In the example given by in the top answer of Simulation based proofs: Simple examples claims that this is insecure the semi-honest, and I need assistance in where I am failing to reason why that this is true.
By definition (at least put simply to my understanding), a protocol is secure under this simulation based definition if the output views of the simulator and the adversary are computationally indistinguishable.
In the case where $x_2$ is 0, the output of the computation is always 0. And in this case, the simulator has to guess the value of $x_1$, namely it can take either value of 0 or 1. In the case that the simulator guesses $x_1$ correctly, it's obvious that these views are identical. However, what I fail to understand is in the case the simulator guesses $x_1$ wrong, how are these views distinguishable? As far as I can see, no distinguisher should be able to tell if the view of $(1,0,0)$ with the message $1$ from $P_1$ was the real/ideal interaction, or if $(0,0,0)$ with the message $0$ from $P_1$ was the real/ideal interaction.