Re: Unpartial Halt Decider 4.0

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.