emuc said:
B and A are local definitions as lambda and delta are local. A(delta, lambda) is defined by equations 2-5. Using M5 B(delta, lambda) is defined by equations 8-11.
Photon 1 and photon 2 of a pair share the same lambda.
One locality part in the "local realism" premise of the theorem is that the choice of measurement setting made at A should be independent from B. The fact that the theorem makes a prediction for the case where there
happens to be a certain relation between a and b and you have perfect anti-correlation, does not mean that this relation must be confused with a dependence.
As far as I can see from what you did, it seems to me that you tried to transform away the a and b, by delta which is is determined by a and gamma, OR by b and gamma, thus you have two different deltas, don't you?
$$\delta_a = w(a,\lambda)$$
$$\delta_b = v(b,\lambda)$$
So what you should have at this point is
$$A=f(\delta_a, \lambda)= f(\delta_a, w(a,\lambda)) = f'(a,\lambda) $$
$$B=g(\delta_b, \lambda)= g(\delta_b, w(b,\lambda)) = g'(b,\lambda) $$
So far, all you did was a change of variables, but you still are stuck in the original form.
Now the premise of the theorem is that a and b are independent. In your case you seem to inappropriate link them together so that the "accidental case where the free choices conincide" becomes forced in your case. This is also why you have dependence between delta_a and delta_b.
This effectively means that you have,
$$A= f'(a,\lambda) $$
$$B= g''(a,\lambda) $$
So IMO what you did is replace the premises that normally consistutes what is "local realism".
I think is somewhat disguised with your mixing in the \delta, though. This is is the reason why i tried to note the general form in the other post to see the big picture before the details. Perhaps i got something wrong in your paper?
/Fredrik