Solution
= Solution
For every $a$, both $aRa$ and $aSa$ hold, so the <intersection of equivalence relations> is <reflexive>. If $a(R\cap S)b$, <symmetry> of both component <binary relations> gives $bRa$ and $bSa$, hence $b(R\cap S)a$. Finally, if $a(R\cap S)b$ and $b(R\cap S)c$, <transitivity> in each component gives $aRc$ and $aSc$, hence $a(R\cap S)c$. These are precisely the three conditions for an <equivalence relation>.