[proofplan]
We first fix notation for the augmented network: $D_\nu=(V,A_\nu)$ is obtained from $D$ by adjoining a new arc $a_\nu:t\to s$, and $l_\nu,u_\nu$ are the capacities extending $l,u$ with value $\nu$ on $a_\nu$. Since the terminals are distinct, this new arc contributes only as an incoming arc at $s$ and only as an outgoing arc at $t$. A feasible integer $s$-$t$ flow of value $\nu$ becomes a zero-balance circulation by assigning exactly $\nu$ units to $a_\nu$. Conversely, a feasible integer circulation in the augmented graph must assign $\nu$ to $a_\nu$, and deleting that arc leaves precisely the required imbalance at $s$ and $t$ while preserving all capacity bounds on the original arcs.
[/proofplan]
[step:Extend a feasible $s$-$t$ flow by putting $\nu$ units on the artificial arc]
Let $a_\nu$ denote a new directed arc with tail $t$ and head $s$, and define the augmented arc set $A_\nu:=A\cup\{a_\nu\}$. Define the augmented directed graph $D_\nu:=(V,A_\nu)$. Define the augmented lower and upper capacity maps $l_\nu,u_\nu:A_\nu\to\mathbb Z$ by setting $l_\nu(a)=l(a)$ and $u_\nu(a)=u(a)$ for every original arc $a\in A$, and by setting
\begin{align*}
l_\nu(a_\nu)=\nu=u_\nu(a_\nu).
\end{align*}
For each vertex $v\in V$, let $\delta^+(v)$ and $\delta^-(v)$ denote the sets of arcs in $A$ leaving and entering $v$, respectively, and let $\delta_\nu^+(v)$ and $\delta_\nu^-(v)$ denote the corresponding sets of arcs in $A_\nu$. Because $s\neq t$, the arc $a_\nu$ enters $s$ but does not leave $s$, and leaves $t$ but does not enter $t$.
Assume first that $f:A\to \mathbb Z$ is a feasible integer [lower-bounded flow](/page/Flow) of value $\nu$ on $D$. We use the balance convention for an $s$-$t$ flow of value $\nu$: outgoing flow minus incoming flow is $\nu$ at $s$, is $-\nu$ at $t$, and is $0$ at every vertex in $V\setminus\{s,t\}$. Define $F:A_\nu\to \mathbb Z$ by setting $F(a)=f(a)$ for every original arc $a\in A$ and setting $F(a_\nu)=\nu$ on the artificial arc. For every original arc $a\in A$, the inequalities $l_\nu(a)\leq F(a)\leq u_\nu(a)$ are exactly the inequalities $l(a)\leq f(a)\leq u(a)$. For the new arc,
\begin{align*}
l_\nu(a_\nu)=\nu=F(a_\nu)=u_\nu(a_\nu),
\end{align*}
so $F$ satisfies all capacity bounds on $A_\nu$.
We now verify zero node balance. For $v\in V\setminus\{s,t\}$, the new arc $a_\nu:t\to s$ is neither outgoing from nor incoming to $v$, hence
\begin{align*}
\sum_{a\in \delta_\nu^+(v)}F(a)-\sum_{a\in \delta_\nu^-(v)}F(a)
=
\sum_{a\in \delta^+(v)}f(a)-\sum_{a\in \delta^-(v)}f(a)
=0.
\end{align*}
At $s$, the new arc enters $s$ and does not leave $s$, so
\begin{align*}
\sum_{a\in \delta_\nu^+(s)}F(a)-\sum_{a\in \delta_\nu^-(s)}F(a)=\sum_{a\in \delta^+(s)}f(a)-\left(\sum_{a\in \delta^-(s)}f(a)+F(a_\nu)\right)=\nu-\nu=0.
\end{align*}
At $t$, the new arc leaves $t$ and does not enter $t$, so
\begin{align*}
\sum_{a\in \delta_\nu^+(t)}F(a)-\sum_{a\in \delta_\nu^-(t)}F(a)=\left(\sum_{a\in \delta^+(t)}f(a)+F(a_\nu)\right)-\sum_{a\in \delta^-(t)}f(a)=-\nu+\nu=0.
\end{align*}
Thus $F$ is a feasible [circulation](/page/Circulation) on $D_\nu$.
[/step]
[step:Delete the artificial arc from a feasible circulation]
Conversely, assume that $F:A_\nu\to \mathbb Z$ is a feasible integer [circulation](/page/Circulation) on $D_\nu$. Since the new arc has equal lower and upper capacity,
\begin{align*}
F(a_\nu)=\nu.
\end{align*}
Define the restriction to the original arcs by $f:A\to \mathbb Z$, where $f(a)=F(a)$ for every $a\in A$. For every $a\in A$, the augmented capacities agree with the original capacities, so
\begin{align*}
l(a)=l_\nu(a)\leq F(a)=f(a)\leq u_\nu(a)=u(a).
\end{align*}
Hence $f$ satisfies the lower and upper bounds on the original network.
[guided]
We start with a circulation $F:A_\nu\to \mathbb Z$ on the augmented graph. The special feature of the artificial arc is that its lower and upper capacities are both equal to $\nu$. Therefore feasibility forces its value:
\begin{align*}
l_\nu(a_\nu)\leq F(a_\nu)\leq u_\nu(a_\nu)\quad\Longrightarrow\quad \nu\leq F(a_\nu)\leq \nu\quad\Longrightarrow\quad F(a_\nu)=\nu.
