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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.
Les messages affichés proviennent d'usenet.