For a <counterexample> to closure under <union of antisymmetric relations>, take $A=\{0,1\}$, $R=\{(0,1)\}$, and $S=\{(1,0)\}$. Each is an <antisymmetric relation>: neither contains an off-diagonal pair together with its reverse. But $R\cup S$ contains both $(0,1)$ and $(1,0)$ while $0\ne1$. Therefore \b[$R\cup S$ is not an <antisymmetric relation>].

Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics, and probability / Discrete mathematics / 3 / b / Solution

 Solution (source code)