\end{align*}
Now remove the artificial arc by restricting $F$ to the original arc set. Define $f:A\to \mathbb Z$ by $f(a)=F(a)$ for every $a\in A$. This map is integer-valued because $F$ is integer-valued. For an original arc $a\in A$, the augmented construction did not change the lower or upper capacity, so $l_\nu(a)=l(a)$ and $u_\nu(a)=u(a)$. Since $F$ is feasible on $A_\nu$, we have
\begin{align*}
l(a)=l_\nu(a)\leq F(a)=f(a)\leq u_\nu(a)=u(a).
\end{align*}
Thus deleting the artificial arc preserves exactly the capacity constraints on every original arc.
[/guided]
[/step]
[step:Compute the remaining node balances after deleting the artificial arc]
It remains to compute the balance of $f$ at each vertex. For $v\in V\setminus\{s,t\}$, the artificial arc is incident to neither $v$, so the zero-balance equation for $F$ gives
\begin{align*}
\sum_{a\in \delta^+(v)}f(a)-\sum_{a\in \delta^-(v)}f(a)
=
\sum_{a\in \delta_\nu^+(v)}F(a)-\sum_{a\in \delta_\nu^-(v)}F(a)
=0.
\end{align*}
At $s$, the arc $a_\nu:t\to s$ contributes only to the incoming arcs in $D_\nu$, so
\begin{align*}
0=\sum_{a\in \delta_\nu^+(s)}F(a)-\sum_{a\in \delta_\nu^-(s)}F(a)=\sum_{a\in \delta^+(s)}f(a)-\left(\sum_{a\in \delta^-(s)}f(a)+F(a_\nu)\right)=\sum_{a\in \delta^+(s)}f(a)-\sum_{a\in \delta^-(s)}f(a)-\nu.
\end{align*}
Therefore
\begin{align*}
\sum_{a\in \delta^+(s)}f(a)-\sum_{a\in \delta^-(s)}f(a)=\nu.
\end{align*}
At $t$, the arc $a_\nu:t\to s$ contributes only to the outgoing arcs in $D_\nu$, so
\begin{align*}
0=\sum_{a\in \delta_\nu^+(t)}F(a)-\sum_{a\in \delta_\nu^-(t)}F(a)=\left(\sum_{a\in \delta^+(t)}f(a)+F(a_\nu)\right)-\sum_{a\in \delta^-(t)}f(a)=\sum_{a\in \delta^+(t)}f(a)-\sum_{a\in \delta^-(t)}f(a)+\nu.
\end{align*}
Therefore
\begin{align*}
\sum_{a\in \delta^+(t)}f(a)-\sum_{a\in \delta^-(t)}f(a)=-\nu.
\end{align*}
[guided]
We compute the node balances after deleting $a_\nu$. The important point is that $s\neq t$, so the artificial arc has exactly one incidence at $s$ and exactly one incidence at $t$: it enters $s$ and leaves $t$. Thus it affects the balance equation at $s$ with a negative sign and the balance equation at $t$ with a positive sign. Since the artificial arc has equal lower and upper capacities, feasibility gives
\begin{align*}
F(a_\nu)=\nu.
\end{align*}
For any vertex $v\in V\setminus\{s,t\}$, the artificial arc is neither incoming nor outgoing at $v$. Since $F$ has zero node balance on $D_\nu$ and $f$ is the restriction of $F$ to $A$, we obtain
\begin{align*}
\sum_{a\in \delta^+(v)}f(a)-\sum_{a\in \delta^-(v)}f(a)
=
\sum_{a\in \delta_\nu^+(v)}F(a)-\sum_{a\in \delta_\nu^-(v)}F(a).
\end{align*}
The zero-balance condition for $F$ at $v$ says that the final expression is $0$, hence
\begin{align*}
\sum_{a\in \delta^+(v)}f(a)-\sum_{a\in \delta^-(v)}f(a)=0.
\end{align*}
At $s$, the arc $a_\nu:t\to s$ appears in $\delta_\nu^-(s)$ and not in $\delta_\nu^+(s)$. Using the zero-balance equation for $F$ at $s$ and the identity $F(a_\nu)=\nu$, we get
\begin{align*}
0=\sum_{a\in \delta_\nu^+(s)}F(a)-\sum_{a\in \delta_\nu^-(s)}F(a)=\sum_{a\in \delta^+(s)}f(a)-\left(\sum_{a\in \delta^-(s)}f(a)+F(a_\nu)\right)=\sum_{a\in \delta^+(s)}f(a)-\sum_{a\in \delta^-(s)}f(a)-\nu.
\end{align*}
Rearranging gives
\begin{align*}
\sum_{a\in \delta^+(s)}f(a)-\sum_{a\in \delta^-(s)}f(a)=\nu.
\end{align*}
At $t$, the arc $a_\nu:t\to s$ appears in $\delta_\nu^+(t)$ and not in $\delta_\nu^-(t)$. Again using zero balance for $F$ and $F(a_\nu)=\nu$, we obtain
\begin{align*}
0=\sum_{a\in \delta_\nu^+(t)}F(a)-\sum_{a\in \delta_\nu^-(t)}F(a)=\left(\sum_{a\in \delta^+(t)}f(a)+F(a_\nu)\right)-\sum_{a\in \delta^-(t)}f(a)=\sum_{a\in \delta^+(t)}f(a)-\sum_{a\in \delta^-(t)}f(a)+\nu.
\end{align*}
Therefore
\begin{align*}
\sum_{a\in \delta^+(t)}f(a)-\sum_{a\in \delta^-(t)}f(a)=-\nu.
\end{align*}
These are exactly the node-balance equations for an integer lower-bounded flow from $s$ to $t$ of value $\nu$.
[/guided]
Thus $f$ is a feasible integer [lower-bounded flow](/page/Flow) from $s$ to $t$ of value $\nu$. This proves the equivalence.
[/step]
