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.