Does Bell's inequalities also rule out non-computable local hidden variable theories? I have beenn reading different articles on Bell's assumptions and interpretations, including superdeterminsm. I always end up dizzy when I try tho think about this specific question, so any hints would be greatly appreciated.
Let us imagine that the randomnes in quantum mechanics arises not because of some chaotic hidden variable process but by an actually non-computable local process. For instance, an underlying 3D local cellular automata with an oracle that can solve the halting problem. In this case there is no way, even in principle, to predict the behaviour of the hidden variables, they will behave trully randomly (not just pseudorandomly). 
So the questions are two:
1) Could such an inherent randomness from a non-computable law be the source of the "true" irreducible randomness of quantum mechanics?
2)Woud such kind of source bypass Bell's assumptions, in a way that you can still mantain local determinism but are at the same time barred from ever using it to make any exact predictions (because of it not being computable) ?
Update:
First, this is not my personal theory, it is only a question about the foundations of physics. It is mainstream physics, and not some sort of vague delirium.
The two questions seem unrelated but I believe they are, and I will try to explain my train of thought about why. 
First, when we usually talk about the randomness being an irreducible property of nature, my interpretation is that we are somehow saying that there are no laws that determine a particular measurement. This in turn implies that there is no hidden variables theory. Bell inequelities are supossed to distinguish between these two positions (I MIGHT BE WRONG HERE). 
If randomness is inherent but because of laws, even if there are not hidden variables (I am not sure if we could talk of a hidden variables theory if such theory is not computable, because there nothing to predict), then perhaps this assumption was not considered as a possibility on bell's theorem. My intuition is that perhaps this assumption is not included in Bells's assumptions. But I am not sure how the option of hidden variables was formalized. Perhaps it is completely irrelevant if there are hidden variables or not. That is why I posted this question.I am confussed.
For instance, equations (1) and (2) of this article , that tries to dissect Bell's assumptions, would not make sense to me.
 A: It shouldn't make a difference whether the hidden variables are generated by a computable or non-computable rule, so yes, Bell's proof should rule out non-computable local hidden variables too. Here's a simple toy model of Bell inequality violations I wrote up a while ago:

