(a) that Y 1 (1) = Y 2 (2) = 1 or (b) that Y 1 (1) = Y 2 (2) = − 1. Suppose that (a) holds: then M (0 , 2) = 0 and Y 2 (2) = 1 imply Y 1 (0) = − 1. But, M (1 , 0) = 0 and Y 1 (1) = 1 imply Y 2 (0) = − 1 and thus,by M (0 , 0) = 0, that Y 1 (0) = 1, a contradiction. 

Suppose instead that (b) holds. Then M (0 , 2) = 0 and Y 2 (2) = − 1 implies Y 1 (0) = 1. But M (1 , 0) = 0 and Y 1 (1) = − 1 implies Y 2 (0) = 1 and thus, by M (0 , 0) = 0, that Y 1 (0) = − 1, again a contradiction. Thus it cannot be the case that Y$_{i}$ ( x 1 ,x 2 ) = Y$_{i}$ ( x$_{i}$ ) ,i = 1 , 2 . 

The next result is given in VanderWeele (2010) in the context of testing for a causal interaction, sometimes referred to as ”epistasis” in genetics. It relates the empirical data E [ M ( x 1 ,x 2 )] to the existence of a unit satisfying M (1 , 2) = 1 ,M (0 , 2) = M (1 , 0) = M (0 , 0) = 0. Within the counterfactual framework, this would constitute a causal interaction for the variable M . Since the proof of the result relating observed data E [ M ( x 1 ,x 2 )] to units such that M (1 , 2) = 1 ,M (0 , 2) = M (1 , 0) = M (0 , 0) = 0 is essentially one line, we give it here also for completeness. 

Theorem 2 . If E [ M (1 , 2)] − E [ M (0 , 2)] − E [ M (1 , 0)] − E [ M (0 , 0)] > 0, then there must exist a unit with M (1 , 2) = 1 ,M (0 , 2) = M (1 , 0) = M (0 , 0) = 0. 

Proof. By contradiction. Suppose there were no unit with M (1 , 2) = 1 ,M (0 , 2) = M (1 , 0) = M (0 , 0) = 0. Then, for all units, M (1 , 2) − M (0 , 2) − M (1 , 0) − M (0 , 0) ≤ 0 which implies E [ M (1 , 2)] − E [ M (0 , 2)] − E [ M (1 , 0)] − E [ M (0 , 0)] ≤ 0, a contradiction. 

An immediate corollary of Theorems 1 and 2 is then: 

Corollary . If E [ M (1 , 2)] − E [ M (0 , 2)] − E [ M (1 , 0)] − E [ M (0 , 0)] > 0, then the the hypothesis of ’local hidden variables’ is false. 

This corollary is referred to as Bell’s theorem in the physics literature. Its premise is referred to as Bell’s inequality. As noted above, from the quantum mechanics of the 2- particle system, and confirmed by experiment, E [ M ( x 1 = i,x 1 = j )] = sin $^{2}$(Δ$_{ij}$ / 2). Thus we have that: 

