an infinite set
one at a time because they know this takes literally forever.
Les messages affichés proviennent d'usenet.