Let $a > b$. Then by definition of indexed summation, both sides are $0$. Let $a \le b$. === Reduction to $a=1$ === By Indexed Summation over Translated Interval: :$\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{j \mathop = 1}^{b - a + 1} \map f {j - a + 1}$ :$\displaystyle \sum_{i \mathop = a}^b \map f {\map \sigma i} = \sum_{j \mathop = 1}^{b - a + 1} \map f {\map \sigma {j - a + 1} }$ Let $n = b - a + 1$. Because $a \le b$, we have $n \ge 1$. By Translation of Integer Interval is Bijection, the mapping $T : \closedint 1 n \to \closedint a b$ defined by: :$\map T i = i + a - 1$ is a bijection. We have: :$\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{j \mathop = 1}^n \map f {\map T j}$ :$\displaystyle \sum_{i \mathop = a}^b \map f {\map \sigma i} = \sum_{j \mathop = 1}^n \map f {\map \sigma {\map T j} }$ By Inverse of Bijection is Bijection and Composite of Bijections is Bijection, the composition $T^{-1} \circ \sigma \circ T$ is a permutation of $\closedint 1 n$. Suppose we have settled the case $a = 1$. Then: :$\displaystyle \sum_{j \mathop = 1}^n \map f {\map T j} = \sum_{j \mathop = 1}^n \map {\paren {f \circ T} \circ \paren {T^{-1} \circ \sigma \circ T} } j$ By Composition of Mappings is Associative, this equals: :$\displaystyle \sum_{j \mathop = 1}^n \paren {f \circ \sigma \circ T} j$ Thus: :$\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{i \mathop = a}^b \map f {\map \sigma i}$ {{qed|lemma}} It remains to investigate the case $a = 1$. === The case $a = 1$ === It remains to show that  :$\displaystyle \sum_{i \mathop = 1}^n \map f i = \sum_{i \mathop = 1}^n \map f {\map \sigma i}$ for any permutation $\sigma$ of $\closedint 1 n$. Let $S_n$ be the symmetric group on $n$ letters, that is, the group of permutations of $\closedint 1 n$. Let $U \subset S_n$ be the subset of all permutations $\sigma$ satisfying :$\displaystyle \sum_{i \mathop = 1}^n \map f i = \sum_{i \mathop = 1}^n \map f {\map \sigma i}$ for all mappings $f: \closedint 1 n \to \mathbb A$. {{qed|lemma}} ==== Summation-preserving permutations form subgroup ==== We prove that $U$ is a subgroup of $S_n$, using Two-Step Subgroup Test. We have to show that $U$ is non-empty. By Identity Mapping is Permutation the identity mapping $I$ on $\closedint 1 n$ is a permutation. By Identity Mapping is Right Identity, $f \circ I = f$. Thus: :$\displaystyle \sum_{i \mathop = 1}^n \map f i = \sum_{i \mathop = 1}^n \map f {\map I i}$ Thus $I \in U$. Let $\sigma, \tau \in U$. We have to show that $\sigma \circ \tau$ is in $U$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \map {f \circ \paren {\sigma \circ \tau} } i       | r = \sum_{i \mathop = 1}^n \map {\paren {\paren {f \circ \sigma} \circ \tau} } i       | c = Composition of Mappings is Associative }} {{eqn | r = \sum_{i \mathop = 1}^n \map {\paren {f \circ \sigma} } i       | c = $\tau \in U$ }} {{eqn | r = \sum_{i \mathop = 1}^n \map f i       | c = $\sigma \in U$ }} {{end-eqn}} Thus $\sigma \circ \tau \in U$. We show that $\sigma^{-1} \in U$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \map {\paren {f \circ \sigma^{-1} } } i       | r = \sum_{i \mathop = 1}^n \map {\paren {\paren {f \circ \sigma^{-1} } \circ \sigma} } i       | c = $\sigma \in U$ }} {{eqn | r = \sum_{i \mathop = 1}^n \map {\paren {f \circ \paren {\sigma^{-1} \circ \sigma} } } i       | c = Composition of Mappings is Associative }} {{eqn | r = \sum_{i \mathop = 1}^n \map {\paren {f \circ I} } i       | c = Identity Mapping is Identity Element in Group of Permutations }} {{eqn | r = \sum_{i \mathop = 1}^n \map f i       | c = $I \in U$ }} {{end-eqn}} Thus $\sigma^{-1} \in U$. Thus $U$ is a subgroup of $S_n$. {{qed|lemma}} ==== Last adjacent transposition preserves summations ==== If $n = 1$, then by First Symmetric Group is Trivial Group and Subgroups of Trivial Group, $U = S_n$. Let $n \ge 2$. By Symmetric Group on n Letters is Generated by Standard Cycle and Adjacent Transposition, it suffices to show that $U$ contains an adjacent transposition and the standard cycle. Let $\sigma \in S_n$ be the adjacent transposition $\begin {bmatrix} n - 1 & n \end {bmatrix}$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \map f {\map \sigma i}       | r = \sum_{i \mathop = 1}^{n - 1} \map f {\map \sigma i} + \map f {\map \sigma n}       | c = {{Defof|Indexed Summation}} }} {{eqn | r = \sum_{i \mathop = 1}^{n - 2} \map f {\map \sigma i} + \map f {\map \sigma {n - 1} } + \map f {\map \sigma n}       | c = {{Defof|Indexed Summation}} }} {{eqn | r = \sum_{i \mathop = 1}^{n - 2} \map f i + \map f n + \map f {n - 1}       | c = {{Defof|Adjacent Transposition}} }} {{eqn | r = \sum_{i \mathop = 1}^{n - 2} \map f i + \map f {n - 1} + \map f n       | c = Commutative Law of Addition }} {{eqn | r = \sum_{i \mathop = 1}^{n - 1} \map f i + \map f n       | c = {{Defof|Indexed Summation}} }} {{eqn | r = \sum_{i \mathop = 1}^n \map f i       | c = {{Defof|Indexed Summation}} }} {{end-eqn}} Thus $\begin {bmatrix} n - 1 & n \end {bmatrix} \in U$. {{qed|lemma}} ==== Standard cycle preserves summations ==== Let $\sigma$ be the standard cycle $\begin {bmatrix} 1 & \ldots & n \end {bmatrix}$. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \map f {\map \sigma i}       | r = \sum_{i \mathop = 1}^{n - 1} \map f {\map \sigma i} + \map f {\map \sigma n}       | c = {{Defof|Indexed Summation}}, $n \ge 1$ }} {{eqn | r = \sum_{i \mathop = 1}^{n - 1} \map f {i + 1} + \map f 1       | c = {{Defof|Standard Cycle}} }} {{eqn | r = \sum_{i \mathop = 2}^n \map f i + \map f 1       | c = Indexed Summation over Translated Interval }} {{eqn | r = \map f 1 + \sum_{i \mathop = 2}^n \map f i       | c = Commutative Law of Addition }} {{eqn | r = \sum_{i \mathop = 1}^n \map f i       | c = Indexed Summation without First Term }} {{end-eqn}} Thus $\begin {bmatrix} 1 & \ldots & n \end {bmatrix} \in U$. {{qed}}