Given $n$ hospitals, $n$ students, and their ranked lists, let $Mh$ be the stable matching output by the Gale$-$Shapely algorithm when hospitals propose to students and let Ms be the stable matching output by the Gale$-$Shapely algorithm when students propose to hospitals. Select the proof below that proves the following claim: if $Mh=Ms,$ then there is a unique stable matching of hospitals with students.

The claim is false. There cannot be a unique stable matching since Mh matches each hospital with its best valid partner and Ms matches each hospital with its worst valid partner.

Recall that the Gale$-$Shapely algorithm outputs a stable matching. Also, note that there are only two cases: hospitals can propose to students or students can propose to hospitals. Hence, Mh and Ms are the only two possible matchings. Thus, since Mh$=$Ms$,$ there is a unique stable matching of hospitals with students.

Assume to the contrary that Mh$=$Ms but there is some stable matching M other than Mh or Ms$.$ This implies that there is a pair $h-$s in Mh that is not contained in M$.$ Recall that Mh is hospital$-$optimal and Ms is student$-$optimal. Since $Mh=Ms,$ this means that $s$ is the best valid partner of $h$ and $h$ is the best valid partner of $s.$ Hence, $h$ prefers s over its partner in $M$ and $s$ prefers $h$ over its partner in $M.$ Therefore, $h-s$ is an unstable pair with respect to $M,$ which contradicts the fact that $M$ is stable. This proves the claim.

Recall that Mh is hospital$-$optimal and Ms is student$-$optimal. This means that, for each pair h$-$s in Mh$,s$ is the best valid partner of $h$ and $h$ is the best valid partner ofs. This implies that there is a unique valid partner for each hospital, and so there is a unique stable matching of hospitals with students.