case $ZZ$, where both modes are 1 and thus observable $Z$ is measured on the left and the right. By , the conditional probability $P\big[(+1,+1){\:\vert\:}ZZ\big]$ that we have $+1$ on the left and the right is $1/3$. Hence $+1\in f_l(Z)\cap f_r(Z)$. In particular, $+1\in f_l(Z)$.

By , in the case $XX$, where both modes are 2 and thus observable $X$ is measured on the left and the right, the conditional probability ${\mathcal P}\big[(+1,+1){\:\vert\:}XX\big]$ that $+1$ is produced on the left and the right is positive (namely $1/12$) and therefore $+1\in f_r(X)$.

Now let’s consider the case $ZX$, where the left mode is 1 and so $Z$ is measured on the left and where right mode is 2 and so $X$ is measured on the right. By noncontextuality, it must be possible to have $+1$ on the left and on the right in the same trial. But this does not happen. For, by , the conditional probabilities in the $ZX$ case are as follows. $${\mathcal P}\big[(+1,-1){\:\vert\:}ZX\big] = 2/3,\quad {\mathcal P}\big[(-1,+1){\:\vert\:}ZX\big]= 1/6,\quad  {\mathcal P}\big[(-1,-1){\:\vert\:}ZX\big] = 1/6,$$ so that ${\mathcal P}\big[(+1,+1){\:\vert\:}ZX\big]=0$ which gives the desired contradiction.

[Observation space.]{} The sample space $\Omega$ consists of 16 outcomes for the combination of the random settings 