stylistic:tone, linguistic:paraphrasing
Proof that the algorithm makes progress

We must show that as long as the matching is not of maximum possible size, the algorithm is always able to make progress — that is, to either increase the number of matched edges, or tighten at least one edge. It suffices to show that at least one of the following holds at every step:

If M is of maximum possible size, we are of course finished. Otherwise, by Berge's lemma, there must exist an augmenting path P with respect to M in the underlying graph G. However, this path may not exist in G y {\displaystyle G_{y}}: Although every even-numbered edge in P is tight by the definition of M, odd-numbered edges may be loose and thus absent from G y {\displaystyle G_{y}}. One endpoint of P is in R S {\displaystyle R_{S}}, the other in R T {\displaystyle R_{T}}; w.l.o.g., suppose it begins in R S {\displaystyle R_{S}}. If every edge on P is tight, then it remains an augmenting path in G y {\displaystyle G_{y}} and we are done. Otherwise, let u v {\displaystyle uv} be the first loose edge on P. If v ∉ Z {\displaystyle v\notin Z} then we have found a loose-tailed path and we are done. Otherwise, v is reachable from some other path Q of tight edges from a vertex in R S {\displaystyle R_{S}}. Let P v {\displaystyle P_{v}} be the subpath of P beginning at v and continuing to the end, and let P ′ {\displaystyle P'} be the path formed by traveling along Q until a vertex on P v {\displaystyle P_{v}} is reached, and then continuing to the end of P v {\displaystyle P_{v}}. Observe that P ′ {\displaystyle P'} is an augmenting path in G with at least one fewer loose edge than P. P can be replaced with P ′ {\displaystyle P'} and this reasoning process iterated (formally, using induction on the number of loose edges) until either an augmenting path in G y {\displaystyle G_{y}} or a loose-tailed path in G is found.