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 $b$:
 * $\displaystyle\sum_{k \mathop \in \left[{a \,.\,.\, b}\right]} f(k) = \sum_{k \mathop = a}^b f(k)$

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:
 * $T(k) = k+a$

is a bijection.

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

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