Next, we show the equivalence.(1). P2⇒P1: Assume an optimal solutionf*ijexists for the max-flow problem. Then constructthe set of path flowsfmfor allmin such a way that the following holdsXm: (i,j)∈Pmfm=f?ij,for all (i, j)∈ E.(4)If such a construction offmis possible, then we can proceed by proving the equivalence. It is easyto see that this selection offmfor allmis greater than or equal to 0, and satisfies the capacityconstraint in (1). Therefore, that construction gives a feasible point in (1). Note that each pathends at the destination, and therefore for eachPm, there existsisuch that (i, d)∈Pm. Therefore,the objective function can be expressed asXfm=Xi:(i,d)∈EXm:(i,d)∈Pmfm=Xi:(i,d)∈Ef?id=f?dswhere the second equality follows from (2) and last equality follows from (3). What remains is totry to prove that such a construction offmwith condition (4) is possible. One way to constructsuch a solution is using the following algorithm1. Initializeˆfij=f?ijfor all (i, j)∈ Eandfm= 0 for all pathsm.2. Select a pathPm.For pathPm, select the edge (i, j)min∈Pmsuch that (i, j)min=argmin(i,j)∈Pmˆfij(the minimum edge flow across the path).Set the flow of the path to beequal to that edge flow,i.e.,fm=ˆf(i,j)min.3. Subtractˆf(i,j)minfromˆfijfor all (i, j)∈Pm(i.e., subtract it from all edge flows of all edgesin the path).4. Repeat the previous steps until in every path there is at least one edge withˆfij= 0.