Gödel's Incompleteness Theorem and the complexity of arithmetic In How complicated can structures be? Jouko Väänänen says:

The guiding result of mathematical logic is the Incompleteness Theorem of Gödel,
  which says that the logical structure of number theory is so complicated that it cannot be
  effectively axiomatized in its entirety. In other
  words, the theory is non-recursive, i.e. there
  is no Turing machine that could tell whether
  a sentence of number theory is true or not.

I've never seen Gödel's Incompleteness Theorem this way: that it's a matter of the overall complexity of the structure of the natural numbers that there are facts about them that cannot be proved. 
So I wonder whether I can take the quote above literally:

Can Gödel's Theorem be rigorously stated in terms
  of complexity? 

Somehow like this: "Every
system which exceeds complexity
threshold X is undecidable."
Or is it just a vague paraphrase, not to be taken too seriously?
 A: I agree with your statement
"Every system which exceeds complexity threshold X is undecidable".
Let us focus on the specific case where we consider the the first order theory of a fixed
structure.
Once the structure is complicated enough to simulate (or to express) computation,
the theory becomes undecidable. 
This is based on the following easy proof of a weak form of the incompleteness theorem:
In the language of number theory, you can write down a formula $\varphi(x)$
such that the natural numbers satisfy $\varphi(t_n)$ for a natural number $n$ iff
$n$ is the Goedel number of a Turing machine that halts on an empty tape.
Here $t_n$ denotes the term for the $n$-th successor of $0$.
If the theory of the natural numbers was decidable, then you could decide the halting 
problem.
A: Yes, this line of thought is perfectly fine.
A set is decidable if and only if it has complexity
$\Delta_1$ in the arithmetic
hiearchy,
which provides a way to measure the complexity of a
definable set in terms of the complexity of its defining
formulas. In particular, a set is decidable when both it
and its complement can be characterized by an existential
statement $\exists n\ \varphi(x,n)$, where $\varphi$ has
only bounded quantifiers.
Thus, if you have a mathematical structure whose set of
truths exceeds this level of complexity, then the theory
cannot be decidable.
To show that the true theory of arithmetic has this level
of complexity amounts to showing that the arithmetic
hierarchy does not collapse. For every $n$, there are sets
of complexity $\Sigma_n$ not arising earlier in the
hierarchy. This follows inductively, starting with a
universal $\Sigma_1$ set.
Tarski's theorem on the non-definability of
truth
goes somewhat beyond the statement you quote, since he
shows that the collection of true statements of arithmetic
is not only undecidable, but is not even definable---it
does not appear at any finite level of the arithmetic
hiearchy.
Finally, it may be worth remarking on the fact that there
are two distinct uses of the word undecidable in this
context. On the one hand, an assertion $\sigma$ is not
decided by a theory $T$, if $T$ neither proves nor refutes
$\sigma$. On the other hand, a set of numbers (or strings,
or statements, etc.) is undecidable, if there is no Turing
machine program that correctly computes membership in the
set. The connection between the two notions is that if a
(computably axiomatizable) theory $T$ is complete, then its
set of theorems is decidable, since given any statement
$\sigma$, we can search for a proof of $\sigma$ or a proof
of $\neg\sigma$, and eventually we will find one or the
other. Another way to say this is that every computably
axiomatization of arithmetic must have an undecidable
sentence, for otherwise arithmetic truth would be
decidable, which is impossible by the halting problem (or
because the arithmetic hierarchy does not collapse, or any
number of other ways).
A: Hmm, nobody has mentioned Chaitin's incompleteness theorem.