Suppose we have a machine that generates pairs of scratch lotto cards,
  each of which has three boxes that, when scratched, can reveal either
  a cherry or a lemon. We give one card to Alice and one to Bob, and
  each scratches only one of the three boxes. When we repeat this many
  times, we find that whenever they both pick the same box to scratch,
  they always get the same result--if Bob scratches box A and finds a
  cherry, and Alice scratches box A on her card, she's guaranteed to
  find a cherry too.
Classically, we might explain this by supposing that there is
  definitely either a cherry or a lemon in each box, even though we
  don't reveal it until we scratch it, and that the machine prints pairs
  of cards in such a way that the "hidden" fruit in a given box of one
  card always matches the hidden fruit in the same box of the other
  card. If we represent cherries as + and lemons as -, so that a B+ card
  would represent one where box B's hidden fruit is a cherry, then the
  classical assumption is that each card's +'s and -'s are the same as
  the other--if the first card was created with hidden fruits A+,B+,C-,
  then the other card must also have been created with the hidden fruits
  A+,B+,C-.
The problem is that if this were true, it would force you to the
  conclusion that if Alice and Bob are picking randomly which box to
  scratch on each trial (with a 1/3 chance of A, B, or C each time),
  then if they do this a large number of times, we should expect that in
  the subset of trials where Alice and Bob happened to pick different
  boxes to scratch, they should find the same fruit at least 1/3 of the
  time. For example, if we imagine Bob and Alice's cards each have the
  hidden fruits A+,B-,C+, then we can look at each possible way that
  Alice and Bob can randomly choose different boxes to scratch, and what
  the results would be:
Bob picks A, Alice picks B: opposite results (Bob gets a cherry,
  Alice gets a lemon) 
Bob picks A, Alice picks C: same results (Bob gets a cherry, Alice
  gets a cherry)
Bob picks B, Alice picks A: opposite (Bob gets a lemon, Alice gets a
  cherry)
Bob picks B, Alice picks C: opposite results (Bob gets a lemon,
  Alice gets a cherry)
Bob picks C, Alice picks A: same results (Bob gets a cherry, Alice
  gets a cherry)
Bob picks C, Alice picks picks B: opposite results (Bob gets a
  cherry, Alice gets a lemon) 
In this case, you can see that that if they are equally likely to pick
  each combination of boxes, then 2 times out of 6 when they choose
  different boxes, they will get the same fruit (i.e. a 1/3 chance of
  the same result). You'd get the same answer if you assumed any other
  preexisting state where there are two fruits of one type and one of
  the other, like A+,B+,C- or A+,B-,C-. On the other hand, if you assume
  a state where each card has the same fruit behind all three boxes, so
  either they're both getting A+,B+,C+ or they're both getting A-,B-,C-,
  then of course even if Alice and Bob pick different boxes to scratch
  they're guaranteed to get the same fruits with probability 1. So if
  you imagine that when multiple pairs of cards are generated by the
  machine, some fraction of pairs are created in inhomogoneous
  preexisting states like A+,B-,C- while other pairs are created in
  homogoneous preexisting states like A+,B+,C+, then the probability of
  getting the same fruits when you scratch different boxes should be
  somewhere between 1/3 and 1. 1/3 is the lower bound, though--even if
  100% of all the pairs were created in inhomogoneous preexisting
  states, it wouldn't make sense for you to get the same answers in less
  than 1/3 of trials where you scratch different boxes, provided you
  assume that each card has such a preexisting state with "hidden
  fruits" in each box.
But now suppose Alice and Bob look at all the trials where they picked
  different boxes, and found that they only got the same fruits 1/4 of
  the time! That would be the violation of Bell's inequality, and
  something equivalent actually can happen when you measure the spin of
  entangled photons along one of three different possible axes. So in
  this example, it seems we can't resolve the mystery by just assuming
  the machine creates two cards with definite "hidden fruits" behind
  each box, such that the two cards always have the same fruits in a
  given box.
Something mathematically analogous is predicted in certain experiments
  where entangled photons are passed through polarizers at different
  angles, in which the probability that both photons will have the same
  result (both pass through their respective polarizers, or both are
  blocked) is $ cos^2 (\theta) $, where $ \theta $ is the angle between
  the two polarizers. Suppose the experimenters decide in advance that
  on each trial, they will choose randomly between one of three angles
  for their polarizer: 0 degrees from vertical, 60 degrees, or 120
  degrees. On a given trial, if both experimenters choose the same
  angle, then $ \theta = 0 $ so they are guaranteed to get the same
  result with probability 1 (both photons pass or both are blocked), but
  if they choose different settings the probability of getting the same
  result would be $ cos^2 (\pm 60) $ or $ cos^2 (\pm 120) $, which in
  both cases gives a probability of 1/4. But the Bell inequality whose
  derivation I sketched says that if you want to explain the perfect
  match when both experimenters make the same choice using local hidden
  variables, the probability of a match when they make different choices
  should be no lower than 1/3.

Note that nothing in the argument depends on how the device printing up pairs of lotto cards decides what hidden fruits to put in the boxes. All that matters is that there must be some fraction of trials where it prints up a "homogeneous" pair and some fraction where it prints up an "inhomogenous" pair, and that the probability that Alice and Bob choose to scratch the same box or different boxes on a given trial is statistically uncorrelated with whether the device printed up a homogenous or inhomogenous pair on that trial (this last assumption would be violated with superdeterminism, but I don't think you were talking about that).
A: Your toy theory is i) local and ii) sounds like it's counter-factual definite, that is, every measurement that could be or could have been performed would result in a single, definite outcome (roughly speaking, an object's properties are pre-existing or real).
You want to know whether you can evade Bell's inequalities [that permit i) or ii) but not both], because of the microscopic details, here a non-computable process?
Bell's inequalities are robust; the assumptions are so reasonable that they are almost akin to common sense and it is difficult to imagine a scientific programme without them. They are implicit in most experiments. 


*

*One assumption is no superdeterminism, that is, that we can chose which measurements to perform, independently of the properties of the objects we are looking at. 

*Another assumption is causality. The outcomes of future experiments can't affect the measurements I previously chose to make or the previous settings of my measurement devices.


I can't see a way in which a non-computable process (or any similar microscopic detail) evade the  Bell's powerful theorem by breaking an assumption. 

I think a possible source of confusion is the CFD wiki page, which in the first line makes reference to computability. I think this is misleading and ought to be corrected.
