Equality of Natural Numbers

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

Then $$\mathbb{N}_m \sim \mathbb{N}_n \iff m = n$$, where $$\sim$$ is as defined as in set equivalence.

Proof

 * By Set Equivalence an Equivalence Relation, we have that $$\mathbb{N}_m \sim \mathbb{N}_n \Longrightarrow m = n$$.


 * Suppose $$m = n$$. We need to show that this implies that $$\mathbb{N}_m \sim \mathbb{N}_n$$.

Let $$S = \left\{{n \in \mathbb{N}: \forall m \in \mathbb{N}_n: \mathbb{N}_m \nsim \mathbb{N}_n}\right\}$$.

That is, $$S$$ is the set of all the natural numbers $$n$$ such that $$\mathbb{N}_m \nsim \mathbb{N}_n$$ for all $$m \in \mathbb{N}_n$$.


 * It is clear that $$0 \in S$$, as $$\mathbb{N}_0 = \varnothing$$ from Consecutive Subsets of N.


 * Let $$n \in S$$.

Let $$m \in \mathbb{N}_{n+1}$$.

If $$m = 0$$, then $$\mathbb{N}_m \nsim \mathbb{N}_{n+1}$$ because $$\mathbb{N}_0 = \varnothing$$ and $$\mathbb{N}_{n+1} \ne \varnothing$$.

If $$m > 0$$ and $$\mathbb{N}_m \sim \mathbb{N}_{n+1}$$, then by Set Equivalence Less One Element that means $$\mathbb{N}_{m-1} \sim \mathbb{N}_n$$.

Thus $$m - 1 < n$$ which contradicts our supposition that $$n \in S$$.

Thus $$n + 1 \in S$$.


 * So, by the Principle of Finite Induction, $$S = \mathbb{N}$$.

Thus $$\mathbb{N}_n \nsim \mathbb{N}_m$$ whenever $$m < n$$.

The result follows from the fact that Set Equivalence an Equivalence Relation, in particular the symmetry clause.