[ExI] Trilemma of Consciousness

[Stuart LaForge]

>On Thu, May 25, 2017 at 7:33 AM, Stuart LaForge <avant at sollegro.com>
wrote:
>> Axiom 1: Let F(n) be the nth computable function with n being an
>> admissible numbering of all possible computable functions.
>> Axiom 2: Let K be the subset of F(n) such that all K share a semantic
>> property k.
>> Definition 1: Let k be called trivial if all F(n) have property k.
>> Definition 2. Let k be called null if no F(n) has property k.
>> Axiom 3: Let Dt be the decision problem as to whether a given F(n) belongs
>> in K.
>>
>> Theorem: By axioms 1-3, definitions 1-2, and Rice's Theorem, Dt is
>> decidable if, and only if, k is trivial or null.

I made an error above in Axiom 1. It should read

Axiom 1: Let F(n) be the nth partial computable function with n being an
admissible numbering of all possible partial computable functions.

Adrian Tymes wrote:

>Counterproof:
>
>Let k be, "Given the numbering in use, is n > 1?"  (This is only null
>if n is at most 1, though in that case all ks will be either trivial
>or null anyway.)  This appears to be computable.

Well by definition of an "admissable numbering", n must be computable. But
"is n > 1?" does not qualify as a *semantic* property of a partial
computable function as per Axiom 2.

Since my understanding of a semantic property of a function is one related
to "meaning" that manifests in the behavior of the function. For example,
"always halts on any input" would be a semantic property.

If "is n > 1?" were a semantic property of F(n), then F(n) would have to
reference n somewhere in the function for it to influence the behavior of
that function. That means that "is n > 1?" would be a syntactical property
of F(n) as well.

But in such a case, F(n) would then violate Axiom 1 because F(n) would no
longer be an indexed list of all possible partial computable functions but
would instead be a list restricted to functions that explicitly referenced
their own indices. Therefore, my theorem and Rice's also, would no longer
apply.

>Let n be some integer greater than 1.  (I am assuming the numberings
>only considers integer numberings starting with 1, but this proof
>works with only slight modification to the examples if not.)
>
>k is neither trivial (since F(1) does not have this property) nor null
>(since F(2) has this property, as are others if n > 2).
>
>However, Dt is decidable: F(1) is not in K.  The rest, which consists
>of at least F(2), are in K.

Your counter-proof only holds if given an infinite list of all possible
partial computable functions F indexed on n, n is a semantic property of
F(n). Can you prove that? It seems to me to be more like some kind of meta
property.

Stuart LaForge