= Solution
Check each <partial order> axiom. For every $(a,b)$, $a\leq a$ and $b\geq b$, so $S$ is a <reflexive relation>. If both $(a,b)S(c,d)$ and $(c,d)S(a,b)$, the first coordinates satisfy $a\leq c\leq a$ and the second coordinates satisfy $b\geq d\geq b$. Therefore $a=c$ and $b=d$, which proves $S$ is an <antisymmetric relation>.
Finally, if $(a,b)S(c,d)$ and $(c,d)S(e,f)$, then $a\leq c\leq e$ and $b\geq d\geq f$. Hence $(a,b)S(e,f)$, making $S$ a <transitive relation>. All three axioms hold, so \b[$S$ is a partial order]. It is a <product order>, with the usual order on the first coordinate and the reversed order on the second.
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