Weak duality ($\leq$). For any flow $f$ and cut $(S,T)$: $|f| = \sum_{v\in S} (\text{net flow out of } v) = \sum_{\substack{u\in S, v\in T}} f(u,v) - \sum_{\substack{u\in T, v\in S}} f(u,v) \leq \sum_{\substack{u\in S, v\in T}} c(u,v) = c(S,T)$.

Strong duality ($\geq$). We show that when no augmenting path exists in $G_f$, the current flow achieves the bound.

Let $f^*$ be a flow with no augmenting $s$-$t$ path in the residual graph $G_{f^*}$. Define $S = \{v \in V : v \text{ is reachable from } s \text{ in } G_{f^*}\}$ and $T = V \setminus S$.

Since no augmenting path exists, $t \notin S$, so $(S, T)$ is a valid $s$-$t$ cut.

Claim: $|f^*| = c(S, T)$.

For any edge $(u, v) \in E$ with $u \in S, v \in T$: if $f^*(u,v) < c(u,v)$, then $c_{f^*}(u,v) > 0$, so $v$ would be reachable from $s$ in $G_{f^*}$, contradicting $v \in T$. Thus $f^*(u,v) = c(u,v)$.

For any edge $(v, u) \in E$ with $u \in S, v \in T$: if $f^*(v,u) > 0$, then $c_{f^*}(u,v) = f^*(v,u) > 0$, so $v$ would be reachable, contradicting $v \in T$. Thus $f^*(v,u) = 0$.

Therefore: $|f^*| = \sum_{\substack{u\in S,v\in T}} f^*(u,v) - \sum_{\substack{v\in T,u\in S}} f^*(v,u) = \sum_{\substack{u\in S,v\in T}} c(u,v) - 0 = c(S,T)$.

Combined with weak duality: $\max|f| = \min c(S,T)$. $\square$