Proof: By double implication: If there is an augmenting path, then M is not maximum: OK. Suppose there is no augmenting path for M but Mopt is maximum. For each node in M Mopt (where is a xor), the degree is either 0, 1 or 2, thus M Mopt consists of disjoint paths or cycles. • The cycles must be even (otherwise M or Mopt is not a matching). • Suppose there is an odd path. Since #Mopt is maximum...