Equality of Natural Numbers

Theorem
Let $m, n \in \N$.

Then:
 * $\N_m \sim \N_n \iff m = n$

where $\sim$ is as defined as in set equivalence.

Proof

 * By Set Equivalence an Equivalence Relation, we have that $\N_m \sim \N_n \implies m = n$.


 * Suppose $m = n$. We need to show that this implies that $\N_m \sim \N_n$.

Let $S = \left\{{n \in \N: \forall m \in \N_n: \N_m \nsim \N_n}\right\}$.

That is, $S$ is the set of all the natural numbers $n$ such that $\N_m \nsim \N_n$ for all $m \in \N_n$.


 * It is clear that $0 \in S$, as $\N_0 = \varnothing$ from Consecutive Subsets of N.


 * Let $n \in S$.

Let $m \in \N_{n+1}$.

If $m = 0$, then $\N_m \nsim \N_{n+1}$ because $\N_0 = \varnothing$ and $\N_{n+1} \ne \varnothing$.

If $m > 0$ and $\N_m \sim \N_{n+1}$, then by Set Equivalence Less One Element that means $\N_{m-1} \sim \N_n$.

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

Thus $n + 1 \in S$.


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

Thus $\N_n \nsim \N_m$ whenever $m < n$.

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