\b[True.] Fix $a\in A$. <Reflexivity> supplies $aR_1a$ and $aR_2a$. In the definition of <composition of relations>, choose the intermediate element to be $a$ itself. Then $a(R_2\circ R_1)a$. This holds for every $a\in A$, so the composite <binary relation> is <reflexive>.

Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 1 / c / i / Solution

 Solution (source code)