Isomorphism to Closed Interval

Theorem
Let $$m, n \in \mathbb{N}$$ such that $$m < n$$.

Then $$\left|{\left[{m + 1 \,. \, . \, n}\right]}\right| = n - m$$.

Let $$h: \mathbb{N}_{n - m} \to \left[{m + 1 \,. \, . \, n}\right]$$ be the mapping defined as:

$$\forall x \in \mathbb{N}_{n - m}: h \left({x}\right) = x + m + 1$$

Then $$h$$ is the unique isomorphism as defined in Unique Isomorphism between Finite Totally Ordered Sets, where the orderings on $$\left[{m + 1 \,. \, . \, n}\right]$$ and $$\mathbb{N}_{n - m}$$ are those induced by the ordering of $$\mathbb{N}$$.