XPost: sci.logic
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its
input D until H correctly determines that its simulated D
*would never stop running unless aborted then*
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
To make this easier to understand we replace line three
with the more conventional terminology this line:
"cannot possibly reach its own simulated final halt state then"
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
I don't think that is the shell game. PO really /has/ an H (it's
trivial to do for this one case) that correctly determines that P(P)
*would* never stop running *unless* aborted. He knows and accepts that
P(P) actually does stop. The wrong answer is justified by what would
happen if H (and hence a different P) where not what they actually are.
Saying that H is required report on the behavior of
machine M is a category error.
Turing machines cannot directly report on the behavior
of other Turing machines they can at best indirectly
report on the behavior of Turing machines through the
proxy of finite string machine descriptions such as ⟨M⟩.
Thus the behavior specified by the input finite string
overrules and supersedes the behavior of the direct
execution.
DDD correctly simulated by HHH cannot possibly
reach its own simulated "return" instruction
final halt state, thus overruling and superseding
the behavior of the directly executed DDD().
--
Copyright 2024 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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