XPost: comp.theory, sci.logic, sci.math
Correctly answering this single question refutes the Linz proof:
Which state does Ĥ applied to ⟨Ĥ⟩ transition to correctly ?
The following simplifies the syntax for the definition of the Linz
Turing machine Ĥ, it is now a single machine with a single start state.
A copy of Linz H is embedded at Ĥ.qx.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Because there are no finite number of steps of the pure simulation of
⟨Ĥ⟩ applied to ⟨Ĥ⟩ by embedded_H such that this simulated input meets the Linz definition of halting:
computation that halts … the Turing machine will halt whenever it enters
a final state. (Linz:1990:234)
Therefore it is correct to say that the input to embedded_H specifies a sequence of configurations that never halts.
https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3
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Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer
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