Max-Flow Min-Cut Theorem

Weak duality first. Take any $s$-$t$ cut $S$,$T$ with $s$ in $S$ and $t$ in $T$, and any feasible flow $|f|$. Every unit of flow leaving $s$ must eventually cross from $S$ to $T$ to reach $t$, and by conservation the net flow across the cut equals the total flow value. Since edges from $T$ back to $S$ can only reduce this net amount, the flow value can never exceed the sum of capacities of edges leaving $S$, i.e. $|f|$ is at most $\mathrm{cap}(S,T)$ for every cut.

Now run the Ford-Fulkerson procedure to completion: repeatedly find a path from $s$ to $t$ in the residual graph (remaining capacity plus reversible flow already sent) and push as much flow as the tightest edge on that path allows. Because capacities are assumed integer or rational and strictly decrease along the path used, this process terminates after finitely many augmentations.

At termination no augmenting path exists. Let $S$ be the set of nodes reachable from $s$ in the final residual graph, and $T$ the rest (so $t$ is in $T$ since no path reaches it). Every edge from $S$ to $T$ in the original network must be fully saturated (otherwise it would still have residual capacity, extending $S$), and every edge from $T$ back to $S$ must carry zero flow (otherwise its reverse residual edge would extend $S$).

Summing flow conservation over all nodes in $S$ shows the total flow value equals exactly the capacity of the saturated forward edges minus the zero backward flow, which is precisely $\mathrm{cap}(S,T)$ for this particular cut. Combined with the upper bound from the first paragraph, this cut achieves the minimum possible capacity, so the maximum flow value equals the minimum cut capacity.