[proofplan] Write the directed cycle as a finite cyclic sequence of vertices and arcs. Expanding the definition of the reduced cost separates the sum into the original cycle cost and a sum of potential differences. The potential differences telescope around the closed cycle, and the closing condition returns the final vertex to the initial vertex, so the telescoping contribution is zero. [/proofplan]

[step:Represent the directed cycle by a closed vertex sequence] Let $k \in \mathbb{N}$ be the length of the directed cycle $C$. Choose vertices $v_0,\dots,v_k \in V$ such that $v_k = v_0$ and, for each index $i \in \{0,\dots,k-1\}$, the arc \begin{align*} a_i := (v_i,v_{i+1}) \end{align*} belongs to $A$. Then the arcs of $C$ are precisely $a_0,\dots,a_{k-1}$, counted once in the cyclic order of the cycle. [/step]

[step:Expand the reduced cost sum along the cycle]By the definition of $c^\pi: A \to \mathbb{R}$, for every $i \in \{0,\dots,k-1\}$, \begin{align*} c^\pi(a_i) = c(a_i) + \pi(v_i) - \pi(v_{i+1}). \end{align*} Summing this identity over $i \in \{0,\dots,k-1\}$ first gives \begin{align*} \sum_{a \in C} c^\pi(a)=\sum_{i=0}^{k-1} c^\pi(a_i). \end{align*} Substituting the displayed formula for each $c^\pi(a_i)$ and using finite additivity of sums gives \begin{align*} \sum_{a \in C} c^\pi(a)=\sum_{i=0}^{k-1} c(a_i)+\sum_{i=0}^{k-1}\bigl(\pi(v_i)-\pi(v_{i+1})\bigr). \end{align*}[/step]

[guided]Let $k \in \mathbb{N}$ be the length of the directed cycle, and write $C$ as the closed vertex sequence $v_0,\dots,v_k \in V$ with $v_k=v_0$. For each index $i \in \{0,\dots,k-1\}$, define the corresponding arc $a_i := (v_i,v_{i+1}) \in A$. The arcs of $C$ are exactly $a_0,\dots,a_{k-1}$, counted once in cyclic order. The reduced cost of an arc is the original cost plus the change in potential from the head back to the tail. Therefore, for each index $i \in \{0,\dots,k-1\}$, the definition of $c^\pi: A \to \mathbb{R}$ gives \begin{align*} c^\pi(a_i)=c(a_i)+\pi(v_i)-\pi(v_{i+1}). \end{align*} Now we sum this equality over all arcs in the directed cycle. Because the list $a_0,\dots,a_{k-1}$ is exactly the cycle $C$ counted once in cyclic order, the sum over $a \in C$ is the same as the indexed sum over $i \in \{0,\dots,k-1\}$. Thus \begin{align*} \sum_{a \in C} c^\pi(a)=\sum_{i=0}^{k-1} c^\pi(a_i). \end{align*} Substituting the displayed formula for each $c^\pi(a_i)$ and using finite additivity of sums gives \begin{align*} \sum_{a \in C} c^\pi(a)=\sum_{i=0}^{k-1} c(a_i)+\sum_{i=0}^{k-1}\bigl(\pi(v_i)-\pi(v_{i+1})\bigr). \end{align*} The second sum is the part introduced by the potential. It telescopes because every intermediate potential value appears once with sign $+$ and once with sign $-$. More formally, \begin{align*} \sum_{i=0}^{k-1}\bigl(\pi(v_i)-\pi(v_{i+1})\bigr)=\pi(v_0)-\pi(v_k). \end{align*} Since the vertex sequence is closed, $v_k=v_0$, so $\pi(v_k)=\pi(v_0)$ and hence \begin{align*} \sum_{i=0}^{k-1}\bigl(\pi(v_i)-\pi(v_{i+1})\bigr)=0. \end{align*} Substituting this into the expanded sum gives \begin{align*} \sum_{a \in C} c^\pi(a)=\sum_{i=0}^{k-1} c(a_i). \end{align*} Finally, because $a_0,\dots,a_{k-1}$ are exactly the arcs of $C$ counted once, we have \begin{align*} \sum_{i=0}^{k-1} c(a_i)=\sum_{a \in C} c(a). \end{align*} Combining the last two identities proves \begin{align*} \sum_{a \in C} c^\pi(a)=\sum_{a \in C} c(a). \end{align*}[/guided]

[step:Show the potential contribution telescopes to zero] The potential contribution is a finite telescoping sum: \begin{align*} \sum_{i=0}^{k-1} \bigl(\pi(v_i)-\pi(v_{i+1})\bigr)=\pi(v_0)-\pi(v_k). \end{align*} Since $v_k = v_0$, we have $\pi(v_k)=\pi(v_0)$, and hence \begin{align*} \pi(v_0)-\pi(v_k)=0. \end{align*} Thus \begin{align*} \sum_{i=0}^{k-1} \bigl(\pi(v_i)-\pi(v_{i+1})\bigr)=0. \end{align*} [/step]

[step:Conclude that the original and reduced cycle costs agree] Substituting the telescoping identity into the expanded reduced cost sum gives \begin{align*} \sum_{a \in C} c^\pi(a)=\sum_{i=0}^{k-1} c(a_i)+0. \end{align*} Therefore \begin{align*} \sum_{a \in C} c^\pi(a)=\sum_{i=0}^{k-1} c(a_i). \end{align*} Since $a_0,\dots,a_{k-1}$ are precisely the arcs of $C$ counted once, this is \begin{align*} \sum_{a \in C} c^\pi(a)=\sum_{a \in C} c(a). \end{align*} This proves that the total cost of a directed cycle is invariant under the potential transformation. [/step]