# 2 party AND computation under passive perfect security

In the book written by Ivan Damagard titled "Secure Multiparty Computation and Secret Sharing", at the end of the third chapter he provides a proof for why it is impossible to securely compute 2 party AND function under perfect and passive security with one corrupt party.

Can someone help me provide an intuitive explanation for why the above is true. I was unable to grasp the solution provided by Damagard. I could not find any other resource regarding the same either.

The proof proceeds by arguing that the intersection between these sets of transcript actually has to be empty (since you can't output both 0 and 1). However, I actually think that the proof is already complete when they show that $T(1,0) \cap T(0,1) \subset T(1,1)$. In order to see this, let $t\in T(1,0)$; by what we have stated now it also follows that $t\in T(1,1)$. Since P1's output is determined by its input and transcript, it follows that it outputs the same bit in both cases. This contradicts correctness.