Edwin, by way of welcoming your return, here's another brief comment on Bell's mental entanglement of physical entanglement.
In the lead-in section of equation (X) -- see my Jun. 19, 2013 @ 13:55 GMT post above -- the word "crazy" appears twice. And we understand that those "crazies" apply only to that particular Bellian context: ie, where particular sets of pristine particle-pairs have been subjected to their once-and-only-once EPRB test.
BUT NO inherent problem would arise had we re-allocated the sets of particle-pairs and instead tested P(b,c) via A(b, L1)A(c,L1) or via A(b, L2)A(c,L2).
Because there is NOTHING special about particular particle-pair sets: so long as we remember that each pristine particle-pair can only be EPRB-tested ONCE. Thus either of the above revised allocations would have given the correct P(b,c)!
However, we'd then need a consequent re-allocation of pairs to test for P(a,b) or P(a,c).
So, to save that hassle, let's do the smart thing and test for P(b,c) under the condition A(b,L3)A(c,L3); ie, from a new set of pristine pairs.
The condition that Bell, you, and I then need satisfied in accord with our earlier agreement -- see my Jun. 19, 2013 @ 13:55 GMT post above -- is this:
A(b,L3)A(c,L3) =?= A(a,L1)A(b,L1)A(a,L2)A(c,L2). ---(Y)
And that satisfaction is so easily delivered: Just take the simple case of a = b!
For (Y) then becomes:
A(b,L3)A(c,L3) = A(b,L1)A(b,L1)A(b,L2)A(c,L2) = A(b,L2)A(c,L2); ---(Z)
no question mark in sight. So it's QED; no problem at all, and hardly a challenge.
But for Bell and his followers: It's a pity about the new super-restrictive boundary condition: a = b!
So, Edwin, when you return, please make these equations your own. DO NOT be mislead by any error on my part. For if I have a minus that should be a plus, or vice versa, we're back to square zero.
PS: In that we've employed L1, L2 and L3, we sure need a neat notation that matches the Essay and fits the approach here.
What do you think about the simplicity of these sets?
{(pi, p'i) -- (pn, p'n)} = p0,
{(pn+i, p'n+i) -- (p2n, p'2n)} = p1,
{(p2n+i, p'2n+i) -- (p3n, p'3n)} = p2, etc;
where the shorthand code-number on RHS is the w-value of the twin-set. For all are happy members of that interesting family of twins
{(pwn+i, p'wn+i)} = pw;
all their co-identified lambdas readily discerned; set shorthand-identifier p1 not the same as particle p1; etc.
Gordon