stevendaryl said:
Okay, I've looked at that paper several times, but there were no equations labeled 14a and 14b, so I assumed you meant another paper.
Bell is assuming that ##A(\overrightarrow{a}, \lambda)## is a function returning ##\pm 1##, and the interpretation is that if a particle has the hypothesized hidden variable ##\lambda##, and you measure the spin along direction ##\overrightarrow{a}## (or polarization), then you will get the result ##A(\overrightarrow{a}, \lambda)##.
So ##A(\overrightarrow{a}, \lambda) A(\overrightarrow{b}, \lambda) - A(\overrightarrow{a}, \lambda) A(\overrightarrow{c}, \lambda)## can be written:
1. ##A(\overrightarrow{a}, \lambda) A(\overrightarrow{b}, \lambda) - A(\overrightarrow{a}, \lambda) A(\overrightarrow{c}, \lambda)##
##= A(\overrightarrow{a}, \lambda) (A(\overrightarrow{b}, \lambda) - A(\overrightarrow{c}, \lambda))##
At this point, we can use that ##A(\overrightarrow{b}, \lambda) = \pm 1##, which implies that ##A(\overrightarrow{b}, \lambda) A (\overrightarrow{b}, \lambda) = 1##. So we can rewrite
##A(\overrightarrow{c}, \lambda)) = A(\overrightarrow{b}, \lambda) A (\overrightarrow{b}, \lambda) A(\overrightarrow{c}, \lambda)##
Since the first two factors multiplied together yield +1. So we can plug this in for ##A(\overrightarrow{c}, \lambda))## into the right-side of equation 1 to get:
2. ##A(\overrightarrow{a}, \lambda) A(\overrightarrow{b}, \lambda) - A(\overrightarrow{a}, \lambda) A(\overrightarrow{c}, \lambda)##
##= A(\overrightarrow{a}, \lambda) (A(\overrightarrow{b}, \lambda) - A(\overrightarrow{b}, \lambda) A (\overrightarrow{b}, \lambda) A(\overrightarrow{c}, \lambda))##
##= A(\overrightarrow{a}, \lambda) A(\overrightarrow{b}, \lambda)(1 - A(\overrightarrow{b}, \lambda) A(\overrightarrow{c}, \lambda))##