 Solution

For local dichotomic outcomes define $A(a,\lambda ),B(b,\lambda )=\pm 1$ and a setting-independent normalized distribution $\rho (\lambda )$. At each $\lambda$,because exactly one bracket vanishes and the other is $\pm 2$. Integrating and using the triangle inequality yieldsFor stochastic local response functions the same bound follows by replacing outcomes with conditional means in $[-1,1]$, or by including local randomness in $\lambda$. The important locality condition is factorization of joint conditional outcome probabilities, together with independence of the setting distribution and $\lambda$.

The spin singlet instead predicts $E(a,b)=-a\cdot b$. Two orthogonal directions at one wing and directions halfway between them at the other give $|S|=2\sqrt{2}$. Therefore reproducing the quantum predictions requires giving up at least one of the stated premises. In the perfect-anticorrelation version, $B(a,\lambda )=-A(a,\lambda )$ gives Bell's three-setting form $|E(a,b)-E(a,c)|\le 1+E(b,c)$ by the same algebra.

This does not establish that all deterministic or hidden-variable theories are impossible: nonlocal guidance is a counterexample to that reading. Nor does it equate Bell locality with no-signalling; averaging quantum outcomes can preserve setting-independent local marginals despite nonfactorizing correlations. The theorem is significant because it turns a local-completeness dispute into incompatible quantitative correlation bounds, not because it refutes every possible meaning of physical reality.^[ref]