let f be any s-t flow, and (A, B) any s-T cut then V(f) = Fout(a) - fin(a)
proofs of why this is maximum flow on 347
max flow = min cut
7.9 - if f is an s-t flow sucht that there is no s-t path in the residual graph Gf, then there is an s-t cut (a*, b*) in G for which v(F) = c(A*,B*) consequently f has maximum value of any flow in G, and (A*, B*) has the minimum capacity of any s-t cut in G.
7.10 - the flow f returned by the ford - fulkerson algorithm is a maximum flow.
7.11 given a flow f of maximum value, we can compute as s-t cut of minimum capacity in O(m) time.
7.13 - in every flow network, the maximum value of an s-t flow is equal to the minimum capacity of an s-t cut. 7.14 if all capacities in the flow network are integers, then there is a maximum flow f for which every flow value f(e) is an integer.