$$ \begin{array}{l}{\text { Let } M \text { be a partial matching in } G=(V, E) . \text { Prove that the following two conditions are }} \\ {\text { equivalent. }}\end{array} $$
$$ \text { (i) There is no partial matching } M^{\prime} \text { containing } M \text { such that }\left|M^{\prime}\right|>|M|$$
$$\text { (ii) } M \text { does not induce any augmenting path of length } 1 $$
I have the (i) implies (ii) proof but not in the other direction. I feel like the "partial matching containing m implies that there is a path of length 1 in m" is the easiest way to prove this but not sure how to proceed. Any help would be appreciated
As I understood by a partial matching you mean a set of mutually non-adjacent edges and by induced by $M$ augmenting path you mean an (M-)alternating path with $M$-free endvertices. Then the claim follows from the observation that augmenting paths of length $1$ are exactly $M$-free edges of $G$.