See Fundamentals of Mathematics, Volume 1 page 101, for context.
This is one of those facts that is so obvious that I find it difficult to prove. My question regards part of the proof of the principle of induction beginning with the natural number $k>1$. That principle can therefore not be used to prove the following claim:
Square brackets $\left[\dots\right]$ enclose function parameter lists. Let $\mathbb{N}_{n}\equiv\left\{ x\in\mathbb{N}\backepsilon x\ge n\right\},$ $k=h+1,$ and the mapping $f:\mathbb{N}_{1}\to\mathbb{N}_{k}$ be $f\left[x\right]\equiv x+h.$
From the cancellation law of addition, $a+c=b+c\implies a=b,$ it follows that the inverse mapping $f^{-1}:\mathbb{N}_{k}\to\mathcal{N}\supseteq\mathbb{N}_{1}$ exists.
It is trivial to show by induction that $f$ is one-to-one, and therefore invertable. But I don't know how best to express this in terms of the cancellation law. What comes to mind is to write a cancellation function $g:\mathbb{N}_{k}\to\mathbb{N}_{1}$ as
$$g\left[a+h\right]=g\left[b+h\right]=a=b.$$
But I'm not sure how to prove the uniqueness and existence of $g$ in a way that doesn't use induction beginning with $k$, and actually makes use of the cancellation law.
What is a good way to show that the mapping $f^{-1}$ follows from the cancellation law?