Summation over Interval equals Indexed Summation

Theorem
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.

Let $a, b \in \Z$ be integers.

Let $\left[{a \,.\,.\, b}\right]$ be the integer interval between $a$ and $b$.

Let $f: \left[{a \,.\,.\, b}\right] \to \mathbb A$ be a mapping.

Then the summation over the finite set $\left[{a \,.\,.\, b}\right]$ equals the indexed summation from $a$ to $b$:
 * $\displaystyle\sum_{k \mathop \in \left[{a \,.\,.\, b}\right]} f \left({k}\right) = \sum_{k \mathop = a}^b f \left({k}\right)$

Proof
By Cardinality of Integer Interval, $\left[{a \,.\,.\, b}\right]$ has cardinality $b - a + 1$.

By Translation of Integer Interval is Bijection, the mapping $T : \left[{0 \,.\,.\, b - a}\right] \to \left[{a \,.\,.\, b}\right]$ defined as:
 * $T \left({k}\right) = k + a$

is a bijection.

By definition of summation over finite set:
 * $\displaystyle \sum_{k \mathop \in \left[{a \,.\,.\, b}\right]} f \left({k}\right) = \sum_{k \mathop = 0}^{b - a} f \left({k + a}\right)$

By Indexed Summation over Translated Interval:
 * $\displaystyle \sum_{k \mathop = 0}^{b - a} f \left({k + a}\right) = \sum_{k \mathop = a}^b f \left({k}\right)$