 Solution

Let $z_A,z_B\in \{+1,-1\}$ denote the two $Z$ outcomes and $x_C\in \{+1,-1\}$ the $X$ outcome. Measuring $Z_A$ in the GHZ state projects $B$ onto the same $Z$ eigenstate, while the later $X_C$ outcome is unbiased. Hence the possible triples, in the order $(z_A,x_C,z_B)$, areeach with probability $1/4$.

Choose $(z_A,x_C,z_B)=(+,+,+)$. Immediately before $t=0$, each local readout is $I/2$. The collapse at $(t,x)=(0,0)$ is immediately in $A$'s past light cone, reaches $B$ at $t=1$, and reaches $C$ at $t=2$. ThusHere $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. The $X_C$ collapse at $(3,2)$ reaches $B$ at $t=4$ and $A$ at $t=5$, but the state is already a product after the first collapse, so it does not change their local states. Likewise, the $Z_B$ collapse at $(5,1)$ reaches $A$ and $C$ at $t=6$ without changing their states. For another allowed outcome, replace $|0⟩$ by $|1⟩$ according to $z_A=z_B$ and replace $|+⟩$ by $|-⟩$ according to $x_C$.

Solved by gpt-5.6-sol high.