[proofplan] We decompose the pooled rank of each $X_i$ into two contributions: the number of $X$-observations not exceeding it and the number of $Y$-observations not exceeding it. The first contribution sums to $1+\cdots+m$ because the $X$-observations are distinct. The second contribution counts exactly the cross-pairs with $Y_j<X_i$. Since every cross-pair satisfies exactly one of $Y_j<X_i$ or $X_i<Y_j$, subtracting from the total number $mn$ of cross-pairs gives the desired formula. [/proofplan] [step:Decompose each pooled rank into its $X$ and $Y$ contributions] For each $i \in \{1,\dots,m\}$, define $A_i$ and $B_i$ by \begin{align*} A_i &:= \#\{k \in \{1,\dots,m\} : X_k \le X_i\}. \end{align*} Also define \begin{align*} B_i &:= \#\{j \in \{1,\dots,n\} : Y_j \le X_i\}. \end{align*} By the definition of the pooled rank, \begin{align*} R_i = A_i + B_i. \end{align*} Because the observations are pairwise distinct, $Y_j \le X_i$ is equivalent to $Y_j < X_i$ for every pair $(i,j)$. Hence \begin{align*} B_i = \#\{j \in \{1,\dots,n\} : Y_j < X_i\}. \end{align*} [/step] [step:Sum the within-sample rank contributions] Since $X_1,\dots,X_m$ are pairwise distinct, the values $A_1,\dots,A_m$ are exactly the integers $1,\dots,m$ in some order. Therefore \begin{align*} \sum_{i=1}^m A_i = \sum_{r=1}^m r = \frac{m(m+1)}{2}. \end{align*} [guided] The number $A_i$ is the rank of $X_i$ inside the $X$-sample alone. Because the $X$-observations are pairwise distinct, no two of them can have the same within-sample rank: for each $r \in \{1,\dots,m\}$, there is exactly one $X_i$ that is the $r$-th smallest among $X_1,\dots,X_m$. Thus the list $(A_1,\dots,A_m)$ is a permutation of $(1,\dots,m)$. Summing a permutation does not change the sum, so \begin{align*} \sum_{i=1}^m A_i = 1+2+\cdots+m = \sum_{r=1}^m r. \end{align*} The finite arithmetic sum is \begin{align*} \sum_{r=1}^m r = \frac{m(m+1)}{2}. \end{align*} Therefore \begin{align*} \sum_{i=1}^m A_i = \frac{m(m+1)}{2}. \end{align*} [/guided] [/step] [step:Identify the summed $Y$ contributions with cross-pairs below the $X$ observations] Summing the decomposition $R_i=A_i+B_i$ over $i \in \{1,\dots,m\}$ gives \begin{align*} S_X = \sum_{i=1}^m R_i = \sum_{i=1}^m A_i + \sum_{i=1}^m B_i = \frac{m(m+1)}{2} + \sum_{i=1}^m B_i. \end{align*} By the definition of $B_i$, \begin{align*} \sum_{i=1}^m B_i = \#\{(i,j) \in \{1,\dots,m\} \times \{1,\dots,n\} : Y_j < X_i\}. \end{align*} Hence \begin{align*} \#\{(i,j) : Y_j < X_i\} = S_X-\frac{m(m+1)}{2}. \end{align*} [/step] [step:Pass from pairs with $Y_j<X_i$ to pairs with $X_i<Y_j$] Let $C_-$ be the set \begin{align*} C_- &:= \{(i,j) \in \{1,\dots,m\} \times \{1,\dots,n\} : Y_j < X_i\}. \end{align*} Let $C_+$ be the set \begin{align*} C_+ &:= \{(i,j) \in \{1,\dots,m\} \times \{1,\dots,n\} : X_i < Y_j\}. \end{align*} For each cross-pair $(i,j)$, pairwise distinctness excludes $X_i=Y_j$, so exactly one of $Y_j<X_i$ and $X_i<Y_j$ holds. Therefore $C_-$ and $C_+$ partition the Cartesian product $\{1,\dots,m\}\times\{1,\dots,n\}$, whose cardinality is $mn$. Thus \begin{align*} \#C_+ = mn-\#C_-. \end{align*} Using the previous step and the definition $U_{m,n}=\#C_+$, \begin{align*} U_{m,n} = mn-\left(S_X-\frac{m(m+1)}{2}\right) = mn+\frac{m(m+1)}{2}-S_X. \end{align*} Since $S_X=\sum_{i=1}^m R_i$, this is precisely \begin{align*} U_{m,n}=mn+\frac{m(m+1)}{2}-\sum_{i=1}^m R_i. \end{align*} The same affine identity also gives \begin{align*} S_X = mn+\frac{m(m+1)}{2}-U_{m,n}, \end{align*} so $U_{m,n}$ and $S_X$ determine each other uniquely. [/step]