Here's something I don't quite understand in the proof of Godel's first Incompleteness Theorem:
Why do you have to introduce some kind of 'minimal theory' (like Q or PA or Robinson Arithmetic) T for which you then show that every recursive function is representable in that theory, i.e. that for every recursive function f(x) there is some formula phi_f(x,y) such that for all x,y: f(x) = y iff in T |- Ay (phi_f(x,y) <-> y = y)?
Why can't you just show that for any recursive function there is a formula phi_f(x,y) such that for all x,y: f(x) = y iff in N |= Ay (Phi_f(x,y) <-> y = y)?
Likewise for the Fixed Point or Diagonal Theorem: why do you have to show that for any formula phi(x) there exists some G such that T |- G <-> phi(g) where g is the godel number for G, instead of just showing that N|= G <-> phi(g)?
In short, why do you need to go through all the trouble of proving all kinds of representability results to show that the corresponding statement is provable in T, rather than just being true under the standard interpretation N? (indeed, why not just stick with definability?)
For if you did the latter, couldn't you then just say that as long as some theory T is recursive, you have a proof predicate phi_proof(x) such that N|=phi_proof(x) iff x is the godel number of a statement that can be proven in T (i.e. element in T), that the Diagonal Lemma gives you the existence of a godel sentence G such that N|=G <-> ~phi_proof(g), and so, as long as T is consistent: if G is provable in T, then N|=G, so N|=phi_proof(g), so G is provable in T, so contradiction, so G is not provable in T? I.e. that any consistent and recursive theory is incomplete? Why wouldn't that all you need? I must be missing something. Please help!