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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Unpartial Halt Decider 4.0
Date: Sat, 19 Apr 2025 19:02:05 -0400
Organization: i2pn2 (i2pn.org)
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On 4/19/25 4:07 PM, Mr Flibble wrote:
> On Sat, 19 Apr 2025 13:00:42 -0700, Keith Thompson wrote:
> 
>> Mr Flibble <flibble@red-dwarf.jmc.corp> writes:
>> [...]
>>> Theorem (Flibble’s Model-Theoretic Parity Principle):
>>> In any theoretical system that permits infinite computational behavior,
>>> a decider analyzing that system may be equipped with equivalent
>>> infinite resources, so long as both reside in a consistent meta-model.
>>
>> If that's a theorem, what's the proof?
> 
> Proof of Flibble’s Model-Theoretic Parity Principle:
> 
> Theorem (Flibble’s Model-Theoretic Parity Principle)
> In any theoretical system S that permits computational entities M to
> exhibit unbounded or infinite behavior (e.g., infinite tape, unbounded
> runtime), it is logically consistent to define an analyzer (decider) D
> within an extended system S' with equally unbounded or infinite resources,
> such that D analyzes M's behavior within the constraints of S, without
> contradiction — provided S' ⊇ S and D is not subject to stricter
> constraints than M.
> Proof
> Let S be a computational system
> S allows machines M ∈ S with infinite computational behaviors, such as
> unbounded tape or unbounded execution time.
> Let D be a proposed decider for machines in S
> D is designed to determine properties such as halting behavior by
> simulating M.
> Let S' be an extension of S
> S' includes all descriptions and behaviors of S and additionally permits
> unbounded computational analysis (e.g., infinite simulation time and
> memory).
> Construction of D
> Define D ∈ S' such that D(M, x) simulates M on input x. D may take
> infinite steps but is allowed to do so in S'. D returns 'halts' or
> 'doesn’t halt' based on complete simulation.
> Consistency of D
> D does not exist in S and does not violate Turing’s Halting Theorem since
> it operates outside the constraints of S. Turing’s proof applies only to
> deciders within the same system as the machine they analyze.
> Conclusion
> Therefore, defining a decider D with equivalent or greater resources than
> machines M ∈ S is logically consistent, provided D exists in a model S'
> that extends S and permits such analysis.
> 
> /Flibble

So, your "proof" is based on the idea that you must be able to compute 
all the mappings.

Note, the decider *IS* given eqquivalent RESOURCES, it just has a 
requirement to answer in FINIT TIME.

Note, Machines that don't answer in finite time are just considered to 
not answer, which just isn't a option for a machine defined to answer 
for all inputs.

So sorry, that isn't a valid proof.

Note, "Logically Consistant" is not the same thing as True.

So, all you are doing is proving that you "Compuatational System" must 
be something very different than what we consider to actually be usable.