Right, which isn't G itself, but something derived from it in Meta-F
Read the proof you have posted, keep track of what system he is talking
about.
Your just PROVING that you don't actually understand what the proof is
about.
>
> Thus Gödel's G is simplified to this:
> G = ¬(F ⊢ G)
Nope, it says that G is true if and only if it is not true that F proves
G, as proven in Meta-F
>
> Translated into Prolog like this:
> ?- G = not(provable(F, G)).
Nop,e as that ISN'T G, only the statment PROVEN to have and equivalent
truth value ot G.
Note, Prolog is incapable of handling this level of Logic.
Can you use Prolog to prove the Pythagorean Theorem?
>
> Found to be incorrect by this:
> ?- unify_with_occurs_check(G, not(provable(F, G))).
> false
Which just means that it is beyond Prolog
Also, since you LIED to Prolog, doesn't mean anything.
>
> Because the Prolog G has an “uninstantiated subterm of itself” we can
> know that unification will fail because it specifies “some kind of
> infinite structure.”The quotes come from: (Clocksin and Mellish 2003:255)
Right, because the logic of the system exceeds the capabilities of Prolog.
>
> So G is unprovable in F because G is incorrect, thus not because F is
> incomplete.
Nope, Since you have shown you don't understand what G actually is, your
logic is incorret.
IF G is incorrect, then there must exist a number that matches the
Primative Recursive Relationship, and thus from the proof in Meta-F, we
know that G is provable, so by your logic, you logic system can prove an
incorrect statement, and thus is shown to be inconsistent.
Of course, since you don't understand what G is, even though you have
presented the paper (translated) of the proof, you are showing that this
is above your head, just shows how little you understand about logic.