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On 7/31/2024 2:32 AM, Mikko wrote:None of above indicates any disagreement by any authority.On 2024-07-30 14:16:20 +0000, olcott said:Computable functions are the formalized analogue of the
On 7/30/2024 1:37 AM, Mikko wrote:Which dictionary (or other authority) disagrees?On 2024-07-29 16:16:13 +0000, olcott said:No you are wrong.
On 7/28/2024 3:02 AM, Mikko wrote:The meaning of "correct" in this context is that if the transition ofOn 2024-07-27 14:08:10 +0000, olcott said:When we compute the mapping from the input to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
On 7/27/2024 2:21 AM, Mikko wrote:And even more simplified semantics.On 2024-07-26 14:08:11 +0000, olcott said:The above is merely simplified syntax for the top of page 3
When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
https://www.liarparadox.org/Linz_Proof.pdf
The above is the whole original Linz proof.
(a) Ĥ copies its input ⟨Ĥ⟩The above is an obvious tight loop of (d), (e), (f), and (g).
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation
You are supposed to evaluate the above as a contiguous
sequence of moves such that non-halting behavior is
identified.
Its relevance (it any) to the topic of the discussion is not
obvious.
to the behavior specified by this input we know that embedded_H
is correct to transition to Ĥ.qn.
embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn is correct if H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to H.qn but
incorrect otherwise.
intuitive notion of algorithms, in the sense that a function
is computable if there exists an algorithm that can do the
job of the function, i.e. *given an input of the function*
*domain it can return the corresponding output*
https://en.wikipedia.org/wiki/Computable_function
The common knowledge that a decider computes the mapping
from its input finite string...
This is almost always the same as the direct execution of
the machine represented by this finite string.
The one rare exception is shown above where Ĥ ⟨Ĥ⟩ haltsThat is not supported by any anuthority.
and the input to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach
its own final state of ⟨Ĥ.qn⟩ when embedded_H acts as if
it was a UTM.
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