How can we derive this lower bound of a special cut in a graph? I have another question about this paper. There the authors prove a special version of the maximal flow-minimal cut theorem for uniform exactly-$k$-splittable $s$-$t$-flows. They define the cut in this situation on page 9.

If you have a partition $C\cup (V\setminus C)$ with $s\in C$ and $t\in V\setminus C$, the $k$-uniform cut capacity $c_k(C)$ is the maximal volume of a packing of $k$ identically sized packages into the bins with size corresponding to the edges from $C$ to $V\setminus C$. The minimum $k$-uniform $s$-$t$ cut, $c_k(s, t)$, is defined by $c_k(s,t):=\min\{c_k(C)|s\in C\subset V\setminus \{t\}\}$ 

My question is then about Theorem 7. How can we derive this lower bound of $k$ for a standard minimum $s$-$t$ cut in the graph $G'$?
Consider a cut $C$ in $G^\prime$. Let $c_k(C)$ be its uniform cut capacity in $G$. Then $c_k(C) \geq c_k(s,t)$. By definition of $c_k(C)$, exactly $k$ packets, each of size $c_k(C)/k$ can be packed into the cut $C$. So sum of $\lfloor ku_e /c_k(C) \rfloor$ over the cut is at least $k$. As $c_k(C) \geq c_k(s,t)$, $ku_e /c_k(C) \leq ku_e /c_k(s,t)$. So 

$$
\sum_C u_e^\prime = \sum_C \lfloor ku_e /c_k(s,t)\rfloor \geq \sum_C \lfloor ku_e /c_k(C)\rfloor \geq k
$$